solution_manual_vme_9e_chapter_13 - · pdf filechapter 13. 3 proprietary material

CHAPTER 13CHAPTER 13

3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

PROBLEM 13.1

A 1300-kg small hybrid car is traveling at 108 km/h. Determine (a) the kinetic energy of the vehicle, (b) the speed required for a 9000-kg truck to have the same kinetic energy as the car.

SOLUTION

v = 108 km/h = 30 m/s

(a) T m vccar J= = = ¥1

2

1

21300 30 585 102 2 3( ) ( ) Tcar = 585 kJ b

(b) T m v

vT

m

truck truck truck

trucktruck

truck

=

= =¥

1

2

2 2 585 10

90

2

23( )( )

000130= m /s2 2

vtruck = 11.40 m/s vtruck = 41.0 km/h b

4


PROBLEM 13.2

An 450 kg satellite is placed in a circular orbit 6400 km above the surface of the earth. At this elevation, the acceleration of gravity is 2.4 m/s2. Determine the kinetic energy of the satellite, knowing that its orbital speed is 20,200 km/h.

SOLUTION

v

v

m

T mv

===

=

=

20 200

5611 11

1

21

2

2

,

.

(

km /h

m/s

Mass of satellite 450 kg

4450 5611 11

7 08402 10

2

9

) ( . )

.T = ¥ J T = 7.08 ¥ 109 J b

Note: Acceleration of gravity has no effect on the mass of the satellite.

5


PROBLEM 13.3

A 1 kg stone is dropped from a height h and strikes the ground with a velocity of 25 m/s. (a) Find the kinetic energy of the stone as it strikes the ground and the height h from which it was dropped, (b) Solve Part a assuming that the same stone is dropped on the moon. (Acceleration of gravity on the moon = 1.62 m/s2.)

SOLUTION

For the stone, m = 1 kg

(a) T mv22 21

2

1

21 25 312 5= = =( ) ( ) . J T2 = 313 J b

On the earth. g = 9.81 m/s2

T1 + U1Æ2 = T2 or 0 + Wh = T2

hT

W= = =2 312 5

1 9 8131 855

.

( ) ( . ). m h = 31.9 m b

On the moon. g = 1.62 m/s2

W = mg = 1.62 N

(b) T mv22 21

212

1 25= = ( ) ( ) ( remains same)T T2 = 313 J b

hT

W= = =2 312 5

1 62192 901

.

.. m h = 192.9 m b

6


PROBLEM 13.4

A 4-kg stone is dropped from a height h and strikes the ground with a velocity of 25 m/s. (a) Find the kinetic energy of the stone as it strikes the ground and the height h from which it was dropped. (b) From what height h1 must a 1 kg stone be dropped so it has the same kinetic energy.

SOLUTION

(a) On the earth.

T mv= = = ◊1

2

1

24 25 12502 2( ) ( )kg m/s N m T = 1250 J b

W mg

T U T T U Wh T

= = =

+ = = = =- -

( ) ( . ) .

.

4 9 81 39 240

0 39 241 1 2 2 1 1 2 2

kg m/s N2

00

1250

39 24031 8552

N

N m

Nmh

T

W= =

◊=

( )

( . ). h = 31.9 m b

(b) K.E. of 1 kg stone = 1250 J = T2

T T U h

hT

W

1 1 2 2 1 2

2

1 9 81

1250

1 9 81127 421

+ = =

= = =

- -U , ( ) ( . )

( ) ( . ). m h = 127.4 m b

7


PROBLEM 13.5

Determine the maximum theoretical speed that may be achieved over a distance of 100 m by a car starting from rest, assuming there is no slipping. The coefficient of static friction between the tires and pavement is 0.75, and 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume (a) front-wheel drive, (b) rear-wheel drive.

SOLUTION

Let W be the weight and m the mass. W mg=

(a) Front wheel drive. N W

s

==

0 60

0 75

.

.m

Maximum friction force without slipping:

F N W

U Fd

W

T T mv

s= ====

= =

Æ

m 0 45

0 45 10045

012

1 2

1 2 22

.

( . ) ( )

,

W J

Principle of work and energy:

T U T

W

gv

v

1 1 2 2

22

22

2

0 451

2

2 9 81 45

882 9

+ =

+ =

==

Æ

W

( )( . )( )

. m /s2

v2 29 714= . m/s v2 107 0= . km/h b

(b) Rear wheel drive. N W

s

==

0 40

0 75

.

.m

Maximum friction force without slipping:

F N W

U Fd

W

T

T mv

s= =====

=

Æ

m 0 30

0 30 1003001

2

1 2

2 22

.

( . )( )W J

1

8


PROBLEM 13.5 (continued)

Principle of work and energy:

T U T

W

gv

v

1 1 2 2

22

22

0 301

2

2 9 81 30

588 6

+ =

+ =

=

=

Æ

( )( . )( )

.

W

m /s2 2

v2 24 261= . m/s v2 87 3= . km/h b

Note: The car is treated as a particle in this problem. The weight distribution is assumed to be the same for static and dynamic conditions. Compare with Sample Problem 16.1 where the vehicle is treated as a rigid body.

9


SOLUTION

(a) First 20 m: Rear wheels skid to generate the maximum force. Since all the weight is on the rear wheels, this force is:

F N

W

T

TW

gv

k===

=

m( . )( )0 60

0

1

2

1

2 202

For first 20 m,

U F

W

W

T U T

WW

gv

v

1 2

1 1 2 2

202

202

20

0 6 20

12

121

2

-

-

===

+ =

=

=

( )( )

( . )( )( )

m

22 9 81 12 235 44( . )( ) .=

v20 15 344= . m/s v20 55 2= . km/h b

(b) For 450 m: Rear wheels skid for first 20 m and roll with sliding impending for remaining 430 m with 60% of the weight on the rear (drive) wheels. The maximum force generated is:

First 20 m: F W a1 0 6= ( . )( ) ( )as in

Remaining 430 m: F N W Ws2 0 85 0 60 0 510= = =m ( . )( . )( ) .

PROBLEM 13.6

Skid marks on a drag race track indicate that the rear (drive) wheels of a car slip for the first 20 m of the 450-m track. (a) Knowing that the coefficient of kinetic friction is 0.60, determine the speed of the car at the end of the first 20-m portion of the track if it starts from rest and the front wheels are just off the ground. (b) What is the maximum theoretical speed for the car at the finish line if, after skidding for 20 m, it is driven without the wheels slipping for the remainder of the race? Assume that while the car is rolling without slipping, 60 percent of the weight of the car is on the rear wheels and the coefficient of static friction is 0.85. Ignore air resistance and rolling resistance.

10



T

TW

gv

U W W

W

1

22

1 2

0

1

2

0 6 20 0 510 430

12 219 3

231

=

=

= += +=

- ( . ) ( ) ( . )( )

( . )

..

.

.

3

0 231 31

2

4538 106

2

2

W

WW

gv

v

f

f

+ =

=

v f = 67 3655. m/s v1320 243= km/h b

11


PROBLEM 13.7

In an ore-mixing operation, a bucket full of ore is suspended from a traveling crane which moves along a stationary bridge. The bucket is to swing no more than 4 m horizontally when the crane is brought to a sudden stop. Determine the maximum allowable speed v of the crane.

SOLUTION

v v

v

T mv

T

1

2

12

2

0

1

20

==

=

=

U mgh d

AB d y y

y y

h

1 2

2 2 2 2 2 2

2

4

10 4

100 16 84 84

10

- = - =

= = + = +

= - = =

=

m

m m( ) ( )

-- = - == - = -

+ =-

-

y

U m

T U T

10 84 0 8349

9 81 0 8349 0 81901 2

1 1 2 2

.

( . )( . ) .

m

m

1

20 8190 0

2 0 8190 16 38

2

2

mv

v

- =

= =

.

( )( . ) .

m

v = 4 05. m/s b

12


SOLUTION

Refer to free body diagram in Problem 13.13.

v v T mv m1 12 23

1

2

1

23 4 5= = = = =m/s m m( ) .

T

U mgh2

1 2

0== --

T U T mgh1 1 2 2 4 5 0+ = - =- . m

h

AB d y d

d

= =

= = + = + -

= +

4 5

9 810 4587

10 10 0 4587

100 91 0

2 2 2 2 2 2

2

.

..

( ) ( . )

. 44

d2 8 96= . d = 2 99. m b

PROBLEM 13.8

In an ore-mixing operation, a bucket full of ore is suspended from a traveling crane which moves along a stationary bridge. The crane is traveling at a speed of 3 m/s when it is brought to a sudden stop. Determine the maximum horizontal distance through which the bucket will swing.

13


PROBLEM 13.9

A package is projected 10 m up a 15º incline so that it just reaches the top of the incline with zero velocity. Knowing that the coefficient of kinetic friction between the package and the incline is 0.12, determine (a) the initial velocity of the package at A, (b) the velocity of the package as it returns to its original position.

SOLUTION

(a) Up the plane, from A to C, - =vC 0.

T mv T

U W F

A A C

A C

= =

= - ∞ --

1

20

15 10

2 ,

( sin )( )m

^SF N W= - ∞ =0 15 0: cos

N W= ∞cos15

F N W

U W

T U T

k

A C

A A C

= = ∞= - ∞ + ∞

+ =

-

-

m 0 12 15

15 0 12 15 10

. cos

(sin . cos )( )m

CC A

A

W

gv W

v

12

15 0 12 15 10

2 9 81 15 0

2

2

- ∞ + ∞

= ∞ +

(sin . cos )( )

( )( . )(sin .

m

112 15 10

73 52

cos )( )

.

∞

=

m

vA

vA = 8 57. m/s 15∞ b

(b) Down the plane from C to A.

T T mv U W F

F

T U

C A A C A

C C

= = = ∞ -

+

-

-

012

15 102 ( sin )

( )reverses direction.

AA A A

A

T W mv

v

= + ∞ - ∞ =

=

0 15 0 12 15 1012

2 9 81 15

2

2

(sin . cos )( )

( )( . )(sin

m

∞∞ - ∞

=

0 12 15 10

28 0392

. cos )( )

.

m

vA

vA = 5 30. m/s 15∞ b

14


SOLUTION

(a) Up the plane from A to B.

T mv

W

g

W

gT

U W F d F N

A A B

A B k

= = = =

= - ∞ - = =-

12

12

8 32 0

15 0 12

2 (

( sin ) .

m/s)2

m NN

^Â = - ∞ = = ∞F N W N W0 15 0 15cos cos

U W d Wd

T U TW

gWd

A B

A A B B

-

-

= - ∞ + ∞ = -

+ = -

(sin . cos ) ( . )

(

15 0 12 15 0 3747

32 0.. )3743 0=

d = 32

9 81 0 3747( . )( . ) d = 8 70. m b

(b) Down the plane from B to A (F reverses direction).

TW

gv T d

U W F d

W

A A B

B A

= = =

= ∞ -

= ∞ --

1

20 8 72

15

15 0 12 1

2 .

( sin )

(sin . cos

m/s

55 8 70

1 245

0 1 2451

22

∞)( . )

.

.

m/s

U W

T U WW

gv

B A

B B A A A

-

-

=

+ = + =T

v

v

A

A

2 2 9 81 1 245

253 9

4 94

===

( )( . )( . )

.

. m/s vA = 4 94. m/s 15∞ b

PROBLEM 13.10

A package is projected up a 15º incline at A with an initial velocity of 8 m/s. Knowing that the coefficient of kinetic friction between the package and the incline is 0.12, determine (a) the maximum distance d that the package will move up the incline, (b) the velocity of the package as it returns to its original position.

15


PROBLEM 13.11

Boxes are transported by a conveyor belt with a velocity v0 to a fixed incline at A, where they slide and eventually fall off at B. Knowing that mk = 0 40. , determine the velocity of the conveyor belt if the boxes leave the incline at B with a velocity of 2.5 m/s.

SOLUTION

T mv

T mv m

T m

U W N

A

B B

B

A B k

=

= =

== ∞ --

1212

12

2 5

3 125

15

02

2 2( . )

.

( sin )

m/s

m (( )6 m

^Â = - ∞ =

= ∞F N W

N W

0 15 0

15

cos

cos

U W

U mg

T U T

mv

A B

A B

A A B B

-

-

-

= ∞ - ∞

= -

+ =

(sin . cos )( )

.

15 0 40 15 6

0 76531

12

m

002 0 76531 3 125- =. .mg m

v

v

02

02 2 2

2 3 125 0 76531

21 265

= +

=

( )( . . )

. /

g

m s v0 4 61= . m/s 15∞ b

16


SOLUTION

T mv T

U W N

A B

A B k

= =

= ∞ --

1

20

15 6

02

( sin )( )m m

^Â = - ∞ =F N0 15 0cos

N W= ∞cos15

U W

U mg

T U T

A B

A B

A A B B

-

-

-

= ∞ - ∞

= -

+ =

(sin . cos )( )

.

15 0 40 15 6

0 76531

m

1

20 76531 00

2mv mg- =.

v

v

02

02 2 2

2 0 76531 9 81

15 0154

=

=

( . )( . )

. /m s v0 3 87= . m/s 15° b

PROBLEM 13.12

Boxes are transported by a conveyor belt with a velocity v0 to a fixed incline at A, where they slide and eventually fall off at B. Knowing that mk = 0 40. , determine the velocity of the conveyor belt if the boxes are to have zero velocity at B.

17


PROBLEM 13.13

Packages are thrown down an incline at A with a velocity of 1 m/s. The packages slide along the surface ABC to a conveyor belt which moves with a velocity of 2 m/s. Knowing that mk = 0 25. between the packages and the surface ABC, determine the distance d if the packages are to arrive at C with a velocity of 2 m/s.

SOLUTION

On incline AB: N mg

F N mg

U mg d F d

mg

AB

AB k AB

A B AB

= ∞= = ∞= ∞ -=

Æ

cos

. cos

sin

30

0 25 30

30

m

(sin cos )d k30 30∞ - ∞m

On level surface BC: N mg x

F mg

U mg x

BC BC

BC k

B C k BC

= === -Æ

7 m

mm

At A, T mv vA A A= =1

212 and m/s

At C, T mv vC C C= =1

222 and m/s

Assume that no energy is lost at the corner B.

Work and energy. T U U TA A B B C C+ + =Æ Æ

1

230 30

1

22

02mv mg d mg x mvA k k BC+ ∞ - ∞ - =(sin cos )m m

Dividing by m and solving for d,

dv g x v gC k BC A

k

=+ -ÈÎ ˘̊

∞ - ∞

=+

2 2

2

2 2

30 30

2 2 9 81

/ /m

m(sin cos )

( ) /( )( . ) (( . )( ) ( ) /( )( . )

sin . cos

0 25 7 1 2 9 81

30 0 25 30

2-∞ - ∞

d = 6 71. m b

18


PROBLEM 13.14

Packages are thrown down an incline at A with a velocity of 1 m/s. The packages slide along the surface ABC to a conveyor belt which moves with a velocity of 2 m/s. Knowing that d = 7 5. m and mk = 0 25. between the packages and all surfaces, determine (a) the speed of the package at C, (b) the distance a package will slide on the conveyor belt before it comes to rest relative to the belt.

SOLUTION

(a) On incline AB: N mg

F N mg

U mg d F d

AB

AB k AB

A B AB

= ∞= = ∞= ∞ -Æ

cos

.

sin

30

0 25 30

30

m cos

== ∞ - ∞mg d k(sin cos )30 30m

On level surface BC: N mg x

F mg

U mg x

BC BC

BC k

B C k BC

= === -Æ

7 m

mm

At A, T mv vA A A= =1

212 and m/s

At C, T mv vC C C= =1

222 and m/s

Assume that no energy is lost at the corner B.

Work and energy. T U U TA A B B C C+ + =Æ Æ

1

230 30

1

22

02mv mg d mg x mvA k k BC+ ∞ - ∞ - =(sin cos )m m

Solving for vC2 ,

v v gd g xC A k k BC

2 2

2

2 30 30 2

1 2 9 81 7 5

= + ∞ - ∞ -

= +

(sin cos )

( ) ( )( . )( . )

m m

((sin . cos ) ( )( . )( . )( )30 0 25 30 2 0 25 9 81 7∞ - ∞ -

= 8 3811. m /s2 2 vC = 2 90. m/s b

(b) Box on belt: Let xbelt be the distance moved by a package as it slides on the belt.

+ Â = - = =

= =

F ma N mg N mg

F N mg

y y

x k k

0

m m

At the end of sliding, v v= =belt m/s2

19



Principle of work and energy.

1

2

1

2

2

8 3811 2

2 2

2 2

mv mg x mv

xv v

g

C k

C

k

- =

=-

=-

m

m

. (

belt belt

beltbelt

))

( )( . )( . )

2

2 0 25 9 81 xbelt m= 0 893. b

20


PROBLEM 13.15

The subway train shown is traveling at a speed of 50 km/h when the brakes are fully applied on the wheels of cars B and C, causing them to slide on the track, but are not applied on the wheels of car A. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the distance required to bring the train to a stop, (b) the force in each coupling.

SOLUTION

mk B

C

F

F

= = ==

0 35 0 35 50 000 9 81 171 675

0 35 40 000

. ( . )( , )( . ) .

( . )( , )(

kN

99 81 137 34. ) .= kN

v1 50125

9= = ¨km/h m/s v T2 20 0= =

(a) Entire train: T U T1 1 2 2+ =-

1

240 000 50 000 40 000

125

9171675 137340 0

2

( , , , ) ( )+ + ÊËÁ

ˆ¯̃

- + =x

x = 40 576. m x = 40 576. m b

(b) Force in each coupling: Recall that x = 40 576. m

Car A: Assume FAB to be in tension.

T V T

F

F

AB

AB

1 1 2 2

21

240 000

125

940 576 0

95081 4

+ =

ÊËÁ

ˆ¯̃

- =

= +

-

( , ) ( . )

. N

FAB = 95 1. kN (tension) b

Car C: T U T1 1 2 2+ =-

1

240 000

125

9137340 40 576 0

137340 95

2

( , ) ( )( . )ÊËÁ

ˆ¯̃

+ - =

- = -

F

F

BC

BC 0081 4.

FBC = + 42258 6. N FBC = 42 3. )kN (tension b

21


PROBLEM 13.16

Solve Problem 13.15, assuming that the brakes are applied only on the wheels of car A.

PROBLEM 13.15 The subway train shown is traveling at a speed of 50 km/h when the brakes are fully applied on the wheels of cars B and C, causing them to slide on the track, but are not applied on the wheels of car A. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the distance required to bring the train to a stop, (b) the force in each coupling.

SOLUTION

(a) Entire train:

F N

v

A A= = =

= =

m ( . )( , )( . ) ,0 35 40 000 9 81 137 340

50125

91

N

km/h m/s v T2 20 0= =

T v T

x

1 1 2 2

21

240 000 50 000 40 000

125

9137 340 0

+ =

+ + ÊËÁ

ˆ¯̃

- =

-

( , , , ) ( , )

x = 91 296. m x = 91 3. m b

(b) Force in each coupling:

Car A: Assume FAB to be in tension.

T v T1 1 2 2+ =-

1

240 000

125

9137340 91 296 0

2

( , ) ( )( . )ÊËÁ

ˆ¯̃

- + =FAB

137340 42 258 4+ =FAB , .

FAB = -95 081 6, . N FAB = 95 1. kN (compression) b

Car C: T v T1 1 2 2+ =-

1

240 000

125

991 296 0

2

, ( . )ÊËÁ

ˆ¯̃

+ =FBC

FBC = - 42 258 4, . N FBC = 42 3. kN (compression) b

22


SOLUTION

Initial speed: v1 108 30= =km/h m/s

Final speed: v2 72 20= =km/h m/s

Vertical drop: h = =( . )( ) .0 02 300 6 00 m

Braking distance: d = 300 m

(a) Braking force. Use cab and trailer as a free body.

m

W mg

= +===

= ¥

1800 5400

7200

7200 9 81

70 632 103

kg

N

( )( . )

.

Work and energy: T U T1 1 2 2+ =Æ

1

2

1

2

1 1

2

1

2

1

300

1

2

12

22

12

22

mv Wh F d v

Fd

mv Wh mv

b

b

+ - =

= + -ÈÎÍ

˘˚̇

= ÊËÁ

ˆ¯̃̄

+ ¥ÈÎÍ

- ˘˚̇

( )( ) ( . )( . ) ( )( )7200 30 70 632 10 6 001

27200 202 3 2

= ¥7 4126 103. N Fb = 7 41. kN b

PROBLEM 13.17

A trailer truck enters a 2 percent downhill grade traveling at 108 km/h and must slow down to 72 km/h in 300 m. The cab has a mass of 1800 kg and the trailer 5400 kg. Determine (a) the average braking force that must be applied, (b) the average force exerted on the coupling between cab and trailer if 70 percent of the braking force is supplied by the trailer and 30 percent by the cab.

23



(b) Coupling force Fc . Use the trailer alone as a free body.

Braking force: ¢ = ¥

= ¥== =

F

m

W mg

b ( . )( . )

.

( )( .

0 70 7 4126 10

5 1888 10

5400

5400 9 81

3

3 N

kg

))

.= ¥52 974 103 N

Work and energy: T U T

mv Wh F d F d mvb c

1 1 2 2

12

221

2

1

2

+ =

+ - ¢ + =

Æ

The plus sign before Fc means that we have assumed that the coupling is in tension.

F Fd

mv Wh mvc b= ¢- + -ÈÎÍ

˘˚̇

= ¥ -

1 1

2

1

2

5 1888 101

300

1

25400 30

12

12

3. ( )( )) ( . )( . ) ( )( )2 3 252 974 10 6 001

25400 20+ ¥ - Ê

ËÁˆ¯̃

ÈÎÍ

˘˚̇

= - ¥3 7068 103. N Fc = 3 71. ( )kN compression b

24


SOLUTION

Initial speed: v1 72 20= =km/h m/s

Final speed: v2 108 30= =km/h m/s

Vertical rise: h = =( . )( ) .0 02 300 6 00 m

Distance traveled: d = 300 m

(a) Traction force. Use cab and trailer as a free body.

m

W mg

= +== =

= ¥

1800 5400

7200

7200 9 81

70 632 103

( )( . )

.

kg

N


1

2

1

21 1

2

1

300

1

27200

12

22

12

12

mv Wh F d mv

Fd

mv Wh mv

t

t

- + =

= + -ÈÎÍ

˘˚̇

= ( )(( ) ( . )( . ) ( )( )30 70 632 10 6 001

27200 202 3 2+ ¥ -È

ÎÍ˘˚̇

= ¥7 4126 103. N Ft = 7 41. kN b

(b) Coupling force Fc . Use the trailer alone as a free body.

m

W mg

===

= ¥

5400

5400 9 81

52 974 103

kg

N

( )( . )

.

PROBLEM 13.18

A trailer truck enters a 2 percent uphill grade traveling at 72 km/h and reaches a speed of 108 km/h in 300 m. The cab has a mass of 1800 kg and the trailer 5400 kg. Determine (a) the average force at the wheels of the cab, (b) the average force in the coupling between the cab and the trailer.

25



Assume that the tangential force at the trailer wheels is zero.


1

2

1

212

22mv Wh F d mvc- + =

The plus sign before Fc means that we have assumed that the coupling is in tension.

Fd

mv Wh mvc = + -ÈÎÍ

˘˚̇

= + ¥

1 1

2

1

2

1

300

1

25400 30 52 974 10

22

12

2 3( )( ) ( . )(( . ) ( )( )6 001

25400 20 2-È

ÎÍ˘˚̇

= ¥5 5595 103. N Fc = 5 56. ( )kN tension b

26


PROBLEM 13.19

Two identical blocks are released from rest. Neglecting the mass of the pulleys and the effect of friction, determine (a) the velocity of Block B after it has moved 2 m, (b) the tension in the cable.

SOLUTION

(a) Kinematics: x xB A= 2

v vB A= 2

A and B. Assume B moves down.

v

T

T m v m v

vv

T v

U

A A B B

BB

B

1

1

22 2

22

22

0

0

1

2

1

2

1

22

4

5

4

==

= +

= +Ê

ËÁˆ

¯̃

=

( )kg

11 2

1 2

30 30

2

1

2 9 81

-

-

= - ∞ + ∞==

=

m g x m g x

x

x

U

A A B B

B

A

(cos )( ) (cos )

( )( .

m

m

)) [ ]

.

3

21 2

16 991 2

Ê

ËÁˆ

¯̃- +

=-U J

Since work is positive, block B does move down.

T U T

v

v

B

B

1 1 2 2

2

2

0 16 995

4

13 59

+ =

+ =

=

-

.

.

vB = 3 69. m/s 60° b

27



(b) B alone.

v

T

v a

1

1

2

0

0

3 69

=== . , ( ( ))m/s from

T m v

U m g x T

B

B B

2 22 2

1 2

1

2

1

22 3 69 13 59

30

= = =

= ∞ --

( )( . ) .

( )(cos )( ) ( )(

J

xx

U T

U

B )

( )( . ) ( ) ( )

.

1 2

1 2

2 9 813

22

33 98

-

-

=Ê

ËÁˆ

¯̃-

È

ÎÍÍ

˘

˚˙˙

=

kg m/s m2

-- 2T

T U T T

T1 1 2 2 0 33 98 2 13 59

2 33 98 13 59 20 39

+ = + - == - =

- . .

. . . T = 10 19. N b

28


PROBLEM 13.20

Two identical blocks are released from rest. Neglecting the mass of the pulleys and knowing that the coefficients of static and kinetic friction are ms = 0 30. and mk = 0 20. , determine (a) the velocity of block B after it has moved 2 m, (b) the tension in the cable.

SOLUTION

Check at 1 to see if blocks move. With motion impending at B downward, determine required friction force at A for equilibrium.

Block B: ZÂ = - ∞ =F N m gB B( ) (sin )30 0

N g gB = Ê

ËÁˆ¯̃

=( )21

2

^Â = - ∞ + =F T m g FB B f( ) (cos ) ( )30 0

( ) ( . )( )

( ) ( . )

. ( )

F N g

T g g

T g

B f s B= =

= ( ) -

= -( )

m 0 30

2 3 2 0 30

3 0 30

/

1

Block A:

^Â = - ∞ =

= ÊËÁ

ˆ¯̃

=

F N m g

N g g

A A

A

( )(sin )

( )

30 0

21

2

ZÂ = - ∞ - =F T m g FA A f2 30 0( ) (cos ) ( )

( ) ( )F T gA f = - ( )2 2 3 2/ 2

Substituting T from 1 into 2

( ) ( ) . ( )F g gA f = -( ) -2 3 0 30 3

Requirement for equilibrium ( ) . .F g gA f = -( ) =3 0 60 1 132

Maximum friction that can be developed at A N gs A= =m 0 3.

Since 0 3 1 132. . ,g g� blocks move.

29



(a) A and B.

( ) ( . )

( ) ( . )

F N g

F N g

A f k B

A f k A

= =

= =

m

m

0 20

0 20

Kinematics: x x

v vB A

B A

==

2

2

v

T

T m v m vv

v

T

A A B BB

B

1

1

22 2

22

2

0

0

1

2

1

2

1

22

4

==

= + = ÊËÁ

ˆ¯̃

+Ê

ËÁˆ

¯̃

=

( )kg

55

42vB

U m g x m g x

F x F x

x

A A B B

A f A B f B

B

1 2 30 30- = - ∞ + ∞- -

=

(cos ) ( ) (cos )

( ) ( ) ( ) ( )

22 1

2 3

21 2 3

22

0 20

1 2

m m

kg m kg m

,

[ ( ) ( ) ( ) ( )

( . )

x

U

A =

= - ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

-

-

(( ) ( . )( )] [ . ]

[( . ) ( . )] [ . ] .

1 0 20 2 9 81

1 732 0 6 9 81 111 2

m m m/s2-= - =-U 1105

1 1 2 2

J

T U T+ =-

0 11 105 1 25

8 88

2

2

+ =

=

. .

.

v

v

B

B

vB = 2 98. m/s 60° b

(b) B alone.

v T

v a

T m vB B

1 1

2

22 2

0 0

2 98

1

2

1

22 2 98

= ==

= = ÊËÁ

ˆ¯̃

. ( ( ))

( )( . )

m/s from

TT

N m g

U m g T FB B

B B f

2

1 2

8 88

30 9

30 2 2 2

== ∞ == ∞ - --

.

sin

(cos )( ) ( )( ) ( ) (

J

N

))

( )( . ) ( ) ( . )( ( )U T

U

1 2

1 2

2 9 813

22 2 0 2 9 2

2 3

-

-

=Ê

ËÁˆ

¯̃- -

=

kg m/s m N) m2

gg T g

T U T g T g

T g

- -

+ = + - - =

= - - =-

2 0 6

0 2 3 2 0 4 8 88

2 2 3 0 4 8 88

1 1 2 2

.

. .

( . )( ) . 221 179. T = 10 59. N b

30


SOLUTION

Kinematics: v vA B= 2

m W m g

WA A A

A

= = ==

3 3

29 43

kg kg) (9.81 m/s

N

2( )

.

(a) 150–N force acts through entire 0.6 m motion of B.

U

T T m vA A

1 2

1 2 22

150 0 6 29 43 1 2 54 68

01

2

- = - =

= =

( )( . ) ( . ) ( . ) .

( )

N m N m J

++ 1

2 22m vB B( )

= ÈÎ ˘̊ + =1

23 2

1

28 102

222

22( ) ( ) ( ) ( ) ( )kg kgv v vB B B

T U T vB1 1 2 2 220 54 68 10+ = + =- : . ( )

( ) .vB 2 2 338= m/s vB = 2 34. m/s b

(b) Initial and final velocities are zero.

T T1 20 0= =

Remove 150-N force after B moves distance d.

U d

d1 2 150 29 43 1 2

150 35 316- = -

= -( ) ( . ) ( . )

.

N N m

J

T U T d1 1 2 2 0 150 31 316 0+ = + - =- : . d = 0 2354. m

d = 235 mm b

PROBLEM 13.21

The system shown is at rest when a constant 150-N force is applied to collar B. (a) If the force acts through the entire motion, determine the speed of collar B as it strikes the support at C. (b) After what distance d should the 150-N force be removed if the collar is to reach support C with zero velocity?

31


PROBLEM 13.22

Blocks A and B have masses of 11 kg and 5 kg, respectively, and they are both at a height h = 2 m above the ground when the system is released from rest. Just before hitting the ground, Block A is moving at a speed of 3 m/s. Determine (a) the amount of energy dissipated in friction by the pulley, (b) the tension in each portion of the cord during the motion.

SOLUTION

Energy dissipated.

(a) v T

v v v

T m m v

T

A B

A B

1 1

2

2 22

2

0 0

3

1

216

23

= == = =

= +

= ÊËÁ

ˆ¯̃

m /s

kg m/s

( )

( )22 72= J

U m g m g E

U E

U

A B p

p

1 2

1 22

1 2

2 2

6 9 81 2

117 7

-

-

-

= - -

= -

=

( ) ( )

( )( . )( )

.

kg m/s m

22 - Ep

T U T

Ep

1 1 2 2

0 117 72 72

+ =+ - =

-

.

Ep = -117 72 72. Ep = 45 7. J b

(b) Block A:

T T m v

U m g T

A

A A

1 2 22 2

1 2

01

2

11

23 49 5

2

= = = ÊËÁ

ˆ¯̃

=

= - =-

( ) .

( )( )

kg m/s J

[[( )( . ) ] [ ]

.

11 9 81 2

215 82 2

2

1 2

kg m/s m-= --

T

U TA

A

T U T TA1 1 2 2 0 215 82 2 49 5+ = + - =- . . TA = 83 2. N b

Block B:

T U T1 1 2 2+ =-

T T m v

U m g T

B

B B

1 2 22 2

1 2

01

2

5

23 22 5

2 2

= = = ÊËÁ

ˆ¯̃

=

= - + =-

kg m/s J( ) .

( ) ( ) -- += - +-

( ) ( . ) ( )

.

5 9 81 2 2

98 1 2

2

1 2

kg m/s m T

U TB

B

0 98 1 2 22 5- + =. .TB TB = 60 3. N b

32


PROBLEM 13.23

The system shown, consisting of a 20-kg collar A and a 10 kg counterweight B, is at rest when a constant 500 N force is applied to collar A. (a) Determine the speed of A just before it hits the support at C. (b) Solve Part a assuming that the counterweight B is replaced by a 98.1 N downward force. Ignore friction and the mass of the pulleys.

SOLUTION

Kinematics:

x x

v vB A

B A

==

2

2

(a) Blocks A and B.

T

T m v m v

T v v

T v

B B A A

A A

A

1

22 2

22 2

22

0

1

2

1

21

210 2

1

220

30

=

= +

= +

=

( ) ( ) ( ) ( )

U x W x W x

U

A A A B B1 2

1 2

500

500 1 20 9 81 1

-

-

= + -

= +

( )( ) ( )( ) ( )( )

( ) ( ) ( ) ( . ) ( )) ( ) ( . ) ( )

. .

-

= + - = ◊-

10 9 81 2

500 196 2 196 2 5001 2U N m

T U T v

v

A

A

1 1 2 22

2

0 500 30

50

3

+ = + =

=

-

vA = Ø4 08. m/s b

33



(b) Since the 10 kg weight at B is replaced by a 98.1 N force, the kinetic energy at 2 is

T m v v v

T

A A A A22 2 2

1

1

2

1

220 10

0

= = =

=

( )

The work done is the same as in Part (a).

U1 2 500- = ◊N m

T U T1 1 2 2+ =-

0 500 10

50

2

2

+ =

=

v

v

A

A vA = Ø7 07. m/s b

34


PROBLEM 13.24

Four packages, each weighing 3 kg, are held in place by friction on a conveyor, which is disengaged from its drive motor. When the system is released from rest, Package 1 leaves the belt at A just as Package 4 comes onto the inclined portion of the belt at B. Determine (a) the speed of Package 2 as it leaves the belt at A, (b) the velocity of Package 3 as it leaves the belt at A. Neglect the mass of the belt and rollers.

SOLUTION

Slope angle: sinm

mb b= = ∞

3

7 523 6

..

(a) Package 1 falls off the belt and 2, 3, 4 move down.

Vertical distance between packets m R= =

= ÈÎÍ

˘˚̇

=

33

1

312

3

2 22

=

T mv

m kgg T vfi =2 229

2

U W R

U1 2

1 2

3 3 3 1 3 3 9 81 1

88 29-

-

= = = ¥= ◊

( )( )( ) ( )( )( ) ( )( . )( )

.

g m

N m

T U T

v

v

1 1 2 2

22

22

0 88 29 4 5

19 62

+ =

+ =

=

-

. .

.

v2 4 43= . m/s 23.6° b

(b) Package 2 falls off the belt and its energy is lost to the system, and 3 and 4 move down 2 ft.

¢= ÈÎÍ

˘˚̇

¢=¢=

= ÈÎÍ

T mv

T

T

T mv

2 22

2

2

3 32

21

2

3 19 62

58 86

21

2

( )

( . )

.

( )

J

˘̆˚̇

=T v3 323

35



U g

U

T U T

2 3

2 3

2 2 3 3

2 3 1 6

58 86-

-

-

= == ◊

+ =

( )( )( )

.

g

N m

58 86 58 86 3

39 24

32

32

. .

.

+ =

=

v

v v3 6 26= . m/s 23.6° b

36


PROBLEM 13.25

Two blocks A and B of mass 4 kg and 5 kg, respectively, are connected by a cord which passes over pulleys as shown. A 3-kg collar C is placed on block A and the system is released from rest. After the blocks have moved 0.9 m, collar C is removed and blocks A and B continue to move. Determine the speed of block A just before it strikes the ground.

SOLUTION

Position 1 to Position 2 .

v T1 10 0= =

At 2 before C is removed from the system

T m m m v

T v v

U m m m g

A B C

A C B

2 22

2 22

22

1 2

1

21

212 6

0 9

= + +

= =

= + --

( )

(

( ) ( .

kg)

m))

( )( )( . )

( )( . )( . )

.

U g

U

1 2

2

1 2

4 3 5 0 9

2 9 81 0 9

17 658

-

-

= + -

==

m

kg m/s m

J

TT U T1 1 2 2+ =-

0 17 658 6

2 943

22

22

+ =

=

.

.

v

v

At Position 2 , collar C is removed from the system.

Position 2 to Position 3 .

¢= + = ÊËÁ

ˆ¯̃

¢=

= +

T m m v

T

T m m

A B

A B

2 22

2

3

1

2

9

22 943

13 244

1

2

( ) ( . )

.

( )

kg

J

(( )v v32

329

2=

37



U m m g

U

A B¢-

¢-

= -

= -= -

2 3

2

2 3

0 7

1 9 81 0 7

6 867

( )( )( . )

( )( . )( . )

.

m

kg m/s m

JJ

m/s

¢+ =

- =

== =

-T U T

v

v

v vA

2 2 3 3

32

32

3

13 244 6 867 4 5

1 417

1 190

. . .

.

.

vA = 1 190. m/s b

38


SOLUTION

(a) Find initial position of the block at 1x0 d .

k

F F kx x

x

s s

== = =

=

2400

100 100 2400

1

24

0 0

0

N/m

At 1 N N N/m

m

d, ( )

T T v

T v

1 2 22

2 22

01

25

2 5

= =

=

, ( )

.

U F dx F x F kx x

F N g

s f k sx

f k k

1 2 0

02400

0 4 5

0- = - + - = =

= = =

Ú ( ) ( );

( ) ( . ) ( )m 119 62

1200 19 62

1200124

1 22 0

0

1 2

2

0

.

[ ] ( . )( )

( )

N

U x x

U

x-

-

= - + -

= ÊËÁ

ˆ¯̃ == - Ê

ËÁˆ¯̃

= ◊-

( .

.

19 62124

1 26581 2

)

N mU

T U T v

v

v

1 1 2 2 22

22

2

0 1 2658 2 5

0 50632

0 71156

+ = + =

==

- . .

.

. m/s

At original position, v = 0 712. m/s b

PROBLEM 13.26

A 5 kg block is attached to an unstretched spring of constant k = 2400 N/m. The coefficients of static and kinetic friction between the block and the plane are 0.60 and 0.40, respectively. If a force F is slowly applied to the block until the tension in the spring reaches 100 N and then suddenly removed, determine (a) the speed of the block as it returns to its initial position, (b) the maximum speed achieved by the block.

39



(b) For any position at a distance to the right of the o2¢¢d

x rrignal positiond

2 .

T T v v1 2 22

2012

2 52

= = = ¢¢( )( ) .m

U F dx F dx x

U x

sx

x

f kx

x

xx

1 2 0

1 22

0 0

0

1

24

1200 19 62

-

-

= - + =

= - +

Ú Ú¢

¢

( )

[ ] .

m

(( )x x- 0

T U T x x1 1 2 2

220 1200

1

2419 62

1

24+ = + Ê

ËÁˆ¯̃

-È

ÎÍÍ

˘

˚˙˙

+ -ÊËÁ

ˆ¯̃

=- ¢ ( ) . 22 5 22. v

Max v, when dv

dxx

2 0

2400 19 62 0

¢ =

- + =.

Max v, when x = ¥ -8 175 10 3. m

2 5 1200

124

8 175 10 19 62 8 752

3 2. ( ) ( . ) . .max¢ = ÊËÁ

ˆ¯̃ - ¥

È

ÎÍÍ

˘

˚˙˙

+-v2

¥¥ -ÊËÁ

ˆ¯̃

= -

-10124

2 5 2 00314 0 645825

3

2. . .maxv

0 1553 1 3336 0 4445 0 8891

0 542926

2

2

. . . .

.

v

v

max

max

= - =

= vmax m/s= 0 737. b

40


PROBLEM 13.27

A 5 kg block is attached to an unstretched spring of constant k = 2400 N/m. The coefficients of static and kinetic friction between the block and the plane are 0.60 and 0.40, respectively. If a force F is applied to the block until the tension in the spring reaches 100 N and then suddenly removed, determine (a) how far the block will move to the left before coming to a stop, (b) whether the block will then move back to the right.

SOLUTION

(a) x0 is initial position at 1d ..

Block has a velocity to the left as it reaches its original position (see Problem 13.26).

k

F x

F N

F g

F

S

f k k

f k

f k

===

=

=

2400

2400

0 4 5

19 62

N/m

N

( )

( ) ( . )( )

( ) .

m

T T

U F dx F dxsx

x

f kx

x

1 2

1 2

0 0

0 0

= =

= - +-

- -

Ú Ú ( )

U x F x x

U x x x x

x

x

f k1 22

0

1 22

02

2400

2

1200 19 62

0

-

-

-

=-

+ - -

= - - - +

( ) ( )

( ) . ( 00 )

T U T x x x x x x

x x1 1 2 2 0 0 0

0

0 1200 19 62 0

1200 19 6

+ = - - + - + =- - -

- ( )( ) . ( )

( ) . 22 0= - = -1200 4 1200 0x x

At 1 , F

F k x x

x x

s

s

== =

= = =

100

2400

100

2400

1

24

23

600

0 0

0

N

m m

41



Total distance moved to the left = +x x0 .

x x01

24

23

6000 08+ = + = . m x x0 0 0800+ = . m b

(b) If Fs at 2 is larger than the maximum possible static friction force, then block will move to the right.

N g= =5 49 05. m

From (a) with m

N

x

F

F N

s

f s s

=

= ÊËÁ

ˆ¯̃

==

==

23

600

240023

600

92

0 60 49 05

29

( )

( . )( . )

.

m

443 N

Since F Fs f s� ( ) , block moves to the right. b

42


PROBLEM 13.28

A 3-kg block rests on top of a 2-kg block supported by, but not attached to a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.

SOLUTION

(a) At the initial Position 1 , the force in the spring equals the weight of both blocks, i.e., 5g N, thus at a distance x, the force in the spring is

F g kx

F g xs

s

= -= -

5

5 40

Maximum velocity of the 2-kg block occurs while the spring is still in contact with the block.

T T mv v v

U g x dx gx gx

1 22 2 2

1 2

01

2

1

22

5 40 2 3 2

= = = ÊËÁ

ˆ¯̃

=

= - - = --

( )( )

( )

kg

00 2

0x

x

Ú

T U T gx x v1 1 2 22 20 3 20+ = + - =- (1)

Maximum v when dv

dxg x

x vg

=

= -

=

0

3 40

3

40( )max m

Substituting in (1) x v2 0 7358(max ) .= m

vmax2 23 9 81 0 7358 20 0 7358

21 65 10 83

10 82

= -= -=

( )( . )( . ) ( )( . )

. .

.

vmax m/s= 3 29. b

43



(b) x

xg g g

F g x

Ts

0

02 3

40 85 40

=

= + =

= -

initial compression of the spring

m( )

11 30 0= =T

U g x dx gh

Ug g

gh

T U T

g

1 3 0

8

1 3

2 2

1 1 3 3

5 40 2

5

8

20

642

020

-

-

-

= - -

= - -

+ = +

Ú ( )/

gggh

hg

2

642 0

10

64

10 9 81

64

- =

= = ( )( . )

h = 1 533. m b

44


SOLUTION

(a) See solution to (a) of Problem 13.28. vmax m/s= 3 29. b

(b) Refer to figure in (b) of Problem 13.28.

T T

U g x dx ghh

1 3

1 3 0

0 0

5 40 2

= =

= - -- Ú ( )

Since the spring remains attached to the 2-kg block, the integration must be carried out throughout the total distance h.

T U T gh h gh

hg

1 1 3 220 5 20 2 0

3

20

3 9 81

20

+ = + - - =

= =

-

( )( . )

h = 1 472. m b

PROBLEM 13.29

Solve Problem 13.28, assuming that the 2-kg block is attached to the spring.

PROBLEM 13.28 A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.

45


SOLUTION

(a) k

kB

A

==

2400

3600

N/m

N/m

Since the collar C leaves the spring at B and there is no friction, it must engage the spring at A.

T T

U k xdx k dx

U

A B

A B B A

y

A B

= =

= - ¥

= ÊËÁ

ˆ¯̃

-

-

Ú Ú0 0

2400

20 05

0

0 05

0

( . )

.

22 23600

2- Ê

ËÁˆ¯̃

y

T U T y

y y

A A B B+ = + - =

= fi = =

- 0 3 1800 0

1

6000 04082 40 82

2

2 . .m mm

Total distance d = + + -50 400 150 40 82( . ) mm

d = 559 mm b

(b) Assume that C does not reach the spring at B because of friction.

N W

F

T T

f

A D

= == ¥ =

= =

4

0 35 4 9 81 13 734

0

g

) N( . )( . .

U xdx F y yA D f- = - = -Ú 2400 3 13 7340

0 05( ) .

.

T U T y

yA A D D+ = + - =

= =- 0 3 13 734 0

0 2184 218 4

.

. .m mm

The collar must travel 400 150 50 300- + = mm before it engages the spring at B. Since y = 218 4. mm it stops before engaging the spring at B.

Total distance d = 218 mm b

PROBLEM 13.30

A 4 kg collar C slides on a horizontal rod between springs A and B. If the collar is pushed to the right until spring B is compressed 50 mm and released, determine the distance through which the collar will travel, assuming (a) no friction between the collar and the rod, (b) a coefficient of friction mk = 0 35. .

46


PROBLEM 13.31

A 3 kg block is attached to a cable and to a spring as shown. The constant of the spring is k = 1600 N/m and the tension in the cable is 15 N. If the cable is cut, determine (a) the maximum displacement of the block, (b) the maximum speed of the block.

SOLUTION

k = 1600 N/m

Â = = - =

= = = =

F F g C

v T T v v

y s0 3 15 14 43

0 01

23 1 5

1

1 1 2 22

22

: ( ) .

, : ( ) .

N

For weight: U g x x1 2 3 29 43- = =( .) 3x g=

For spring: U dx x xx

1 22

014 43 14 43 800- = - = - -Ú ( . ) .+1600x

T U T x x x v

x x v

1 1 2 22

22

222

0 29 43 14 43 800 1 5

15 800 1 5

+ = + - - =

- =- : . . .

. (1)

(a) For x vm , :2 0= 15 800 02x x- =

x xm m= = =015

8000 01875, .m mm xm = Ø18 75. mm b

47



(b) For vm , we seek maximum of v T x x1 2 2215 800- = = -maximum of

dT

dxx x2 15 1600 0

15

1600

3

3209 375= - = = = =m m mm.

Eq. (1): 153

320800

3

3201 5

22Ê

ËÁˆ¯̃

- ÊËÁ

ˆ¯̃

= . vm

v vm m2 3 64 0 2165= =/ . m/s vm = 0 217. m/sb b

Note: U1 2- for the spring may be computed using F xs - curve

U

x x

1 2

214 431

2

- =

= +

area

(1600).

48


PROBLEM 13.32

An uncontrolled automobile traveling at 100 km/h strikes squarely a highway crash cushion of the type shown in which the automobile is brought to rest by successively crushing steel barrels. The magnitude F of the force required to crush the barrels is shown as a function of the distance x the automobile has moved into the cushion. Knowing that the weight of the automobile is 1000 kg and neglecting the effect of friction, determine (a) the distance the automobile will move into the cushion before it comes to rest, (b) the maximum deceleration of the automobile.

SOLUTION

(a) 100250

9km/h m/s=

Assume auto stops in m m.1 5 4. � �d

v

T mv

T

1

1 12

2

1

250

9

1

2

1

21000

250

9

385 802 5

385 8

=

= = ÊËÁ

ˆ¯̃

==

m/s

J

( )

, .

. 0025

0

0

80 1 5 120 1 5

120

2

2

1 2

kJ

kN m kN

v

T

U d

=== + -{ }= - +

- ( )( . ) ( )( . )

(

-1120 180

120 60

385 8025 120 60

3 7

1 1 2 2

d

d

T U T

d

d

-= - ◊

+ == -=

-

)

( )

.

.

- kN m

115m d = 3 72. m b

Assumption that d � 4 m is ok.

(b) Maximum deceleration occurs when F is largest. For d F= =3 715 120. m, kN . Thus, F maD=

120 000 1000, = aD aD = 120 2m/s b

49


SOLUTION

Pressures vary inversely as the volume

P

P

Aa

AxP

Pa

xP

P

Aa

A a xP

Pa

a x

LL

RR

= =

=-

=-( ) ( )2 2

Initially at 1 , v xa

T

= =

=

02

01

At 2 , x a T mv= =, 221

2

U P P Adx PaAx a x

dx

U paA x

L R a

a

a

a

1 2 22

1 2

1 1

2-

-

= - = --

È

ÎÍ

˘

˚˙

= +

ÚÚ ( )

[

//

ln lln

ln ln

( )]

ln ln

/2

2

3

2

2

1 2

a x

U paA a aa a

aa-

= + - ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚̇

-

UU paA aa

paA

T U T paA

1 22

2

1 1 2 2

3

4

4

3

0

-

-

= -È

ÎÍ

˘

˚˙ = Ê

ËÁˆ¯̃

+ = +

ln ln ln

lnn4

3

1

22Ê

ËÁˆ¯̃

= mv

vpaA

m

paA

m2

243

0 5754=

ÊËÁ

ˆ¯̃

=ln

. vpaA

m= 0 759. b

PROBLEM 13.33

A piston of mass m and cross-sectional area A is equilibrium under the pressure p at the center of a cylinder closed at both ends. Assuming that the piston is moved to the left a distance a/2 and released, and knowing that the pressure on each side of the piston varies inversely with the volume, determine the velocity of the piston as it again reaches the center of the cylinder. Neglect friction between the piston and the cylinder and express your answer in terms of m, a, p, and A.

50


PROBLEM 13.34

Express the acceleration of gravity gh at an altitude h above the surface of the earth in terms of the acceleration of gravity g0 at the surface of the earth, the altitude h and the radius R of the earth. Determine the percent error if the weight that an object has on the surface of earth is used as its weight at an altitude of (a) 1 km, (b) 1000 km.

SOLUTION

FGM m

h R

GM m R

hR

mgE Eh=

+=

+ÊËÁ

ˆ¯̃

( )2

2

2

1

/

At earth’s surface, (h = 0) GM m

RmgE

2 0=

GM

Rg gE

h

GME

R

hR

2 02

12

= =+

ÊËÁ

ˆ¯̃

Thus, gg

R

hhR

=

=

+ÊËÁ

ˆ¯̃

0

12

6370 km

At altitude h, “true” weight F mg Wh T= =

Assume weight W mg0 0=

Error = =-

=-

=-

=+Ê

ËÁˆ¯̃

=

-

EW W

W

mg mg

mg

g g

g

gg

hR

E

g

T h h

h

g

0

0

0

0

0

0

02

00

1

1 ++( )= -

+( )È

ÎÍÍ

˘

˚˙˙

hR

hRg

2

02

11

1

(a) h = 1 km: P E= =-

+ÊËÁ

ˆ¯̃

È

Î

ÍÍÍ

˘

˚

˙˙˙

100 1001

1

11

6370

2 P = 0.0314%b

(b) h = 1000 km: P E= =-

+ÊËÁ

ˆ¯̃

È

Î

ÍÍÍ

˘

˚

˙˙˙

100 1001

1

110006370

2 P = 25.3% b

51


PROBLEM 13.35

A rocket is fired vertically from the surface of the moon with a speed v0 . Derive a formula for the ratio h hn u/ of heights reached with a speed v, if Newton’s law of gravitation is used to calculate hn and a uniform gravitational field is used to calculate hu . Express your answer in terms of the acceleration of gravity gm on the surface of the moon, the radius Rm of the moon, and the speeds v and v0 .

SOLUTION

Newton’s law of gravitation

T mv T mv

u F dr Fmg R

r

u mg R

n nm m

R

R h

m

m

m n

1 02

22

1 2

2

2

1 2

1

2

1

2= =

= - =

= -

-+

-

Ú ( )

mm R

R h

m mm m n

dr

r

u mg RR R h

T u T

mv

m

m n22

1 22

1 1 2 2

02

1 1

1

2

+

-

-

Ú

= -+

ÊËÁ

ˆ¯̃

+ =

++ -+

ÊËÁ

ˆ¯̃

=mg RR

R hmvm m

m

m n

1

22

hv v

g

R

Rn

m

m

m

v v

gm

=-( )

-

È

Î

ÍÍÍ

˘

˚

˙˙˙

-( )02 2

02 2

22

(1)

Uniform gravitational field

T mv T mv

u F dr mg R h R mgh

T

u m m u m uR

R h

m

m n

1 02

22

1 2

1

1

2= =

= - = - + - = --+

Ú ( ) ( )

++ = - =-u T mv mg h mvm u1 2 2 02 21

2

1

2

hv v

gum

=-( )0

2 2

2 (2)

Dividing (1) by (2) h

hn

uv v

g Rm m

=-

-( )1

1 02 2

2( )

b

52


SOLUTION

Solve for hm .

At maximum height, the total velocity is the horizontal component of the velocity, which is constant and the same in both cases.

T mv T mv

U mg h

mg h

H

e e

m m

12

22

1 2

1 2

12

12

= =

= -= -

-

-

Earth

MoonU

Earth 1

2

1

22 2mv mg h mve e H- =

Moon 1

2

1

22 2mv mg h mvm m H- =

Subtracting - + = =

=ÊËÁ

ˆ¯̃

=

g h g hh

h

g

g

hg

g

e e m mm

e

e

m

me

e

0

600 165

363 636

( ).

.

m

m

PROBLEM 13.36

A golf ball struck on the earth rises to a maximum height of 60 m and hits the ground 225 m away. How far will the same golf ball travel on the moon if the magnitude and direction of its velocity are the same as they were on the earth immediately after the ball was hit? Assume that the ball is hit and lands at the same elevation in both cases and that the effect of the atmosphere on the earth is neglected, so that the trajectory in both cases is a parabola. The acceleration of gravity on the moon is 0.165 times that on earth.

53



Equation of a parabola /( ) ( )y h C x R- = - - 2 2

( )

( )

y h C xR

y h C xR

e ee

m mm

- = - -ÊËÁ

ˆ¯̃

- = - -ÊËÁ

ˆ¯̃

2

2

2

2

Earth

Moon

At is the same, thus is the same.x v dydx= 0,

dy

dxC R C R

C

C

R

R

xe e m m

e

m

m

e

== =

=

0

At 0:x y= =0, h CR

h CR

h

h

C R

C R

R

R

e ee

m mm

m

e

m m

e e

m

e

= =

= =

2 2

2

2

4 4

h

h

g

g

R

RR

g

gm

e

e

m

m

em

e

e

= = =ÊËÁ

ˆ¯̃

=

0 165225

1363 636

.( )

.

m

m

Rm = 1364 m b

54


PROBLEM 13.37

A 300-g brass (nonmagnetic) block A and a 200-g steel magnet B are in equilibrium in a brass tube under the magnetic repelling force of another steel magnet C located at a distance x = 4 mm from B. The force is inversely proportional to the square of the distance between B and C. If block A is suddenly removed, determine (a) the maximum velocity of B, (b) the maximum acceleration of B. Assume that air resistance and friction are negligible.

SOLUTION

(a) Calculating K.

Â = + - ¥

= ¥

= ¥

-

-

-

F m m g k

k g

k

A B( ) ( )

( ) ( . )( )

/ m

m kg m/s

4 10

4 10 0 5

8 10

3 2

3 2 2

6 gg N m◊

v T

v v

T m v v

U F m g dx

U

B

B

1 1

2

22 2

1 2 1

2

1 2

0 0

12

0 1

8 10

= ==

= = ◊

= -

= ¥

-

-

Ú

.

( )

N m

d

d

--

¥

-

-

-Ê

ËÁˆ

¯̃

+ = + ¥ -Ê

ËÁˆ

-Ú6

24 10

1 1 2 2

6

2

0 2

08 10

0 2

3

g

xg dx

T U Tg

xg

x.

.¯̃̄

=¥ -Ú4 10

23

0 1x

dx v.

For maximum v, d v

dx

( . )0 10

2

=

Thus, 8 10

0 2 06

2

¥- =

- g

xg.

55



At vmax, x = 0 00632. m

08 10

0 2 0 1

08 10

6

22

0 004

0 00632

6

+¥

-Ê

ËÁˆ

¯̃=

+- ¥

-

-

Úg

xg dx v

x

. .

( )

.

.

max

(( . ) . ( . ) ..

.

9 81 0 2 9 81 0 10 004

0 006322-

È

ÎÍ

˘

˚˙ =k v

m

m

max

vmax m/s= 0 1628.

vm = 162 8. mm/s b

(b) Maximum acceleration at x = 0.004 m where SF is maximum.

Â = - =

¥ --

F k x W m aB B/ 2

6 28 10 9 81 0 004 0 2 9( )( . )/( . ) ( . )( .881 0 2) ( . )= am

am = ≠14 72 2. m/s b

56


PROBLEM 13.38

Nonlinear springs are classified as hard or soft, depending upon the curvature of their force-deflection curve (see figure). If a delicate instrument having a mass of 5 kg is placed on a spring of length l so that its base is just touching the undeformed spring and then inadvertently released from that position, determine the maximum deflection xm of the spring and the maximum force Fm exerted by the spring, assuming (a) a linear spring of constant k = 3 kN/m, (b) a hard, nonlinear spring, for which F x x= +( ( ).3 160 3kN/m)

SOLUTION

W mg g

W

= ==

( )

.

5

49 05

kg

N

Since T T T U T U1 2 1 1 2 2 1 20 0= = + = =- -, yields

U Wx Fdx x Fdxm m

xx mm

1 2 0049 05 0- = - = - =ÚÚ . (1)

(a) For F kx x= = ( )300 N/m

Eq. (1): 49 05 3000 00

. x x dxm

xm- =Ú 49 05 1500 02. x xm m- = xm = ¥ =-32 7 10 32 73. .m m b

F xm m= = ¥ -3000 3000 32 7 10 3( . ) Fm = ≠98 1. N b

(b) For F x x= +( ) ( )3000 1 160 2N/m

Eq. (1) 49 05 3000 160 03

0. ( )x x x dxm

xm- + =Ú

49 05 30001

240 02 4. x x xm m m- +Ê

ËÁˆ¯̃

= (2)

Solving by trial: xm = ¥ -30 44 10 3. m xm = 30 4. mm b

Fm = ¥ + ¥- -( )( . )[ ( . ) ]3000 30 44 10 1 160 30 44 103 3 2 Fm = ≠104 9. N b

57


PROBLEM 13.39

The sphere at A is given a downward velocity v0 and swings in a vertical circle of radius l and center O. Determine the smallest velocity v0 for which the sphere will reach Point B as it swings about Point O (a) if AO is a rope, (b) if AO is a slender rod of negligible mass.

SOLUTION

T mv

T mv

U mgl

T U T mv mgl mv

v v

1 02

22

1 2

1 1 2 2 02 2

02 2

1

21

2

1

2

1

2

=

=

= -

+ = - =

=

-

-

++ 2gl

Newton’s law at 2

(a) For minimum v, tension in the cord must be zero.

Thus, v gl2 =

v v gl gl02 2 2 3= + = v gl0 3= b

(b) Force in the rod can support the weight so that v can be zero.

Thus, v gl02 0 2= + v gl0 2= b

58


PROBLEM 13.40

The sphere at A is given a downward velocity v0 of magnitude 5 m/s and swings in a vertical plane at the end of a rope of length l = 2 m attached to a support at O. Determine the angle q at which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere.

SOLutiOn

T mv m

T m

T mv

U mg

1 02 2

1

22

1 2

1

2

1

25

12 5

1

22

= =

=

=

=-

( )

.

( )sinq

T U T m mg mv1 1 2 2212 5 2

1

2+ = + =- . sinq

25 4 2+ =g vsinq (1)

Newton’s law at 2 .

+ 22

2 2

mg mg mv

mv- = =sinq

l

v g g2 4 2= - sinq (2)

Substituting for v2 from Eq. (2) into Eq. (1)

2 4 4 2

4 9 81 25

6 9 810 2419

J g g g+ = -

= - =

sin sin

sin( )( . )

( )( . ).

q q

q q = ∞14 00. b

59


PROBLEM 13.41

A section of track for a roller coaster consists of two circular arcs AB and CD joined by a straight portion BC. The radius of AB is 30 m and the radius of CD is 80 m. The car and its occupants, of total weight 300 kg, reach Point A with practically no velocity and then drop freely along the track. Determine the normal force exerted by the track on the car as the car reaches Point B. Ignore air resistance and rolling resistance.

SOLutiOn

v T T mv v vA A B B B B= = = = =0 01

2

1

2300 1502 2 2( )

U

U

U

AB

AB

AB

= - ∞

==

mg

kg m/s ) m

( )( cos )

( )( . ( )( . )

30 1 40

300 9 81 30 0 234

2

2

00655 9. N.m

T U T vA A B B B+ = + =- 0 20655 9 150 2.

vB2 20655 9

150= .

vB2 2 2137 706= . m /s

Newton’s law at B

+ N Wmv

RvB

B- ∞ = - =cos ; .40 137 7062

2

N = -( ( )( . )

(300 300

137 706

30)(9.81)(cos40 )

)∞

N = - =2254 5 1337 06 877 44. . . N N = 877 Nb

60


PROBLEM 13.42

A section of track for a roller coaster consists of two circular arcs AB and CD joined by a straight portion BC. The radius of AB is 30 m and the radius of CD is 80 m. The car and its occupants, of total weight 300 kg, reach Point A with practically no velocity and then drop freely along the track. Determine the maximum and minimum values of the normal force exerted by the track on the car as the car travels from A to D. Ignore air resistance and rolling resistance.

SOLutiOn

Normal force at B.

See solution to Problem 13.41, NB = 877 N.

Newton’s law

From B to C (car moves in a straight line).

+ ¢ - =N WB cos40 0

¢ =NB 300 9 81 40¥ ¥ ∞. cos

¢ =NB 2254 5. N

At C and D (car in the curve at C).

At C:

+ N W mv

RCC- =cosq2

N m gv

RCC= +

Ê

ËÁˆ

¯̃cosq

2

At D:

+ N Wv

RDD- = + m2

N m gv

RDD= +

Ê

ËÁˆ

¯̃

2

61



Since v v N ND C D C� � �and cos ,q 1

Work and energy from A to D.

v T T m v vA A D D D= = = =0 012

1502 2

U mgA B- = + =( ) ( )( . )(30 20 300 9 81 50)

UA B- = ◊147150 N m

T U T vA A B B D+ = + =- 0 147150 150 2

vD2 147150

150981= =

ND = +ÊËÁ

ˆ¯̃

=300 9 81981

806621 75. . N

N NBmin = = 877N N NDmax = = 6620 N b

62


PROBLEM 13.43

A small sphere B of mass m is released from rest in the position shown and swings freely in a vertical plane, first about O and then about the peg A after the cord comes in contact with the peg. Determine the tension in the cord (a) just before the sphere comes in contact with the peg, (b) just after it comes in contact with the peg.

SOLutiOn

Velocity of the sphere as the cord contacts A

v T

T mv

U mg

T U T

B B

C C

B C

B B C C

= =

=

=+ =

-

-

0 0

1

20 4

2

( )( . )

0 0 41

22+ =. mg mvC

v gC2 0 8= ( . )( )

Newton’s law

(a) Cord rotates about Point O R L( )=

+ T mg mv

LC- ∞ =(cos )602

T mgm g= ∞ +(cos )

( . )

.60

0 8

0 8

T mg= 3

2 T mg=1 5. b

63



(b) Cord rotates about A RL=Ê

ËÁˆ¯̃2

T mgmvC

L- ∞ =(cos )60

2

2

Tmg m g

= +2

0 8

0 4

( . )( )

.

T mg mg= +ÊËÁ

ˆ¯̃

=1

22

5

2 T mg= 2 5. b

64


PROBLEM 13.44

A small block slides at a speed v = 2 5. m/s on a horizontal surface at a height h = 1m above the ground. Determine (a) the angle q at which it will leave the cylindrical surface BCD, (b) the distance x at which it will hit the ground. Neglect friction and air resistance.

SOLutiOn

Block leaves surface at C when the normal force N = 0.

+ mg ma

gv

n

n

C

cos

cos

q

q

=

=2

v gh gyC2 = =cosq (1)

Work-energy principle.

(a) T mvB = = =1

2

1

22 5 3 1252 2m m( . ) .

T mv U W h y mg h y

T U T

C C B C C

B B C C

= = - = -

+ =

-

-

1

22 ( ) ( )

3 1251

22. ( )m + - =mg h y mvC

Using Eq. (1) 3 1251

2. ( )+ - =g h y gyC C (2)

3 1253

23 125

3 125 9 81 1

9 81

32

32

.

( . )

( . . )

( . )

+ =

= +( )

= +

gh gy

ygh

g

y

C

C

C¥

65



y mC = 0 87903. (3)

y hy

hCC= = = =cos cos

..q q 0 87903

10 87903 q = ∞-28 5. b

(b) From (1) and (3)

v gy

v

v

C

C

C

=

==

( . )( . )

.

9 81 0 87903

2 9365 m/s

At C: ( ) cos ( . )(cos . ) .v vC x C= = ∞ =q 2 9365 28 5 2 5806m/s

( ) sin ( . ) sin( . )v vC y C= - = - ∞ =q 2 9365 28 5 -1.40118 m/s

y y v t gt t tC C y= + - = - -( ) . . .1

20 87903 1 40118 4 9052 2

At E: y t tE = + - =0 4 905 1 40118 0 87903 02. . .

neglecting the –ve root: t = 0 3039. s

At E: x h v tC x= + = ∞ +(sin ) ( ) ( ) sin( . ) ( . )( . )q 1 28 5 2 5806 0 3039

x = + =0 47716 0 78424 1 2614. . . m

x = 1 261. m b

66


PROBLEM 13.45

A small block slides at a speed v on a horizontal surface. Knowing that h = 2 5. m, determine the required speed of the block if it is to leave the cylindrical surface BCD when q = ∞40 .

SOLutiOn

See above; block leaves the surface

when N gv

hC= =02

cos q

h = = ∞2 5 40. ,m q

Thus, v gh

v

c

c

2

2

9 81 2 5 40

18 79

= = ∞

=

cos ( . )( . )(cos )

.

q

Work-energy principle.

T mv

T mv m

T m

B

C C

C

=

= =

=

1

21

2

1

218 79

9 395

2

2 ( . )

.

U mghB C- = -( cos )1 q

T U TB B C C+ =-

1

2

1

22 2mv mgh mvC+ (1- =cos )q

v gh

v

2

2

2 1 18 79

2 9 81 2 5

= - - +

= - - ∞

( cos ) .

( . )( .

q

m/s m)(1 cos 40 ) + 18.2 7792v = 7 315.

vC = 2 70. m/s b

67


PROBLEM 13.46

(a) A 50 kg woman rides a 7.5 kg bicycle up a 3-percent slope at a constant speed of 1.5 m/s. How much power must be developed by the woman? (b) A 75-kg man on a 9 kg bicycle starts down the same slope and maintains a constant speed of 6 m/s by braking. How much power is dissipated by the brakes? Ignore air resistance and rolling resistance.

SOLutiOn

tan .q q= = ∞3

1001 718

(a) m m mB W= + = + =50 7 5 57 5. . kg

W = 57 5. ¥ 9.81 = 564.075N

P W vW = ◊ =W v ( sin )( )q

PW = ∞( . )(sin . )( . )564 075 1 718 1 5

PW = 25 367. W

PW = 25 4. W b

(b) W W W gB m= + = + =( ) ( )( . )75 9 84 9 81

W = 824 04. N

Brakes must dissipate the power generated by the bike and the man going down the slope at 6 m/s.

P W vB = ◊ =W v ( sin )( )q

PB =

=( . )(sin . )( )

.

824 04 1 718 6

148 23W

PB = 148 2. W b

(a)

(b)

68


SOLutiOn

(a) Material is lifted to a height b at a rate, ( )( ) [ ( )]m g mgkg/h m/s N/h2 =

Thus,

DDU

t

mg b m mgb= = ÊËÁ

ˆ¯̃

◊[ ( )][ ( )]

( )

N/h

s/hN m/s

3600 3600

1000 1N m/s kw◊ =

Thus, including motor efficiency, h,

Pmgb

( )( )

( ) ( )kw

N m/sN m/s

kw

=◊

◊ÊËÁ

ˆ¯̃3600

1000 h

Pmgb

( ) .kw = ¥ -0 278 10 6

h b

PROBLEM 13.47

A power specification formula is to be derived for electric motors which drive conveyor belts moving solid material at different rates to different heights and distances. Denoting the efficiency of the motors by h and neglecting the power needed to drive the belt itself, derive a formula (a) in the SI system of units for the power P in kW, in terms of the mass flow rate m in kg/h, the height b and horizontal distance l in meters.

69


PROBLEM 13.48

A chair-lift is designed to transport 900 skiers per hour from the base A to the summit B. The average weight of a skier is 80 kg and the average speed of the lift is 75 m/min. Determine (a) the average power required, (b) the required capacity of the motor if the mechanical efficiency is 85 percent and if a 300 percent overload is to be allowed.

SOLutiOn

Note: Solution is independent of speed.

(a) Average power = = =DDu

t

( )(,

900 80

360058 860

2kg)(9.81m/s )(300 m)

sW

Average power = 58 9. kw b

(b) Maximum power required with 300% overload

= + =100 300

) W

235kW100

58 860 235 440( , ,

=

Required motor capacity (85% efficient)

Motor capacity = =235440

0 85276 988

., W

277kw=

b

70


SOLutiOn

(a) First 20 m. (Calculating velocity at 20 m.) Force generated by rear wheels = mk w, since car skids.

Thus, F gs = ( . )( )( )0 68 1000

Fs = =( . )( )( . ) .0 68 1000 9 81 6670 8kg m/s N2

Work and energy. T T mv v1 2 202

2020

1

2500= = =,

T U T1 1 2 2+ =-

U Fs1 2 20 20- = =( ) (m)( m)(6670.8 N)

U1 2 133 420- = , J

0 133 420 500 202+ =, v

v202 133 420

500266 83= =

,.

v20 16 335= . m/s

Power = =( )( ) ( .F vs 20 6670 8 N)(16.335 m/s)

Power = =108 970, J/s 108.97 kJ/s

1 1kJ/s kW=

1 0 7457hp kW= . Power = =109 0 109 0. . kJ/s kW b

Power = =( .

( ..

109 0

0 7457146 2

kW)

kW/hp) hp. b

PROBLEM 13.49

In an automobile drag race, the rear (drive) wheels of a 1000-kg car skid for the first 20 m and roll with sliding impending during the remaining 380 m. The front wheels of the car are just off the ground for the first 20 m and for the remainder of the race, 80 percent of the weight is on the rear wheels. Knowing that the coefficients of friction are ms = 0 90. and mk = 0 68. , determine the power developed by the car at the drive wheels (a) at the end of the 20-m portion of the race, (b) at the end of the race. Give your answer in kW and in hp. Ignore the effect of air resistance and rolling friction.

71



(b) End of race. (Calculating velocity at 400 m.) For remaining 380 m, with 80% of weight on rear wheels, the force generated at impending sliding is ( )ms mg(0.80)( )

FI = ( . )( . )( )0 90 0 80 1000 kg)(9.81 m/s2

FI = 7063 2. N

Work and energy, from 20 m 2 to 28 m 3 .

v2 16 335= . m/s [from Part (a)]

T221

21000= ( ) kg)(16.335 m/s

T2 133 420= , J

T mv v3 3802

38021

2500= =

U FI2 3 380 7063 2- = =( )( ( . m) N)(380 m)

U2 3 2 684 000- = , , J

T U T2 2 3 3+ =-

( , ) ( , ,133 420 2 684 000 500 302 J J)+ = v

v30 75 066= . m/s

Power = =( )( ) ( .F vI 30 7063 2 N) (75.066 m/s)

= 530 200, J

kW Power = =530 200 530, J kW b

hp Power = =530711

kW

(0.7457 kW/hp)hp b

72


PROBLEM 13.50

It takes 15 s to raise a 1200-kg car and the supporting 300-kg hydraulic car-lift platform to a height of 2.8 m. Determine (a) the average output power delivered by the hydraulic pump to lift the system, (b) the average electric power required, knowing that the overall conversion efficiency from electric to mechanical power for the system is 82 percent.

SOLutiOn

(a) ( ) ( )( ) ( )( )( )P F v m m g vP A A C L A= = +

vA = = =s/t m s m/s( . ) / ( ) .2 8 15 0 18667

( ) [( ) ( )]( . .PP A = +1200 300 9 81 0 18667 3kg kg m/s )( m/s)2

( ) .PP A = 274 7 kJ/s ( ) .PP A = 2 75 kW b

(b) ( ) ( ) ( .P PE A P= =/ kW)/(0.82)h 2 75 ( ) .PE A = 3 35 kW b

73


PROBLEM 13.51

The velocity of the lift of Problem 13.50 increases uniformly from zero to its maximum value at mid-height in 7.5 s and then decreases uniformly to zero in 7.5 s. Knowing that the peak power output of the hydraulic pump is 6 kW when the velocity is maximum, determine the maximum lift force provided by the pump.

SOLutiOn

Newton’s law

Mg M M g g

Mg gC L= + = +

=( ) ( )1200 300

1500

+ SF F g a= - =1500 1500 (1)

Since motion is uniformly accelerated, a = constant.

Thus, from (1), F is constant and peak power occurs when the velocity is a maximum at 7.5 s.

av

= max

.7 5 s

P F v

v F

= ==

( ) ( )( )

( )max

max

6000

6000

W

/

Thus, a F= ( ) ( . )( )6000 7 5/ (2)

Substituting (2) into (1)

F g F- =1500 1500 6000 7 5( )( ) ( . )( )/

F F2 15001500

7 50- -

◊=( )

(

( . kg)(9.81 m/s

kg)(6000 N m/s)

s)2

F F

F

2 614 715 1 2 10 0

14 800

- - ¥ ==

, .

, N F = 14 8. kN b

74


PROBLEM 13.52

A 100,000 kg train traveling on a horizontal track requires 300 kW to maintain a constant speed of 80 km/h. Determine (a) the total force needed to overcome axle friction, rolling resistance, and air resistance, (b) the additional horsepower required if the train is to maintain the same speed going up a 1-percent grade.

SOLutiOn

(a) P

v

= =

= =

300 300 000

802009

kW W

km/h m/s

,

P F v

F P v

R

R

= ◊

= = =/ N300 000

200 913500

,

( / )

FR = 13 5. kN b

(b) W

W

==

( , )( . )

,

100 000 9 81

981 000

N

N

DP W v= ◊sinq

D ∞

D

P

P

= ÊËÁ

ˆ¯̃

=

( , )(sin . )

,

981 000 0 573200

9

218 012W

DP = 218kW b

75


PROBLEM 13.53

The frictional resistance of a ship is known to vary directly as the 1.75 power of the speed v of the ship. A single tugboat at full power can tow the ship at a constant speed of 4.5 km/h by exerting a constant force of 300 kN. Determine (a) the power developed by the tugboat, (b) the maximum speed at which two tugboats, capable of delivering the same power, can tow the ship.

SOLutiOn

(a) Power developed by tugboat at 4.5 km/h.

v

F0

0

4 5 1 25

300

= ==

. .km/h m/s

kN

P F v= =0 0 300( kN)(1.25 m/s) P = 375 kW b

(b) Maximum speed.

Power required to tow ship at speed v:

F Fv

vP Fv F

v

vv F v

v

v=

ÊËÁ

ˆ¯̃

= =ÊËÁ

ˆ¯̃

=ÊËÁ

ˆ¯̃0

0

1 75

00

1 75

0 00

1 75. . .

(1)

Since we have two tugboats, the available power is twice maximum power F v0 0 developed by one tugboat.

2

2 2 1 2

0 0 0 00

3 50

0

3 50

01

0

F v F vv

v

v

vv v v

=ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

= = =

.

.

( ) ( ./3.50 8867)

Recalling that v0 4 5= . km/h

v = =( . .4 5 5 7902km/h)(1.2867) km/h v = 5 79. km/h b

76


PROBLEM 13.54

The elevator E has a mass of 3000 kg when fully loaded and is connected as shown to a counterweight W of mass 1000 kg. Determine the power in kW delivered by the motor (a) when the elevator is moving down at a constant speed of 3 m/s, (b) when it has an upward velocity of 3 m/s and a deceleration of 0 5. .m/s2

SOLutiOn

(a) Acceleration = 0

Counterweight Elevator Motor

+

SF T M gW W= - = 0 +

SF T T M gC W E= + - =2 0

TW = (1000 kg)(9.81 m/s )2 2 9810 3000 9 81TC = - +( ) ( )( .N kg m/s )2

TW = 9810 N TC = 9810 N

Kinematics: 2 2 2 6x x x x v vE C E C C E= = = =& & m/s

P T vC C= ◊ = =( ,9810 6 58 860 N)( m/s) J/s P = 58 9. kW b

(b) aE = ≠0 5. m/s2 , vE = Ø3 m/s Counterweight

Counterweight: SF Ma=

SF T M g M a

T

T

W W W W

W

W

= - =

= +=

( )[( . ) ( . )]

,

1000 9 81 0 5

10 310

kg m/s m/s

N

2 2

77



Elevator: SF Ma= : Elevator

+ SF T T M g M aC W E E E= + - = -2

2 3000 9 81 0 5 10 310T

T vC

C C

= - -= =

( [( . ) ( . )] , kg) m/s m/s N

8810 N 6 m/

2 2

ss [see ( )]a

P T vC C= ◊ = (8810 N)( m/s)6

P = = =52 860 52 860 52 86, . .J/s kJ/s kW P = 52 9. kW b

78


PROBLEM 13.55

A force P is slowly applied to a plate that is attached to two springs and causes a deflection x0 . In each of the two cases shown, derive an expression for the constant ke , in terms of k1 and k2 ,of the single spring equivalent to the given system, that is, of the single spring which will undergo the same deflection x0 when subjected to the same force P.

SOLutiOn

System is in equilibrium in deflected x0 position.

Case (a) Force in both springs is the same = P

x x x

xP

ke

0 1 2

0

= +

=

xP

k11

= xP

k22

=

Thus, P

k

P

k

P

ke

= +1 2

1 1 1

1 2k k ke

= + kk k

k ke =+1 2

1 2

b

Case (b) Deflection in both springs is the same = x0

P k x k x

P k k x

P k xe

= += +=

1 0 2 0

1 2 0

0

( )

Equating the two expressions for

P k k x k xe= + =( )1 2 0 0 k k ke = +1 2 b

79


PROBLEM 13.56

A block of mass m is attached to two springs as shown. Knowing that in each case the block is pulled through a distance x0 from its equilibrium position and released, determine the maximum speed of the block in the subsequent motion.

SOLutiOn

We will use an equivalent spring constant ke (see Problem 13.55).

Choosing 1 at initial undeflected position

V T mv1 1 120

1

2= =

Choosing 2 at x0 where v = 0

V k x Te2 02

21

20= =

T V T V mv k xe1 1 2 2 12

020

1

2

1

20+ = + + = +

Thus, v v xk

me

1 0= =max

Case (a) kk k

k ke =+1 2

1 2

v xk k

m k kmax ( )=

+01 2

1 2

b

Case (b) k k ke = +1 2 v xk k

mmax =+

01 2 b

80


PROBLEM 13.57

A 1.2-kg collar C may slide without friction along a horizontal rod. It is attached to three springs, each of constant k = 400 N/m and 150-mm undeformed length. Knowing that the collar is released from rest in the position shown, determine the maximum speed it will reach in the ensuing motion.

SOLutiOn

Maximum velocity occurs at E where collar is passing through position of equilibrium.

Position 1

T1 0=

Note: Undeformed length of springs is 0.150 m

Spring AC: L = 335 4. mm

D = - =0 3354 0 150 0 1854. . . m

Spring CD: L == - =

212 1

0 2121 0 150 0 0621

.

. . .

mm

mD

Spring BD: L = =150 mm, 0D

Potential energy. (k = 400 N/m for each spring)

V k k

V

12 2 2

1

1

2

1

2

1

2400 0 0621 0

7 6

= = = + +

=

S D SD ( ( . ) ]

.

N/m)[(0.1854 m) m2

4467 J

Position 2

m T mv v= = =1 21

2

1

21 22 2

222. ( .kg; kg)

Spring AC: L == - =

212 1

0 2121 0 150 0 0621

.

. . .

mm

mD

Spring CD: L ==

150

0

mm

D

Spring BC: L =

= - =212 1

0 2121 0 150 0 0621

.

. . .

mm

mD

81



Potential energy. V k k

V

22 2

2

1

2

1

21

2400 0 0 0621 1 5

= =

= + + =

S D SD

( ( . ] .N/m)[(0.0621 m) m)2 2 4426 J

Conservation of energy. T V T V v1 1 2 2 220 7 6467

1

21 2 1 5426+ = + + = +: J J. ( . ) .

v22 10 1736= . v2 3 19= ´. m/s b

82


PROBLEM 13.58

A 5-kg collar B can slide without friction along a horizontal rod and is in equilibrium at A when it is pushed 100 mm to the right and released. The underformed length of each spring is 300 mm and the constant of each spring is k = 320 N/m. Determine (a) the maximum speed of the collar, (b) the maximum acceleration of the collar.

SOLutiOn

(a) Maximum velocity occurs at A where the collar is passing through its equilibrium position.

Position 1 .

T1 0= k

L

LOC

OC

=

= + == - =

320

0 3 0 1 0 31623

0 31623 0 3 0 01623

2 2

N m

m

/

( . ) ( . ) .

. . .D mm

mm mDLAC = =100 0 1.

V k L k L

V

OC AC12 2

2 2

1

1

2

1

21

2320 0 01623

1

2320 0 1

= +

= +

( ) ( )

( )( . ) ( )( . )

D D

== 1 64215. J

Position 2 .

T mv v v21

2

1

2(5) 2.5= = =2

222 2

max

V2 (Both springs are unstretched.)= 0

T V T V v1 2 220 1 64215 2 5 0+ = + + = +1 . . max

vm

smax .2

2

20 65686= vmax .= 0 810 m/s b

83



(b) Maximum acceleration occurs at C where the horizontal force on the collar is at maximum.

SD D

F ma F F ma

k L k L maOC AC

= + =+ =

1 2

320 0 01623100

31

cos

cos

( . )

max

max

qq

66 230 1 5

.. max

ÊËÁ

ˆ¯̃

+ÈÎÍ

˘˚̇

= a

amax.= 33 6423

5 amax .= 6 73m/s2 b

84


PROBLEM 13.59

An elastic cord is stretched between two Points A and B, located 400 mm apart in the same horizontal plane. When stretched directly between A and B, the tension is 50 N. The cord is then stretched as shown until its midpoint C has moved through 150 mm to ¢C ; a force of 300 N is required to hold the cord at ¢C . A 100 gm pellet is placed at ¢C , and the cord is released. Determine the speed of the pellet as it passes through C.

SOLutiOn

Let l = undeformed length of cord.

Position 1 . Length mm 0.5 m; ElongationAC B x¢ = = = -500 0 51= . l

SF F Fx = ÊËÁ

ˆ¯̃

- = =0 300 0 2501 1: 23

5N N

Position 2 . Length m; Elongation

Given: N

ACB x

F

= = = -=

0 4 0 4

502

2

. . l

F k x F k x1 1 2 2= =

F F k x x

k k1 2 1 2

250 50 0 5 0 4 0 1

- = -- = - - - =

( )

( ) [( . ) ( . )] .l l

k = =200

0 12000

.N/m

xF

k11 250

2000

1

8= = = m

xF

k22 50 1

40= = =

2000m

Position 1 . T V k x1 1 120

1

2

1

22000

1

8

125

8= = = Ê

ËÁˆ¯̃

=( N/m) m J2

Position 2 . m = 0 1. kg

T mv v v

V k x

2 22

22

22

2 22

1

2

1

20 1 0 05

1

2

1

22000

1

40

= = =

= = ÊËÁ

ˆ¯̃

( . ) .

( N/m)22

J= 5

8

85



Conservation of energy.

T V T V

v

1 1 2 2

220

125

80 05

5

8

+ = +

+ = +.

15 0 05 17 32122

2= =. .v vfi m/s v2 17 32= . m/s b

Note: The horizontal force applied at the midpoint of the cord is not proportional to the horizontal distance ¢C C. A solution based on the work of the horizontal force would be rather involved.

86


PROBLEM 13.60

A 10-kg collar slides without friction along a vertical rod as shown. The spring attached to the collar has an undeformed length of 100 mm and a constant of 600 N/m. If the collar is released from rest in position 1, determine its velocity after it has moved 150 mm to position 2.

SOLutiOn

Position 1. Potential Energy. The elongation of the spring is

x1 200= mm 100mm 100mm 0.1 m- = =

V kxe = = =1

2

1

2600 0 1 31

2 2( )( . )m J

Choosing the datum as shown, we have Vg = 0. Therefore,

V V Ve g1 3= + = J

Kinetic Energy. T1 = 0

Position 2. Potential Energy. The elongation of the spring is

x2 250 100 150 0 15= - =mm mm mm m= .

V kxe = = =1

2

1

2600 0 15 6 752

2 2( )( . ) .m J

V Wyg = = =( )( . )( . ) .10 9 81 0 15 14 715- -m J

V V Ve g2 6 75= + = . -14.715 = -7.965 J

T mv v v2 22

22

221

2

1

210 5= = =( )

T V T V1 1 2 2+ = +

0 3 5 7 965

1 480922

2

+ = -=

v

v

.

. m/s v2 1 481= Ø. m/s b

87


PROBLEM 13.61

A 500-g collar can slide without friction on the curved rod BC in a horizontal plane. Knowing that the undeformed length of the spring is 80 mm and that k = 400 kN/m, determine (a) the velocity that the collar should be given at A to reach B with zero velocity, (b) the velocity of the collar when it eventually reaches C.

SOLutiOn

(a) Velocity at A.

T mv v

T v

L

A A A

A A

A

= = ÊËÁ

ˆ¯̃

== -

1

2

0 5

2

0 25

0 150 0 080

2 2

2

.

( . )

. .

kg

m mDDLL

V k L

V

V

A

A A

A

A

=

=

= ¥

=

0 070

1

21

2400 10 0 070

980

2

3 2

.

( )

( )( . )

m

N/m m

D

JJ

m m m

v T

L

V k L

B B

B

B B

= == - =

= =

0 0

0 200 0 080 0 120

1

22

D

D

. . .

( )11

2400 10 0 120

2880

3 2( )( . )¥

=

N/m m

JVB

T V T V v

v

v

A A B B A

A

A

+ = + + = +

= -

=

0 25 980 0 2880

2880 9800 25

7600

2

2

2

.

( )( . )

mm /s2 2

vA = 87 2. m/s b

88



(b) Velocity at C.

Since slope at B is positive, the component of the spring force FP, parallel to the rod, causes the block to move back toward A.

T V a

T mv v v

B B

C C C C

= =

= = =

0 2880

1

2

0 50 252 2

, [

( . ).

J from Part ( )]

kg

222

2 3

0 100 0 080 0 020

1

2

1

2400 10

D

D

L

V k L

C

C C

= - =

= = ¥

. . .

( ) (

m m m

N/m)(0..020 m) J

m

2 =

+ = + + = +

=

80 0

0 2880 0 25 80 0

11 200

2

2

.

. .

,

T V T V v

v

B B C C C

C22 2/s

vC = 105 8. m/s b

89


PROBLEM 13.62

A 3-kg collar can slide without friction on a vertical rod and is resting in equilibrium on a spring. It is pushed down, compressing the spring 150 mm, and released. Knowing that the spring constant is k = 2 6. kN/m, determine (a) the maximum height h reached by the collar above its equilibrium position, (b) the maximum speed of the collar.

SOLutiOn

(a) Maximum height is reached when

v2 0=

Thus, T T1 2 0= =

V V Vg e= +

Position 1 . ( )Vg 1 0=

Total spring deflection from undeflected spring position x1

x mg k

x mg k

1

1 3

0 150

0 1503 9 81

2 6 100

= +

= + =¥

+

/

/ kg m/s

N/m)

2

.

.( )( . )

( ...

. . .

( ) ( .

150

0 01132 0 150 0 1613

1

2

1

22 6 10

1

1 12 3

m

m

N/m

x

V k xe

= + =

= = ¥ ))( . ) .0 1613 33 832 m J=

Position 2 .

V

V mg h g hg

1

2

0 33 83 33 83

0 150 3 0 150

= + == + = +

. .

( ) ( . ) ( . )

J

( )Ve 2 0= (spring is not attached to the collar)

T V T V V V V V

g

g e g e1 1 2 2 1 1 2 20 0

0 0 33 83 0 3 0 150

+ = + + + = + +

+ + = +

( ) ( ) ( ) ( )

. ( . ++ +

= -

=

h

h

) 0

J

(3 kg)(9.81 m/s m)

m

2

33 830 150

0 9995

.

)( .

.

h = 1000 mm b

90



(b) Maximum velocity occurs when the acceleration = 0, i.e., at equilibrium.

At Position 3

T mv v v

V V V mg kg e

3 32 2 2

3 3 3

1

2

1

23 1 5

0 1501

2

= = =

= + = +

( ) .

( ) ( ) ( . ) (

max max

xx

V

12

33

0 150

3 9 81 0 1501

22 6 10

-

= + ¥

. )

( )( . )( . ) ( . kg m/s m N/m)(02 ..1613 0.150 m)

J J J

2-

= + =

+ = + +

V

T V T V

3

1 1 3 3

4 415 0 1660 4 581

0 33

. . .

.. . .

( . ) . .

max

max

83 1 5 4 581

29 249 1 5 19 50

2

2

= +

= =

v

v / m /s2 2 vmax .= 4 42 m/s b

91


SOLutiOn

Denote by k an equivalent spring constant.

Static deflection of beam is then yW

kst = (1)

Drop W from height h.

T V Wh T V Wy ky

T V T V Wh Wy ky

m m

m m

1 1 2 22

1 1 2 22

0 01

2

0 01

2

= = = = - +

+ = + + = - +

From Equation (1), W ky

ky h y kym m

=

+ =

st

st ( )1

22

y y y y hm m2 2 2 0- - =st st y y

h

ym = +Ê

ËÁˆ

¯̃stst

12

b

PROBLEM 13.63

It is shown in mechanics of materials that when an elastic beam AB supports a block of weight W at a given Point B, the deflection yst (called the static deflection) is proportional to W. Show that if the same block is dropped from a height h onto the end B of a cantilever beam AB and does not bounce off, the maximum deflection ym in the ensuing motion can be

expressed as y y h ym = + +st st/( ).1 1 2 Note that this formula is approximate, since it is based on the assumption that there is no energy dissipated in the impact and that the weight of the beam is small compared to the weight of the block.

92


Solution

(a) Maximum height.

Above B is reached when the velocity at E is zero.

T

T

V V V

C

E

e g

=== +

0

0

Point C. D

D

L

L

BC

BC

= ÊËÁ

ˆ¯̃

=

( . )0 36

20

m rad

m

p

p

q p= ∞ =306

rad

R = =300 mm 0.3 m

( ) ( )

( ) ( ) .

V k L

V

C e BC

C e

C e

=

= ÊËÁ

ˆ¯̃ =

=

12

12

5020

0 61685

2

2

D

N/m m J

( ) m

p

V gg kg m/s m2R

VC g

( )( . )( . )( . )( cos )

( ) .

1 0 25 9 81 0 3 1 30

0 09857

- - ∞cosq= 22J

Point E. ( )VE e = 0 (spring is unattached)

( ) ( . )( . )( ) .

. .

V WH H

T V T V

E g

C C E E

= = =

+ = ++ +

0 25 9 81 2 4525

0 0 61685 0 098

H

5572 0 0 2 4525= + + . H H = 0 292. m b

PRoBlEM 13.64

A thin circular rod is supported in a vertical plane by a bracket at A. Attached to the bracket and loosely wound around the rod is a spring of constant k = 50 N/m and undeformed length equal to the arc of circle AB. A 250-gm collar C, not attached to the spring, can slide without friction along the rod. Knowing that the collar is released from rest when q = 30o , determine (a) the maximum height above Point B reached by the collar, (b) the maximum speed of the collar.

93



(b) The maximum velocity is at B where the potential energy is zero, v vB = max.

T V

T mv v

T

C C

B B

B

= = + =

= =

0 0 61685 0 098572 0 715422

1

2

1

20 252 2

. . .

( . ) max

J

===

+ = + =

=

0 125

0

0 715422 0 125

5 723

2

2

2

.

. .

.

max

max

max

v

V

T V T V v

v

B

C C B B

442

2

m

s

vmax . /= 2 39 m s

94


SOLutiOn

(a) Smallest angle q occurs when the velocity at D is close to zero.

v v

T T

V V V

C D

C D

e g

= == == +

0 0

0 0

Point C.

DL

V k L

V

BC

C e BC

C e

= =

= D

= =

( . ) ( . )

( ) ( )

( ) ( )( . ) .

0 3 0 3

1212

50 0 3 2 2

2

2

m mq

q 55 2q R = =300 0 3mm m.

( ) ( cos )

( ) ( . )( . )( . ) )

V WR

V

C g

C g

= -

= -

1

0 25 9 81 0 3

q

q(1 cos

( ) . ( cos )

( ) ( ) . . ( cos )

V

V V V

C g

C C e C g

= -

= + = + -

0 73575 1

2 25 0 7375 12

q

q q

Point D.

( ) (

( ) ( ) ( . )( . )( . )

V

V W RD e

D g

== =

0

2 0 25 9 81 0 6

spring is unattached)

==

+ = +

1 4715

2 25 0 7375 12

.

. . ( cosT V T VC C D D q q- - ) =1.4715

2 25 0 73752. . cosq q- = 0.734

By trial, q = 0 7548. rad q = ∞43 2. b

PROBLEM 13.65

A thin circular rod is supported in a vertical plane by a bracket at A. Attached to the bracket and loosely wound around the rod is a spring of constant k = 50 N/m and undeformed length equal to the arc of circle AB. A 250 gm collar C, not attached to the spring, can slide without friction along the rod. Knowing that the collar is released from rest at an angle q with the vertical, determine (a) the smallest value of q for which the collar will pass through D and reach Point A, (b) the velocity of the collar as it reaches Point A.

95



(b) Velocity at A.

Point D.

V T VD D D= = =0 0 1 4715.

Point A.

T mv v

T v

V V W R

A A A

A A

A A g

= =

== = =

1

2

1

20 25

0 125

0 25 9 81

2 2

2

( . )

.

( ) ( ) ( . )( . )(( . ) .

. . .

.

0 3 0 73575

0 125 0 73575 1 4715

5 886

2

2

=

+ = +

+ =

=

T V T V

v

v

A A D D

A

A m22 2/s vA = Ø2 43. m/s b

96


SOLutiOn

L

L

L

L

AF

AF

BF

= + +

= =

= +

( . ) ( . ) ( . )

( . ) ( . )

0 5 0 4 0 3

22

0 4 0 3

2 2 2

2 2

m 0.07071 m

BBF

FE

FE

e g

L

L

V V V

=

= +

= =

= +

0 5

0 5 0 3

3410

0 5831

2 2

.

( . ) ( . )

.

m

m

(a) Speed at B. v TA A= =0 0,

Point A.

( ) ( ) . .V k L L L LA e AF AF AF= = - = -1

20 7071 0 252

0D D

DLAF = 0 4571. m

( ) ( )( . )

( ) .

( ) ( )( . )

( )( .

V

V

V

A e

A e

A g

=

==

=

1

280 0 4571

8 3576

0 4

1 9 81

2

J

mg

))( . )

.

( ) ( )

. .

.

0 4

3 924

8 3576 3 924

12 2816

== +

= +=

J

J

V V VA A e A g

PROBLEM 13.66

A 2.1 kg collar can slide along the rod shown. It is attached to an elastic cord anchored at F, which has an undeformed length of 250 mm and a spring constant of 80 N/m. Knowing that the collar is released from rest at A and neglecting friction, determine the speed of the collar (a) at B, (b) at E.

97



Point B. T mv v

T v

B B B

B B

= =

=

1

2

1

21

0 5

2 2

2

( )

.

( ) ( ) . .V k L L L LB e BF BF BF= = - = -1

20 5 0 252

0D D

DLBF = 0 25. m

( ) ( )( . .

( ) ( . ) .

( ) ( )

V

V

V v V

B e

B g

B B e B g

= =

= =

= +

1

280 0 25 2 5

0 4 3 924

) J

mg J

2

== 2 5. J+3.924J=6.424J

T V T V

v

v

A A B B

B

B

+ = +

+ = +

=

0 12 2816 0 5 6 424

11 7152

2

2

. . .

. m/s2 vB = 3 42. m/s b

(b) Speed at E.

Point A. T VA A= =0 12 2816. J [from Part (a)]

Point E.

T mv v

V k L L L L

E E E

E e FE FE FE

= =

= D D = - = -

12

0 5

12

0 5831 0 25

2 2

20

.

( ) ( ) . .

D =LFE 0 3331. m

( ) ( .

( ) .

V

V V

T V T V

E e

E g E

A A E E

= =

= =

+ = +

1

280 4 38244

0 4 38244

)(0.331) J

J

2

112 2816 0 5 4 382442. . .= +vE

vE2 = 15.79832 m /s2 2 vE = 3.97 m/s b

98


PROBLEM 13.67

The system shown is in equilibrium when f = 0. Knowing that initially f = 90o and that block C is given a slight nudge when the system is in that position, determine the speed of the block as it passes through the equilibrium position f = 0. Neglect the weight of the rod.

SOLutiOn

Find the unstretched length of the spring.

q

q

=

= ∞

= +

= =

-tan.

..

( . ) ( . )

.

1

2 2

0 3

0 171 565

0 3 0 1

1

100 31623

L

L

BD

BD m

Equilibrium SM F

F

A s

s

= - =

=

( . )( sin ) ( )( . )( . )

( )( . )( . )

( . )

0 1 12 9 81 0 6 0

12 9 81 0 6

0 1

q

((sin . )

.

71 565

744 527

∞== D

N

F k Ls BD

744 527 10 000

0 074453

. ( , )( )

.

= DD =

L

LBD

BD m

Unstretched length L L L

LBD BD0

0 0 31623 0 074453

0 241777

= -= -=

D. .

. m

Spring elongation, D ¢LBD, when f = ∞90 .

DD

¢ = + -¢ = -

=

L L

LBD

BD

( .

.

.

0 3

0 4

0 158223

0m 0.1m)

m 0.241777m

m

99



At 1 , ( )f = ∞90 v T

V V Ve g

1 1

1 1 1

0 0= == +

,

( ) ( )

( ) ( )

( ) ( , )( . )

( ) .

V k L

V

V

e BD

e

e

12

12

1

1

21

210 000 0 158223

125 173

= ¢

=

=

D

J

(( ) ( )( . )( . ) .

. .

.

V

V

g1

1

12 9 81 0 6 70 632

125 173 70 632

54 541

= - = -

==

J

J

-

At 2 , ( )f = ∞0

( ) ( ) ( , .

( ) .

( )

V k L

V

V

e BD

e

g

22 2

2

2

1

2

1

210 000 0 074453

27 7162

= =

=

D )( )

J

== =

= = =

+ = +

+

0 27 7162

1

2

1

212 6

0 54 541

2

2 22

22

22

1 1 2 2

V

T mv v v

T V T V

.

( )

.

J

== +

=

6 27 7162

4 4708

22

22

v

v

.

. m /s2 2

v2 2 11= . m/s b

100


PROBLEM 13.68

A spring is used to stop a 50-kg package, which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held by cables so that it is initially compressed 50 mm. Knowing that the velocity of the package is 2 m/s when it is 8 m from the spring and neglecting friction, determine the maximum additional deformation of the spring in bringing the package to rest.

SOLutiOn

Let Position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by the cable. Let Position 2 be the position of maximum compression. Let x be the additional compression of the spring. Use the principle of conservation of energy. T V T V1 1 2 2+ = + .

Position 1: T mv

V mghg

12

1 1

1

2

1

250 2

50 9 81 8 20

= = =

= = ∞ =

12 100 J

134

( )( )

( )( . )( sin ) 22.09 J

37.5 JV kee1 12 3 21

2

1

230 10 0 050= = ¥ =( )( . )

Position 2: T mv v

V mgh x xg

2 2

2

12

0 0

50 9 81 20

= = =

= = - ∞ =

22

2 167.76

since

( )( . )( sin ) -

37.5 150022V ke x x xe2

3 2 212

12

30 10 0 05 15 000= = ¥ + = +( )( . ) ,+

Principle of conservation of energy.

100 1342.09 37.5 167.61 37.5 1500 15,000

15,000 133

2

2

+ + = - + +

+

x x x

x

+

22.24 1442.09 0x - =

Solving for x,

x = -0 26882 0 35764. .and x = 0 269. m b

101


PROBLEM 13.69

Solve Problem 13.68, assuming the kinetic coefficient of friction between the package and the incline is 0.2.

PROBLEM 13.68 A spring is used to stop a 50-kg package, which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held by cables so that it is initially compressed 50 mm. Knowing that the velocity of the package is 2 m/s when it is 8 m from the spring and neglecting friction, determine the maximum additional deformation of the spring in bringing the package to rest.

SOLutiOn

Let Position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by the cable. Let Position 2 be the position of maximum compression. Let x be the addition compression of the spring. Use the principle of work and energy. T V U T V1 1 1 2 2 2+ + = +Æ

Position 1: T mv

V mghg

12

1

1

2

1

250 2

50 9 81 8 20

= = =

= = ∞ =

12

1

100 J

134

( )( )

( )( . )( sin ) 22.09 J

37.5 J12V kee1

3 21

2

1

230 10 0 05= = ¥ =( )( . )

Position 2: T mv v

V mgh xg

2 2

2

1

20 0

50 9 81 20

= = =

= = - ∞ =

22

2 167.76

since .

( )( . )( sin ) - xx

V ke x x xe

37.5 15002 22 3 2 21

2

1

230 10 0 05 15 000= = ¥ + = +( )( . ) ,+

Work of the friction force.

+ SFn = 0

N mg

N mg

F Nf k

N

- = cos 20∞= ∞= ∞==

0

20

50 9 81 20

460 92

cos

( )( . )cos

.

m

==== -

= - += - -

Æ

( . )( . )

.

. ( )

. .

0 2 460 92

92 184

92 184 8

737 47 92 18

1 2U F d

x

f

44x

102



Principle of work and energy.

100 +1342.09 + 37.5 - 737.47 - 92.184x

= - + + +167 76 37 5 1500 15 000 2. . ,x x x

15,000 1424.42 704.62 0x x2 + - =

Solving for x,

x = -0.17440 and 0.26936 x = 0.1744 m b

103


PROBLEM 13.70

A 300-g pellet is released from rest at A and slides with friction along the surface shown. Determine the force exerted on the pellet by the surface (a) just before the pellet reaches B, (b) immediately after it has passed through B.

SOLutiOn

Velocity at 2 .

v T V mg

V

1 1 1

12

0 0 (1.2)(sin 30 )

(3 kg)(9.81 m/s )(1.2 m)

= = =

= ÊËÁ

ˆ

∞1

2 ¯̃̄=

= = ( ) =

=

+ = +

V

T mv v v

v

T V T

1

2 22

22

22

2

1 1

(1.766 J)

1

2

1

2(3 kg) 0.15

0

2 VV v

v

2 22

22

0 1 766 0 15 0+ = +

=

. .

11.77 m /s2 2

(a) SF N g

N

= - ∞=

=Ê

ËÁˆ

¯̃=

3 30

0

m s

N

cos

( . )( . )

.

0 3 9 813

2

2 55

2kg

(b) S ∞F N g

mv

r

= -

=

=

3 30

(0.3 kg)(117.72 m /s )

(0.8 m)

2

2 2

cos

N = +=

2.55 4.41

6.96 N N = 6.96 N b

104


PROBLEM 13.71

A 300-g pellet is released from rest at A and slides without friction along the surface shown. Determine the force exerted on the pellet by the surface (a) just before the pellet reaches C, (b) immediately after it has passed through C.

SOLutiOn

Velocity at C.

v TA A= =0 0

V

V

T mv

T v

A

A

C C

C C

==

=

=

( . )( . )( . )0 3 9 81 0 7072 kg m/s m

2.081 J

1

2

0.15

2

2

22

2

2 2 2

0

0 2.081 0.15 0

13.873 m /s

V

T V T V

v

v

C

A A C C

C

C

=+ = +

+ = +

=

(a) Normal force just before C.

Pellet is in the curve a v rn C=( )2 / .

+

SF mv rC= 2 /

N mgmv

r

N m gv

r

N

C

C

- =

= +Ê

ËÁˆ

¯̃

=

2

22 2

(0.3 kg) 9.81 m/s +13.873 m /s

(

2

00.8 m)

Ê

ËÁˆ

¯̃

N = +(0.3)(9.81 17.34) N = 8.15 N b

(b) Normal force just after C.

Pellet is on the straight section of the surface ( 0).an =

+ SF N mg= - = 0

N g= 0 3. N = 2.94 N b

105


PROBLEM 13.72

A 600 gm collar can slide without friction along the semicircular rod BCD. The spring is of constant 1.360 N/m and its undeformed length is 200 mm. Knowing that the collar is released from rest at B, determine (a) the speed of the collar as it passes through C, (b) the force exerted by the rod on the collar at C.

SOLutiOn

L

L

AB

AC

= + + =

= + =

(0.3) (0.15) (0.075) 0.34369m

(0.075) 0.

2 2 2

2 2( . )0 3 330923m

N/mk = 360

(a) Speed at C.

At B: v TB B= =0 0

V V V

L

L

V k L

B B e B g

AB

AB

B e AB

= +

= -=

=

( ) ( )

.

( ) ( )

DD

D

0.34369 0.2

m0 14369

1

222

21

2360 0 14369

3 71643

( ) ( )( . )

( ) .

V

V

B e

B e

=

= J

( )

( ) ( ) ( )

V Wr

V V V

B g

B B e B g

= = =

= + =

(0.6)(9.81)(0.15) 0. J

3.71643

8829

++ =0.8829 J4 59933.

At C: T mv v v

V k L

L

C C C C

C e AC

AC

= = =

=

= -

1

2

1

22 2 2( . ) .

( ) ( )

. .

0 6 0 3

1

20 30923 0

2D

D 22

0 10923= . m

106



( ) =1

2(360)(0.10923) . J

( ) 0

( ) ( ) ( ) .

2V

V

V V V

C e

C g

C C e C g

=

=

= + =

214761

2114761

0

8 172422

2

J

4.59933 0.3 2.14761

m

s

2

T V T V

v

v

B B C C

C

C

+ = +

+ = +

= . vC = 2.86m/s b

(b) Force of rod on collar at C.

F

F F

k L

z

x y

e AC

== +

= =

=

-

0 (no friction)

tan300

14.04

( )(c

F i j

F

q 1 75 ∞

D oos sin )

( )( )(cos 14.04 sin 14.04 )

38.

q qi k

F i k

F

+= ∞ + ∞=

e

e

360 0 10923.

11481 9.5397 (N)i k+

SF i j k

j k

= + + - +

= +

+ =

( 38.1481) ( 5.886) 9.5397

38.1481 0

2

F F

mv

rma

F

x y

x FF

F

F

y

x

y

= +

= -=

5.886(0.6)(8.1724)

N

38.5756 N

( . )

.

0 15

38 148

F i j= - +(38.1N) (38.6 N) b

107


PROBLEM 13.73

A 500 g collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 125 mm and a constant k = 150 N/m . Knowing that the collar is released from being held at A, determine the speed of the collar and the normal force between the collar and the rod as the collar passes through B.

SOLutiOn

For the collar, m = 0 5. kg

For the spring, k O= =150 N/m ml 0 125.

At A: l

l l

A

A O

= + =- = =

0 175 0 25 0 425

0 425 0 3

. . .

. .

m

m 0.125m m-

At B: l

l l

B

B O

= + + =- = =

( . . ) ( . ) .

. . .

0 175 0 125 0 125 0 325

0 325 0 125 0 2

2 2 m

m-

Velocity of the collar at B.

Use the principle of conservation of energy.

T V T VA A B B+ = +

where T mvA A= =1

202

V k W

T mv

A A O

B B

= - +

= + =

= =

12

( )

J

2l l ( )

( )( . ) .

( . )

0

12

150 0 3 0 6 75

12

12

0 5

2

2 vv v

V k Wh

B B

B B O

2 20 25=

= - +

= +

.

12

( )

12

(150)(0.2) (0.5)(9.81)( 0

2

2

l l

- ..125)

J

0 6.75 0.25 2.38688

17.4525m

s

2

22

2

=

+ = =

=

2 38688.

v

v

B

B vB = ¨4 18. m/s b

108



Forces at B.

F k

mamv

s B O

nB

= - = =

=

=

=

=

( ) ( )( . )

.

. (

l l 150 0 2 30

0 125

0 5 17

2

N

sin5

13m

a

r

r.. )

..

4525

0 12569 81= N

+ SF ma F W N may y s n= - + =: sin b

N ma W F

N

n s= + -

= + - ÊËÁ

ˆ¯̃

=

sin

.

a

69.81 (0.5)(9.81) 305

13

N63 1765 N = ≠63 2. N b

109


PROBLEM 13.74

A 200-g package is projected upward with a velocity v0 by a spring at A; it moves around a frictionless loop and is deposited at C. For each of the two loops shown, determine (a) the smallest velocity v0 for which the package will reach C, (b) the corresponding force exerted by the package on the loop just before the package leaves the loop at C.

SOLutiOn

(1) (a) The smallest velocity at B will occur when the force exerted by the tube on the package is zero.

m/s m

m /s

2

2 2

SF mgmv

r

v gr

v

B

B

B

= + =

= =

=

0

9 81 0 5

4 905

2

2

2

( . )( . )

.

At A: T mv VA A= =1

200

2

At B: T mv mB B= = =1

24 905 2 4532 1

2m( . ) .

V mg mg

T V T V

mv mg

v

B

A A B B

= + =+ = +

+ = +

=

12

0 2.453 m

2[(2.

( . . )2 5 0 5 3

302

02 4453 J 63.77) ( . )+ =3 9 81 v0 7 99= . m/s b

At C: T mv V mg

T V T V

mv mv mg

v

C C C

A A C C

C

C

= =

+ = +

+ = +

=

12

2 5

12

012

2 5

6

2

02 2

2

( . )

.

[

m

33 77 5 0 9 81

14 722

. ( . )( . )]

.

-

=vC m /s2 2

(b) + kg m /s

mSF N

mv

rCC= = =2 2 20 2 14 72

0 5

( . )( . )

( . )

Package on tube, NC = ¨5 89. N b

+

110



(2) (a) The velocity at B can be nearly equal to zero since the weight of the package is supported by the tube.

Thus, v TB B= =0 0,

V mg

V mg

T mv V

T V T V mg mv

B

B

A A

B B A A

= +=

= =

+ = + + =

( . . )2 5 0 5

3

12

0

0 312

02

m m

002

02

0

6

+

=v g

v0 7 67= . m/s b

(b) T mv V mg

T V T V mv mv mg

v

C C C

A A C C C

C

= =

+ = + + = +

=

12

2 5

12

012

2 5

6

2

02 2

2

( . )

.

m

gg g- =5 9 81. m /s2 2

+ SF Nmv

rCC= =2

NC = ( . )( . ) ( . )0 2 9 81 0 5 kg m/s / m2

Package on tube, NC = ¨3 92. N b

111


PROBLEM 13.75

If the package of Problem 13.74 is not to hit the horizontal surface at C with a speed greater than 3.5 m/s, (a) show that this requirement can be satisfied only by the second loop, (b) determine the largest allowable initial velocity v0 when the second loop is used.

SOLutiOn

(a) Loop (1). The smallest allowable velocity at B will occur when the force exerted by the tube on the package is zero

+ SF mg mv r

v gr

v

B

B

B

= + =

= = ==

0

9 81 0 5 4 905

2 21

2

2 2 2 2

/

m/s m m /s( . )( . ) .

. 55 m/s

The velocity at B cannot be less than 2.215 m/s if the package is to maintain contact with the tube. For vC to be as small as possible, vB must be as small as possible; that is, vB = 2 215. m/s.

At B: T mvB B= =1

2

1

22 2152 2m( . )

T

V mgB

B

== + =

2 453

2 5 0 5

.

( . . )

m

3 mg

At C: T mv

V mg

T V T V

mg mv mg

v

C C

C

B B C C

C

C

=

=+ = +

+ = +

=

122 5

2 453 312

2 5

2 2

2

2

2

.

. .

[

m

.. . ( . )]

.

453 0 5 9 81

14 72

2

2

+

=

m/s

m /s2 2vC

vC = 3 836 3 5. . m/s m/s� b

Thus, Loop (1) cannot meet the requirement.

112



(b) Loop (2).

At A: T mv

V

A

A

=

=

1

20

02

At C: v

T

T

v mg

TA

T V

mv

C

C

C

C

A C C

=

=

==

+ = +

+

3 5

12

3 5

6 125

2 5

1

12

2

02

.

( . )

.

.

m/s

m

m

00 6 125 2 5= +. .m mg

v g02 2 6 125 2 5 61 3= + =( . . ) . m /s2 2 v0 7 83= . m/s b

Note: A larger velocity at A would result in a velocity at C greater than 3.5 m/s.

113


PROBLEM 13.76

The 1 kg ball at A is suspended by an inextensible cord and given an initial horizontal velocity of 5 m/s. If l = 600 mm and xB = 0, determine yB so that the ball will enter the basket.

SOLutiOn

Let Position 1 be at A. v v1 0=

Let Position 2 be the point described by the angle where the path of the ball changes from circular to parabolic. At Position 2, the tension Q in the cord is zero.

Relationship between v2 and q is based on Q = 0. Draw the free body diagram.

+

SF Q mg mamv

ln= + = =0 22

: sinq

With Q = 0, v gl v gl22

2= =sin sinq qor (1)

Relationship among v v0 , 2 and q is based on conservation of energy.

T V T V

mv mgl mv mgl

1 1 2 2

02

221

2

1

2

+ = +

- = + sinq

v v mgl02

22 2 1- = +( sin )q (2)

114



Eliminating v2 from Eqs. (1) and (2),

v gl mgl

v

gl

02

02 2

2 1

3 25

9 81 0 62

48

- = +

= - = -

=

sin ( sin )

sin( )

( . )( . )

q q

q

q ..514∞

From Eq. (1), v

v22

2

9 81 0 6 48 514

2 0998

= ∞=

( . )( . )sin( . )

. m/s

x and y coordinates at Position 2.

x l

y l2 0 6 48 514 0 39746

0 6 48 514

= = ∞ == = ∞ =

cos . cos( . ) .

sin . sin( . )

qq

m

2 00 44947. m

Let t2 be the time when the ball is at Position 2.

Motion on the parabolic path. The horizontal motion is

&x v

x x t t

= - = -= -

= - -

2

2 2

2 0998 48 514

1 5730

1 573

sin . sin( . )

.

. ( )

qm/s

At Point B, xB = 0

0 0 39746 1 573 0 252682 2= - - - =. . ( ) .t t t tB B s

The vertical motion is

y y v t t g t t= + - - -2 2 2 221

2cos ( ) ( )q

At Point B, y y v t t g t tB B B= + - - -2 2 2 221

2cos ( ) ( )q

yB = +

=

0 44947 2 0998 48 514 0 252681

29 81 0 25268

0

2. . cos( . )( . ) ( . )( . )-

..48777m yB = 0 488. m b

115


SOLutiOn

Let Position 1 be at A. v v1 0=

Let Position 2 be the point described by the angle where the path of the ball changes from circular to

parabolic. At Position 2, the tension Q in the cord is zero.

Relationship between v2 and q is based on Q = 0. Draw the free body diagram.

+

SF Q mg mamv

n= + = =0 22

: sinql

With Q = 0, v g v g22

2= =l lsin sinq qor (1)

Relationship among v v0 2, , and q is based on conservation of energy.

T V T V

mv mg mv mg

1 1 2 2

02

221

2

1

2

+ = +

- = +l lsinq

v v g02

22 2 1= + +l( sin )q (2)

PROBLEM 13.77*

The 1-kg ball at A is suspended by an inextensible cord and given an initial horizontal velocity of v0. If l = 600 mm xB = 90 mm and yB = 120 mm, determine the initial velocity v0 so that the ball will enter the basket.

116


PROBLEM 13.77* (continued)

x and y coordinates at Position 2.

x2 = lcosq (3)

y2 = lsinq (4)

Let t2 be the time when the ball is in Position 2.

Motion on the parabolic path. The horizontal motion is

&x v= - 2 sinq

x x v t t= - -2 2 2( sin )( )q (5)

At Point B, x x t tB B= =and . From Eq. (5),

( )cos

sint t

x

vBB- =

-2

2

l qq

(6)

Vertical motion: &y v g t t= - -2 2cos ( )q

y y v t t g t t= + - - -2 2 2 221

2( cos )( ) ( )q

At Point B,

y v t t g t tB B B= + - - -lsin ( cos )( ) ( )q q2 2 221

2 (7)

Data: l = = = =0 6 0 09 0 12 9 81. . . .m, m, m, m/s2x y gB B

With the numerical data,

Eq. (1) becomes v2 9 81 0 6 5 886= =( . )( . )sin . sinq q (1)¢

Eq. (6) becomes t tvB - = -

22

0 6 0 09. cos .

sin

qq

(6)¢

Eq. (7) becomes y v t t t tB B B= + - - -0 6 4 9052 2 22. sin ( cos )( ) . ( )q q (7)¢

Method of solution. From a trial value of q, calculate v2 from Eq. (1)¢, t tB - 2 from Eq. (6)¢, and yB from Eq. (7)¢. Repeat until yB = 0.12 m as required.

Try q = ∞30 .

117



Use trial and error (it may be useful and convenient to use an excel worksheet) we get q = 33.552°

Substituting q

q

==

= + +

=

33 552

1 80365

2 1

1 803

2

02

22

.

.

( sin )

( .

∞V

V V gl

m/s in Eq. (2)

665 2 9 81 0 6 1 33 552

21 5315

2) . . ( sin . )

.

+ +

=

¥ ¥ ∞

m /s2 2 V0 4 64= . m/s

118


SOLutiOn

(a) At Point A: v T

V mghA A

A

= ==

0 0

At any Point P: T mvP = 1

22

V Wy mgy

T V T V

mgh mv mgy

v g h y

P

A A P P

= =+ = +

+ = +

= -

01

2

2

2

2 ( )

en along principal normal, horizontal and directed toward y axis

et tangent to centerline of the chute

eD along binormal

bp p

b= = = ∞

=

- -tan tan .1 1

2

613 427

0

h

R

mab

m

2 (4m)

since ab = 0

Note: Friction is zero.

SS

S

F ma mg ma a g

F ma N W N W

F ma N

t t t t

b b b B

n n

= = == - = =

=

sin sin

cos cos

b bb b0

nnmv m g h y

Wh y= = - = -2 2

2r r r

( ) ( )

PROBLEM 13.78*

Packages are moved from Point A on the upper floor of a warehouse to Point B on the lower floor, 6 m directly below A, by means of a chute, the centerline of which is in the shape of a helix of vertical axis y and radius R = 4 m. The cross section of the chute is to be banked in such a way that each package, after being released at A with no velocity, will slide along the centerline of the chute without ever touching its edges. Neglecting friction, (a) express as a function of the elevation y of a given Point P of the centerline the angle f formed by the normal to the surface of the chute at P and the principal normal of the centerline at that point, (b) determine the magnitude and direction of the force exerted by the chute on a 10 kg package as it reaches Point B. Hint: The principal normal to the helix at any Point P is horizontal and directed toward the y axis, and the radius of curvature of the helix is r = R[1 + (h/2pR)2].

119



The total normal force is the result of Nb and Nn, lies in the b–m plane, and forms angle f with m axis.

tan

tan cos( ( ))

tan ( ( ))cos

f

f br

f r b

=

= -

= -

N N

WW h y

h y

b n/

/

2

2

Given: r = + ÊËÁ

ˆ¯̃

È

ÎÍÍ

˘

˚˙˙

= + =Rh

RR

R1

21

22

2pb

b( tan )

cos

Thus, tan( )

cos( )cos

tan( )cos .

.

f bb

f

=-

=-

=- ∞

=

r2 2

42 6 14 327

2 0642

h y

R

h y

y

m

66 - y

or f =ÊËÁ

ˆ¯̃

tan.-

-1 2 0642

6 y b

(b) At Point B: y = 0. For x, y, z axes, we write, with W = log = 98.1 N,

N N W Nx b x= = = ∞ ∞ =sin cos sin ( . )cos . sin . .b b b 98 1 14 327 14 327 23 5204N N

NN N W N

N N Wh

y b y

z n

= = = ∞ =

= - = -

cos cos ( . )cos . .b b2 298 1 14 327 92 0929

2

N N

-- = - -

= - - ∞ = -

yW

h y

R

N Nz z

r b2

2 98 16

14 327 276 2

2

2

/cos

( . )( )

cos ( . ) .N0

47787N

N = + + - =( . ) ( . ) ( . ) .23 5204 92 0929 276 2787 292 17162 2 2 N N = 292N b

cos.

.qx

xN

N= = 23 5204

292 1716 qx = ∞85 4. b

cos.

.q y

yN

N= = 92 0929

292 1716 q y = ∞71 6. b

cos.

.q z

zN

N= = - 276 2787

292 1716 q z = ∞161 0. b

120


PROBLEM 13.79*

Prove that a force F(x, y, z) is conservative if, and only if, the following relations are satisfied:

∂F

y

F

x

F

z

F

y

F

x

F

zx y y z z x

∂∂∂

∂∂

∂∂

∂∂

∂∂

= = =

SOLutiOn

For a conservative force, Equation (13.22) must be satisfied.

FV

xF

V

yF

V

zx y z= - = - = -∂∂

∂∂

∂∂

We now write ∂∂

∂∂ ∂

∂∂

∂∂ ∂

F

y

V

x y

F

x

V

y xx y= - = -

2 2

Since ∂∂ ∂

∂∂ ∂

2 2V

x y

V

y x= :

∂∂

∂∂

F

y

F

xx y= b

We obtain in a similar way

∂∂

∂∂

∂∂

∂∂

F

z

F

y

F

x

F

zy z z x= = b

121


PROBLEM 13.80

The force F i j k= + +( )yz zx xy xyz/ acts on the particle P x y z( , , ), which moves in space. (a) Using the relation derived in Problem 13.79, show that this force is a conservative force. (b) Determine the potential function associated with F.

SOLutiOn

(a) Fyz

xyzF

zx

xyzx y= =

∂∂

∂∂

∂∂

∂

∂F

y y

F

x xx x y y=

( )= =

( )=

1 1

0 0

Thus, ∂∂

∂∂

F

y

F

xx y=

The other two equations derived in Problem 13.79 are checked in a similar way.

(b) Recall that Fv

xF

v

yF

v

zx y z= - = - = -∂∂

∂∂

∂∂

, ,

Fx

v

xV x f y zx = = - = - +1 ∂

∂ ln ( , ) (1)

Fy

v

yV y g z xy = = - = - +1 ∂

∂ ln ( , ) (2)

Fz

v

zV z h x yz = = - = - +1 ∂

∂ ln ( , ) (3)

Equating (1) and (2)

- + = - +ln ( , ) ln ( , )x f y z y g z x

Thus, f y z y k z( , ) ln ( )= - + (4)

g z x x k z( , ) ln ( )= - + (5)

Equating (2) and (3)

- + = - +ln ( , ) ln ( , )z h x y y g z x

g z x z l x( , ) ln ( )= - +

From (5),

g z x x k z( , ) ln ( )= - +

122



Thus,

k z z( ) ln= -

l x x( ) ln= -

From (4),

f y z y z( , ) ln ln= - -

Substituting for f y z( , ) in (1)

V x y z= - - -ln ln ln

V xyz= - ln b

123


SOLUTION

(a) F ky FF

yk

F

xx yx y= = = =0 0

∂∂

∂∂

Thus, ∂∂

∂∂

F

y

F

xx y= F is not conservative.

U d ky dy ky dx dy ky dxABCAABCA

A

B

B

C

B

A= ◊ = ◊ + ◊ + + ◊Ú Ú Ú ÚF r i j i i j i j( )

=Ú 0A

B, F is perpendicular to the path.

ky dx dy ky dxB

C

B

Ci i j◊ + =Ú Ú( )

From B to C, the path is a quarter circle with origin at A.

Thus, x y a

y a x

2 2 2

2 2

+ =

= -

Along BC, kydx k a x dxka

B

C a

Ú Ú= - =2 22

0 4

p

ky dx y CAC

Ai jÚ ◊ = =0 0( )on

Uka

ABCA A

B

B

C

C

A= + + = + +Ú Ú Ú 0

40

2p U

kaABCA = p 2

4 b

(b) F k F kxF

yk

F

xkx y y

x y= = = =∂∂

∂∂

∂∂

∂∂

F

y

F

xx y= F is conservative.

Since ABCA is a closed loop and F is conservative, U ABCA = 0 b

PROBLEM 13.81*

A force F acts on a particle P(x, y) which moves in the xy plane. Determine whether F is a conservative force and compute the work of F when P describes in a clockwise sense the path A, B, C, A, including the quarter circle x y a2 2 2+ = , if (a) F i= ky , (b) F i j= +k y x( ).

124


SOLUTION

(a) PV

x

x y z

xx x y zx = - = -

- + += + + -∂

∂∂

∂( )

( )2 2 2 1 2

2 2 2 1 2/

/ b

PV

y

x y z

yy x y zy = - = -

- + += + + -∂

∂∂

∂( )

( )2 2 2 1 2

2 2 2 1 2/

/ b

PV

z

x y z

zz x y zz = - = -

- + += + + -∂

∂∂

∂( )

( )2 2 2 1 2

2 2 2 1 2/

/ b

(b) U U U UOABD OA AB BD= + +

O–A: Py and Px are perpendicular to O–A and do no work.

Also, on O–A x y Pz= = =0 1and

Thus, U P dz dz aO A z

a a

- = = =Ú Ú0 0

A–B: Pz and Py are perpendicular to A–B and do no work.

Also, on A–B y z a= =0, and Px

x ax =

+( ) /2 2 1 2

Thus, Uxdx

x a

a

A B

a

- =+

= -( )Ú ( )2 2 1 20

2 1

/

B–D: Px and Pz are perpendicular to B–D and do no work.

On B D- , x a

z a

Py

y ay

==

=+( ) /2 2 1 22

PROBLEM 13.82*

The potential function associated with a force P in space is known to be V x y z x y z( , , ) ( ) .= - + +2 2 2 1 2/ (a) Determine the x, y, and z components of P. (b) Calculate the work done by P from O to D by integrating along the path OABD, and show that it is equal to the negative of the change in potential from O to D.

125



Thus, Uy

y ady y a

U a a a

BD

a a

BD

=+

= +

= + -

Ú ( )( )

( ) ( )

2 2 1 20

2 2 1 2

0

2 2 1 2 2 1

22

2 2

//

/ /22 3 2= -( )= + +- - -

a

U U U UOABD O A A B B D

= + -( ) + -( )a a a2 1 3 2 U aOABD = 3 b

D = -

= - + + -

V V a a a V O O O

a a a O

OD ( , , ) ( , , )

( )2 2 2 1 2/ D = -V aOD 3 b

Thus, U VOABD OD= - D

126


PROBLEM 13.83*

(a) Calculate the work done from D to O by the force P of Problem 13.82 by integrating along the diagonal of the cube. (b) Using the result obtained and the answer to Part b of Problem 13.82, verify that the work done by a conservative force around the closed path OABDO is zero.

PROBLEM 13.82* The potential function associated with a force P in space is known to be V(x, y, z) = - + +( ) .x y z2 2 2 1 2/ (a) Determine the x, y, and z components of P. (b) Calculate the work done by P from O to D by integrating along the path OABD, and show that it is equal to the negative of the change in potential from O to D.

SOLUTION

From solution to (a) of Problem 13.82

Pi j k

=+ +

+ +x y z

x y z( )2 2 2 1 2/

(a) U dOD O

D= ◊Ú P r

r i j k

r i j k

Pi j k

= + += + +

=+ +

+ +

x y z

d dx dy dz

x y z

x y z( )2 2 2 1 2/

Along the diagonal. x y z= =

Thus, P ◊ = =drx

x

3

33

2 1 2( ) /

U dx aO D

a

- = =Ú 3 30

U aOD = 3 b

(b) U U UOABDO OABD DO= +

From Problem 13.82

U aOABD = 3 at left

The work done from D to O along the diagonal is the negative of the work done from O to D.

U U aDO OD= - = - 3 [see Part (a)]

Thus,

U a aOABDO = - =3 3 0 b

127


PROBLEM 13.84*

The force F i j k= + + + +( )x y z x y z/ ( ) /2 2 2 3 2 acts on the particle P x y z( , , ) , which moves in space. (a) Using the relations derived in Problem 13.79, prove that F is a conservative force. (b) Determine the potential function V(x, y, z) associated with F.

SOLUTION

(a) Fx

x y z

Fy

x y z

x

y

=+ +

=+ +

( )

( )

/

/

2 2 2 3 2

2 2 2 12

∂∂

∂∂

F

y

x y

x y z

F

x

y y

x y z

x

y

=-( )

+ +

=-( )

+ +

32

2 2 2 5 2

32

2 2 2 5 2

2

2

( )

( )

( )

/

/

Thus, ∂∂

∂∂

F

y

F

xx y=

The other two equations derived in Problem 13.79 are checked in a similar fashion.

(b) Recalling that FV

xF

V

yF

V

z

FV

xV

x

x y zdx

V x

x y z

x

= - = - = -

= - = -+ +

=

Ú

∂∂

∂∂

∂∂

∂∂

, ,

( )

(

/2 2 2 3 2

22 2 2 12+ + +-y z f y z) ( , )/

Similarly integrating ∂ ∂V y/ and ∂ ∂V z/ shows that the unknown function f x y( , ) is a constant.

Vx y z

=+ +

12 2 2 12( ) /

b

128


PROBLEM 13.85

While describing a circular orbit 300 km above the earth, a space vehicle launches a 3600-kg communications satellite. Determine (a) the additional energy required to place the satellite in a geosynchronous orbit at an altitude of 35,770 km above the surface of the earth, (b) the energy required to place the satellite in the same orbit by launching it from the surface of the earth, excluding the energy needed to overcome air resistance. (A geosynchronous orbit is a circular orbit in which the satellite appears stationary with respect to the ground.)

SOLUTION

For any circular orbit of radius r, the total energy

E T V mvGMm

r= + = -1

22

M = mass of the earth

m = 3600 kg = satellite mass

Newton’s second law

F maGMm

r

mv

rv

GM

rn= = =:2

22

T mv mGM

rV

GMm

r

E T VGMm

r

GMm

r

GMm

r

GM gR Eg

E

= = = -

= + = - = -

= = -

1

2 21

2

1

2

1

2

2

2 RR m

r

Er

Er

E2

2 3 2

15

1

2

9 81 6370 10 3600

716 15 10

= -¥

= -¥

( . )( ) ( )

.(

m/s m kg

NN m◊ )

For a geosynchronous orbit. ( . )

.

.

.

r

EGS

26

15

6

9

42 140 10

716 10

42 140 10

17 003 10

17 0

= ¥

= -¥

¥

= - ¥= -

m

J

003 GJ

129



(a) At 300 km ( . ).r166 67 10= ¥ m

E300

15

6

9

716 10

6 67 10

107 42 10

107 42

= -¥¥

= - ¥= -

.

.

.

J

GJ

Additional energy D = -E E EGS300 300

D = - +E300 17 003 107 42. . D =E300 90 4. GJ b

(b) Launch from the earth ( ).RE = 6370 km

At launch pad, E VGMm

R

gR m

R

E

E

EE

E

E

E

E

= = - = -

= - ¥

= -

2

2 39 81 6370 10 3600

2

( . )( )( )m/s m kg

224 96 10 224 969. .¥ = -J GJ

Additional energy D = -E E EE GS S

D = - +EE 17 003 224 96. . D =EE 208 GJ b

130


SOLUTION

Total energy per unit mass for the orbit.

E T V

T V

E

m

v GM

r

v GM

r

A A

P P

A

A

P

P

0

02 2

2 2

= += +

= - = - (1)

vv

vGM

r rAP

A A P

22

21 2

1 1-Ê

ËÁˆ

¯̃= -

ÊËÁ

ˆ¯̃

v

v

r

rA

P

P

A

= (given)

vr

rGM

r r

r r

vr r r r

r

AA

P

P A

A P

AP A P A

P

22

2

22

1 2-Ê

ËÁˆ

¯̃=

-ÊËÁ

ˆ¯̃

- +=

( )( )22GM

r r

r rP A

A P

( )-

v GMr

r r rAP

A P A

2 21=+

ÊËÁ

ˆ¯̃

(2)

PROBLEM 13.86

A satellite is to be placed in an elliptic orbit about the earth. Knowing that the ratio v vA P/ of the velocity at the apogee A to the velocity at the perigee P is equal to the ratio r rP A/ of the distance to the center of the earth at P to that at A, and that the distance between A and P is 80,000 km, determine the energy per unit mass required to place the satellite in its orbit by launching it from the surface of the earth. Exclude the additional energy needed to overcome the weight of the booster rocket, air resistance, and maneuvering.

131



Substituting vA in (2) in (1)

E

mGM

r

r r r

GM

r

E

mGM

r

r r r

r r

P

A P A A

A

P P A

P A

0

0

1

1

=+

ÊËÁ

ˆ¯̃

-

=- +

+È

ÎÍ

˘

˚˙ =

( )--

+

= = ¥

GM

r r

GM gR

P A

E2 2 39 81 6370 10( . )( )m/s m

r rP A+ = ¥80 000 103, m (given)

E

m

E

m

02 3 2

3

0 6

9 81 6370 10

80 000 10

4 9765 10

= -¥

¥

= ¥

( . )( )

,

.

m/s m

m

N-m/kg == - 4 9765. MJ/kg

Total energy per unit mass on the earth.

E T V v T VmGM

R

E

m

gR

R

E E E E E EE

E E

E

= + = = = -

= - = - ¥

0 0

9 81 6370 102

2( . )(m/s 33

662 490 10 62 49

m

N-m/kg MJ/kg

)

. .E

mE = - ¥ = -

Energy per unit mass needed for propulsion,

E

m

E

m

E

mE

m

P E

P

= -

= - +

0

4 9765 62 490. .MJ/kg MJ/kg E

mP = 57 5. MJ/kg b

132


PROBLEM 13.87

Knowing that the velocity of an experimental space probe fired from the earth has a magnitude vA = ¥36 103 km/h at Point A, determine the velocity of the probe as it passes through Point B.

SOLUTION

At A: vA = ¥ =36 10 10 0003 km/h m/s,

T m

m

VGMm

r

gR m

r

r

A

AA A

A

=

= ¥

= - = -

= = ¥

12

10 000

50 10

10370 10 37 1

2

6

2

( )

km

,

. 00

6370 6 37 10

9 81 6 37 10

10 37 10

6

6

2 6 2

m

km m

m/s m

R

Vm

A

= = ¥

= - ¥¥

.

( . )( . )

( . 66

638 3857 10

m)

.= - ¥ m

At B: T mvB B= 1

22

VGMm

r

gR m

r

r

V

BB B

B

B

=-

=-

= = ¥

=-

2

6

2

18 870 18 87 10

9 81 6 37

, .

( . )( .

km m

m/s ¥¥¥

= - ¥

10

18 87 10

21 0948 10

6 2

6

6

m

m

)

( . )

.

m

V mB

133



T V T V

m m mv

v

A A B B

B

B

+ = +

¥ - ¥ = - ¥

=

50 10 38 3857 101

221 0948 10

2 50

6 6 2 6

2

. .

[ ¥¥ - ¥ + ¥

= ¥ - +

=

10 38 3857 10 21 0948 10

2 10 50 38 3857 21 0948

6

6 6 6

6

. . ]

( . . )

55 4182 10

8088 152 29 117 10

62

2

3

.

. .

¥

= = ¥

m

s

m/s km/hvB vB = ¥29 1 103. km/h b

134


PROBLEM 13.88

A lunar excursion module (LEM) was used in the Apollo moon-landing missions to save fuel by making it unnecessary to launch the entire Apollo spacecraft from the moon’s surface on its return trip to earth. Check the effectiveness of this approach by computing the energy per kilogram required for a spacecraft (as weighed on the earth) to escape the moon’s gravitational field if the spacecraft starts from (a) the moon’s surface, (b) a circular orbit 80 km above the moon’s surface. Neglect the effect of the earth’s gravitational field. (The radius of the moon is 1740 km and its mass is 0.0123 times the mass of the earth.)

SOLUTION

Note: GM GMmoon earth= 0 0123.

By Eq. 12.30, GM gREmoon = 0 0123 2.

At • distance from moon: r v2 2 0= • =, assume

E T V

GM mM

2 2 2

0

0 0

0

= +

= -•

= -=

(a) On surface of moon. RM = = ¥1740 1 74 106km m.

v T1 10 0= = RE = = ¥6370 6 37 106km m.

VGM m

RM

M1 = - E T V

gR m

RE

M1 1 1

2

00 0123

= + = -.

Em

1

2 6 2

6

0 0123 9 81 6 37 10

1 74 10=

- ¥¥

( . )( . )( . )

( . )

m/s m

WE = Weight of LEM on the earth

E m

E E E

16

2

2

2 1

62

2

2 81387 10

2 81387 10

= - ¥Ê

ËÁˆ

¯̃D = -

= + ¥Ê

ËÁˆ

¯̃

.

.

m

s

m

sm

Energy per kg: DE

m= ¥2 81 106.

Jkg

b

135



(b) r R

rM1

1

6

80

1740 80 1820

1 82 10

= += + =

= ¥

km

km km

m

( )

.

Newton’s second law:

F maGM m

rm

v

rnM= =:

12

12

1

vGMm

rT mv m

GM

r

VGM m

r

E T VGM m

r

GM

M

M

M M

12

11 1

2

1

11

1 1 11

1

2

1

2

1

2

= = =

= -

= + = -mm

r

EGM m

r

gR m

r

E

M E

1

11

2

1

1

2

1

2

1

2

0 0123

1

2

0 0123 9 81

= - = -

= -

.

( . )( . )m/s (( . )

( .

( )

( .

6 37 10

1 82 10

0 1 3451

6 2

6

2 1

¥¥

¥

D = - = +

m

) m

m

m

E E E

= -1.3451 106

¥¥ 106 m) J

Energy per kg: D = ¥E

m1 35 106.

J

kg b

136


PROBLEM 13.89

A satellite of mass m describes a circular orbit of radius r about the earth. Express (a) its potential energy, (b) its kinetic energy, (c) its total energy, as a function of r. Denote the radius of the earth by R and the acceleration of gravity at the surface of the earth by g, and assume that the potential energy of the satellite is zero on its launching pad.

SOLUTION

(a) Potential energy VGMm

r

gR m

r= - = - +

2

constant

(cf. Equation 13.17)

Choosing the constant so that V r R= =0 for :

V mgRR

r= -Ê

ËÁˆ¯̃

1 b

(b) Kinetic energy


F maGMm

rm

v

rn= =:2

2

vGM

r

gR

r

T mv

2

2

21

2

=

=

= TmgR

r= 1

2

2

b

Energy

(c) Total energy

E T V mgR

rmg

R

r= + = + -Ê

ËÁˆ¯̃

1

21

2

E mgRR

r= -Ê

ËÁˆ¯̃

12

b

137


PROBLEM 13.90

How much energy per kilogram should be imparted to a satellite in order to place it in a circular orbit at an altitude of (a) 600 km, (b) 6000 km?

SOLUTION

Before launching: r R v

E T VGMm

R

gR m

RmgR

16

1

1 1 1

2

6 37 10 0

0

= = ¥ =

= + = - = - = -

. ;m

In circular orbit of radius r2. [cf. Eq. 12.30]


F maGMm

rm

v

rn= =:22

22

2

vGM

r

gR

r

E T V mvGMm

r

E mgR

r

gR m

r

22

2

2

2

2 2 2 22

2

2

2

2

2

2

1

2

1

2

1

= =

= + = -

= - = -22

2

2

gR m

r

Energy imparted is D = -

= - - -

= -ÊËÁ

ˆ¯̃

E E E

gR m

rmgR

RmgR

r

2 1

2

2

2

1

2

12

( )

Energy per kg is D = -ÊËÁ

ˆ¯̃

E

mRg

R

r1

2 2

(a) r2 6370 600 6970= + = km

D = ¥ -ÊËÁ

ˆ¯̃

E

m( . )( . )

( )

( )6 37 10 9 81 1

6370

2 69706 D =E

m33 9. MJ/kg b

(b) r2 6370 6000 12 370= + = , km

D = ¥ -ÊËÁ

ˆ¯̃

E

m( . )( . )

( , )6 37 10 9 81 1

6370

2 12 3706 D =E

m.46 4 MJ/kg b

138


PROBLEM 13.91

(a) Show that, by setting r R y= + in the right-hand member of Eq. (13.17¢) and expanding that member in a power series in y/R, the expression in Eq. (13.16) for the potential energy Vg due to gravity is a first-order approximation for the expression given in Eq. (13.17¢). (b) Using the same expansion, derive a second-order approximation for Vg.

SOLUTION

VWR

rr R y V

WR

R y

WR

V WRy

R

g g yR

g

= - = + = -+

= -+

= - +ÊËÁ

ˆ¯̃

=-

2 2

1

1

1

setting :

-- +-

+- -

◊ÊËÁ

ˆ¯̃

+È

ÎÍÍ

˘

˚˙˙

WRy

R

y

R1

1

1

1 2

1 2

2( ) ( )( )L

We add the constant WR, which is equivalent to changing the datum from r r R= • =to :

V WRy

R

y

Rg = - ÊËÁ

ˆ¯̃

+È

ÎÍÍ

˘

˚˙˙

2

L

(a) First order approximation:

V WRy

RWyg = Ê

ËÁˆ¯̃

= b

[ ]Equation 13.16

(b) Second order approximation: V WRy

R

y

Rg = - ÊËÁ

ˆ¯̃

È

ÎÍÍ

˘

˚˙˙

2

V WyWy

Rg = -2

b

139


PROBLEM 13.92

Observations show that a celestial body traveling at 2 106¥ km/h appears to be describing about Point B a circle of radius equal to 60 light years. Point B is suspected of being a very dense concentration of mass called a black hole. Determine the ratio MB/MS of the mass at B to the mass of the sun. (The mass of the sun is 330,000 times the mass of the earth, and a light year is the distance traveled by light in one year at a speed of 3 ¥ 105 km/s.)

SOLUTION

One light year is the distance traveled by light in one year.

Speed of light = 3 ¥ 108 m/s

r

r

= ¥ ¥ ¥ ¥ ¥

= ¥

( ) ( )

.

60 365 24 3600 3 10

5 67648 10

8

17

s m/s

m


FGM m

rm

v

r

Mrv

G

GM gR

B

B

= =

=

=

= ¥

2

2

2

2

69 81 6 37 10

earth earth

2m/s m( . )( . )22

143 98059 10= ¥. ( )m /s3 2

M M GM GMEsun sun earth= =330 000 330 000, : ,

GM

G

sun

m /s

= ¥

= ¥

=¥

( , )( . )

.

.

330 000 3 98059 10

1 313596 10

1 31359

14

20 3 2

110

1 31359 10

20

2 2

20

M

Mrv

G

rv MB

sun

sun= =¥.

M

MB

sun

=¥ ¥

¥ ¥( . )( )

( ) .

5 67648 10 5 10

81 1 31359 10

17 6 2

20

M

MB

sun

= ¥1 334 109. b

140


PROBLEM 13.93

A 200-g ball may slide on a horizontal frictionless surface and is attached to a fixed Point O by means of an elastic cord of constant k = 150 N/m and undeformed length equal to 600 mm. The ball is placed at Point A, 900 mm from O, and is given an initial velocity vA in a direction perpendicular to OA. Knowing that the ball passes at a distance d = 100 m from O, determine (a) the initial speed vA of the ball, (b) its speed v after the cord has become slack.

SOLUTION

Position A. Elongation of cord

x

V k x

T mv

A

A A

A

= - =

= = =

=

0 9 0 6 0 3

1

2

1

2150 0 3 6 75

1

2

2 2

. . .

( )( . ) .

m m m

N/m m J

AA A Av v2 2 21

20 2 0 1= =( . ) .kg

Position B. Elongation of cord = 0

V k x

T mv v v

B B

B B B B

= =

= = =

1

20

1

2

1

20 2 0 1

2

2 2 2( . ) . kg


T V T V v vA A B B A B+ = + + =: ( . ) . ( . )1

20 2 6 75

1

20 22 2 (1)

Conservation of angular momentum. d = 100 mm

mv r mv d v vA A B A B( . ) ( . )= =0 9 0 1m m (2)

Substituting for vB from (2) into (1)

1

20 2 6 75

1

20 2 92 2( . ) . ( . )( )v vA A+ =

(a) 6 75 8 2. = vA vA = 0 919. m/s b

(b) Eq. (2) v vB A= =9 9 0 919( . ) vB = 8 27. m/s b

141


PROBLEM 13.94

For the ball of Problem 13.93, determine (a) the smallest magnitude of the initial velocity vA for which the elastic cord remains taut at all times, (b) the corresponding maximum speed reached by the ball.

PROBLEM 13.93 A 200-g ball may slide on a horizontal frictionless surface and is attached to a fixed Point O by means of an elastic cord of constant k = 150 N/m and undeformed length equal to 600 mm. The ball is placed at Point A, 900 mm from O, and is given an initial velocity vA in a direction perpendicular to OA. Knowing that the ball passes at a distance d = 100 m from O, determine (a) the initial speed vA of the ball, (b) its speed v after the cord has become slack.

SOLUTION

Position A. Elongation of cord

x

V k x

T mv

A

A A

A A

= - =

= = =

=

0 9 0 6 0 3

1

2

1

2150 0 3 6 75

1

2

2 2

. . .

( )( . ) .

m m m

N/m m J

22 2 21

20 2 0 1= =( . ) .kg v vA A

Position B. Elongation of cord = 0

V k x

T mv v v

B B

B B B B

= =

= = =

1

20

1

2

1

20 2 0 1

2

2 2 2( . ) . kg


T V T V v vA A B B A B+ = + + =: ( . ) . ( . )1

20 2 6 75

1

20 22 2 (1)

(a) For smallest vA for which cord stays taut,

d = =600 0 6mm m.

Eq. (2) mv mv v vA B B A( . ) ( . ) .0 9 0 6 1 5m m= =

Substituting v vB A= 1 5. into Eq. (1)

1

20 2 6 75

1

20 2 1 5

6 750 2

21 5 1 0 125

2 2

2 2

( . ) . ( . )( . )

..

( . ) .

v v

v

A A

A

+ =

= - = vvA2

vA = 7 348. m/s vA = 7 35. m/s b

(b) Corresponding maximum speed: v vB A= =1 5 11 02. . m/s b

142


PROBLEM 13.95

Collar A weighs 5 kg and is attached to a spring of constant 800 N/m and of undeformed length equal to 450 mm. The system is set in motion with r = 300 mm, vq = 5 m/s, and vr = 0. Neglecting the mass of the rod and the effect of friction, determine the radial and transverse components of the velocity of the collar when r = 525 mm.

SOLutiOn

r

v

r

1

1

2

300 0 3

5

525

===

mm m

m/s

mm 0.525 m

=

=

.

Conservation of angular momentum.

r mv r m v

v

v

1 1 2 2

2

2

0 3 5 0 525

20

7

( )

( . ) ( ) ( . ) ( )

( )

==

=

q

q

q

m m

m/s ( )vq 2 3= m/s b


T mv1 12 21

212

5 5 62 5

12

12

= = = D -

D =

( ) ( ) .

(

J; 0.45 0.3 0.15m

V k( )

1

1 12

= =

= 8800 0 15

9

2) ( . )

= J

T m v v vr r2 22

22

22

2

1

2

1

25

20

7 2= +ÈÎ ˘̊ = Ê

ËÁˆ¯̃

+È

ÎÍÍ

˘

˚˙˙

D =

( ) ( ) ( ) ( )

(

q

00 45 0 525 0 075

1

2

1

2800 0 0752 2

2 2

. . ) ( . )

( ) ( . )

- = - fi

= D = =

m extension

V k 22 25. J

T V T V vr1 1 2 2

2262 5 9 2 5

20

72 25

2+ = + + = Ê

ËÁˆ¯̃

+È

ÎÍÍ

˘

˚˙˙

+: . . ( ) .

20

727 7

2

22Ê

ËÁˆ¯̃

+ =( ) .vr

( ) . /v mr 22 2 219 5367= s ( ) .vr 2 4 42= m/s b

143


PROBLEM 13.96

For the motion described in Problem 13.95, determine (a) the maximum distance between the origin and the collar, (b) the corresponding speed. (Hint: Solve by trial and error the equation obtained for r.)

PROBLEM 13.95 Collar B weighs 5 kg and is attached to a spring of constant 800 N/m and of undeformed length equal to 450 mm. The system is set in motion with r = 300 mm, vq = 5 m/s, and vr = 0. Neglecting the mass of the rod and the effect of friction, determine the radial and transverse components of the velocity of the collar when r = 525 mm.

SOLUTION

r

v1

1

0 3

5

==

. m

m/s


At position of maximum distance r2 0=

r mv r mv

r mv1 1 2 2

20 3 5

==( . ) ( )m m/s max

vr21 5= .

max

(1)


T mv1 12 21

2

1

25 5 62 5= = =( ) ( ) . J

Spring: D = = D = =1 1 12 20 15

12

12

800 0 15 9. ; ( ) ( . )m JV k

T mv v2 22

221

22 5= = .

Spring: D = -

= D = -

+ = +

2

2 22 2

1 1 2 2

0 45

1

2

1

2800 0 45

( . )

( ) ( . )

:

max

max

r

V k r

T V T V

62 5 9 2 5 400 0 4522 2. . ( . )+ = + -v rmax (2)

144



(a) Substituting for v2 from (1) into (2):

71 5 2 51 5

400 0 452

2. ..

( . )=ÊËÁ

ˆ¯̃

+ -r

rmax

max

71 55 625

400 0 45 2..

( . )- = -r

rmax

max2

Solving by trial: rmax m= 0 849052. Error in given data. Let length of rod be 1 m.b

(b) Substituting into (1):

v21 5

1 7666= =..

0.849052m/s v2 1 767= . m/s b

145


PROBLEM 13.97

Solve Sample Problem 13.8, assuming that the elastic cord is replaced by a central force F of magnitude (80/r2) N directed toward O.

SOLUTION

(a) The force exerted on the sphere passes through O. Angular momentum about O is conserved.

Minimum velocity is at B, where the distance from O is maximum.

Maximum velocity is at C, where distance from O is minimum.

r mv r mvA A m msin 60 =

( . )( . )( )sin ( . )0 5 0 6 20 60 0 6m kg m/s kg= r vm m

vrmm

= 8 66. (1)


At Point A, T mv

V Fdrr

drr

V

A A

A

= = =

= = =-

=-

ÚÚ

1

2

1

20 6 20 120

80 80

8

2 2

2

( . )( )

,

kg m/s J

00

0 5160

.= - J

At Point B, T mv v vB m m m= = =1

2

1

20 6 0 32 2 2( . ) . kg

(and Point C) Vr

T V T V

Bm

A A B B

=-

+ = +

80

120 160 0 3802- = -. vrmm

(2)

Substitute (1) into (2) - =ÊËÁ

ˆ¯̃

-40 0 38 66 80

2

( . ).

r rm m

r r

r rm m

m m

2 2 0 5625 0

0 339 1 661

- + =¢ = =

.

. .m and m

rmax m= 1 661. b

rmin m= 0 339. b

146



(b) Substitute ¢rm and rm from results of Part (a) into (1) to get corresponding maximum and minimum values of the speed.

¢ = =vm8 66

0 33925 6

.

.. m/s vmax .= 25 6 m/s b

vm = =8 66

1 6615 21

.

.. m/s vmin .= 5 21 m/s b

147


PROBLEM 13.98

A 1.8-kg collar A and a 0.7-kg collar B can slide without friction on a frame, consisting of the horizontal rod OE and the vertical rod CD, which is free to rotate about CD. The two collars are connected by a cord running over a pulley that is attached to the frame at O. At the instant shown, the velocity vA of collar A has a magnitude of 2.1 m/s and a stop prevents collar B from moving. If the stop is suddenly removed, determine (a) the velocity of collar A when it is 0.2 m from O, (b) the velocity of collar A when collar B comes to rest. (Assume that collar B does not hit O, that collar A does not come off rod OE, and that the mass of the frame is negligible.)

SOLUTION

(a) Conservation of angular momentum about D, C.

( . )( )( ) ( . )( )( )

( ).

.( . )

0 1 0 2

0 1

0 22 1

m m

m/s

m v m v

v

A A A A T

A T

= ¢

¢ = ÊËÁ

ˆ¯̃

== 1 05. m/s


1 v T

v

A

B

= = =

=

2 11

21 8 2 1 3 969

0

12. ( . )( . ) .m/s kg m/s J

Choose datum for B at its initial position and note that the potential energy of A does not change. Thus, we take V1 0= .

2 ( ) . ( )¢ = ¢ = ¢v v vA T A R B1 050 m/s (kinematics)

T m v v m v

T

A A T A R B B22 2 2

2

1

2

1

21

21 8 1 050

= ¢ + ¢ÈÎ ˘̊ + ¢

=

( ) ( ) ( )

( . ) ( .kg m//s kg) ( ) ( . )( )

. . ( )

2 2 2

22

2

1

20 7

0 9923 1 25

+ ¢ÈÎ ˘̊ + ¢

= + ¢

=

v v

T v

V

A R A R

A R

mm gB ( . ) ( . )( . )( . ) .0 1 0 7 9 81 0 1 0 6867m kg m/s m J2= =

T V T V vA R1 1 2 223 969 0 0 9923 1 25 0 6867+ = + + = + ¢ +. . . ( ) .

( ) . ;

( ) .

¢ =¢ =

v

vA R

A R

2 2 21 832

1 354

m /s

m/s

148



¢ = ¢ + ¢

= +=

= -

v v vA A T A R( ) ( )

[( . ) ( . ) ]

.

tan

/

2 2

2 2 1 2

1

1 05 1 354

1 713 m/s

q(( )

( )

tan.

..

¢¢

=

= ∞

-

v

vA T

A R

1 1 05

1 35437 8 ¢ =vA 1 713. m/s 37.8∞ b

(b) When B comes to rest, the distance x moved by A and B is unknown.

Conservation of angular momentum about D, C.

( . )( )( ) ( . )( )( )0 1 0 1m m mm v x m vA A A A T= + ¢

Kinematics: ( ) ( ) ,¢ = ¢ =v vA R B 0

Thus, ( )

.

( . )( . ) ( . )

..

¢ = ¢== + ¢

=¢

-

v v

v

x v

xv

A T A

A

A

A

2 1

0 1 2 1 0 1

0 210 1

m/s


At 1 , v

v

T m v

T

V

A

B

A A

==

=

=

==

21

0

1

21

21 8 2 1

3 969

0

12

2

1

1

m/s

kg

J

( . )( . )

.

At 2 , ¢ = ¢ =v vB A R0 0, ( )

T m v v

V m gx x

V x

A A A

B

22 2

2

2

1

20 9

0 7 9 81

6 867

= ¢ = ¢

= ==

.

( . )( . )

.

kg m/s2

T V T V

v xA

1 1 2 2

23 969 0 0 9 6 867

+ = +

+ = ¢ +. . .

149



From conservation of angular momentum.

xvA

=¢

-0 210 1

..

Thus, 3 969 0 9 6 8670 21 0 1

3 969 0 9 1 4

2

3

. . ( . )( . . )

. . .

= ¢ +- ¢

¢

¢ = ¢ +

vv

v

v v

AA

A

A A 442 0 6867

4 6557 0 9 1 442

5 173 1 602

3

3

- ¢

¢ = +

¢ = ¢ +

.

. . .

. .

v

v v

v v

A

A A

A A

By trial: ¢ =vA 0 316. m/s b

150


PROBLEM 13.99

Using the principles of conservation of energy and conservation of angular momentum, solve Part a of Sample Problem 12.9.

SOLUTION

R

r

r

v

== +

= = ¥=

=

6370

500 6370

6870 6 87 10

36 900

36

0

06

0

km

km km

km m

km/h

.

,

..

.

.

9 10

3 6 10

10 25 10

6

3

3

¥¥

= ¥

m

s

m/s


r mv r mv r r r r

Vr

rv

r

A

A

0 0 1 0 1

0

10

6

1

6 870 10

= = =

=ÊËÁ

ˆ¯̃

=¥Ê

ËÁ¢

, ,

.

min max

ˆ̂

¯̃¥( . )10 25 103

VrA¢ =

¥70 418 109

1

. (1)


Point A:

v

T mv m

T m

A

A

03

02 3 2

6

10 25 10

1

2

1

210 25 10

52 53 10

= ¥

= = ¥

= ¥

.

( . )

( )( . )(

m/s

JJ

m/s m

m /s

2

3 2

)

( . )( . )

VGMm

r

GM gR

GM

r

A = -

= = ¥

= ¥

0

2 6 2

12

9 81 6 37 10

398 10

006

12

6

6

6 87 10

398 10

6 87 10

57 93 10

= ¥

= -¥

¥

= - ¥

.

( )

( . )

.

m

m /s

m

m (J

3 2

Vm

A

))

151



Point A¢:

T mv

VGMm

r

m

r

T V T V

A A

A

A A A A

¢ ¢

¢

¢ ¢

=

= -

= -¥

+ = +

¥

1

2

398 10

52 53 10

2

1

12

1

( )

.

J

66 6 212

1

57 93 101

2

398 10m m mv

m

rA- ¥ = -¥

¢.

Substituting for vA¢ from (1)

- ¥ =¥

-¥

- ¥ =

5 402 1070 418 10

2

398 10

5 402 102 4

69 2

12

12

1

6

.( . )

( )( )

.( .

r r

7793 10 398 1021

12

12

1

¥-

¥)

r r

( . ) ( ) .5 402 10 398 10 2 4793 10 0612 12

121¥ - ¥ + ¥ =r r

r1666 7 10= ¥ ¥. m, 6.87 10 m6 rmax ,= 66 700 km b

152


PROBLEM 13.100

A spacecraft traveling along a parabolic path toward the planet Jupiter is expected to reach Point A with a velocity vA of magnitude 26.9 km/s. Its engines will then be fired to slow it down, placing it into an elliptic orbit which will bring it to within 100 ¥ 103 km of Jupiter. Determine the decrease in speed Dv at Point A, which will place the spacecraft into the required orbit. The mass of Jupiter is 319 times the mass of the earth.

SOLUTION


Point A:

T m v v

VGM m

r

GM GM gR

R

GM

A A A

AJ

A

J E E

E

= -

=-

= =

= ¥

1

2

319 319

6 37 10

2

2

6

( )

.

D

m

JJ

J

A

GM

r

= ¥

= ¥

= ¥

( )( . )( . )

.

319 9 81 6 37 10

126 98 10

350 1

2 6 2

15 3 2

m/s m

m /s

00

126 98 10

350 10

362 8 10

6

15 3 2

6

6

m

m /s

mV

m

V m

A

A

=- ¥

¥

= - ¥

( . )

( )

( . )

Point B:

T mv

VGM m

r

m

V

B B

BJ

B

B

=

=-

=- ¥

¥

= -

1

2

126 98 10

100 10

126

2

15 3 2

6

( . )

( )

(

m /s

m

99 8 10

1

2362 8 10

1

21269 8

6

2 6 2

. )

( ) . .

¥+ = +

- D - ¥ = -

m

T V T V

m v v m mv

A A B B

A A B ¥¥ 106 m

( )v v vA A B- - = - ¥D 2 2 61814 10 (1)

153




r

r

r m v v r mv

A

B

A A A B B

= ¥

= ¥- =

350 10

100 10

6

6

m

m

( )D

vr

rv v

v v

BA

BA A

A A

=ÊËÁ

ˆ¯̃

-

= ÊËÁ

ˆ¯̃

-

( )

( )

D

D350

100 (2)

Substitute vB in (2) into (1)

( ) [ ( . ) ]

( ) .

( )

v v

v v

v v

A A

A A

A A

- - = - ¥

- = ¥

- =

D

D

D

2 2 6

2 6

1 3 5 1814 10

1612 4 10

mm12.698 10 m/s3¥

(Take positive root; negative root reverses flight direction.)

vA = ¥26 9 103. m/s (given)

D = ¥ - ¥vA ( . . )26 9 10 12 698 103 3m/s m/s

D =vA 14 20. km/s b

154


PROBLEM 13.101

After completing their moon-exploration mission, the two astronauts forming the crew of an Apollo lunar excursion module (LEM) would prepare to rejoin the command module which was orbiting the moon at an altitude of 140 km. They would fire the LEM’s engine, bring it along a curved path to a Point A, 8 km above the moon’s surface, and shut off the engine. Knowing that the LEM was moving at that time in a direction parallel to the moon’s surface and that it then coasted along an elliptic path to a rendezvous at B with the command module, determine (a) the speed of the LEM at engine shutoff, (b) the relative velocity with which the command module approached the LEM at B. (The radius of the moon is 1740 km and its mass is 0.01230 times the mass of the earth.)

SOLUTION


mr v mr v

vr

rv

v

A A B B

BA

BA

A

=

=

= 1748

1880

v vB A= 0 9298. (1)


At Point A: T mv

VGM m

r

A A

AA

=

= --

1

22

moon

M M

GM GM gR

GM

moon earth

moon earth earth

moo

=

= =

0 0123

0 0123 0 0123 2

.

. .

nn

moon3

m/s m

m /s

= ¥

= ¥

=

( . )( . )( . )

.

0 0123 9 81 6 37 10

4 896 10

2 6 2

12 2GM

rA 11748

4 896 10

2 801 10

12

6

¥

= - ¥¥

= - ¥

10 m

m /s

(1748 10 m)

m

3

3 2

3Vm

A( . )

.

155



At Point B: T mv r

VGM m

r

m

B B B

BB

= = ¥

= - = -¥

1

21880 10

4 896 10

188

2 3

12

m

m /smoon3 2( . )

( 00 102 604 10

1

22 801 10

1

22

36

2 6 2

¥= - ¥

+ = + - ¥ = -

m).

; .

m

T V T V mv m mvA A B B A B ..604 106¥ m

v vA B2 2 3393 3 10= + ¥. )(m /s2 2 (2)

(a) Speed at A:

Substitute vB in (1) into (2)

v

v

v

v

A

A

A

A

2 2 3

2 6

3

1 0 9298 393 3 10

2 903 10

1 704 10

1

[ ( . ) ] .

.

.

- = ¥

= ¥

= ¥=

m/s

7704 m/s

(b) At Point B:

From (1) and result in (a) v

vB

B

==

( . )( )

.

0 9298 1704

1584 0 m/s

Command module is in circular orbit, rB = ¥1 88 106. m (Eq. 12.44)

vGM

rBcirc

moon

m/s

=

=¥¥

=

4 896 10

1 88 10

1613 8

12

6

.

.

.

Relative velocity = - = -v vBcirc 1613 8 1584 0. . vrel m/s= 29 8. b

156


PROBLEM 13.102

The optimal way of transferring a space vehicle from an inner circular orbit to an outer coplanar circular orbit is to fire its engines as it passes through A to increase its speed and place it in an elliptic transfer orbit. Another increase in speed as it passes through B will place it in the desired circular orbit. For a vehicle in a circular orbit about the earth at an altitude h1 300= km, which is to be transferred to a circular orbit at an altitude h2 800= km, determine (a) the required increases in speed at A and at B, (b) the total energy per unit mass required to execute the transfer.

SOLUTION

r

r

A

A

= + =

= ¥

6370 300 6670

6 670 106

km

m.

r

r

R

B

B

= + =

= ¥

= = ¥

6370 800 7170

7 17 10

6370 6 37 10

6

6

km

m

m

.

( ) .km

GM gR

GM

= = ¥

= ¥

2 6 2

14 2

9 81 6 37 10

3 98059 10

( . )( . )

.

m

m /s3

Elliptical orbit between A and B.



mr v mr v

vr

rv v v v

A A B B

AB

AB B A B

=

= = fi =7 17

6 671 07496

.

..

(1)

Point A:

T mv

VGMm

r

m

V

A A

AA

A

=

= - = - ¥¥

= - ¥

1

2

3 98059 10

59 679 10

2

14( . )

.

(6.67 10 )6

66 m

Point B: T mv

VGMm

r

m

B B

BB

=

= - = - ¥¥

= - ¥

1

2

3 98059 10

7 17 1055 5173 10

2

14

66( . )

( . ). mm

157



T V T V

mv m mv

v v

A B B

A B

A B

A + = +

- ¥ = - ¥

- =

1

259 679 10

1

255 5173 10

8

2 6 2 6

2 2

. .

.33234 106¥

From (1) N vA B= 1 07496.

v

v

v

B

B

B

2 2 6

2 6

1 07496 1 8 3234 10

53 5133 10

7315 28

[( . ) ] .

.

.

- = ¥

= ¥=

m /s2 2

mm/s

m/s

v

vA

A

==

( . )( . )

.

1 07496 7315 28

7863 63

Circular orbit at A and B.

(Eq. 12.44) ( ).

..

).

vGM

r

vGM

r

A CA

B CB

= = ¥¥

=

= =

3 98059 10

6 67 107137 61

3 980

14

6m/s

(559 10

7 17 107450 99

14

6

¥¥

=.

. m/s

(a) Increases in speed at A and at B.

Dv v vA A A C= - = - =( ) . . .7863 63 7137 61 726 02m/s DVA = 726m/s b

Dv v vB B C B= - =( ) .135 71m/s DVB = 135 7. m/s b

(b) Total energy per unit mass.

E

mv v v v

E

m

A A C B C B= - + -ÈÎ ˘̊

= -

1

21

27863 63 7137 61

2 2 2 2

2 2

( ) ( ) ( )

( . ) ( . ) ++ -ÈÎ ˘̊( . ) ( . )7450 99 7315 282 2

E

m= ¥12 90 106 2. m /s2 b

158


PROBLEM 13.103

A spacecraft approaching the planet Saturn reaches Point A with a velocity vA of magnitude 20 103¥ m/s. It is to be placed in an elliptic orbit about Saturn so that it will be able to periodically examine Tethys, one of Saturn’s moons. Tethys is in a circular orbit of radius 300 103¥ about the center of Saturn, traveling at a speed of 11 1 103. ¥ . Determine (a) the decrease in speed required by the spacecraft at A to achieve the desired orbit, (b) the speed of the spacecraft when it reaches the orbit of Tethys at B.

SOLUTION

(a)

r

r

A

B

= ¥

= ¥

185 10

300 10

6

6

m

m

¢ =vA speed of spacecraft in the elliptical orbit after its speed has been decreased

Elliptical orbit between A and B.


Point A: T mv

VGM m

r

A A

AA

= ¢

=-

1

22

sat

Msa = mass of Saturn; determine GMsa from the speed of Tethys in its circular orbit.

(Eq. 12.44) vGM

rGM r vBcirc

satsat circ= = 2

GM

VA

sat

3

m/s

m /s

= ¥ ¥

= ¥

= -¥

( )( . )

.

( .

300 10 11 1 10

3 6963 10

3 6963

6 3 2

16 2

110

199 8 10

16 2

6

m /s

(185 10 m)

m

3

6

)

.

m

¥

= - ¥

159



Point B: T mv VGM m

r

m

V

B B BB

B

= =-

= -¥¥

= - ¥

1

2

3 6963 10

300 10

123 21 1

216

6sat ( . )

( )

. 006 m

T V T V

mv m mv m

A A B B

A B

+ = +

¢ - ¥ = - ¥

;

. .

1

22 6 2 6199 8 10

1

2123 21 10

¢ - = ¥v vA B2 2 676 59 10.


r mv r mv vr

rv v v

v

A A B B BA

BA A A

A

¢ = = ¢ = ¥¥

¢ = ¢

¢

185 10

300 100 616667

1

6

6

2

.

[ -- = ¥

¢ = ¥¢ =

( . ) ] .

.

,

0 616667 76 59 10

123 5877 10

11 117

2 6

2 6v

vA

A m/s

(a) Dv v vA A A= - ¢ = -( , , )20 000 11 117 DvA = 8 880, m/s b

(b) vr

rvB

A

BA= ¢ = Ê

ËÁˆ¯̃

37

6011 117( , ) vB = 6860 m/s b

160


PROBLEM 13.104

A spacecraft is describing an elliptic orbit of minimum altitude hA = 2400 km and maximum altitude hB = 9600 km above the surface of the earth. Determine the speed of the spacecraft at A.

SOLUTION

r

r

r

A

A

B

= +== +=

6370 2400

8770

6370 9600

15 970

km km

km

km km

km,

Conservation of momentum. r mv r mvA A B B=

vr

rv v vB

A

BA A A= = =8770

15 9700 5492

,. (1)

Conservation of energy. T mv VGMm

rT mv V

GMm

rA A AA

B B BB

= = - = = -1

2

1

22 2

GM gR

VA

= = ¥ = ¥

=- ¥

2 2 3 2 129 81 6370 10 398 1 10

398 1 1

( . )( ) .

( .

m/s m m /s3 2

00

8770 1045 39 10

398 1 10

15 970 102

12

36

12

3

).

( . )

( , )

mm

Vm

B

¥= - ¥

=- ¥

¥= - 44 93. m

T V T V mv m

mv m

A A B B A

B

+ = + - ¥

= - ¥

1

22 6

2 6

45 39 10

1

224 93 10

.

. (2)

Substituting for vB in (2) from (1)

v

v

A

A

2 2 6

2 6 2

1 0 5492 40 92 10

58 59 10

[ ( . ) ] .

.

- = ¥

= ¥ m /s2

vA = ¥7 65 103. m/s vA = ¥27 6 103. km/h b

161


PROBLEM 13.105

A spacecraft describing an elliptic orbit about the earth has a maximum speed vA = ¥26 3 103. km/h at A and a minimum speed vB = ¥18 5 103. km/h at B. Determine the altitude of the spacecraft at B.

SOLUTION

v

v

v

A

A

B

= ¥

= ¥

= ¥

= ¥

26 3 10

7 31 10

18 5 10

5 14 10

3

3

3

3

.

.

.

.

km/h

m/s

km/h

m/ss

Conservation of momentum. r mv r mvA A B B=

r v r v rv

vr r

r r

A A B B AB

AB B

A B

= = =

=

18 5

26 3

0 7034

.

.

. (1)

Conservation of energy. T mv T m m

T mv T m

A A A

B B B

= = ¥ = ¥

= = ¥

1

2

1

27 31 10 26 69 10

1

2

1

25 14

2 3 2 6

2

( . ) .

( . 110 13 20 10

9 81 6370 10

3 2 6

2 3 2

) .

( . )( )

= ¥

= - = = ¥

m

VGMm

rGM gRA

A

m/s

GM = ¥398 1 1012 3 2. m /s

Vr

VGMm

r r

T V T V

AA

BB B

A A B B

= - ¥

= - = - ¥

+ = +

398 1 10

398 1 10

12

12

.

.

26 69 10398 1 10

13 20 10398 1 106

126

12

..

..¥ - ¥ = ¥ - ¥

mr

m mrA B

m

162



Substituting for rA from (1)

398 1 10 1

0 70341 13 49 10

180 37 10

12

126

9

.

( . ).

.

¥ -È

ÎÍ

˘

˚˙ = ¥

= ¥

=

-

r

r

r

B

B

B ..

,

442 10

12 442

6¥=

m

km

h r RR B= -= -12 442 6370, km km hB = 6070 km b

163


PROBLEM 13.106

Upon the LEM’s return to the command module, the Apollo spacecraft or Problem 13.101 was turned around so that the LEM faced to the rear. The LEM was then cast adrift with a velocity of 200 m/s relative to the command module. Determine the magnitude and direction (angle f formed with the vertical OC) of the velocity vC of the LEM just before it crashed at C on the moon’s surface.

SOLUTION

Command module in circular orbit

r

GM GM gR

B = + = = ¥

= =

1740 140 1880 1 88 10

0 0123 0 0123

6km m

moon earth

.

. . 22

6 2

12 3 2

0 0123 9 81 6 37 10

4 896 10

= ¥

= ¥

. ( . )( . )

. m /s R = 1740 km

SF maGM m

r

mv

r

vGMm

r

nm

B B

B

= =

= = ¥¥

202

0

12

6

4 896 10

1 88 10

.

.

v

vB

0 1614

1614 200 1414

== - =

m/s

m/s GMm

rB0

Conservation of energy between B and C.

1

2

1

2

21

2 2

2 2

mvGM m

rmv

GM m

rr R

v vGMm

r

r

R

v

Bm

BC

m

CC

C BB

B

C

- = - =

= + -ÊËÁ

ˆ¯̃

22 212 3 2

6

6

1414 24 896 10

1 88 10

1 88 10

1 74 1= + ¥

¥¥¥

( )( . )

( . )

.

. m/s

m /s

m 001

1 999 10 0 4191 10 2 418 10

6

2 6 6 6 2 2

-Ê

ËÁˆ

¯̃

= ¥ + ¥ = ¥vC . . . m /s

vC = 1555 m/s b

164




r mv Rmv

r v

r v

B B C

B B

C C

=

=

=¥¥

sin

sin

( . )( )

( .

f

f

1 88 10 1414

1 74 10

6

6

m m/s

mm m/s)( )

.

1555

0 98249=

f = ∞79 26. f = ∞79 3. b

165


PROBLEM 13.107

A satellite is projected into space with a velocity vo at a distance r0 from the center of the earth by the last stage of its launching rocket. The velocity v0 was designed to send the satellite into a circular orbit of radius r0. However, owing to a malfunction of control, the satellite is not projected horizontally, but at an angle a with the horizontal and, as a result, is propelled into an elliptic orbit. Determine the maximum and minimum values of the distance from the center of the earth to the satellite.

SOLUTION

For circular orbit of radius r0

F ma

GMm

rm

v

r

vGM

r

n= =

=

02

02

0

02

0

But v0 forms an angle a with the intended circular path.

For elliptic orbit:


r mv r mvA A0 0 cosa =

vr

rvA

A

=ÊËÁ

ˆ¯̃

00cosa (1)


1

2

1

2

21

02

0

2

02 2

0

0

mvGMm

rmv

GMm

r

v vGM

r

r

r

AA

AA

- = -

- = -ÊËÁ

ˆ¯̃

Substitute for vA from (1)

vr

r

GM

r

r

rA A02 0

22

0

012

1-ÊËÁ

ˆ¯̃

È

ÎÍÍ

˘

˚˙˙

= -ÊËÁ

ˆ¯̃

cos a

But vGM

r02

0

= , thus 1 2 102

2 0-ÊËÁ

ˆ¯̃

= -ÊËÁ

ˆ¯̃

r

r

r

rA A

cos a

cos2 02

02 1 0ar

r

r

rA A

ÊËÁ

ˆ¯̃

-ÊËÁ

ˆ¯̃

+ =

166



Solving for r

rA

0

r

r

r

A

A

02

2 2

2 4 4

2

1

1

1 1

1

=+ ± -

= ±-

= + -±

cos

cos

sin

sin

( sin )( sin )

s

aa

aa

a aiin

( sin )a

ar r0 01= m

also valid for Point ¢A

Thus, r rmax ( sin )= +1 0a r rmin ( sin )= -1 0a b

167


PROBLEM 13.108

A space platform is in a circular orbit about the earth at an altitude of 300 km. As the platform passes through A, a rocket carrying a communications satellite is launched from the platform with a relative velocity of magnitude 3.44 km/s in a direction tangent to the orbit of the platform. This was intended to place the rocket in an elliptic transfer orbit bringing it to Point B, where the rocket would again be fired to place the satellite in a geosynchronous orbit of radius 42,140 km. After launching, it was discovered that the relative velocity imparted to the rocket was too large. Determine the angle g at which the rocket will cross the intended orbit at Point C.

SOLUTION

R

r

r

r

GM gR

GM

A

A

C

== +

= ¥

= ¥

=

=

6370

6370 300

6 67 10

42 14 10

6

6

2

km

km km

m

m

.

.

(99 81 6 37 10

398 1 10

6 2

12 3 2

. )( . )

.

m/s m

m /s

2 ¥

= ¥GM

For any circular orbit:

F mamv

rn n= = circ2

F

GMm

rm

v

r

vGM

r

n = =

=

2circ2

circ

Velocity at A.

( )

( . )

( . ).v

GM

r

v

AA

A

circm /s

mm/s= =

¥¥

= ¥

=

398 1 10

6 67 107 726 10

12 3 3

63

(( ) ( ) . . .v vA A Rcirc m/s+ = ¥ + ¥ = ¥7 726 10 3 44 10 11 165 103 3 3

168



Velocity at C.

Conservation of energy. T V T V

mvGM m

rmv

GM m

r

A A C C

AA

CC

+ = +

- = -1

2

1

22 2

v v GMr rC AC A

2 2

3 2 12

21 1

11 165 10 2 398 1 101

42 14

= + -ÊËÁ

ˆ¯̃

= ¥ + ¥( . ) ( . ). ¥¥

-¥

ÊËÁ

ˆ¯̃

= ¥ - ¥

= ¥

10

1

6 67 10

124 67 10 100 48 10

24 19 10

6 6

2 6 6

6 2

.

. .

.

vC

m //s

m/s

2

34 919 10vC = ¥.


r mv r mv

r v

r v

A A C C

A A

C C

=

=

= ¥ ¥¥

cos

cos

( . )( . )

( .

g

g

6 67 10 11 165 10

42 14

6 3

110 4 919 10

0 35926

6 3)( . )

cos .

¥=g

g = ∞68 9. b

169


SOLUTION

(a) r

r

r

A

B

B

= + == + =

= ¥

6370 360 6730

6370 6430

6 43 10

6

6

km m

60 km

= 6.73 ¥ 10

. mm

m

m

m /s

R

GM gR

GM

= ¥

= = ¥ ¥

= ¥

6 37 10

9 81 6 37 10

3 9806 10

6

2 6 2

14 3 2

.

( . ) ( . )

.


T mv

VGMm

r

T m

A A

AA

B

=

= -

= - ¥¥

= - ¥

=

1

2

3 9806 10

6 73 10

59 147 10

1

2

2

14

6

6

.

.

. m

vv

VGMm

r

T V T V

B

BB

A A B B

2

14

6

6

3 9806 10

6 43 10

61 907 10

= -

= - ¥¥

= - ¥+ = +

.

.

. m

11

259 147 10

1

261 907 102 6 2 6mv mvA B- ¥ = - ¥. .m m

v vA B2 2 62 76 10= - ¥. (1)

PROBLEM 13.109

A space vehicle is in a circular orbit at an altitude of 360 km above the earth. To return to the earth, it decreases its speed as it passes through A by firing its engine for a short interval of time in a direction opposite to the direction of its motion. Knowing that the velocity of the space vehicle should form an angle fB = 60o with the vertical as it reaches Point B at an altitude of 60 km, determine (a) the required speed of the vehicle as it leaves its circular orbit at A, (b) its speed at Point B.

170




r mv r mv

vr v

r

A A B B B

BA A

B B

=

= =∞

ÊËÁ

ˆ¯̃

sin

( )

( )(sin ) sin

f

f6730

6430

1

60vvA

v vB A= 1 2086. (2)

Substituting vB from (2) in (1)

v v

v

v

A A

A

A

2 2 6

2 2 6

2

1 2086 2 76 10

1 2086 1 2 76 10

5 990

= - ¥

- = ¥

=

( . ) .

[( . ) ] .

. 77 10

2447 6

6 2 2¥=

m /s

m/sVA .

(a) vA = 2450 m/s b

(b) From (2)

v vB A===

1 2086

1 2086 2447 6

2958 16

.

. ( . )

. m/s

vB = 2960 m/s b

171


PROBLEM 13.110*

In Problem 13.109, the speed of the space vehicle was decreased as it passed through A by firing its engine in a direction opposite to the direction of motion. An alternative strategy for taking the space vehicle out of its circular orbit would be to turn it around so that its engine would point away from the earth and then give it an incremental velocity DvA toward the center O of the earth. This would likely require a smaller expenditure of energy when firing the engine at A, but might result in too fast a descent at B. Assuming this strategy is used with only 50 percent of the energy expenditure used in Problem 13.109, determine the resulting values of fB and υB .

SOLUTION

r

r

r

A

B

B

= + == + =

= ¥

6370 360 6730

6370 6430

6 43 10

6

6

km m

60 km

= 6.73 ¥ 10

. mm

m

m

m /s

R

GM gR

GM

= ¥

= = ¥ ¥

= ¥

6 37 10

9 81 6 37 10

3 9806 10

6

2 6 2

14 3 2

.

( . ) ( . )

.

Velocity in circular orbit at 360 km altitude.


F maGMm

r

m v

r

vGM

r

nA

A

A

AA

= =

=

= ¥¥

:( )

( )

.

.

2

2

14

6

3 9806 10

6 73 10

circ

circ

== 7690 72. m/s

Energy expenditure.

From Problem 13.109, vA = 2447 6. m/s

172



Energy, D

D

D

E m v mv

E m m

E

A A1092 2

1092 2

1

2

1

21

27690 72

1

224476

= -

= -

( )

( . ) ( )

circ

11096

110 109

6

26 5782 10

0 5026 5782 10

2

= ¥

= = ¥Ê

ËÁˆ

¯̃

( . )

( . ).

m

E E

J

JD D m

Thus, additional kinetic energy at A is

1

213 2891 102

1106m v E mA( ) ( . )D D= = ¥ J (1)

Conservation of energy between A and B.

T m v v VGMm

rA A A AA

= + = -1

22 2[( ) ( ) ]circ D

T mv VGMm

rB B BA

= = -1

22

T V T VA A B B+ = +

1

27690 72 13 2891 10

1

261 9072 6 6 2 6m m m mvB( . ) ( . ) .+ =¥ - 59.147 ¥10 - ¥10 mm

v

vB

B

2 691 2454 10

9552 2

= ¥=

.

.

m /s

m/s

2 2

vB = 9560 m/s b

Conservation of angular momentum between A and B.

r m v r mvA A B B B( ) sincirc = f

sin( )

( )

( )

( )

( . )

( . )fB

A

B

A

B

r

r

v

v=

ÊËÁ

ˆ¯̃

=circ 6730

6430

7690 72

9552 2== 0 8427.

fB = ∞57 4. b

173


PROBLEM 13.111

When the lunar excursion module (LEM) was set adrift after returning two of the Apollo astronauts to the command module, which was orbiting the moon at an altitude of 140 km, its speed was reduced to let it crash on the moon’s surface. Determine (a) the smallest amount by which the speed of the LEM should have been reduced to make sure that it would crash on the moon’s surface, (b) the amount by which its speed should have been reduced to cause it to hit the moon’s surface at a 45° angle. (Hint: Point A is at the apogee of the elliptic crash trajectory. Recall also that the mass of the moon is 0.0123 times the mass of the earth.)

SOLUTION

r

r r R

GM

A

B C

= +

= ¥

= = = = ¥=

1740 140

1880 10

1740 1740 10

0

3

3

km km

m

km m

moon .00123

0 0123

0 0123 9 81 6 37 10

4

2

6 2

GM

gR

GM

E

E=

= ¥

=

.

( . )( . )( . )m/s m2

moon ..896 1012 2¥ m /s2

Velocity in a circular orbit at 140 km altitude.

vGM

rAcirc

moon12 3

3

4.896 10 m /s

1880 10 m

m/s

= ¥¥

= ¥

2

31 6138 10.

(a) An elliptic trajectory between A and C, where the LEM is just tangent to the surface of the moon, will give the smallest reduction of speed at A, which will cause impact.

Conservation of energy (A and C).

T mv VGM m

r

T m

A A AM

A

C

= = - = -¥¥

= - ¥

=

1

2

4 896 10

1880 102 604 10

1

2

212

36.

.m

m

vv VGM m

r

T V T V

C CM

C

A A C C

212

364 896 10

1740 102 814 10

1

2

= - =¥¥

- ¥

+ = +

..

mm

mmv mv

v v

A C

A C

2 6 2 6

2 2 3

2 604 101

22 81410 10

419 1 10

- ¥ = - ¥

= - ¥

. .

.

m m

(1)

174



Conservation of angular momentum (A and C).

r mv r mvA A C C=

vr

rv v vC

A

CA A A= = =1880

17401 0805. (2)

Replace vC in (1) by (2)

v v

v v

A A

A A

2 2 3

2 2 3 2

1 0806 419 1 10

1 0805 1 419 1 10 2 5

= - ¥

- = ¥ =

( . ) .

[( . ) ] . . 002 10

1582

6¥=vA m

Dv v vA A A= - = -( )circ 1619 1582 DvA = 31 5. m/s b

(b) Conservation of energy (A and B).

Since r rB C= , conservation of energy is the same as between A and C.

Thus, from (1) v vA B2 2 3419 1 10= - ¥. (1¢)

Conservation of angular momentum (A and B).

r mv r mv

vr v

r

vv

A A B B

BA A

B

AA

= = ∞

= =∞

=

sin

sin sin.

f f

f

45

1880

1740 451 528

(3)

Replace vB in (1¢) by (3)

v v

v v

A A

A A

2 2 3

2 3

1 528 419 1 10

313 98 10 560

= - ¥

= ¥ =

( . ) .

. m/s

Dv v vA A A= - = -( )circ 1614 560 DvA = 1053 m/s b

175


PROBLEM 13.112*

A space probe describes a circular orbit of radius nR with a velocity v0 about a planet of radius R and center O. Show that (a) in order for the probe to leave its orbit and hit the planet at an angle q with the vertical, its velocity must be reduced to av0, where

a qq

= --

sin( )

sin

2 12 2

n

n

(b) the probe will not hit the planet if a is larger than 2 1/ ( ).+ n

SOLUTION

(a) Conservation of energy.

At A: T m v

VGMm

nR

A

A

=

= -

1

2 02( )a

At B: T mv

VGMm

R

B

B

=

= -

1

22

M = mass of planet

m = mass of probe

T V T VA A B B+ = +

1

2

1

202m v

GMm

nRmv

GMm

R( )2a - = - (1)


nRm v Rmva q0 = sin

vn v

=a

q0

sin (2)

Replacing v in (1) by (2)

( )sin

aa

qv

GM

nR

n v GM

R02 0

22 2- = ÊËÁ

ˆ¯̃

- (3)

176



For any circular orbit:

av

rn =2


- =

=

GMm

r

m v

r

vGM

r

2

2( )circ

circ

For r nR= , v vGM

nR0 = =circ

Substituting for v0 in (3)

a aq

aq

22 2

2

22

2

2 2

1 2 1

GM

nR

GM

nR

n GM

nR

GM

R

n

- = ÊËÁ

ˆ¯̃

-

-È

ÎÍ

˘

˚˙ =

sin

sin( --

= --

= --

n

n

n

n

n

)

( )(sin )

(sin )

( )sin

( sin )a q

qq

q2

2

2 2

2

2 2

2 1 2 1

a qq

= --

sin( )

sin

2 12 2

n

n Q.E.D. b

(b) Probe will just miss the planet if q � 90∞,

a = ∞ -- ∞

=+

sin( )

sin90

2 1

90

2

12 2

n

n n b

Note: n n n2 1 1 1- = - +( )( )

177


SOLUTION


r mv r mvA A P P=

vr

rvA

P

AP= (1)


1

2

1

22 2mv

GMm

rmv

GMm

rPP

AA

- = - (2)

Substituting for vA from (1) into (2)

vGM

r

r

rv

GM

rPP

P

AP

A

22

22 2- =ÊËÁ

ˆ¯̃

-

1 21 1

22-

ÊËÁ

ˆ¯̃

Ê

ËÁÁ

ˆ

¯˜̃ = -

ÊËÁ

ˆ¯̃

r

rv GM

r rP

AP

P A

r r

rv GM

r r

r rA P

AP

A P

A P

2 2

22 2

-=

-

with r r r r r rA P A P A P2 2- = - +( )( )

vGM

r r

r

rPA P

A

P

2 2=+

ÊËÁ

ˆ¯̃

(3) b

Exchanging subscripts P and A

vGM

r r

r

rAA P

P

A

2 2=+

ÊËÁ

ˆ¯̃

Q.E.D. b

PROBLEM 13.113

Show that the values vA and vP of the speed of an earth satellite at the apogee A and the perigee P of an elliptic orbit are defined by the relations

vGM

r r

r

rv

GM

r r

r

rAA P

P

AP

A P

A

P

2 22 2=+

=+

where M is the mass of the earth, and rA and rP represent, respectively, the maximum and minimum distances of the orbit to the center of the earth.

178


PROBLEM 13.114

Show that the total energy E of an earth satellite of mass m describing an elliptic orbit is E GMm r rA P= - +/ ( ), where M is the mass of the earth, and rA and rP represent, respectively, the maximum and minimum distances of the orbit to the center of the earth. (Recall that the gravitational potential energy of a satellite was defined as being zero at an infinite distance from the earth.)

SOLUTION

See solution to Problem 13.113 (above) for derivation of Equation (3).

vGM

r r

r

rPA P

A

P

2 2=+( )

Total energy at Point P is

E T V mvGMm

r

GMm

r r

r

r

GMm

r

GMmr

r

P P PP

A

A

P P

A

P

= + = -

=+

È

ÎÍ

˘

˚˙ -

=

1

2

1

2

2

2

0( )

(( )

( )

( )

r r r

GMmr r r

r r r

A P P

A A P

P A P

+-

È

ÎÍ

˘

˚˙

=- -

+

1

EGMm

r rA P

= -+

b

Note: Recall that gravitational potential of a satellite is defined as being zero at an infinite distance from the earth.

179


PROBLEM 13.115

A spacecraft of mass m describes a circular orbit of radius r1 around the earth. (a) Show that the additional energy DE which must be imparted to the spacecraft to transfer it to a circular orbit of larger radius r2 is

DEGMm r r

r r=

-( )2 1

1 22

where M is the mass of the earth. (b) Further show that if the transfer from one circular orbit to the other is executed by placing the spacecraft on a transitional semielliptic path AB, the amounts of energy DEA and DEB which must be imparted at A and B are, respectively, proportional to r2 and r1 :

D DEr

r rEA =

+2

1 2

D DEr

r rEB =

+1

1 2

SOLUTION

(a) For a circular orbit of radius r

F maGMm

rm

v

rn= =:2

2

vGM

r2 =

E T V mvGMm

r

GMm

r= + = - = -1

2

1

22 (1)

Thus DE required to pass from circular orbit of radius r1 to circular orbit of radius r2 is

DE E EGMm

r

GMm

r= - = - +1 2

1 2

1

2

1

2

DEGMm r r

r r=

-( )2 1

1 22 Q.E.D. (2)

(b) For an elliptic orbit, we recall Equation (3) derived in

Problem 13.113 ( )with v vP = 1

vGm

r r

r

r12

1 2

2

1

2=+( )

At Point A: Initially spacecraft is in a circular orbit of radius r1.

vGM

r

T mv mGM

r

circ

circ circ

2

1

2

1

1

2

1

2

=

= =

180



After the spacecraft engines are fired and it is placed on a semi-elliptic path AB, we recall

vGM

r r

r

r12

1 2

2

1

2=+

◊( )

and T mv mGMr

r r r1 12 2

1 1 2

1

2

1

2

2= =

+( )

At Point A, the increase in energy is

D

D

E T T mGMr

r r rm

GM

r

EGMm r r r

r

A

A

= - =+

-

=- -

12

1 1 2 1

2 1 2

1

2

2 1

2

2

2

circ ( )

( )

11 1 2

2 1

1 1 2

2

1 2

2 1

1 2

2

2

( )

( )

( )

( )

r r

GMm r r

r r r

Er

r r

GMm r r

r rA

+=

-+

=+

-È

ÎD ÍÍ

˘

˚˙

Recalling Equation (2): D DEr

r rEA =

+2

1 2( ) Q.E.D.

A similar derivation at Point B yields,

D DEr

r rEB =

+1

1 2( ) Q.E.D.

181


PROBLEM 13.116

A missile is fired from the ground with an initial velocity v0 forming an angle f 0 with the vertical. If the missile is to reach a maximum altitude equal to aR, where R is the radius of the earth, (a) show that the required angle f 0 is defined by the relation

sin ( )f a aa0

0

2

1 11

= + -+

ÊËÁ

ˆ¯̃

v

vesc

where vesc is the escape velocity, (b) determine the range of allowable values of v0 .

SOLUTION

(a) r RA =


Rmv r mvB B0 0sinf =

r R R RB = + = +a a( )1

vRv

R

vB =

+=

+0 0 0 0

1 1

sin

( )

sin

( )

fa

fa

(1)


T V T V mvGMm

Rmv

GMm

RA A B B B+ = + - = -+

1

2

1

2 102 2

( )a

v vGMm

R

GMm

RB02 2 2

11

1

2

1- = -

+ÊËÁ

ˆ¯̃

=+

ÊËÁ

ˆ¯̃a

aa

Substituting for vB from (1)

vGMm

R02

202

11

2

1-

+

Ê

ËÁˆ

¯̃=

+ÊËÁ

ˆ¯̃

sin

( )

fa

aa

From Equation (12.43): vGM

Resc2 2=

v v02

202

11 1

-+( )

Ê

ËÁ

ˆ

¯˜ =

+ÊËÁ

ˆ¯̃

sin fa

aaesc

2

sin

( )

202

0

2

11

1

fa

aa+

= -ÊËÁ

ˆ¯̃ +

v

vesc (2)

sin ( )f a aa0

0

2

1 11

= + -+

ÊËÁ

ˆ¯̃

v

vesc Q.E.D.

182



(b) Allowable values of v0 (for which maximum altitude = aR)

0 120� �sin f

For sin ,f0 0= from (2)

0 1

1

1

0

2

0

= -ÊËÁ

ˆ¯̃ +

=+

v

v

v v

esc

esc

aa

aa

For sin ,f0 1= from (2)

1

11

1

11

1

1

20

2

0

2

( )+= -

ÊËÁ

ˆ¯̃ +

ÊËÁ

ˆ¯̃

= + -+

ÊËÁ

ˆ¯̃

=

aa

a

aa

a

v

v

v

v

esc

esc 11 2 1

1

2

1

1

2

2

0

+ + -+

= ++

= ++

a aa a

aa

aa

( )

v vesc

v v vesc esca

aaa1

1

20+++

� � b

183


PROBLEM 13.117*

Using the answers obtained in Problem 13.107, show that the intended circular orbit and the resulting elliptic orbit intersect at the ends of the minor axis of the elliptic orbit.

SOLUTION

If the point of intersection P0 of the circular and elliptic orbits is at an end of the minor axis, then v0 is parallel to the major axis. This will be the case only if a q+ ∞ =90 0 , that is if cos sin .q a0 = - We must therefore prove that

cos sinq a0 = - (1)

We recall from Equation (12.39):

1

2r

GM

hC= + cosq (2)

When q = 0, r r r r= = -min min ( sin )and 0 1 a

1

102r

GM

hC

( sin )-= +

a (3)

For q = ∞180 , r r r= = +max ( sin )0 1 a

1

102r

GM

hC

( sin )+= -

a (4)

Adding (3) and (4) and dividing by (2):

GM

h r

r

20

02

1

2

1

1

1

1

1

=-

++

ÊËÁ

ˆ¯̃

=

sin sin

cos

a a

a

Subtracting (4) from (3) and dividing by (2):

Cr

r

Cr

=-

-+

ÊËÁ

ˆ¯̃

=ÊËÁ

ˆ¯̃ -

=

1

2

1

1

1

1

1

2

2

1

0

02

0

sin sin

sin

sin

sin

a a

aa

accos2 a

Substituting for GMh2 and C into Equation (2)

1 1

10

2r r= +

cos( sin cos )

aa q (5)

184



Letting r r= =0 0 and q q in Equation (5), we have

cos sin cos

coscos

sin

sin

sinsin

20

0

2

2

1

1

a a q

q aaaa

a

= +

= -

= -

= -

This proves the validity of Equation (1) and thus P0 is an end of the minor axis of the elliptic orbit.

185


PROBLEM 13.118*

(a) Express in terms of rmin and vmaxthe angular momentum per unit mass, h, and the total energy per unit mass, E/m, of a space vehicle moving under the gravitational attraction of a planet of mass M (Figure 13.15). (b) Eliminating vmax between the equations obtained, derive the formula

1

1 12

2

2

r

GM

h

E

m

h

GMmin

= + + ÊËÁ

ˆ¯̃

È

ÎÍÍ

˘

˚˙˙

(c) Show that the eccentricity e of the trajectory of the vehicle can be expressed as

e = + ÊËÁ

ˆ¯̃

12 2E

m

h

GM

(d) Further show that the trajectory of the vehicle is a hyperbola, an ellipse, or a parabola, depending on whether E is positive, negative, or zero.

SOLUTION

(a) Point A.

Angular momentum per unit mass.

h

H

mr mv

m

=

=

0

min max

h r v= min max (1) b

Energy per unit mass

E

m mT V

E

m mmv

GMm

rv

GM

r

= +

= -ÊËÁ

ˆ¯̃

= -

1

1 1

2

1

22 2

( )

maxmin

maxmin

(2) b

(b) From Eq. (1): v h rmax min= / substituting into (2)

E

m

h

r

GM

r= -1

2

2

2min min

1 2 1

20

2

2 2r

GM

h r

Em

hmin min

ÊËÁ

ˆ¯̃

- ◊ -

ÊËÁ

ˆ¯̃

=

186



Solving the quadratic: 1 2

2 2

2

2r

GM

h

GM

h h

Em

min

= + ÊËÁ

ˆ¯̃

+( )

Rearranging

1

1 12

2

2

r

GM

h

E

m

h

GMmin

= + + ÊËÁ

ˆ¯̃

È

ÎÍÍ

(3) b

(c) Eccentricity of the trajectory.

Eq. (12.39¢) 11

2r

GM

h= +( cos )e q

When q = 0, cos minq = =1 and r r

Thus,

1

12r

GM

hmin

( )= + e (4)

Comparing (3) and (4), e = + ÊËÁ

ˆ¯̃

12 2E

m

h

GM (5)

(d) Recalling discussion on pages 708, 709, and in view of Eq. (5)

1. Hyperbola if e �1, that is, if E � 0 b

2. Parabola if e = 1, that is, if E = 0 b

3. Ellipse if e �1, that is, if E � 0 b

Note: For circular orbit e = 0 and

12

02

2 2

+ ÊËÁ

ˆ¯̃

= = -ÊËÁ

ˆ¯̃

E

m

h

GME

GM

h

mor ,

but for circular orbit vGM

rh v r GMr2 2 2 2= = =and ,

thus E mGM

GMr

GMm

r= - = -1

2

1

2

2( )

187


PROBLEM 13.119

A 1200-kg automobile is moving at a speed of 90 km/h when the brakes are fully applied, causing all four wheels to skid. Determine the time required to stop the automobile (a) on dry pavement (mk = 0.75), (b) on an icy road (mk = 0.10).

SOLUTION

mv Wt

tmv

W

mv

mg

v

g

k

k k k

1

1 1 1

0- =

= = =

m

m m m

(a) For mk = 0 75.

t =25 m/s

(0.75)(9.81 m/s )2 t = 3 4. 0 s b

(b) For mk = 0 10.

t =25 m/s

(0.10)(9.81 m/s )2 t = 25 5. s b

188


PROBLEM 13.120

A 4 ¥ 107 kg ocean liner has an initial velocity of 4 km/h. Neglecting the frictional resistance of the water, determine the time required to bring the liner to rest by using a single tugboat which exerts a constant force of 200 kN.

SOLUTION

mv Ft

t

1

7 3

0

410

9200 10 0

- =

ÊËÁ

ˆ¯̃

=( ) ( )¥10 - ¥

t = 222 22. s t = 3 42min s b

189


PROBLEM 13.121

The initial velocity of the block in position A is 10 m/s. Knowing that the coefficient of kinetic friction between the block and the plane is mk = 0 30. , determine the time it takes for the block to reach B with zero velocity, if (a) q = 0, (b) q = ∞20 .

SOLutiOn

(a) q = 0

mv mg t tv

gA kA

k

- = = =ÊËÁ

ˆ¯̃

mm

010

9 81 0 3m/s

m/s( . )( . ) t = 3 40. s b

(b) q = ∞20

Impulse-momentum in x direction

+ mv mg mg tA k- ∞ - ∞ =m cos sin20 20 0

tv

gA

k

=∞ + ∞

=∞ + ∞

( cos sin )

)( . cos sin )

m 20 20

10

0 3 20 20

m/s

(9.81m/s2 t = 1 634. s b

190


PROBLEM 13.122

A 2-kg particle is acted upon by the force, expressed in newtons, F i j k= - + - + +( ) ( ) ( ) .8 6 4 42t t t Knowing that the velocity of the particle is v = + -( ) ( ( )150 100 250 m/s m/s) m/si j k at t = 0, determine (a) the time at which the velocity of the particle is parallel to the yz plane, (b) the corresponding velocity of the particle.

SOLutiOn

m dt mv F v0 + =Ú (1)

where F i j k

i j

dt t t t

t t t t

t

Ú Ú= - + - + +

= - + -ÊËÁ

ˆ¯̃

+

[( ) ( ) ( ) ]

( )

8 6 4 4

8 3 41

3

0

2

2 3 441

22t t+Ê

ËÁˆ¯̃

k

Substituting m = 2 kg, v0 150 100 250= + -i j k into (1):

( )( ) ( )2 150 100 250 8 3 41

34

1

22 3 2 kg i j k i j+ - + - + -Ê

ËÁˆ¯̃

+ +ÊËÁ

t t t t t t ˆ̂¯̃

=k v( )2 kg

v i j= + -ÊËÁ

ˆ¯̃

+ + -ÊËÁ

ˆ¯̃

+ - + +ÊËÁ

ˆ¯̃

150 43

2100 2

1

6250 2

1

42 3 2t t t t t t kk

(a) v is parallel to yz plane when vx = 0, that is, when

150 43

20 11 4222+ - = =t t t . s t = 11 42. s b

(b) v j= + -ÈÎÍ

˘˚̇

+ - + +

100 2 11 4221

611 422

250 2 11 4221

411 42

3( . ) ( . )

( . ) ( . 22 2)ÈÎÍ

˘˚̇k

v j k= - -( . ) ( .125 5 194 5 m/s m/s) b

191


SOLutiOn

(a) First 20 m

Velocity at 20 m. Rear wheels skid to generate the maximum force resulting in maximum velocity and minimum time since all the weight is on the rear wheel: this force is F N Wk= =m 0 60. .

Work and energy. T U T0 0 20 60+ =-

T U F T mv0 0 20 20 2020 20

12

= = =- ( )( )

0 201

2

2 0 60 20 9 81

15 34

202

202

20

+ =

==

( )( )

( )( . )( )( . )

.

mk mg mv

v

v

m m/s2

44 m/s

Impulse-momentum.

+ 0 15 3440 20 20 20+ = =-mk mgt mv v . m/s

t0 60

15 344

2 607

- =

=

.

)

.

m/s

(0.6)(9.81 m/s

s

2 t0 20 2 61- = . s b

PROBLEM 13.123

Skid marks on a drag race track indicate that the rear (drive) wheels of a car slip for the first 20 m of the 400 m track. (a) Knowing that the coefficient of kinetic friction is 0.60, determine the shortest possible time for the car to travel the initial 20 m portion of the track if it starts from rest with its front wheels just off the ground. (b) Determine the minimum time for the car to run the whole race if, after skidding for 60 ft, the wheels roll without sliding for the remainder of the race. Assume for the rolling portion of the race that 60 percent of the weight is on the rear wheels and that the coefficient of static friction is 0.85. Ignore air resistance and rolling resistance.

192



(b) For the whole race

The maximum force on the wheels for the first 20 m is F W Wk= =m 0 60. . For remaining 400 m, the maximum force, if there is no sliding and 60% of the weight is on the rear (drive) wheels, is

F W Ws= = =m ( . ) ( . )( . ) .0 60 0 85 0 60 0 510W

Velocity at 400 m.

Work and energy. T U U T0 0 20 20 400 400+ + =- -

T U W U W

T mv

0 0 20 20 400

1320 40

0 0 60 20 0 510 380

12

= = =

=

- -( . )( ), ( . )( ) m m

002

0 12 0 510 3801

2 4002+ + =mg mg mv( . )( )

v400 63 544= . m/s

Impulse–momentum.

From 20–400 m

F N W

v

v

s= ===

m 0 510

15 344

63 54420

400

.

.

.

m/s

m/s

m mgt m t( . ) ( . ) ( . ); .15 344 0 510 63 544 9 63420 400 20 400+ = =- - s

t t t0 400 0 20 20 400 63 544- - -= + = m( . ) t0 400 12 24- = . s b

193


PROBLEM 13.124

A truck is traveling on a level road at a speed of 90 km/h when its brakes are applied to slow it down to 30 km/h. An antiskid braking system limits the braking force to a value at which the wheels of the truck are just about to slide. Knowing that the coefficient of static friction between the road and the wheels is 0.65, determine the shortest time needed for the truck to slow down.

SOLutiOn

v

v1

2

90 25

30 8 33

= == =

km/h m/s

km/h m/s.

N mg=

mv Nt mvs1 2- =m

m m t m( ( . ) ( . ) ( .25 0 65 9 81 8 332 m/s) m/s m/s)- =

t = -25 8 33

0 65 9 81

.

( . )( . ) t = 2 61. s b

194


PROBLEM 13.125

A truck is traveling down a road with a 4-percent grade at a speed of 80 km/h when its brakes are applied to slow it down to 30 km/h. An antiskid braking system limits the braking force to a value at which the wheels of the truck are just about to slide. Knowing that the coefficient of static friction between the road and the wheels is 0.60, determine the shortest time needed for the truck to slow down.

SOLutiOn

q = = ∞-tan .1 4

1002 29

m mv v1 2+ =-S imp1 2

+ mv mg t Ft mv1 2+ - =sinq

v N W W mg

v F N mgs s

1

2

80200

9

3025

3

= = = =

= = = =

km/h m/s

km/h m/s

cosq

m m ccosq

( ) ( )( . )(sin . )( ) ( . )( )( . )(cos .m m t m200

99 81 2 29 0 60 9 81 2Ê

ËÁˆ¯̃

+ ∞ - 22925

3∞ = Ê

ËÁˆ¯̃

)( ) ( )t m

t =-

∞ - ∞

253

2009

9 81 2 29 0 6 2 29( . sin . . cos . )¥ 9.81¥ t = 2 53. s b

195


PROBLEM 13.126

Baggage on the floor of the baggage car of a high-speed train is not prevented from moving other than by friction. Determine the smallest allowable value of the coefficient of static friction between a trunk and the floor of the car if the trunk is not to slide when the train decreases its speed at a constant rate from 200 km/h to 90 km/h in a time interval of 12 s.

SOLutiOn

v

v1

2

200 55 56

90 25 0

= == =

km/h m/s

km/h m/s

.

.

+ mv m gt mvs1 1 2 2- =-m

( . ) ( . )( )55 56 9 81 12 252 m/s m/s s m/s- =ms

ms = - =( . . )

( . )( ).

55 56 25 0

9 81 120 2596 ms = 0 260. b

196


PROBLEM 13.127

Solve Problem 13.126, assuming that the train is going down a 5-percent grade.

PROBLEM 13.126 Baggage on the floor of the baggage car of a high-speed train is not prevented from moving other than by friction. Determine the smallest allowable value of the coefficient of static friction between a trunk and the floor of the car if the trunk is not to slide when the train decreases its speed at a constant rate from 200 km/h to 90 km/h in a time interval of 12 s.

SOLutiOn

v

v1

2

200 55 56

90 25 0

= == =

km/h m/s

km/h m/s

.

.

t1 2 12- = s

Nt Wt1 2 1 2- -= cosq

+ mv m gt m gt mvs1 1 2 1 2 2- + =- -m q qcos sin

( . ) ( . )( )(cos . ) ( . )(55 56 9 81 12 2 86 9 81 122 2 m/s m/s s m/s s- ∞ +ms ))(sin . )2 86 25∞ = m/s

ms = - + ∞∞

55 56 25 0 9 81 12 2 86

9 81 12 2 86

. . ( . )( )(sin . )

( . )( )(cos . ) ms = 0 310. b

197


PROBLEM 13.128

A sailboat weighing 500 kg with its occupants is running downwind at 12 km/h when its spinnaker is raised to increase its speed. Determine the net force provided by the spinnaker over the 10-s interval that it takes for the boat to reach a speed of 18 km/h.

SOLutiOn

v t

v

1 1 2

2

1210

310

18 5

= = =

= =

- km/h m/s

km/h m/s

sec

m v mv

m F mn

◊ + =

ÊËÁ

ˆ¯̃

+ =

-1 1 2 2

10

310 5

imp

s )( ) (

Fn =

ÊËÁ

ˆ¯̃

= =( )

.500 5

103

102503

83 33-

N Fn = 83 3. N b

Note: Fn is the net force provided by the sails. The force on the sails is actually greater and includes the force needed to overcome the water resistance on the hull.

198


SOLutiOn

(a) Entire train: v1 72 20= = km/h m/s

m mA B+ = + =( , , ) ,18 000 13 000 31 000 kg

+ :0 21500 21500 31 000 201 2= - + +-( ) ( , )( )t

t1 2

31 000 2043000

14 41865

- =

=

( , )( )( )

.

kg m/sN

s t1 2 14 42- = . s b

(b) Car A: m tA = =-18000 14 41861 2 kg, s.

+

: ( )0

0

1 2 1= - + +

=

F F t m v

F

B C A

C

-

-(21500 + )14.4186 + (18,000)(20)

FC = 3467 7. N FC = 3470N (tension) b

PROBLEM 13.129

A light train made of two cars travels at 72 km/h. Car A weighs 18000 kg, and car B weighs 13000 kg. When the brakes are applied, a constant braking force of 21.5 kN is applied to each car. Determine (a) the time required for the train to stop after the brakes are applied, (b) the force in the coupling between the cars while the train is slowing down.

199


SOLutiOn

(a) Entire train: v1 72 20= = km/h m/s

m mA B+ = 31 000, kg

+ :0 21500 31 000 201 2= - +-( ) ( , )( )N kg m/st

t1 2 28 837- = . s t1 2 28 8- = . s b

(b) Car A:

+ :0 28 83751 2 1 1 2= - + =- -F t m v tC A( ) . s

FC = =( , )(

), .

18 000 2012 483 9

)

(28.837N FC = 12 480, N (tension) b

PROBLEM 13.130

Solve Problem 13.129, assuming that a constant braking force of 21.5 kN is applied to car B, but the brakes on car A are not applied.

PROBLEM 13.129 A light train made of two cars travels at 72 km/h. Car A weighs 18000 kg, and car B weighs 13000 kg. When the brakes are applied, a constant braking force of 21.5 kN is applied to each car. Determine (a) the time required for the train to stop after the brakes are applied, (b) the force in the coupling between the cars while the train is slowing down.

200


SOLutiOn

v = =90 25 km/h m/s

(a) The shortest time for the rig to come to a stop will be when the friction force on the wheels is maximum. The downward force exerted by the trailer on the cab is assumed to be zero. Since the trailer brakes fail, all of the braking force is supplied by the wheels of the cab, which is maximum when the wheels of the cab are at impending sliding.

Ft N t N m g g

Ft gts C C C1 2 1 2

1 2

2000

0 65 2000- -

-

= = ==

m ( )

( . )( )

[( ) ] [( ) ]m m v Ft m m vC T C T+ = - + +2 1

+ 0 0 65 2000 9 81 10 000 2521 2= - =-( . )( )( . )( ) , ( ) kg m/s kg m/st

t1 2 19 60- = . s b

(b) For the trailer:

+ [ ] [ ]m v Qt m vT T2 1 2 1= - +-

From (a), t

Q

Q

1 2 19 60

19 60 8000

10 204

- == - +=

.

( . ) (

,

s

0 s kg)(25 m/s)

N

Q = 10 20. kN (compression) b

PROBLEM 13.131

A trailer truck with a 2000-kg cab and an 8000-kg trailer is traveling on a level road at 90 km/h. The brakes on the trailer fail and the antiskid system of the cab provides the largest possible force which will not cause the wheels of the cab to slide. Knowing that the coefficient of static friction is 0.65, determine (a) the shortest time for the rig to come to a stop, (b) the force in the coupling during that time.

201


SOLutiOn

(a) Blocks A and C

[( ) ] ( ) ( ) [( ) ]m m v T t m m gt m m vA C A C A C+ - + + = +- -1 1 2 1 2 2

0 12 0 8 12+ - =( )( . )g T v (1)

Block B

[ ] ( ) ( )m v T t m gt m vB B B1 1 2 1 2 2+ - =- -

0 4 0 8 4+ - =( )( . )T g v (2)

Adding (1) and (2), (eliminating)

( )( . ) ( )12 4 0 8 12 4g g- = +

v =( )( . )( . )8 9 81 0 8

16

2 kg m/s s

kg v = 3 92. m/s b

(b) Collar A

( ) ( )m v F m gA C A1 0 0= + + (3)

From Eq. (2) with v = 3 92. m/s

Tv

g

T

T

= +

= +

=

4

0 84

4 3 924 9 81

58 8

2

.( )( . )

( )( . )

.

kg m/s

(0.8 s) kg m/s

44 N Solving for FC in (3)

FC = - +( )( .( )( . ) .

4 3 924 9 81 58 84

kg m/s)

(0.8 s) kg m/s N2

FC = 39 2. N b

PROBLEM 13.132

An 8-kg cylinder C rests on a 4-kg platform A supported by a cord which passes over the pulleys D and E and is attached to a 4-kg block B. Knowing that the system is released from rest, determine (a) the velocity of block B after 0.8 s, (b) the force exerted by the cylinder on the platform.

+

+

+

202


SOLutiOn

Kinematics: Dependent motion

Cable length: L x x xA A B= + + + constant

dL

dtv vA B= + =2 0

Here velocities are defined as positive if downward.

v vB A= -2 (1)

Let T be the tension in the cable.

Use the principle of impulse and momentum.

Collar A, components : 2Tt m gt m vA A A- = - (2)

Block B, components : Tt m gt m vB B B- = - (3)

Subtract twice Eq. (3) from Eq. (2) to eliminate Tt.

( )

( )

2 2

2

2

m m gt m v m v

tm v m v

m m g

B A B B A A

B B A A

B A

- = -

=-

-

Data: m m m mA B B A= = - =20 30 2 10 kg kg kg

vA = -1 0. m/s, i.e., 1.0 m/s

From Eq. (1) vB = 2 0. m/s

t = - -[( )( )( . ) ( )( . )]

( )( . )

2 15 2 0 20 1 0

10 9 81 t = 0 815. s b

PROBLEM 13.133

The system shown is released from rest. Determine the time it takes for the velocity of A to reach 1 m/s. Neglect friction and the mass of the pulleys.

203


Solution

The block does not move until P = 2 ¥ 9.81 = 19.62N

From t t P t= = =0 2 25to s, .

Thus, the block starts to move when t = =19 62

250 7848

.. s.

(a) For 0 2� �t s

P t= 25

+ t t v1 2 10 7848 2 0= = =. s s,

mv Pdt mg t t mv

tdt v

t

t

1 2 1 2

20 784

1

2

0 25 19 62 2 0 7848 2

+ - - =

+ - - =

Ú ( )

. ( . ). 88

2

Ú

v22 21

2

25

22= ÈÎ ˘̊Ï

ÌÓ

¸˝˛

( ) - (0.7848) - 23.842

v2 9 2296= . m/s v2 9 23= . m/s b

(b) From t t= =2 3 s to s

v a

P t

t t

2 9 2296

50 2 3

2 3

=== =

. m/s, from ( )

N s s

s s2 3

� �

mv Pdt mg t t mv

t

t

2 3 2 3

3

2

3

2 9 2296 50 3 2 2

+ - - =

+

Ú ( )

( . ) ( )- -19.62(3 - ) = 2v

v3 24 4196= . m/s v3 24 4= . m/s b

PRoBlEM 13.134

A 2 kg collar which can slide on a frictionless vertical rod is acted upon by a force P which varies in magnitude as shown. Knowing that the collar is initially at rest, determine its velocity at (a) t = 2 s, (b) t = 3 s.

204


SOLutiOn

(a) Determine time at which collar starts to move.

P t t= < <257 0 2, s

Collar moves when P g tP= = = = =2 19 6225

19 62

250 7848.

..N or s

mv Pdt Wdt mvt t

1 4 5 4 5 2+ - =Ú Ú/ /

For t < 2 s, P t= 25 ( )N

2 3 50 s s, N

3 s, 0

< < => =

t P

t P

W = 19 62. N

The maximum velocity occurs when the total impulse is maximum.

Area impulse

Ar

ABCD = = +max ( )( . ) ( . )( )12

50 2 0 7848 50 19 62 1-19.62 - -

eea N.s

0 N.s m/s

ABCD

v v

=

+ = =

48 8389

48 8389 2 24 4194

.

. .max maxfi

vmax .= 24 4 m/s b

(b) Velocity is zero when total impulse is zero at t + Dt.

For 0 7848 3 48 8389. . ( ))< < =t a s, impulse N.s (Part

For D Dt t beyond s, impulse3 19 62= - .

Thus,

Total impulse

s

= = -=

0 48 8389 19 62

2 489

. .

.

DD

t

t

Time to zero velocity t = +3 2 489 s s. t = 5 49. s b

PROBLEM 13.135

A 2 kg collar which can slide on a frictionless vertical rod is acted upon by a force P, which varies in magnitude as shown. Knowing that the collar is initially at rest, determine (a) the maximum speed of the collar, (b) the time when the velocity is zero.

205


SOLutiOn

+

0 + - =ÚÚ Pdt Fdt mv

At any time: vm

Pdt Fdt= -ÈÎ

˘˚ÚÚ

1 (1)

(a) Block starts moving at t1.

P F Ws s= = = =m ( . )( )( . ) .0 50 60 9 81 294 3N

t

F

t

s

1 18

500 294 3

8

5004 7088= = =

s

N N

s

Ns;

.. t1 4 71= . s b

(b) Maximum velocity: At t tm=

where P F Wk k= = = =

=

=

m ( . )( )( . ) .

..

0 4 60 9 81 235 44

16235 44

85500

12

1

1

N;

- tm

tm 3333s

Block moves at t = 4 7088. s.

Shaded area is maximum net impulse Pdt F dtK- ÚÚ .

when t t v vm m= =1

PROBLEM 13.136

A 60 kg block initially at rest is acted upon by a force P which varies as shown. Knowing that the coefficients of friction between the block and the horizontal surface are ms = 0 50. and mk = 0 40. , determine (a) the time at which the block will start moving, (b) the maximum speed reached by the block, (c) the time at which the block will stop moving.

206



Eq. (1): vm mm =

È

ÎÍ

˘

˚˙ = + -1 1 1

2294 3 500 235 44 8

shaded

area( . . )(- 235.44 - 4.70088 - 235.44) ( )( . )

( . )

+ÈÎÍ

˘˚̇

=

1

2500 4 333

11105 389

m

vm = =11105 389 18 423

60. . m/s vm = 18 42. m/s b

(c) Block stops moving when Pdt Fdt Qdt Fdt-ÈÎ

˘˚ = =ÚÚ Ú Ú0; or

Assume tm >16 s.

Pdt

Fdt tm

Ú

Ú

= =

= + -

21

2500 4000

1

2294 3 4 7088 235 44

¥ ¥ 8 ¥ N.s

( . )( . ) ( . )( 44 7088. )

Pdt Fdt- ÚÚ tm = 18 755. s

tm > 16 s OK tm = 18 76. s b

207


PROBLEM 13.137

Solve Problem 13.136, assuming that the weight of the block is 90 kg.

PROBLEM 13.136 A 60 kg block initially at rest is acted upon by a force P, which varies as shown. Knowing that the coefficients of friction between the block and the horizontal surface are ms = 0 50. and mk = 0 40. , determine (a) the time at which the block will start moving, (b) the maximum speed reached by the block, (c) the time at which the block will stop moving.

SOLutiOn

See solution of Problem 13.136.

W vm

Pdt Fdt= = = -ÈÎ

˘˚ÚÚ90 9 81 882 9

1¥ . . N (1)

(a) Block starts moving. P F Ws s= = = =m ( . )( . ) .0 50 882 9 441 45N

See first figure of Problem 13.136.

t

F

t

s

1 18

500 441 5

8

500= = s s

N;

. t1 7 0632= . s b

(b) Maximum velocity. P F Wk k= = = =m 0 4 882 9 353 16. ( . ) . N

16

353 16

8

50010 349

-=

=

t

t

m

m

..

s

s

Eq. (1): vm

m

m =È

ÎÍ

˘

˚˙

= + +

1

1 12

88 29 146 84 812

146

shaded

area

( . . )( ) ( .- 7.0632 884 10 349 8

1282 598

)( . )

( . )

-ÈÎÍ

˘˚̇

=m

vm =

=

1282 598

3 139990

( . )

. m/s vm = Æ3 14. m/s b

208


Problem 13.137 (continued)

(c) Block stops moving when net impulse ( )P F dt-ÈÎ

˘˚ =Ú 0

Assume ts �16 s.

Pdt

tt

t

t ss

s

s

0

2

1

2500 8

1

2500 500

16

88

250

8

Ú = + +-È

ÎÍ˘˚̇

-

= -

( )( )( )

( )

( -- +32 128ts )

Fdt tt

ss

0

12

441 5 7 0632 353 16 7 0632Ú = + -( . )( . ) . ( . )

Pdt Fdt tsÚ Ú- = + =0 31 25 3064 761 02fi - 646.84. .ts

Solving for ts , ts = 13 356. s ts = 13 36. s b

(the second rod < 8, hence discarded)

209


SOLutiOn

At t = 0, p p c c

c p

= = -=

0 1 2

1 0

0( )

At t = ¥ -1 6 10 3. s, p

c c

cp

=

= - ¥

=¥

-

-

0

0 1 6 10

1 6 10

1 23

20

3

( . )

.

s

s

m = ¥ -20 10 3 kg

0 2

0

1 6 10 3

+ =¥ -

ÚA pdt mv. s

A

A

=

= ¥

-

-

p ( )

.

10

4

78 54 10

3 2

6 2m

020 10

1 2

3

0

1 6 10 3

+ - = ¥ -¥ -

ÚA c c t dtg

( ). s

( . ) ( )( . )( )( .

(78 54 10 1 6 101 6 10

2206 2

13 2

3

¥ ¥ -¥È

ÎÍ

˘

˚˙ = ¥- -

-m s

s)2

cc

110 7003- kg m/s)( )

1 6 10 1 280 10 178 25 10

1 6 101 280

31

62

3

3 20

. . .

( . )( .

¥ - ¥ = ¥

¥ ◊ -¥

- -

-

c c

pm s110

1 6 10178 25 10

222 8 10

6 2

3 03

06 2

-

-◊

¥= ¥ ◊

= ¥

m s

skg m/s

N/m

2 )

( . ).

.

p

p p0 223= MPa b

PROBLEM 13.138

A simplified model consisting of a single straight line is to be obtained for the variation of pressure inside the 10-mm-diameter barrel of a rifle as a 20-g bullet is fired. Knowing that it takes 1.6 ms for the bullet to travel the length of the barrel and that the velocity of the bullet upon exit is 700 m/s, determine the value of p0.

210


PROBLEM 13.139

The following mathematical model was suggested for the variation in pressure inside the 10-mm-diameter barrel of a rifle as a 25-g bullet was fired:

p t e t( ) ( ) ( . )= -950 0 16 MPa / ms

where t is expressed in ms. Knowing that it took 1.44 ms for the bullet to travel the length of the barrel and that the velocity of the bullet upon exit was measured to be 520 m/s, determine the percent error introduced if the above equation is used to calculate the muzzle velocity of the rifle.

SOLutiOn

A = = ¥-

-p ( ).

10

478 54 10

3 26 2m

0 78 54 10 950 106 2 6 0 16 10

0

1 44 10 33

+ ¥ ¥- - ¥¥ --

Ú( . ) ( / )( )/ ..m N m2s

e dtt == ¥

¥ ¥ - ¥

-

- - -

( )

( . )( )( . )( . / .

25 10

78 54 10 950 10 0 16 10

32

6 6 3 1 44 0

kg v

e 116 321 25 10- = ¥ -) v dt

v2 477 46

477 46 520 42 54

10042 54

5

== - = -

= -

.

. .

%.

m/s

Error m/s

error220

ÊËÁ

ˆ¯̃

percent error = 8 18. % b

211


SOLutiOn

m t m tv P W v1 2 0 18+ - = =( ) .D D s

Vertical components

0 0 18 12 50

12 50

9 81 0 18

+ - = ∞

= +∞

( )( . ) ( )(sin )

( )(sin )

( . )( . )

P WW

g

P W

v

v WW

P Wv = 6 21. b

PROBLEM 13.140

The triple jump is a track-and-field event in which an athlete gets a running start and tries to leap as far as he can with a hop, step, and jump. Shown in the figure is the initial hop of the athlete. Assuming that he approaches the takeoff line from the left with a horizontal velocity of 10 m/s, remains in contact with the ground for 0.18 s, and takes off at a 50° angle with a velocity of 12 m/s, determine the vertical component of the average impulsive force exerted by the ground on his foot. Give your answer in terms of the weight W of the athlete.

212


PROBLEM 13.141

The last segment of the triple jump track-and-field event is the jump, in which the athlete makes a final leap, landing in a sand-filled pit. Assuming that the velocity of a 90 kg athlete just before landing is 9 m/s at an angle of 35° with the horizontal and that the athlete comes to a complete stop in 0.22 s after landing, determine the horizontal component of the average impulsive force exerted on his feet during landing.

SOLutiOn

W

t

=D =

90

0 22

¥ 9.81 = 882.9N. s

m t mv P W v1 2+ - =( )D

Horizontal components

m( )(cos ) ( . )

( )( )cos( . )

.

9 35 0 22 0

90 9 350 22

3015

∞ - =

= ∞ =

P

P

H

Hkg m/s

s997 N PH = 3020 N b

213


PROBLEM 13.142

An estimate of the expected load on over-the-shoulder seat belts is to be made before designing prototype belts that will be evaluated in automobile crash tests. Assuming that an automobile traveling at 72 km/h is brought to a stop in 110 ms, determine (a) the average impulsive force exerted by a 100 kg man on the belt, (b) the maximum force Fm exerted on the belt if the force-time diagram has the shape shown.

SOLutiOn

(a) Force on the belt is opposite to the direction shown.

v

W1 72 20

100 9 81 981

= == =

km/h m/s

N¥ .

m dt mv F v1 2- =Ú

Fdt F t

t

=

= =

Ú ave

s

D

D ¥ -3110 10 0 110.

( )( ) ( . )100 20 0 110 0- =Fave s

Fave N= =( )( )

( . ), .

100 20

0 11018 181 8 Fave kN= 18 18. b

(b) Impulse = area under F t Fm- = diagram s1

20 110( . )

From (a), impulse ave= =F tD 2000

1

20 110 2000

36363 6

F

F

m

m

( . )

.

=

= N Fm = 36 4. kN b

214


PROBLEM 13.143

A 46-g golf ball is hit with a golf club and leaves it with a velocity of 50 m/s. We assume that for 0 � �t t0, where t0 is the duration of the impact, the magnitude F of the force exerted on the ball can be expressed as F = F t tm sin ( ).p / 0 Knowing that t0 0 5= . ms, determine the maximum value Fm of the force exerted on the ball.

SOLutiOn

m

t

= =

= = ¥ -

46 0 046

0 5 0 5 1003

g kg

ms s

.

. .

The impulse applied to the ball is

F dt F

t

tdt

F t t

t

F t

t

m

t mt

m

0 00

0

0 0

0

0 00

0

Ú Ú= = -

= - -

sin cos

(cos cos )

pp

p

pp ==

2 0F tm

p

Principle of impulse and momentum.

m dt mt

v F v1 0 20+ =Ú

With v1 0= ,

02 0

2+ =F t

mvm

p

Solving for Fm, Fmv

tm = = ¥¥

= ¥-

-p p2

0

3

33

2

46 10 50

2 0 5 107 23 10

( )( )

( )( . ). N Fm = 7 23. kN b

215


SOLutiOn

m

F

t

= == == =

200 0 200

2 2000

2 0 002

g kg

kN N

ms save

.

.D

(a) Velocity immediately after impact.

Use principle of impulse and momentum.

v v F t

m p m1 2 1 2

1 1 2 2

0= = =+ =

Æ

Æ

? ( )Imp ave Dv Im v

0 2+ =F t mvave ( )D

vF t

m22000 0 002

0 200= =ave ( ) ( )( . )

.

D v2 20 0= . m/s b

(b) Average resistance to penetration.

Dx

v

v

= ===

1 0 001

20 0

02

3

mm m

ft/s

.

.

Use principle of work and energy.

T U T mv R x2 2 3 3 221

20+ = - =Æ or ave ( )D

Rmv

xave N= = = ¥22 2

3

2

0 200 20 0

2 0 00140 10

( )

( . )( . )

( )( . )D Rave kN= 40 0. b

PROBLEM 13.144

The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur. Assuming the punch applies an average force of 2 kN over a time of 2 ms to the 200 g implant, determine (a) the velocity of the implant immediately after impact, (b) the average resistance of the implant to penetration if the implant moves 1 mm before coming to rest.

216


SOLutiOn

(a) The momentum of the system consisting of the two cars is conserved immediately before and after coupling.

Before coupling After coupling

+

S Smv mv= ¢

0 20 4 20 40+ = + ¢( )( ) ( )( )Mg km/h Mg Mg v

¢ =+

v( )( )

( )

20 4

20 40 ¢ = ¨v 1 333. km/h b

(b) After coupling

The friction force acts only on the 40 Mg car since its wheels are locked. Thus,

F N

F

f k

f

= = ¥

= ¥

m 403

3

0 30 40 10 9 81

117 72 10

( . )( )( . )

.

kg m/s

N

2

From (a), v v1 1 333 0 3704= ¢ = =. . km/h m/s

Impulse momentum

S Smv F dt mv

dt

f

t

1 20

3 3

060 10 0 3704 117 72 10 0

+ =

¥ - ¥ =

Ú( )( . ) ( . )kg m/s N

tt

t

Ú

= ¥¥

( )( . )

( . )

60 10 0 3704

117 72 10

3

3 t = 0 1888. s b

PROBLEM 13.145

A 20-Mg railroad car moving at 4 km/h is to be coupled to a 40-Mg car which is at rest with locked wheels ( . ).mk = 0 30 Determine (a) the velocity of both cars after the coupling is completed, (b) the time it takes for both cars to come to rest.

217


PROBLEM 13.146

At an intersection, car B was traveling south and car A was traveling 30° north of east when they slammed into each other. Upon investigation, it was found that after the crash, the two cars got stuck and skidded off at an angle of 10° north of east. Each driver claimed that he was going at the speed limit of 50 km/h and that he tried to slow down but couldn’t avoid the crash because the other driver was going a lot faster. Knowing that the masses of cars A and B were 1500 kg and 1200 kg, respectively, determine (a) which car was going faster, (b) the speed of the faster of the two cars if the slower car was traveling at the speed limit.

SOLutiOn

(a) Total momentum of the two cars is conserved.

Smv x, : m v m m vA A A Bcos ( ) cos30 10∞ = + ∞ (1)

Smv y, : m v m v m m vA A B B A Bsin ( ) sin30 10∞ - = + ∞ (2)

Dividing (1) into (2),

sin

cos cos

sin

cos

30

30 30

10

10

∞∞

-∞

=∞∞

m v

m vB B

A A

v

v

m

m

v

v

B

A

A

B

B

A

=∞ - ∞ ∞

=

(tan tan )( cos )

( . )( )

( )c

30 10 30

0 40101500

1200oos

. .

30

0 434 2 30

∞

= =v

vv vB

AA B

Thus, A was going faster. b

(b) Since vB was the slower car,

vB = 50 km/h

vA = ( . )( )2 30 50 vA = 115 2. km/h b

218


PROBLEM 13.147

A mother and her child are skiing together, with the mother holding the end of a rope tied to the child’s waist. They are moving at a speed of 7.2 km/h on a flat portion of the ski trail when the mother observes that they are approaching a steep descent. She decides to pull on the rope to decrease the child’s speed. Knowing that this maneuver causes the child’s speed to be cut in half in 3 s and neglecting friction, determine (a) the mother’s speed at the end of the 3-s interval, (b) the average value of the tension in the rope during that time interval.

SOLutiOn

(a) Consider mother and child as a single system. Assuming the friction force on the skis is negligible, momentum is conserved.

m v m v m v m v

v v

v

C C M M C C M M

C M

C

+ = ¢ + ¢= =

¢ =+

7 2

3 6

20 7 2

.

.

( )( . ) (

km/h

km/h

555 7 2 20 3 6 55)( . ) ( . ) ( )( )= + ¢vM

¢ =vm 8 51. km/h b

(b) Child alone.

t

m v F t m v

v

v

C C AV C C

C

C

=- = ¢

= =¢ = =

3

7 2 2

3 6 1

s

km/h m/s

km/h m/s

.

.

( )( ) ( ) ( )( )20 2 3 20 1 kg m/s s kg m/s- =FAV

FAV = = ◊( )( )

( ).

20 1

36 67 2 kg m/s

s kg m/s FAv = 6 67. N b

219


SOLutiOn

Masses:

Bullet: mB = ¥ -15 10 3kg 0.015kg=

Blocks A and C: m mA C= = 2kg

Block C + bullet: m mC B+ = 2 015. kg

Normal forces for sliding blocks from N mg- = 0

Block A: N m gA A= = =( )( . ) .2 9 81 19 62N

Block C + bullet: N m m gC C B= + = =( ) ( . )( . ) .2 015 9 81 19 767N

Let v0 be the initial speed of the bullet;

v1 be the speed of the bullet after it passes through block A;

vA be the speed of block A immediately after the bullet passes through it;

vC be the speed of block C immediately after the bullet becomes embedded in it.

Four separate processes and their governing equations are described below.

1. The bullet hits block A and passes through it. Use the principle of conservation of momentum.

( )vA 0 0=

m v m v m v m v

v vm v

m

B A A B A A

A A

B

0 0 1

0 1

+ = +

= +

( )

(1)

2. The bullet hits block C and becomes embedded in it. Use the principle of conservation of momentum.

( )vC 0 0=

m v m v m m v

vm m v

m

B C C B C C

B C C

B

1 0

1

+ = +

=+

( ) ( )

( ) (2)

PROBLEM 13.148

Bullet B weighs 15 g and blocks A and C both weigh 2 kg. The coefficient of friction between the blocks and the plane is mk = 0 25. . Initially, the bullet is moving at v0, and blocks A and C are at rest (Fig. 1). After the bullet passes through A, it becomes embedded in block C, and all three objects come to stop in the positions shown (Fig. 2). Determine the initial speed of the bullet v0.

220



3. Block A slides on the plane. Use principle of work and energy.

T U T

m v N d vN d

m

d

A A k A A Ak A A

A

A

1 1 2 2

21

20

2

0 15

+ =

- = =

=

Æ

mm

or

m. (3)

4. Block C with embedded bullet slides on the plane. Use principle of work and energy.

dC = =100 0 1mm m.

T U T

m m v N d vN d

m mC B C k C C Ck C C

C B

1 1 2 2

21

20

2

+ =

+ - = =+

Æ

( ) mm

or (4)

Applying the numerical data:

From Eq. (4), vC =

=

( )( . )( . )( . )

..

2 0 25 19 767 0 1

2 0150 70035m/s

From Eq. (3), vA =

=

( )( . )( . )( . )

( )

.

2 0 25 19 62 0 15

2

0 85776m/s

From Eq. (2), v12 015 0 70035

0 015

94 080

=

=

( . )( . )

( . )

. m/s

From Eq. (1), v0 94 0802 0 85776

0 015= +.

( )( . )

( . ) v0 208= m/s b

221


SOLutiOn

(a) For the system consisting of both balls connected by a cord, the total momentum is conserved.

x mv m vB x: cos ( )- = ¢0 q (1)

( ) cos¢ = -

= -

v v

va

L

B x 0

0

q

y mv mv m v yA B: sin ( )0 q = ¢ + ¢ (2)

Since the cord is inextensible,

¢ = ¢v vA B y( ) (3)

Thus, from (2) v vA0 2sinq = ¢ ¢ = -v v L L aA ( / )02 22 b

PROBLEM 13.149

Two identical spheres A and B, each of mass m, are attached to an inextensible inelastic cord of length L, and are resting at a distance a from each other on a frictionless horizontal surface. Sphere B is given a velocity v0 in a direction perpendicular to line AB and moves without friction until it reaches B¢, when the cord becomes taut. Determine (a) the magnitude of the velocity of each sphere immediately after the cord has become taut, (b) the energy lost as the cord becomes taut.

222



From (3): ( )¢ = ¢v vB y A

= -

¢ = ¢ + ¢

= + -

( / )

( ) ( )

( )

v L L a

v v v

va

L

L a

L

B B x B y

C

02 2

2 2

2

2

2 2

2

2

4 ¢ = +v v L L aB ( / )0

2 22 3 b

(b) Initial T mv= 1

2 02

¢ = ¢ + ¢

= - + +

¢ =

T m v m v

m v L L a L a

T mv

A B1

2

1

2

1

22 3

1

2

2

02 2 2 2 2

( ) ( )

( / ) [( ) ( )]

002 2 2 2

02 2 2 2

02

02

4 2 2

4

1

24

/ ( )

/ ( )

/

L L a

mv L L a

T T T

mv mv L

( ) +

= ( ) +

= - ¢

= -

D

22 2 2( ) +( )L a DT mv L L a= ( ) -02 2 2 24/ ( ) b

223


PROBLEM 13.150

Two swimmers A and B of weight 90 kg and 60 kg, respectively, are at diagonally opposite corners of a floating raft when they realize that the raft has broken away from its anchor. Swimmer A immediately starts walking toward B at a speed of 0.6 m/s relative to the raft. Knowing that the raft weighs 150 kg, determine (a) the speed of the raft if B does not move, (b) the speed with which B must walk toward A if the raft is not to move.

SOLutiOn

(a) The system consists of A and B and the raft R.

Momentum is conserved.

( ) ( )S Sm m

m m mA A B B R R

v v

v v v1 2

0

== + + (1)

v v v v = v v

v v v v

A A R R B B R R B R

A R B R

A

v

m

= + + =

= + =

= +

/ / /

.

[ .

0

0 6

0 0 6

m/sZ

Z

A

B

A

Bvv v

v

R B R R R

RA

A B R

m m

m

m m m

]

.

( )

( . )( )

+ +

=-+ +

=-

+

v

0 6 0 6 90

60

m/s kg

(90 kg kgg kg)+150 vR = 0 18. m/s b

(b) From Eq. (1),

0 0 0

0 6

0 6

0

= + + =

= - = + =

= -

m v m v v

vm v

mv v v

v

A A B B R

BA A

BA A R R

B

( )

.

( .

/ Z

m/s

m//s kg

kgm/s

)( )

( ).

90

600 9= vB = 0 900. m/s b

224


SOLutiOn

(a) For the system which is the ball and the plate, momentum is conserved.

All forces are non-impulsive, except the equal and opposite forces between the plate and the ball.

+ ( ) ( ) ( )mv mv mvB B p= - ¢ + ¢

( . ( . ) ( .

.

0 125 0 125 0 250

0 5

kg)(3 m/s) kg)( kg)= - ¢ + ¢

¢ = ¢ +

v v

v v

B p

p B 11 5. (1)

Since there is no energy lost, the kinetic energy of the system is conserved.

Before impact, ¢ =

=

=

T m vB B1

21

20 125

0 563

2

( .

.

kg)(3 m/s)

J

2

After impact, ¢ = ¢( ) + ¢( )T m v m vB B P P1

2

1

22 2

Substituting for ¢vp from (1)

¢ = ¢ + ÊËÁ

ˆ¯̃ ¢ +

¢ =

T v v

T

B B1

20 125

1

20 250 1 5

0

2 2( . ) ( . . ]

.

kg)( kg)[0.5

009375 0 1875 0 2813

0 563 0 09375 0 1875

2

2

( ) . .

. . ( ) .

¢ + +

= ¢ = ¢ +

v v

T T v

B B

B vvB + 0 2813.

¢ + ¢ - =

¢ = - + = - = - Ø + ≠

v v

v

B B

B

2 2 3 0

2 4 12

21 2 3 1

m

m ,

( ¢ = -vB 3 m/s before impact) ¢ = ≠vB 1 000. m/s b

PROBLEM 13.151

A 125-g ball moving at a speed of 3 m/s strikes a 250-g plate supported by springs. Assuming that no energy is lost in the impact, determine (a) the velocity of the ball immediately after impact, (b) the impulse of the force exerted by the plate on the ball.

225



(b) Ball alone.

+ ( . ( .0 125 3 0 125 kg)( m/s) kg)(1 m/s)- + =F tD FDt = ◊ ≠0 500. N s b

226


SOLutiOn

For the system which is the bullet and the block, momentum in the horizontal direction is conserved.

+ mv M m v

vmv

M m

0

0

cos ( )

cos

( )

qq

= + ¢

¢ =-

+

Bullet alone.

+ - + = ¢mv P t mvx0 cosq D

P t mvm

M mxD = -+

ÈÎÍ

˘˚̇0 1cosq P t

mM

M mvxD =

+ 0 cosq b

+ - + =mv P ty0 0sinq D

P t mvyD = 0 sinq b

PROBLEM 13.152

A bullet of mass m is fired with a velocity v0 forming an angle q with the horizontal and gets lodged in a wooden block of mass M. The blocks can roll without friction on a hard floor and is prevented by springs from hitting the wall. Determine the horizontal and vertical components of the impulse of the force exerted by the block on the bullet.

227


SOLutiOn

Velocity of the block just before impact.

T V Wh

T mv V

T V T V

1 1

22

1 1 2 2

0 30 441 45

1

20

0

= = = =

= =

+ = +

+

( .)(9.81)(1.5) J

2

4441 451

230

2 441 45

305 4249

2. ( )

( . )

.

=

=

=

v

v

m/s

(a) Rigid columns.

+ - + = =mv F t F tD D0 30 5 4249( )( . )

F tD = ◊ ≠162 747. N s on the block. F tD = ◊162 7. N s b

All of the kinetic energy of the block is absorbed by the chain.

T =

=

1

230 5 4249

441 45

2( )( . )

. J E = 441 J b

PROBLEM 13.153

In order to test the resistance of a chain to impact, the chain is suspended from a 120-kg rigid beam supported by two columns. A rod attached to the last link is then hit by a 30-kg block dropped from a 1.5 m height. Determine the initial impulse exerted on the chain and the energy absorbed by the chain, assuming that the block does not rebound from the rod and that the columns supporting the beam are (a) perfectly rigid, (b) equivalent to two perfectly elastic springs.

228



(b) Elastic columns.

Momentum of system of block and beam is conserved.

mv M m v

vm

m Mv

= + ¢

¢ =+

=

( )

( )( .

30

1505 4249m/s) ¢ =v 1 085. m/s

Referring to figure in Part (a),

- + = - ¢= - ¢= -= ◊

mv F t mv

F t m v v

DD ( )

( )( . . )

.

30 5 4249 1 085

130 197 N s F tD = ◊130 2. N s b

E mv mv mv= - ¢ - ¢

=

1

2

1

2

1

21

230 5 4249

1

230 1 085

1

212

2 2 2

2 2( )( . ) ( )( . ) (- - 00 1 085

353 151

2)( . )

.= J E = 353J b

229


SOLutiOn

v

m

td

v

= =

=

= = = ( )

140350

90 15

0 15

91

60

km/h m/s

kg

sAV

.

.

+ Æ - + =0 F t mv Fmv

tAV AV

=

ÊËÁ

ˆ¯̃

( )=

(0.15)

N

3509

350

160

b

PROBLEM 13.154

A baseball player catching a ball can soften the impact by pulling his hand back. Assuming that a 150 g ball reaches his glove at 140 km/h and that the player pulls his hand back during the impact at an average speed of 9 m/s over a distance of 150 mm, bringing the ball to a stop, determine the average impulsive force exerted on the player’s hand.

230


SOLutiOn

Let vA and vB be velocity components (positive to the right) of collar A and B. Let ¢vA and ¢vB be the velocities after impact.

Conservation of momentum.

m v m v m v m vA A B B A A B B+ = ¢ + ¢

Data: m m

v vA B

A B

= == = -

5 3

2 1 5

kg, kg

m/s, m/s.

( )( ) ( )( . )5 2 3 1 5 5 3+ - = ¢ + ¢v vA A

or 5 3 5 5¢ + ¢ =v vA B . (1)

Coefficient of restitution.

¢ - ¢ = -= - -

¢ - ¢ =

v v e v v

v v

B A A B

B A

( )

. [ ( . )]

.

0 80 2 1 5

2 8 (2)

(a) Solving Eqs. (1) and (2) simultaneously,

v

vA

B

= -=

0 3625

2 4375

.

.

m/s

m/s ¢ = ¨vA 0 363. m/s b

¢ = ÆvB 2 44. m/s b

(b) Energy loss.

T m v m v

T mv

A A B B

A

12 2

2 2

1

2

1

2

1

25 2

1

23 1 5

13 375

1

2

= +

= + -

=

=

( )( ) ( )( . )

. J

222 2

2 2

1

2

1

25 0 3625

1

23 2 4375

9 2406

+ ¢

= - +

=

m vB B

( )( . ) ( )( . )

. J

T T1 2 4 1344- = . J Energy loss 4.13 J= b

PROBLEM 13.155

The coefficient of restitution between the two collars is known to be 0.80. Determine (a) their velocities after impact, (b) the energy loss during impact.

231


SOLutiOn

(a) m m m

eA B= =

= 0

Conservation of total momentum.

+

mv mv mvA B- = ¢2 ¢ = -v v vA B1

2( ) b

(b) Energy loss.

E T T T T

E m v v m v v

L A B A B

L A B

= + - ¢ + ¢

= +( ) - ¢ + ¢

( )

( )1

2

1

22 2 2 2

From (a),

¢ = -v v vA B1

2( )

E m v v m v v

E m v v m v

L A B A B

L A B A

= +( ) - -ÈÎÍ

˘˚̇

= +( ) - -

1

2

1

2

1

2

1

2

1

4

2 2 2

2 2 2

( )

22

1

42

1

4

2

2 2

2

v v v

E m v v v v

m v v

A B B

L A A B B

A B

+( )= + +ÈÎ ˘̊

= +( ) b

PROBLEM 13.156

Collars A and B, of the same mass m, are moving toward each other with the velocities shown. Knowing that the coefficient of restitution between the collars is 0 (plastic impact), show that after impact (a) the common velocity of the collars is equal to half the difference in their speeds before impact, (b) the loss in kinetic energy is 1

42m A B( ) .υ υ+

232


SOLutiOn

The total momentum is conserved.

+ m v m v m v m vA A B B A A B B+ = ¢ + ¢

( )( ) ( )( ) ( )( ) ( )( .

.

3 3 2 2 3 2 3 1

9 4 6

kg m/s kg m/s kg kg m/s)+ = ¢ +

¢ = + -v

v

A

A22

334

15= m/s


ev v

v vB A

A B

=¢ - ¢

-

=-

-

=

3 13415

3 25

60 8333

.

. e = 0 833. b

PROBLEM 13.157

Two steel blocks are sliding on a frictionless horizontal surface with the velocities shown. Knowing that after impact the velocity of B is observed to be 3.1 m/s to the right, determine the coefficient of restitution between the two blocks.

233


Solution

(a) The total momentum is conserved.


( )( ) ( )( ) ( ) ( )3 3 2 2 3 2

9 4 3 2

kg m/s kg m/s kg kg+ = ¢ + ¢+ = ¢ + ¢

v v

v vA B

A B (1)

Relative velocities.

( ) ( )

( )( . )

.

v v e v v

v v

v v

A B B A

B A

B A

- = ¢ - ¢

- = ¢ - ¢

¢ - ¢ =

3 2 0 75

0 75 (2)

Solving (1) and (2) simultaneously,

¢ = ÆvB 3 05. m/s b

¢ = ÆvA 2 30. m/s b

(b) Energy loss.

E m v m v m v m v

E

L A A B B A A B B

L

= + - ¢ - ¢

= + -

1

2

1

2

1

2

1

21

23 3 2 2

2 2 2 2

2 2[( )( ) ( )( ) (( )( . ) ( )( . ) ]3 2 3 2 3 052 2-

EL = 0 2625. J EL = 0 263. J b

PRoBlEM 13.158

Two steel blocks are sliding on a frictionless horizontal surface with the velocities shown. Knowing that the coefficient of restitution between the two blocks is 0.75, determine (a) the velocity of each block after impact, (b) the loss of kinetic energy due to the impact.

234


SOLutiOn

m m m mA B C= = =

Collision between B and C.


+ mv mv mv mvB C B C¢ + ¢ = +

¢ + ¢ = +v vB C 0 1 5. (1)


( )( ) ( )

( . )( . ) ( )

.

v v e v v

v v

v v

B C BC C B

C B

C B

- = ¢ - ¢- = ¢ - ¢

- = ¢ - ¢1 5 0 8

1 2 (2)


¢ =¢ =

v

vB

C

1 35

0 15

.

.

m/s

m/s ¢ = ¨vC 0 15. 0 m/s b

Since ¢ ¢v vB C� , car B collides with car A.

Collision between A and B.

mv mv mv mv

v vA B A B

A B

¢ + ¢¢ = + ¢¢ + ¢¢ = +0 1 35. (3)

PROBLEM 13.159

Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the same weight, has been pushed by dockworkers and hits car B with a velocity of 1.5 m/s. Knowing that the coefficient of restitution is 0.8 between B and C and 0.5 between A and B, determine the velocity of each car after all collisions have taken place.

235




( ) ( )

( . )( . )

.

v v e v v

v v

v v

A B AB B A

B A

A B

- ¢ = ¢¢ - ¢- = ¢¢ - ¢

¢ - ¢¢ =0 1 35 0 5

0 675 (4)


2 1 35 0 675¢ = +vA . .

¢ = ¨vA 1 013. m/s b

¢¢ = ¨vB 0 338. m/s b

Since ¢ ¢¢ ¢v v vC B A� � , there are no further collisions.

236


SOLutiOn

(a) First collision (between A and B).


mv mv mv mv

v v vA B A B

A B

+ = ¢ + ¢= ¢ + ¢0 (1)


v v e v vA B B A-( ) = ¢ - ¢( ) v e v vB A0 = ¢ - ¢ (2)

Solving Equations (1) and (2) simultaneously,

¢ =-

vv e

A0 1

2

( ) b

¢ =+

vv e

B0 1

2

( ) b

(b) Second collision (between B and C).


mv mv mv mvB C B C¢ + = ¢¢ + ¢

Using the result from (a) for ¢vB

v e

v vB C0 1

20

( )++ = ¢¢ + ¢ (3)


( )¢ - = ¢ - ¢¢v e v vB C B0

PROBLEM 13.160

Three steel spheres of equal weight are suspended from the ceiling by cords of equal length which are spaced at a distance slightly greater than the diameter of the spheres. After being pulled back and released, sphere A hits sphere B, which then hits sphere C. Denoting by e the coefficient of restitution between the spheres and by v0 the velocity of A just before it hits B, determine (a) the velocities of A and B immediately after the first collision, (b) the velocities of B and C immediately after the second collision. (c) Assuming now that n spheres are suspended from the ceiling and that the first sphere is pulled back and released as described above, determine the velocity of the last sphere after it is hit for the first time. (d) Use the result of Part c to obtain the velocity of the last sphere when n = 6 and e = 0 95. .

237



Substituting again for ¢vB from (a)

ve

e v vC B01

2

( )( )

+ = ¢ - ¢¢ (4)

Solving equations (3) and (4) simultaneously,

¢ =+

+ +ÈÎÍ

˘˚̇

vv e

v ee

C1

2

1

21

20

0( )

( )( )

¢ =+

vv e

C0

21

4

( ) b

¢¢ =-

vv e

B0

21

4

( ) b

(c) For n spheres

n balls

n -1th collision,

we note from the answer to Part (b) with n = 3

¢ = ¢ = ¢ =+

v v vv e

n C30

21

4

( )

or ¢ =+ -

-vv e

30

3 1

3 1

1

2

( )( )

( )

Thus, for n balls

¢ =+ -

-vv e

n

n

n0

1

1

1

2

( )( )

( ) b

(d) For n e= =6 0 95, . ,

from the answer to Part (c) with n = 6

¢ =+

=

-

-vv

v

B0

6 1

6 1

05

5

1 0 95

2

1 95

2

( . )

( . )

( )

( )

( )

¢ =v vB 0 881 0. b

238


SOLutiOn

Total momentum conserved.


( ) ( )m v m v m vA B B B0 0 0+ - = + ¢

¢ = -ÊËÁ

ˆ¯̃

vm

mvA

B

1 0 (1)


¢ - ¢ = -¢ =

v v e v v

v evB A A B( )

2 0 (2)

Subtracting Eq. (2) from Eq. (1) and dividing by v0 ,

m

me

m

me

A

B

A

B

- - =

= +

1 2 0

1 2

mm

eBA=

+1 2 (3)

(a) With mA = 3 kg and e = 0 5. ,

mB =+

3

1 2 0 5( )( . ) mB = 1 500. kg b

(b) With mA = 3 kg and e = 1,

m mB B=+

=3

1 2 11 000

( )( ). kg

With mA = 3 kg and e = 0,

m mB B=+

=3

1 2 03 00

( )( ). kg

Range: 1 000 3 00. . kg kg� �mB b

PROBLEM 13.161

Two disks sliding on a frictionless horizontal plane with opposite velocities of the same magnitude v0 hit each other squarely. Disk A is known to have a mass of 3 kg and is observed to have zero velocity after impact. Determine (a) the mass of disk B, knowing that the coefficient of restitution between the two disks is 0.5, (b) the range of possible values of the mass of disk B if the coefficient of restitution between the two disks is unknown.

239


SOLutiOn

(a) Packages A and B.


m v m v m v m v

v vA A B B A A B B

A B

+ = ¢ + ¢+ = ¢ + ¢

=( ( (8 0 8 4

4

kg)(2 m/s) kg) kg)

22 ¢ + ¢v vA B (1)


( ) ( )

( )( . )

v v e v v

v vA B B A

B A

- = ¢ - ¢= ¢ - ¢2 0 3 (2)

Solving Equations (1) and (2) simultaneously,

¢ = ÆvA 1 133. m/s

¢ = ÆvB 1 733. m/s

Packages B and C.

+ m v m v m v m vB B C C B B C C¢ + = ¢¢ + ¢

(4 4 6 kg)(1.733 m/s) 0+ = ¢¢ + ¢v vB c

6 932 4 6. = ¢¢ + ¢v vB C (3)

PROBLEM 13.162

Packages in an automobile parts supply house are transported to the loading dock by pushing them along on a roller track with very little friction. At the instant shown, packages B and C are at rest and package A has a velocity of 2 m/s. Knowing that the coefficient of restitution between the packages is 0.3, determine (a) the velocity of package C after A hits B and B hits C, (b) the velocity of A after it hits B for the second time.

240




( )

( . )( . ) .

¢ - = ¢ - ¢¢= = ¢ - ¢¢

v v e v v

v vB C C B

C B1 733 0 3 0 5199 (4)

Solving equations (3) and (4) simultaneously,

¢ = ÆvC 0 901. m/s b

(b) Packages A and B (second time),


( )( . ) ( )( . )

.

8 1 133 4 0 381 8 4

10 588 8 4

+ = ¢¢ + ¢¢¢= ¢¢ + ¢¢

v v

v vA B

A B (5)


( )

( . . )( . ) .

¢ - ¢¢ = ¢¢¢- ¢¢- = = ¢¢¢- ¢¢

v v e v v

v vA B B A

B A1 133 0 381 0 3 0 2256 (6)


¢¢ =vA 0 807. m/s ¢¢ = ÆvA 0 807. m/s b

241


PROBLEM 13.163

One of the requirements for tennis balls to be used in official competition is that, when dropped onto a rigid surface from a height of 2.5 m the height of the first bounce of the ball must be in the range 1.325 m £ h £ 1.45 m. Determine the range of the coefficients of restitution of the tennis balls satisfying this requirement.

SOLutiOn

Uniform accelerated motion.

v gh

v gh

=

¢ = ¢

2

2

Coefficient of restitution. ev

v

eh

h

= ¢

= ¢

Height of drop h = 2 5. m

Height of bounce 1 325 1 45. .m m� �¢h

Thus, 1 325

2 5

1 45

2 5

.

.

.

.� �e 0 728 0 762. .� �e b

242


SOLutiOn

Momentum in t direction is conserved (no friction).

mv mv

v vt t= ¢

= ¢sin sinq a (1)

Coefficient of restitution (n-direction).

v e v v e vn n= ¢ = ¢(cos )( ) cosq a (2)

Dividing Equation (2) into Equation (1),

tan

tan

qa

= e

Thus,

for 0 1� �e tan tana q a q� � and b

% loss in kinetic energy.

Squaring both sides of (1) and (2) and adding

v e v

T m v v

mv e

2 2 2 2 2

2 2

2 2 2

1

21

21

(sin cos ) ( )

[ ( ) ]

[ (sin c

q q

q

+ = ¢

= - ¢

= - +

D

oos )]

cos ( )

2

2 2 21

21

q

qDT mv e= -

% ( )cos loss = = -100 100 112

22 2DT

mve q b

PROBLEM 13.164

Show that for a ball hitting a frictionless fixed surface, a q� . Also show that the percent loss in kinetic energy due to the impact is 100(1 - e2) cos2 q.

243


PROBLEM 13.165

A 600-g ball A is moving with a velocity of magnitude 6 m/s when it is hit as shown by a 1-kg ball B, which has a velocity of magnitude 4 m/s. Knowing that the coefficient of restitution is 0.8 and assuming no friction, determine the velocity of each ball after impact.

SOLutiOn

Before After

v

v

v

A

A n

A t

== ∞ == - = -

6

6 40 4 596

6

m/s

( m/s

( (sin 40 ) 3.857

) ( )(cos ) .

) ∞ m/s

m/s

(

v v

vB B n

B t

= = -=

( )

)

4

0

t-direction.


m v m v m v m vA A t B B t A B t B B t( ) ( ) ( ) ( )

( . . (

+ = ¢ + ¢- + =0 6 3 857 0 0 kg)( m/s) .. ) ( )

. . ) ( )

6 1

2 314 0 6

kg)( kg)(

m/s (

¢ + ¢- = ¢ + ¢

v v t

v vA t B

A t B t (1)

Ball A alone.

Momentum conserved.

m v m v v

v

A A t A A t A t

A t

( ) = ¢( ) - = ¢

¢ = -

3 857

3 857

. )

( ) .

(

m/s (2)

Replacing ( )¢vA t in (2) in Eq. (1)

- = - + ¢- = - + ¢

¢ =

2 314 0 6 3 857

2 314 2 314

0

. ( . )( . ) ( )

. . ( )

( )

v

v

v

B t

B t

B t

244



n-direction


[( ) ( ) ] ( ) ( )

[( . ) ( )]( . ) ( ) (

v v e v v

v vA n B n B n A n

B n

- = ¢ - ¢- - = ¢ - ¢4 596 4 0 8 AA n

B n A nv v

)

. ( ) ( )6 877 = ¢ - ¢ (3)


m v m v m v m vA A n B B n A A n B B n( ) ( ) ( ) ( )

( . (

+ = ¢ + ¢+0 6 1 kg)(4.596 m/s) kg))( m/s) kg)( kg)(

(

- = ¢ + ¢- = ¢ +

4 1 0 6

1 2424 0 6

( ) ( . )

. ( ) .

v v

vB n A n

B n ¢¢vA n) (4)

Solving Eq. (4) and (3) simultaneously,

( ) .

) .

¢ =¢ =

v

vA n

B n

5 075

1 802

m/s

( m/s

Velocity of A.

tan( )

)

.

.

. .

( .

b

b b

=

=

= ∞ + ∞ = ∞

¢ =

| |

|( |

v

v

v

A t

A n

A

3 857

5 075

37 2 40 77 2

3 857)) ( . )

.

2 25 075

6 37

+

= m/s

¢ =vA 6 37. m/s 77 2. ∞ b

Velocity of B. ¢ =vB 1 802. m/s 40∞ b

245


SOLutiOn

( ) (

( ) (

v

vA n

A t

= == - = -

3

3

m/s) cos 20 2.819 m/s

m/s) sin 20 1.0

∞∞ 2261 m/s

m mA B=

( ) ( .

) (

v

vB n

B t

= - = -=

3 2 819

3

m/s) cos 20 m/s

( m/s) sin 20 =1.0

∞∞ 2261 m/s

t-direction

Momentum of A is conserved.

m v m v v

vA A t A A t A t

A t

( ) ( ) . ( )

( ) .

= ¢ - = ¢¢ = -

1 0261

1 0261 m/s

Momentum of B is conserved.

m v m v t v

vB B t B B B t

B t

( ) ( ) . ( )

( ) .

= ¢ = ¢¢ =

1 0261

1 0261 m/s

n-direction

Total momentum is conserved.

m v m v m v m vA A n B B n A A n B B n( ) ( ) ( ) ( )+ = ¢ + ¢

m mA B=

2 819 2 819. . ( ) ( )

( ) ( )

- = ¢ + ¢¢ = - ¢

v v

v vA n B n

A n B n

PROBLEM 13.166

Two identical hockey pucks are moving on a hockey rink at the same speed of 3 m/s and in parallel and opposite directions when they strike each other as shown. Assuming a coefficient of restitution e = 1, determine the magnitude and direction of the velocity of each puck after impact.

246



Relative velocities (coefficient of restitution).

e = 1 [( ) ( ) ] ( ) ( )

[ . ( . )]( ) ( ) (

v v e v v

v v

A n B n B n A n

B n

- = ¢ - ¢

- - = ¢ - ¢2 819 2 819 1 AA n)

( ) ( )¢ = - ¢v vA n B n ( ) ( ) .

( ) .

¢ - ¢ =¢ = -

v v

vB n A n

A n

5 638

2 5 638

( ) . ) .¢ = - ¢ =v vA n B n2 819 2 819 m/s ( m/s

¢ =vA 3 00. m/s 40∞ b

¢ =vB 3 00. m/s 40∞ b

247


PROBLEM 13.167

Two identical pool balls of 60 mm diameter may move freely on a pool table. Ball B is at rest and ball A has an initial velocity v i= v0 . (a) Knowing that b = 50 mm and e = 0 7. , determine the velocity of each ball after impact. (b) Show that if e = 1, the final velocities of the balls form a right angle for all values of b.

SOLutiOn

Geometry at instant of impact.

sin

.

q

q

= =

= ∞

b

d

50

6056 443

Directions n and t are shown in the figure.

Principle of impulse and momentum.

Ball B

Ball A

Ball A, t-direction: mv m v v vA t A t0 00sin ( ) ( ) sinq q+ = = (1)

Ball B, t-direction: 0 0 0+ = =m v vB t B t( ) ( ) (2)

Balls A and B, n-direction: mv m v m vA n B n0 0cos ( ) ( )q + = +

( ) ( ) cosv v vA n B n+ = 0 q (3)

Coefficient of restitution. ( ) ( ) [ cos ]v v e vB n A n- = 0 q (4)

(a) e = 0 7. . From Eqs. (1) and (2),

( ) .v vA t = 0 83334 0 ( )1 ¢

( )vB t = 0 ( )2 ¢

248



From Eqs. ( )3 and ( )4 ,

( ) ( ) .v v vA n B n+ = 0 55277 0 ( )3 ¢

( ) ( ) ( . )( . )

.

v v v

vB n A n- =

=0 7 0 55277

0 386940

0 ( )4 ¢

Solving Eqs. (5) and (6) simultaneously,

( ) . ( ) .

( ) ( )

( .

v v v v

v v v

A n B n

A A n A t

= =

= +

=

0 082915 0 469855

0 082195

0 0

2 2

vv v

v

v

v

vA n

A t

02

02

0

0

0 83334

0 83738

0 082915

0 833

) ( .

.

tan( )

( )

.

.

+=

= =b334

0 099497

5 682

90

90 56 443 5 682

27 875

0v=

= ∞= ∞ - -= ∞ - ∞ - ∞= ∞

.

.

. .

.

bj q b

vA v= 0 837 0. 29∞ b

vB v= 0 470 0. 56 4. ∞ b

(b) e = 1. Eqs. (3) and (4) become

( ) ( ) cosv v vA n B n+ = 0 q ( )3 ¢¢

( ) ( )v v vB n A n- = 0 cosq ( )4 ¢¢

Solving Eqs. ( )3 ¢¢ and ( )4 ¢¢ simultaneously,

( ) , ( ) cosv v vA n B n= =0 0 q

But ( ) sin , )v v vA t B t= =0 0q and (

vA is in the t-direction and vB is in the n-direction; therefore, the velocity vectors form a right angle.

249


SOLutiOn

(a) Since ¢vB is in the x-direction and assuming no friction, the common tangent between A and B at impact must be parallel to the y-axis.

Thus, tan

tan

.

q

q

=-

=-

= ∞

-

250

150250

150 6070 201

1

D

q = ∞70 2. b

(b) Conservation of momentum in x(n) direction.

mv m v m v mv

v vA B n A n B

An B

cos ( ) ( )

( )cos( . )

.

q + = ¢ + ¢+ = ¢ + ¢1 70 201 0

0 338722 = ¢ + ¢( ) ( )v vA n B (1)

Relative velocities in the n direction.

e = 0 9. ( cos ( ) ) ( )v v e v vA B n B A nq - = ¢ - ¢

( . )( . ) ( )0 33872 0 0 9- = ¢ - ¢v vB A n (2)

( ) ( )1 2+ 2 0 33872¢ =vB ( . ) ¢ = ÆvB 0 322. m/s b

PROBLEM 13.168

The coefficient of restitution is 0.9 between the two 60 mm diameter billiard balls A and B. Ball A is moving in the direction shown with a velocity of 1 m/s when it strikes ball B, which is at rest. Knowing that after impact B is moving in the x direction, determine (a) the angle q , (b) the velocity of B after impact.

250


SOLutiOn

Law of sines.

sin sin

.

.

.

q

qa

R R2

135

20 705

45 20 705

24 295

= ∞

= ∞= ∞ - ∞= ∞

Conservation of momentum for ball in t direction.

- = - ¢v vsin sinq a

Coefficient of restitution in n. v e v(cos ) cosq a= ¢

Dividing, tan

tanq a

e=

e = ∞∞

tan .

tan .

20 705

24 295 e = 0 857. b

PROBLEM 13.169

A boy located at Point A halfway between the center O of a semicircular wall and the wall itself throws a ball at the wall in a direction forming an angle of 45∞ with OA. Knowing that after hitting the wall the ball rebounds in a direction parallel to OA, determine the coefficient of restitution between the ball and the wall.

251


PROBLEM 13.170

A girl throws a ball at an inclined wall from a height of 1.2 m, hitting the wall at A with a horizontal velocity v0 of magnitude 15 m/s. Knowing that the coefficient of restitution between the ball and the wall is 0.9 and neglecting friction, determine the distance d from the foot of the wall to the Point B where the ball will hit the ground after bouncing off the wall.

SOLutiOn

Momentum in t direction is conserved.

mv mv

v t

v

t

t

sin

( )(sin )

.

30

15 30

7 5

∞ = ¢∞ = ¢¢ = m/s

Coefficient of restitution in n-direction.

( cos )

( )(cos )( . )

.

v e v

v

v

n

n

n

30

15 30 0 9

11 69

∞ = ¢∞ = ¢

¢ = m/s

Writing v in terms of x and y components ¢

( ) cos sin

( ) ( . )(cos ) ( . )(sin

¢ = ¢ ∞ - ¢ ∞¢ = ∞ -

v v v

vx n t

x

0

0

30 30

11 69 30 7 5 330 6 374

30 30

11 690

0

∞ =¢ = ¢ ∞ + ¢ ∞

¢ =

) .

( ) sin cos

( ) ( . )(si

m/s

v v v

v

y n t

y nn ) ( . )(cos ) .30 7 5 30 12 340∞ + ∞ = m/s

Motion of a projectile. (origin at 0)

y y v tgt

y tt

y= + ¢ -

= + -

0 0

2

2

2

1 2 12 340 9 812

( )( )

. ( . ( . ) m/s) m/s2

252



Time to reach Point B ( )yB = 0

0 1 2 12 3409 81

22= + - Ê

ËÁˆ¯̃

. ..

t tB B tB = 2 610. s

x x v t

x t

x t

x

x

B B

= + ¢= +==

0 0

0 6 374

6 374

6 374

( )

.

( . )( )

( . m/s)(2.610 s)

BB = 16 63. m

d xB= -=

1 2. cot 60

15.94 m d = 15 94. m b

253


PROBLEM 13.171

A ball hits the ground at A with a velocity v0 of 5 m/s at an angle of 60° with the horizontal. Knowing that e = 0 6. between the ball and the ground and that after rebounding the ball reaches Point B with a horizontal velocity, determine (a) the distances h and d, (b) the velocity of the ball as it reaches B.

SOLutiOn

(a) Rebound at A.

v0 5= m/s

Conservation of momentum in the t-direction.

mv m v

vA t

A t

0 60

5 60

2 5

cos ( )

( ) ( )(cos )

.

q = ¢¢ = ∞

=m/s

m/s

Coefficient of restitution in the n-direction.

( ( ) ) ( )

( )(sin )( . ) ( )

( )

- - = -∞ =

= =

v e v

v

v

A n A n

A n

A n

0 0

5 60 0 6

3 32

¢¢

¢

m/s

2..5981m/s

Projectile motion between A and B.

After rebound

( ) ( ) .

( ) .

v v

v vx A t

A n

0 2 5

2 5981

= - = -= =

¢¢

m/s

( ) m/sy 0

x-direction: x v t t

Ux

x

= = -= -

( ) .

.0 2 5

2 5 m/s

y-direction: y v t gt t t

v v gt t

y

y y

= - = -

= - = -

( ) . .

( ) . .

02 2

0

1

22 5981 4 905

2 5981 9 81

254



At B: ( )

( ) . .

.

v

v t

t

y h

B y

B y A B

A B

B

=

= = -

==

-

-

0

0 2 5981 9 81

0 2648 s

h t t

h

A B A B= -

= -- -( . ) .

( . )( . ) ( . )( .

2 5981 4 905

2 5981 0 2648 4 905 0 2648

2

))2 h = 0.344 m b

x d t

dB A B= - = -

=-2 5

2 5 0 2648 0 662

.

( . )( . ) .= m d = 0 662. m b

(b) v vB x= = -( ) .0 2 5 m/s vB = 2 50. m/s ¨ b

255


PROBLEM 13.172

A sphere rebounds as shown after striking an inclined plane with a vertical velocity v0 of magnitude v0 = 5 m/s. Knowing that a = ∞30 and e = 0 8. between the sphere and the plane, determine the height h reached by the sphere.

SOLutiOn

Rebound at A

Conservation of momentum in the t-direction.

mv m v

vA t

A t

0 30

5 30

2 5

sin ( )

( ) ( )(sin )

.

∞ == ∞=

¢¢ m/s

m/s

Relative velocities in the n-direction.

( cos ) ( )

( ) ( . )( )(cos )

( )

- ∞ = -= ∞

=

v e v

v

v

A n

A n

A n

0 30 0 0

0 8 5 30

2 3

- ¢¢

¢

m/s

mm/s 3.4641m/s=

Projective motion between A and B.

After rebound ( ) ( ) cos ( ) sin

( ) ( . )(cos ) ( .

v v v

vx A t A n

x

0

0

30 30

2 5 30 3 4641

= ∞ + ∞= ∞ +

¢ ¢))sin

( ) .

( ) ( ) sin ( ) cos

(

30

3 8971

30 300

0

∞== - ∞ + ∞

v

v v v

v

x

y A t A n

y

m/s

¢ ¢

)) ( . )(sin ) ( . )cos

( ) .0

0

2 5 30 3 4641 30

1 7500

= - ∞ + ∞

=vy m/s

x-direction: x v t v v

x t v vx x x

x B

= == = =

( ) ( )

. .0 0

3 8971 3 8971 m/s

y-direction: y v t gt

v v gt

y

y y

= -

= -

( )

( )

02

0

1

2

At A: v v gt

t v g

t

y y AB

AB y

A B

= = -=

=

=-

0

1 75

9 80 17839

0

0

( )

( )

.

..

/

m/s

m/ss

2

256



At B:

y h

v tgt

h

y A BA B

=

= -

= -

--( )

( . )( . ) ( . )( . )

0

2

2

2

1 75 0 17839 4 905 0 17839 h = 0 1561. m b

257


PROBLEM 13.173

A sphere rebounds as shown after striking an inclined plane with a vertical velocity v0 of magnitude v0 6= m/s. Determine the value of a that will maximize the horizontal distance the ball travels before reaching its maximum height h, assuming the coefficient of restitution between the ball and the ground is (a) e = 1, (b) e = 0 8. .

SOLutiOn

Directions x, y, n, and t are shown in the sketch.

Analysis of the impact. Use the principle of impulse and momentum for components in the t-direction. mv m vt0 10sin ( )a + = ¢

( ) sinv vt 1 0= a (1)

Coefficient of restitution. ( ) ( )v vn ne1 0= -

( ) cosv evn 1 0= a (2)

x and y components of velocity immediately after impact.

( ) ( ) sin ( ) cos ( )sin cosv v v v ex n t1 1 1 0 1= + = +a a a a

= +1

21 20v e( )sin a (3)

( ) ( ) cos ( ) sin ( cos sin )

[ ( cos )

v v v v e

v e

y n t1 1 1 02 2

01

21 2

= =

= +

a a a a

a

- -

-- -

- -

( cos )]

[( )cos ( )]

1 2

1

21 2 10

a

a= +v e e (4)

Projectile motion. Use coordinates x and y with the origin at the point of impact.

x

y0

0

0

0

==

Vertical motion: v v gt

v v e e gt

y y

y

=

= +

( )

[( )cos ( )]

1

01

21 2 1

-

- - -a

vy = 0 at the position of maximum height where

tv

g

v

ge ey

21 0

21 2 1= = +

( )[( )cos ( )]a - - (5)

258



Horizontal motion: v v v e

x v t

x x

x

= = +

=

( ) ( )sin

( )

1 0

1

1

21 2a

At the point of maximum height,

x v tv

ge e ex2 1 2

02

41 2 1 2 1= = + +( ) ( )sin [( )cos ( )]a a - -

Let q a= 2 . To determine the value of q that maximizes x2 , differentiate the expression v with respect to q.

v e e

dv

de e e

=

= -

sin [( )cos ( )]

cos [( )cos ( ) ( )sin

q q

qq q

1 1

1 1 1 2

+ - -

+ - - + qq

q q q

q

]

( )cos ( )cos ( )( cos )

( )cos ( )

= - - =

- -

1 1 1 1 0

2 1 1

2 2

2

+ - - +

+

e e e

e e ccos ( )q - 1 0+ =e

This is a quadratic equation for cos .q

(a) e = 1 4 2 02cos q - =

cos

cos

2 1

2

2

2

q

q

=

= ±

q = ± ∞45 and ± ∞135

a = ∞22 5. and 67 5. ∞

Reject the negative values which make x2 negative.

Reject a = ∞67 5. since it makes a smaller maximum height.

a = ∞22 5. b

(b) e = 0 8. 3 6 0 2 1 8 02. cos . cos .q q- - =

cos . .

. .

.

qqa

== ± ∞ ± ∞= ± ∞

0 73543 0 67987

42 656 132 833

21 328

and

and

and

-

±± ∞66 417. a = ∞21 3. b

259


PROBLEM 13.174

A 1-kg block B is moving with a velocity v0 of magnitude v0 2= m/s as it hits the 0.5-kg sphere A, which is at rest and hanging from a cord attached at O. Knowing that mk = 0 6. between the block and the horizontal surface and e = 0 8. between the block and the sphere, determine after impact (a) the maximum height h reached by the sphere, (b) the distance x traveled by the block.

SOLutiOn

Velocities just after impact

Total momentum in the horizontal direction is conserved.

m v m v m v m v

v vA A B B A A B B

A B

+ = ¢ + ¢+ = ¢ + ¢0 1 2 0 5 1( )( ) ( . )( ) ( )( )kg m/s kg kg

4 2= ¢ + ¢v vA B (1)


( ) ( )

( )( . )

.

v v e v v

v v

v v

A B B A

B A

B A

- -- -

- -

= ¢ ¢= ¢ ¢= ¢ ¢

0 2 0 8

1 6 (2)


¢ =¢ =

v

vB

A

0 8

2 4

.

.

m/s

m/s


T m v V

T m m

A

A A

1 12

1

12

1

20

1

22 4 2 88

= =

= =( . ) .m/s

T

V m ghA

2

2

0==

T V T V m mA A1 1 2 2 2 88 0 0 9 81+ = + + = +. ( . )

h = 0 294. m b

260



(b) Work and energy.

T m v m m T

U F x Nx m gx

B B B

f k x B

1 12 2

2

1 2

1

2

1

20 8 0 32 0= = = =

= = =

( . ) .m/s

- - - -m m ==

=+ =

-

--

-

-

( . )( )( . )

.

. .

0 6 9 81

5 886

0 32 5 8861 2

1 1 2 2

m x

U m x

T U T m m x

B

B

B B == 0

x = 0 0544. m x = 54 4. mm b

261


SOLutiOn

m m

vA B

A

= ==

1 5

5

. kg

m/s

Phase I impact


+ m v m v m v m v v vA A B B A A B B A B+ ¢ = ¢ + ¢ = ¢ + ¢: 5 (1)


( ) ( )v v e v v v v eA B B A B A- - -= ¢ ¢ fi ¢ ¢ = 5 (2)

Solving Eqs. (1) and (2), ¢ = + ¢ == ¢ = ¢ == ¢ =

v e v e

e v v

e v

B A

B A

B

2 5 1 2 5 1

1 5 0

0 2 5

. ( ), . ( )

:

: .

-m/s,

m/s, ¢¢ =vA 2 5. m/s

(a) Conservation of energy phase II.

k = 80 N/m

e = 1: ¢ = ¢ =

= ¢

=

= =

v v

T m v

V

B A

B B

5 0

1

21

21 5 5

18 75 0

12

2

1

m/s,

kg)( m/s)

J

( )

( .

. ;

PROBLEM 13.175

A 1.5-kg block B is attached to an undeformed spring of constant k = 80 N/m and is resting on a horizontal frictionless surface when it is struck by an identical block A moving at a speed of 5 m/s. Considering successively the cases when the coefficient of restitution between the two blocks is (1) e = 1, (2) e = 0, determine (a) the maximum deflection of the spring, (b) the final velocity of block A.

262



At x x= max , T V k x x x

T V T V

2 22 2 2

1 1 2 2

01

2

1

280 40

18 75 40

= = = =

+ = + =

; ( ) ( )

: .

max max max

xxmax2

xmax .= 0 68465 m e x= =1 0 685, .max m b

e = 0: ¢ = ¢ =

= + ¢ = =

v v

T m m v

B A

A B A

2 5

1

2

1

23 0 9 3751

2 2

.

( )( ) ( . .

m/s

kg)(2.5 m/s) J;;

,

: .

max

max

V

T V x

T V T V x

1

2 22

1 1 2 22

0

0 40

9 375 40

=

= =

+ = + =

xmax .= 0 48412 m e x= =0 0 484, .max m b

(b) e = 1: block B is returned to Position 1 with a velocity of 5 m/s , since energy is conserved, and impacts block A, which is at rest. In the impact, total momentum is conserved and phase I is repeated with the velocities of A and B interchanged. Thus, ¢¢¢=vA 5 m/s and ¢¢¢=vB 0.Since there is no friction, these velocities are the final velocities of A and B.

e A= ¢¢¢ = Æ1 5 00, .v m/s b

e = 0: blocks A and B are returned to Position 1 with the same velocity of 2 5. m/s Æ , since energy is conserved. There is no additional impact, the spring slows block B down, and A and B separate, with A continuing with a velocity of 2 5. m/s to the right.

e A= ¢¢¢ = Æ0 2 50, .v m/s b

263


SOLutiOn

Let the x-direction be positive to the right and the y-direction vertically upward.

Let ( ) , ( ) , ( ) ,v v vA x A y B x and ( )vB y be velocity components just before impact and ( ) , ( ) , ( ) ,¢ ¢ ¢ ¢v v vA x A y B x and ( )¢vB y those just after impact. By inspection,

( ) ( ) ( ) ( )v v v vA y B y A y B y= = ¢ = ¢ = 0

Conservation of momentum for x-direction.

0 0+ = + =m v m v v vA A x B B x B x A x( ) ( ) ( ) ( )-b (1)

0 0+ = ¢ + ¢ ¢ = ¢m v m v v vA A x B B x B x A x( ) ( ) ( ) ( )-b (2)

where b = m mA B/

Conservation of energy during friction, less sliding.

Initial potential energies: m ghA for A, 0 for B.

Potential energy just before impact: V1 0=

Initial kinetic energy: T0 0= (rest)

Kinetic energy just before impact: T m v m vA A B B12 21

2

1

2= +

T V T V

m gh m v m v m m v

m v

A A A B B A B A

A A

0 0 1 1

2 2 2 21

2

1

2

1

21

21

+ = +

= + = +

= +

( )

( )

b

b 22

v vgh

vgh

A A x A2 2 2

1

2

1= =

+=

+( )

b b (3)

PROBLEM 13.176

Block A is released from rest and slides down the frictionless surface of B until it hits a bumper on the right end of B. Block A has a mass of 10 kg, and object B has a mass of 30 kg and can roll freely on the ground. Determine the velocities of A and B immediately after impact when (a) e = 0, (b) e = 0 7. .

264



Velocities just before impact: v

v

A

B

gh

gh

=+

Æ

=+

¨

2

1

2

1

b

bb

Analysis of Impact. Use Eq. (2) together with coefficient of restitution.

( ) ( ) [( ) ( ) ]

( ) ( ) [( ) ( )

¢ ¢ =¢ ¢ = +

v v e v v

v v e v vB x A x A x B x

A x A x A x A x

- -- -b b ]]

( ) ( )¢ =v e vA x A x- (4)

Data: m

m

h

g

A

B

===

=

10

30

0 2

9 81

kg

kg

m

m/s2

.

.

b = =10

300 33333

kg

kg.

From Eq. (3), vA =

=

( )( . )( . )

..

2 9 81 0 2

1 333331 71552 m/s

(a) e = 0: ( ) ( )¢ = ¢ =v vA x B x0 0 ¢ =vA 0 b

¢ =vB 0 b

(b) e = 0 7. : ( ) ( . )( . )

.

( )

¢ -

¢

v x

v x

A

A

=

= -

= -(0.33333)(1.2008

0 7 1 71552

1 20086 m/s

66)

= 0.40029 m/s

¢ = ¨vA 1 201. m/s b

¢ = ÆvB 0 400. m/s b

265


PROBLEM 13.177

A 90-g ball thrown with a horizontal velocity v0 strikes a 720-g plate attached to a vertical wall at a height of 900 mm above the ground. It is observed that after rebounding, the ball hits the ground at a distance of 480 mm from the wall when the plate is rigidly attached to the wall (Fig. 1), and at a distance of 220 mm when a foam-rubber mat is placed between the plate and the wall (Fig. 2). Determine (a) the coefficient of restitution e between the ball and the plate, (b) the initial velocity v0 of the ball.

SOLutiOn

(a) Figure (1), ball alone

Relative velocities. v e vB0 1= ¢( )

Projective motion. t = time for the ball to hit the ground.

0 480 0. m = v e t (1)

Figure (2), ball and plate


+ ( ) ( )v v e v vB P P B- = ¢ + ¢ 2

v v

v

v e v v

B

P

P B

=== ¢ + ¢

0

0 2

0

( ) (2)


+ m v m v m v m vB B P P B B P P+ - ¢ + ¢= ( ) ( )2

( . )( ) ( . )( ) ( . )( )0 09 0 0 09 0 7200 2kg kg kgv v vB P+ = ¢ + ¢-

v v vo B P= - ¢ + ¢( )2 8 (3)

Solving (2) and (3) simultaneously for ( )1 2B /

( )( )

¢ =v ve

B 2 08 1

9

-

266



Projectile motion.

0 2208 1

90.( )

m = ve

t-

(4)

Dividing Eq. (4) by Eq. (3),

0 220

0 480

8 1

94 125 8 1

.

..

=

=

e

ee e

-

- e = 0 258. b

(b) From Figure (1)

Projectile motion.

h gt

t t

=

= =

1

2

0 9009 81

21 80 9 81

2

2 2.( . )

, . . (5)

Equation (1):

0 480

0 480

0 258

1 860

0

0

0

.

( . )

( . )

.

=

=

=

v et

tv

v

Substituting for t in Eq. (5),

1 800 9 811 860

18 855

2

02

02

. ( . )( . )

.

=

=

v

v v0 4 34= . m/s b

267


PROBLEM 13.178

A 600-g sphere A is dropped from a height of 5 m onto a 1.2 kg plate B, which is supported by a nested set of springs and is initially at rest. Knowing that the coefficient of restitution between the sphere and the plate is e = 0 8. , determine (a) the height h reached by the sphere after rebound, (b) the constant k of the single spring equivalent to the given set if the maximum deflection of the plate is observed to be equal to 3h.

SOLutiOn

Velocity of A and B after impact

m

mA

B

==

0 6

1 2

.

.

kg

kg

Sphere A falls. Use conservation of energy to find vA , the speed just before impact. Use the plate surface as the datum.

T V m gh T m v V

T V T V m gh m v

A A A

A A A

1 1 0 22

2

1 1 2 2 02

01

20

01

20

= = = =

+ = + + = +

, , ,

With h

v ghA

0

0

5

2 2 9 81 5

=

= =

m,

( . )( )

vA = Ø9 9045. m/s

Analysis of the impact. Conservation of momentum.

m m m mA A B B A A B Bv v v v+ = ¢ + ¢ with vB = 0

Dividing by mA and using y-components + with ( )m mB A/ = 2

-9 9045 0 2. ( ) ( )+ = ¢ + ¢v vA y B y (1)

Coefficient of restitution. ( ) ( ) [( ) ( ) ]¢ ¢ =v v e v vB y A y A y B y- -

( ) ( ) ( ) .¢ ¢ = =v v e v eB y A y A y- -9 9045 (2)

268



Solving Eqs. (1) and (2) simultaneously with e = 0 8. gives

( ) .

( ) .

.

.

¢ =

¢ =

¢ = ≠

¢ =

v

v

A y

B y

A

B

1 9809

5 9427

1 9809

5 9427

m/s

m/s

m/s

m

-

v

v //s Ø

(a) Sphere A rises. Use conservation of energy to find h.

T m v V T V m gh

T V T V m v

A A A

A A

12

1 2 2

1 1 2 22

1

20 0

1

20 0

= ¢ = = =

+ = + ¢ + = +

( ) , , ,

: ( ) mm ghA

hv

gA=¢

=( ) ( . )

( . )

2 2

2

1 9809

2 9 81 h = 0 200. m b

(b) Plate B falls and compresses the spring. Use conservation of energy.

Let d0 be the initial compression of the spring and D be the additional compression of the spring after impact. In the initial equilibrium state,

+ S F k W k Wy B B= = =0 00 0: d d- or (3)

Just after impact: T m v V kB B12

1 021

2

1

2= ¢ =( ) , d

At maximum deflection of the plate, T2 0=

V V V W kg e B2 2 2 021

2= + = + +( ) ( ) ( )- D Dd

Conservation of energy: T V T V1 1 2 2+ = +

1

2

1

20

1

2

1

22

02

02

02m v k W k k kB B B( )¢ + = + + +d d d- D D D

Invoking the result of Eq. (3) gives

1

2

1

22 2m v kB B( )¢ = D (4)

Data: m vB B= ¢ =1 2 5 9427. .kg, m/s

D = = =3 3 0 200 0 6h ( )( . ) . m

km vB B=

¢=

( ) ( . )( . )

( . )

2

2

2

2

1 2 5 9427

0 6D k = 117 7. N/m b

269


Solution

m

m

k

A

B

===

0 6

1 2

1000

.

.

kg

kg

N/m

Sphere A falls. Use conservation of energy to find vA , the speed just before impact. Use the plate surface as the datum.

T V m gh

T m v V

A

A A

1 1 0

22

2

0

1

20

= =

= =,

With h0 5= m, v ghA = =2 2 9 81 50 ( )( . )( )

vA = Ø9 9045. m/s

Analysis of impact. Conservation of momentum.

m m m m vA A B B A A B Bv v v+ = ¢ + ¢ with vB = 0

Dividing by mA and using y components + with ( )m mB A/ = 2

-9 9045 2. ( ) ( )= ¢ + ¢v vA y B y (1)

Coefficient of restitution. ( ) ( ) [( ) ( ) ]¢ ¢ =v v e v vB y A y A y B y- -

( ) ( ) ( ) .

( ) . ( )

¢ ¢ = =

¢ = + ¢

v v e v e

v e v

B y A y A y

B y A y

- -

-

9 9045

9 9045 (2)

PRoBlEM 13.179

A 600 g sphere A is dropped from a height of 5 m onto a 1.2 kg plate B, which is supported by a nested set of springs and is initially at rest. Knowing that the set of springs is equivalent to a single spring of constant k = 1000 N/m, determine (a) the value of the coefficient of restitution between the sphere and the plate for which the height h reached by the sphere after rebound is maximum, (b) the corresponding value of h, (c) the corresponding value of the maximum deflection of the plate.

270



Substituting into Eq. (1),

- -9

-

9 9045 2 9045

3 3015 2 1

. ( ) ( )[ . ( ) ]

( ) . ( )

= ¢ + + ¢

¢ =

v e v

v e

A y A y

A y (3)

From Eq. (2), ( ) ( )¢ = +v eB y -3.3015 1 (4)

(a) Sphere A rises. Use conservation of energy to find h.

T m v V

T V m gh

T V T V m v m

A A

A

A A

12

1

2 2

1 1 2 22

1

20

0

1

20 0

= ¢ =

= =

+ = + ¢ + = +

( ) ,

,

: ( ) AA

A

gh

hv

g

e=¢

=( ) ( . ) ( )

( )( . )

2 2 2

2

3 3015 2 1

2 9 81

-

Since h is to be maximum, e must be as large as possible.

Coefficient of restitution for maximum h: e = 1 000. b

(b) Corresponding value of h. ( ) . [( )( ) ] .¢ = =vA 3 3015 2 1 1 3 3015- m/s

hv

gA= ¢ = =( ) ( . )

( )( . ).

2 2

23 30152 9 81

0 5556 m h = 0 556. m b

(c) Plate B falls and compresses the spring. Use conservation of energy.

Let d0 be the initial compression of the spring and D be the additional compression of the spring after impact. In the initial equilibrium state,

S F k W k Wy B B= = =0 00 0: d d- or (3)

Just after impact: T m v V kB B12

1 021

2

1

2= ¢ =( ) , d

At maximum deflection of the plate, T2 0=

V V V W kg e B2 2 2 021

2= + = + +( ) ( ) ( )- D Dd


1

2

1

20

1

2

1

22

02

02

02m v k W k k kB B B( )¢ + = + + +d d d- D D D

271



Invoking the result of Eq. (3) gives

1

2

1

22 2m v kB B( )¢ = D

Data: m

vB

B y

=¢ = +

=

1 2

3 3015 1 1

6 603

.

( ) . ( )

.

kg,

m/s.

-

-

¢ = Ø =vB k6 603 1000. ,m/s N/m

D22

21 2 6 603

1000

0 05232

=¢

=

=

m v

kB B( )

( . )( . )

.m

s

2

2 D = 0 229. m b

272


PROBLEM 13.180

Two cars of the same mass run head-on into each other at C. After the collision, the cars skid with their brakes locked and come to a stop in the positions shown in the lower part of the figure. Knowing that the speed of car A just before impact was 8 km/h and that the coefficient of kinetic friction between the pavement and the tires of both cars is 0.30, determine (a) the speed of car B just before impact, (b) the effective coefficient of restitution between the two cars.

SOLutiOn

(a) At C:


m m mA B= =

+

820

9

20

9

km/h m/s=

+ = ¢ + ¢

+ = ¢ + ¢

m v m v m v m v

v v v

A A B B A A B B

B A B-

(1)

Work and energy.

Car A (after impact):

T m v

T

U F

U m g

T U T

m

A A

f

k A

A

12

2

1 2

1 2

1 1 2 2

1

20

4

4

1

2

= ¢

==

=+ =

( )

( )

( )

(

-

-

-

-

-m m

¢¢ =

¢ =

=¢

v m g

v

A k A

A

) ( )

( ) ( )( )( . )( . )

.

2

2

4 0

2 4 0 3 9 81

23 544

- m

m m/s

m/s

2

2

vvA = 4 8522. m/s

273



Car B (after impact):

T m v

T

U m g

T U T

B B

k B

12

2

1 2

1 1 2 2

1

20

1

= ¢

==

+ =

( )

( )-

-

-m

1

21 02m v m gB B k B( ) ( )¢ =- m

¢ =

¢ =¢ =

v

v

v

B

B

2

2

2 1 0 3 9 81

5 886

2 4261

( )( )( . )( . )

( ) .

.

m m/s

m /s

m/s

2

2 2

B

From (1) v v vB A B= + ¢ + ¢

= + +=

20

92 222 5 886 2 4261

10 5343

2

. . .

. m/s

=37.924 km/h

vB = 37 9. km/h b

(b) Relative velocities.

+ ( )- - -v v e v vA B B A= ¢ ¢

( . . ) . .

.

- - -2 222 10 5343 2 4261 4 8522

0 19019

e

e

== e = 0 1902. b

274


PROBLEM 13.181

Blocks A and B each weigh 400-g and block C weighs 1.2 kg. The coefficient of friction between the blocks and the plane is mk = 0 30. . Initially, block A is moving at a speed v0 5= m/s and blocks B and C are at rest (Fig. 1). After A strikes B and B strikes C, all three blocks come to a stop in the positions shown (Fig. 2). Determine (a) the coefficients of restitution between A and B and between B and C, (b) the displacement x of block C.

SOLutiOn

(a) Work and energy.

Velocity of A just before impact with B.

T v T vA1 02

22

2

1

2

1

2= = ( )m mA A

U W

T U Tk A1 2

1 1 2 2

0 3-

-

-=+ =

m ( . m)

12

0 3120

2 2

2m m mA A Av g vk A- m ( . ) = ( )

v v g

v

A k

A

2

2 02

2

2

2 0 3 5 2 0 3 9 81 0 3

23 234

( ) = =

( ) =

- - (m ( . ) ( )( . )( . )( .

.

) )2

22 4 82022m /s m/s2 2, ( ) .vA =

Velocity of A after impact with B. ( )¢vA 2

T v T

U W

T U T v

A

k A

A

22

2 3

2 3

2 2 3 32

2

1

20

0 075

1

2

= ( ) =

=

+ = ( )

m

m

A

A

-

-

-

-

m

m

( . )

(, kk )m gA (0.075) = 0

v

v

A

A

2

2

1

2

2 0 3 9 81 0 075 0 44145

0 6644

( ) = =

¢ =

( . )( . )( . ) .

( ) .

m/s m m /s2 2 2

mm/s

275



Conservation of momentum as A hits B.

( ) .

( ) .

v

vA

A

2

2

4 8202

0 6644

=¢ =

m/s

m/s

+ m v m v m v m v m m

v vA A B B B A B B A B

B B

( ) ( )

. . .2 2

4 8202 0 0 6644 4 1

+ = ¢ + ¢ =+ = + ¢ ¢ = 5558 m/s

Relative velocities (A and B).

+ [( ) ] ( )v v e v vA B AB B A2 2- -= ¢ ¢

( . ) . .4 8202 0 4 1558 0 6644- -eAB = eAB = 0 724. b

Work and energy.

Velocity of B just before impact with C.

T v

T v

v

B B

B

B

2 22 2

4 42

2

1

2

1

24 1558

1

24 1558

= ¢ =

= ¢

¢ =

m m

m

B

B

( ) ( . )

( )

( ) . m/s UU g

F W T U T

k

f k B

2 4

2 2 4 42

0 3

1

24 1558 0 3

-

-

-

-

=

= + = fi

m

m

m

m m

B

B B

( . )

( . ) ( . )(

m

))( . )( . ) ( )

( ) .

9 81 0 31

23 9376

42

4

= ¢

¢ =

mB v

v

B

B m/s

Conservation of momentum as B hits C.

m

mB

C

==

0 4

1 2

.

.

kg

kg

( ) .¢ =vB 4 3 9376 m/s

+ m v m v m v m v

vB B C C B B C C

B

( ) ( )

. ( . ) . ( ) .

¢ + = ¢¢ + ¢+ = ¢¢ + ¢

4 4

40 4 3 9376 0 0 4 1 2vv

v vC

B C3 9376 34. ( )= ¢¢ + ¢

276



Velocity of after hits ,B B C vB( ) .¢¢ =4 0

(Compare Figures (1) and (2).)

Thus, ¢ =vC 1 3125. m/s

Relative velocities (B and C).

(( ) ) ( )

. .

¢ = ¢ ¢¢=

v v e v v

eB C BC C B

BC

4 4

3 9376 0 1 3125 0

- -- -( )

eBC = 0 288. b

(b) Work and energy (Block C).

T m v T U W x

T U T m

C C k C

C

42

5 4 5

4 4 5 52

1

20

1

21 3125 0 3

= ¢ = =

+ =

( ) ( )

( . ) ( .

-

-

-

-

m

)) ( . )( )m xC 9 81 0=

x = =( . )

( . )( . ).

1 3125

2 9 81 0 30 29267

2

m

x = 293 mm b

277


PROBLEM 13.182

The three blocks shown are identical. Blocks B and C are at rest when block B is hit by block A, which is moving with a velocity vA of 1 m/s. After the impact, which is assumed to be perfectly plastic ( ),e = 0 the velocity of blocks A and B decreases due to friction, while block C picks up speed, until all three blocks are moving with the same velocity v. Knowing that the coefficient of kinetic friction between all surfaces ismk = 0 20. , determine (a) the time required for the three blocks to reach the same velocity, (b) the total distance traveled by each block during that time.

SOLutiOn

(a) Impact between A and B; conservation of momentum

mv mv mv mv mv mvA B C A B C+ + = ¢ + ¢ + ¢

1 0 0+ = ¢ + ¢ +v vA B (1)

Relative velocities ( )e = 0

( )v v e v v

v v

v v

A B B A

B A

A B

- --

= ¢ ¢= ¢ ¢

¢ = ¢0

From (1) m/s¢ = ¢ =v vA B 0 5.

v = Final common velocities( )

Block C impulse and momentum

+ m gv F t m v F m gc c f c f k c+ = = m

m gv m gt m v

gt v v gtc c k c c+ =

+ = =m

0 0 2 0 2. ( . ) (2)

278



Block A and B impulse and momentum

2 0 5 4 0 2 2

1 0 8 2

m( . ) ( . )

.

--

mgt mv

gt v

== (3)

Substituting v from Eq. (2) into Eq. (3)

1 0 8 0 4- . .gt gt=

t = 1

1 2 9 81

m/s

m/s2( . )( . ) t = 0 08495. s b

(b) Work and energy.

From Eq. (2) v = =( . )( . )( . ) .0 2 9 81 0 08495 0 16667 m/s

Block C:

T T m v m1 22 20

1

2

1

20 16667= = = ( ) ( . )

U F x Wx Wxf C k C C1 2 0 2- = = =m .

T U T mg x mvC1 1 2 220 0 2

1

2+ = + =- ( . )( )

xC = = ¥( .

( . )( ..

0 16667

2 0 2 9 817 0794 10 3)

)m

2-

xC = 7 08. mm b

Blocks A and B:

T m m T m m12

221

22 0 5 0 25

1

22 0 16667 0 027779= ( ) = = ( ) =( . ) . ( . ) .

U Wgx mgx mxk A A A1 2 4 0 8 7 848- - - -= = =m . .

T U T1 1 2 2+ =-

≠ =0 25 0 027779. .m x mA- 7.848m

xA = 0 02832. m xA = 28 3. mm b

279


PROBLEM 13.183

After having been pushed by an airline employee, an empty 40-kg luggage carrier A hits with a velocity of 5 m/s an identical carrier B containing a 15-kg suitcase equipped with rollers. The impact causes the suitcase to roll into the left wall of carrier B. Knowing that the coefficient of restitution between the two carriers is 0.80 and that the coefficient of restitution between the suitcase and the wall of carrier B is 0.30, determine (a) the velocity of carrier B after the suitcase hits its wall for the first time, (b) the total energy lost in that impact.

SOLutiOn

(a) Impact between A and B.


m v m v m v m v m mA A B B A A B B A B+ = ¢ + ¢ = = 40 kg

+ 5 0 m/s + = ¢ + ¢v vA B (1)


( )v v e v vA B AB B A- -= ¢ ¢

( )( . )5 0 0 80- -= ¢ ¢v vB A (2)

Adding Eqs. (1) and (2)

( )

.

5 80 2

4 5

m/s)(1

m/s

+ = ¢¢ = Æ

v

vB

B

Impact between B and C (after A hits B)


+ m v m v m v m vB B C C B B C C¢ + ¢ = ¢¢ + ¢¢

( ( (40 0 40 15kg)(4.5 m/s) kg) kg)+ = ¢¢ + ¢¢v vB C

4 5 0 375. .= ¢¢ + ¢¢v vB C (3)

280




( )

( . )( . )

¢ ¢ = ¢¢ ¢¢= ¢¢ ¢¢

v v e v v

v vB C BC C B

C B

- -- -4 5 0 0 30 (4)

Adding Eqs. (4) and (3)

( . )( . ) ( . )

.

4 5 1 0 3 1 375

4 2545

+ = ¢¢¢¢ =

v

vC

C m/s

(3) ¢¢ = ¢¢ =v vB B4 5 0 375 4 2545 2 90. . ( . ) .- m/s ¢¢ =vB 2 90. m/s b

(b) D -T T T T T

T m v

L B C B C

B B B

= ¢ + ¢ ¢¢+ ¢¢

¢ = ¢ = ÊËÁ

ˆ¯̃

( ) ( )

( ) ( .1

2

40

24 52 kg m/s)) J

kg J

2 =

¢ = ¢¢ = ¢¢ = ÊËÁ

ˆ¯̃

=

¢

405

01

2

40

22 90 168 722 2T T m vC B B B( ) ( . ) .

¢¢ = ¢¢ = ÊËÁ

ˆ¯̃

=T m vC C C1

2

15

24 2545 135 762( ) ( . . kg m/s) J2

D -TL = + + =( ) ( . . ) .405 0 168 72 135 76 100 5 J DTL = 100 5. J b

281


PROBLEM 13.184

A 20-g bullet fired into a 4-kg wooden block suspended from cords AC and BD penetrates the block at Point E, halfway between C and D, without hitting cord BD. Determine (a) the maximum height h to which the block and the embedded bullet will swing after impact, (b) the total impulse exerted on the block by the two cords during the impact.

SOLutiOn

Total momentum in x is conserved.

+ m v m v m v m v v vbl bl bu bu bl bl bu bu bl bu+ ∞ = ¢ + ¢ ¢ = ¢cos ( )20

0 0 02 600 20 4 02+ ∞ = ¢( . ( cos ) ( . ) kg) m/s)( kg)( bl- v

¢ =vbl m/s2 805.


T m m v

T

T

12

12

1

1

24 02

22 805

15 81

= + ¢

= ÊËÁ

ˆ¯̃

=

( )( )

.( . )

.

bl bu bl

kg m/s

55 J

V

T V m m gh

V h

1

2 2

2

0

0

4 02 9 81 3

== = +

= =

kg) m/s

bl bu

2

( )

( . ( . )( ) 99 44

1 1 2 2

. h

T V T V+ = +

15 815 0 0 39 44

0 401

. .

.

+ = +=

h

h m h = 401 mm b

(b) Refer to figure in Part (a).

Impulse-momentum in y direction.

+ m v F t m m v ybu bu bl bu blsin ( )( )20∞ + = + ¢D

( )v ybl = 0

( . sin )0 02 600 20 0 kg)( m/s)(- D∞ + =F t F tD = ◊4 10. N s b

282


PROBLEM 13.185

A 70-g ball B dropped from a height h0 1 5= . m reaches a height h2 0 25= . m after bouncing twice from identical 210-g plates. Plate A rests directly on hard ground, while plate C rests on a foam-rubber mat. Determine (a) the coefficient of restitution between the ball and the plates, (b) the height h1 of the ball’s first bounce.

SOLutiOn

(a) Plate on hard ground (first rebound):


1

2

2

02

0

0 0

m v m gh

v gh

B B=

=


v e v v e gh0 1 1 02= =

Plate on foam rubber support at C:


Points 1 and 3 : V V

m v m v

v e gh

B B B

B

1 3

12

32

3 0

0

1

2

1

2

2

= =

=

=

( )

( )


+ At 3 : m v m v m v m vB B P P B B P P( ) ( )- -3 3+ = ¢ ¢

m

me gh v vP

BB P= = = ¢ ¢

210

703 2 30 3- -( ) (1)


[( ) ( )] ( )

( )

- - - -v v e v v

e gh v v

B P P B

P B

3 3

20 32 0

= ¢ ¢

+ = ¢ + ¢ (2)

283



Multiplying (2) by 3 and adding to (1)

4 2 33 02( ) ( )¢ =v gh e eB -

Conservation of energy at 3 , ( )¢ =v ghB3 22

Thus, 4 2 2 3

3 4 40 25

1 51 63299

2 02

2 2

0

gh gh e e

e eh

h

=

= = =

( )

.

..

-

-

3 1 633 02e e- - . = e = 0 923. b

(b) Points 1 and 2 :


1

2

1

221

21

20 1m v m gh e gh ghB B= =; ( )

h e h12

020 923 1 5= = ( . ) ( . )

h1 1 278= . m b

284


Solution

Ball A falls

T V T V1

0

1 2 2

0+ = +Z

Z (Put datum at 2)

h

mgh mv

v gh

A

A

=

=

= = =

0 2

1

2

2 2 9 81 0 2 1 9809

2

.

( . )( . ) .

m

m/s

Impact

q = = ∞-sin 1

230

r

r

Impulse-Momentum

Unknowns: ¢ ¢ ¢v v vB At An, ,

x-dir

0 0 30 30+ = ¢ + ¢ ∞ + ¢ ∞m v m v m vB B A An A Atsin cos (1)

Noting that m mA B= and dividing by mA

¢ + ¢ ∞ + ¢ ∞ =v v vB An Atsin cos30 30 0 (1)

PRoBlEM 13.186

Ball B is hanging from an inextensible cord. An identical ball A is released from rest when it is just touching the cord and drops through the vertical distance hA = 200 mm before striking ball B. Assuming e = 0 9. and no friction, determine the resulting maximum vertical displacement hB of ball B.

285


PRoBlEM 13.186 (continued)

Ball A alone:

Momentum in t-direction.

-m v m vA A A Atsin 30 0∞ + =

¢ = ∞ = ∞ =v vAt A- - -0.sin . sin30 1 9809 30 99045 m/s (2)


¢ ¢ =v v e v vBn An An Bn- -( )

¢ ∞ ¢ = ∞v v vB An Asin . ( cos )30 0 9 30 0- - (3)

With known value for vAt, Eqs. (1) and (3) become

¢ + ¢ ∞ = ∞¢ ∞ ¢ =

v v

v vB An

B An

sin . cos

sin ( . )( . )co

30 0 99045 30

30 0 9 1 9809- ss 30∞

Solving the two equations simultaneously,

¢ =

¢ =v

vB

An

1 3038

0 8921

.

.

m/s

m/s-

After the impact, ball B swings upward. Using B as a free body

¢ + ¢ = +T V T VB B

where ¢ = ¢

¢ ==

T m v

V

T

B B

B

1

20

0

2( ) ,

,

and V m ghB B B=

1

22m v m ghB B B B( )¢ =

hv

gBB=¢

=

=

1

2

1

2

1 3038

9 81

0 08664

2

2

( )

( . )

( . )

. m hB = 86 6. mm b

286


PRoBlEM 13.187

A 700-g sphere A moving with a velocity v0 parallel to the ground strikes the inclined face of a 2.1-kg wedge B which can roll freely on the ground and is initially at rest. After impact, the sphere is observed from the ground to be moving straight up. Knowing that the coefficient of restitution between the sphere and the wedge is e = 0 6. , determine (a) the angle q that the inclined face of the wedge makes with the horizontal, (b) the energy lost due to the impact.

Solution

(a) Momentum of sphere A alone is conserved in the t-direction.

m v m v

v vA A A

A

0

0

cos sin

tan

q qq

= ¢= ¢ (1)

Total momentum is conserved in the x-direction.

m v m v m v vB B A B B A x+ = ¢ + ¢0 ( )

v vB A x= ¢ =0 0, ( )

0 0 700 2 1 00

0

+ = ¢ +¢ =

. .v v

v v BB

B / (2)

Relative velocities in the n-direction.

( sin ) sin cos

( )( . ) cot

- - - -v e v v

v v vB A

B A

0

0

0

0 6

q q qq

= ¢ ¢= ¢ + ¢ (3)

Substituting ¢vB from Eq. (2) into Eq. (3)

0 6 0 333

0 2670 0

0

. . cot

. cot

v v v

v vA

A

= + ¢= ¢

qq (4)

Dividing (4) into (1)

1

0 2672

.

tan

cottan= =

qq

q

tan .q = 1 935 q = ∞62 7. b

(b) From (1) v v vA A0 1 935= ¢ = ¢tan ( . )q

¢ = ¢ =v v v v BA B0 5168 0 0. , / (2)

T m v m v m vA A A A B Blost = ¢ +( )1

2

1

22 2 2- ( )

287



T v v v

T

lost

lost

/= +

=

1

20 7

1

20 7 0 5168 2 1 30

20

20

2( . )( ) [( . )( . ) ( . )( ) ]-

11

20 7 0 1870 0 2333

0 1400

02

02

[ . . . ]

.

- - v

T vlost J=

T vlost = 0 1400 02. b

288


PRoBlEM 13.188

When the rope is at an angle of a = ∞30 , the 1-kg sphere A has a speed v0 0 6= . m/s. The coefficient of restitution between A and the 2-kg wedge B is 0.8 and the length of rope l =1m. The spring constant has a value of 1500 N/m and q = ∞20 . Determine the velocity of A and B immediately after the impact.

Solution

Masses: m

mA

B

==

1

2

kg

kg

Analysis of sphere A as it swings down.

Initial state: a a= ∞ = = ∞ =30 1 1 1 30 0 133970, ( cos ) ( )( cos ) .h l - - m

V m gh

T mv

A0 0

0 02 2

1 9 81 0 13397 1 31429

1

2

1

21 0 6

= = =

= = =

( )( . )( . ) .

( )( . )

J

00 18. J

Just before impact: a = = =0 0 01 1, ,h V

T m v v vA A A A12 2 21

2

1

21 0 5= = =( ) .


0 18 1 31429 0 5 0

2 98858

1 72875

2

2

. . .

.

.

+ = +

==

v

v

A

A

A

m /s

m/s

2 2

v

b

Analysis of the impact. Use conservation of momentum together with the coefficient of restitution e = 0 8. .

Note that the ball rebounds horizontally and that an impulse TdtÚ is applied by the rope. Also, an impulse NdtÚ is applied to B through its supports.

289



Both A and B.

Momentum in x-direction:

m v m v m v

v vA A x A A x B B x

A x

( ) ( ) ( )

( )( . ) ( )( ) ( )(

+ = ¢ + ¢+ = ¢ + ¢

0

1 1 72875 0 1 2 BB x) (1)

Coefficient of restitution: ( ) ( ) cosv vA n A x= q

( ) , ( ) ( ) cos , ( ) cos ( )

( ) ( )

v v v v v

v v eB n A n A x B x B n

B n A n

= ¢ = ¢ ¢ ¢¢ ¢ =

0 q q =- [[( ) ( ) ]

( ) cos ( ) cos [( ) cos ]

v v

v v e vA n B n

B x A x A x

--¢ ¢ =q q q

Dividing by cos q and applying e = 0 8. gives

( ) ( ) ( . )( . )¢ ¢ =v vB x A x- 0 8 1 72875 (2)


( ) .

( ) .

¢ =¢ =

v

vA x

B x

-0 34575

1 03725

m/s

m/s

¢ =vA 0 346. m/s b

¢ =vB 1 037. m/s b

290


PRoBlEM 13.189

When the rope is at an angle of a = ∞30 , the 0.5-kg sphere A has a speed v0 1 2= . m/s. The coefficient of restitution between A and the 0.9-kg wedge B is 0.7 and the length of rope l = 0 8. m. The spring constant has a value of 500 N/m and q = ∞20 . Determine the velocity of A and B immediately after the impact.

Solution

Masses: m mA B= =0 5 0 9. . kg, kg

Analysis of sphere A as it swings down.

Initial state: a a= ∞ = = ∞ =30 1 0 8 1 30 0 107180, ( cos ) ( . )( cos ) .h l - - m

V m gh

T m v

A

A

0 0

0 02

0 5 9 81 0 10718 0 52576

1

2

1

20 5

= = =

= =

( . )( . )( . ) .

( . )(

J

11 2 0 362. ) .= J

Just before impact: a = = =0 0 01 1, ,h V

T m v v vA A A A12 2 21

2

1

20 5 0 25= = =( . ) .


0 36 0 52576 0 25 0

3 54304

2

2

. . .

.

+ = +

=

v

v

A

A m /s2 2

vA = 1 8823. m/s b

Analysis of the impact. Use conservation of momentum together with the coefficient of restitution e = 0 7. .

Note that the rope does not apply an impulse since it becomes slack.

291



Sphere A. Momentum in t-direction:

m v m v

v vA A A A t

A t A

sin ( )

( ) sin . sin .

(

qq

+ = ¢¢ = = ∞ =

0

1 8823 20 0 6438 m/s

vAA t) .= ∞0 6438 m/s 70Z

Both A and B. Momentum in x-direction:

m v m v m v m vA A A A n A A t B B+ = ¢ + ¢ + ¢=

0

0 5 1 8823 0 5

( ) cos ( ) sin

( . )( . ) ( . )(

q qvv v

vA n B

A n

) cos ( . )( . ) sin .

( . )( ) cos .

20 0 5 0 6438 20 0 9

0 5 20 0

∞ + ∞ + ¢¢ ∞ + 99 0 83105¢ =vB . (1)

Coefficient of restitution:

( ) ( ) [( ) ( ) ]

cos ( ) [ cos ]

c

¢ ¢ =¢ ¢ =

¢

v v e v v

v v e v

v

B n A n A n B n

B A n A

B

- -- -q q 0

oos ( ) ( . )( . ) cos20 0 7 1 8823 20∞ ¢ = ∞- vA n (2)

Solving Eqs. (1) and (2) simultaneously for ( )¢vA n and ¢vB ,

( ) .

.

¢ =¢ =

v

vA n

B

-0 24853

1 053

m/s

m/s

Resolve vA into horizontal and vertical components.

tan

( )

( )

.

.

. .

( .

b

b b

=¢

- ¢

=

= ∞ + ∞ = ∞

¢ =

v

v

v

A t

A n

A

0 6438

0 24853

68 9 20 88 9

0 64438 0 24853

0 690

2 2) ( . )

.

+

= m/s

¢ =vA 0 690. m/s 88.9° b

¢ =vB 1 053. m/s b

292


PRoBlEM 13.190

A 60-kg pellet shot vertically from a spring-loaded pistol on the surface of the earth rises to a height of 90 m. The same pellet shot from the same pistol on the surface of the moon rises to a height of 570 m Determine the energy dissipated by aerodynamic drag when the pellet is shot on the surface of the earth. (The acceleration of gravity on the surface of the moon is 0.165 times that on the surface of the earth.)

Solution

Since the pellet is shot from the same pistol, the initial velocity v0 is the same on the moon and on the earth Work and energy.

Earth: T mv

U mg E

EE L

L

1 02

1 2

1

290

=

==

- - -( )

(

m

loss of energy due to drag))

T mg EE L1 90 0- - =

Moon: T mv T1 02

21

20= =

U mgM1 2 570- -= ( ) (1)

T

T mgM

2

1

0

570 0

==- (2)

Subtracting (1) from (2) -570 90 0mg mg EM E L+ + =

g g

m

E g g

M E

L M E

===

0 165

0 06

570 0 06 90 0 06

570 0 06 0 16

.

.

( . ) ( . )

( . )( .

kg

-= 55 90 0 06

9 81 0 06 570 0 165 90

2 3838

) ( . )

. . ( . )

.

g gE E-= ¥=

¥ -J

EL = 2 38. J b

293


PRoBlEM 13.191

An elastic cable is to be designed for bungee jumping from a tower 40 m high. The specifications call for the cable to be 25 m long when unstretched, and to stretch to a total length of 30 m when a 300 kg weight is attached to it and dropped from the tower. Determine (a) the required spring constant k of the cable, (b) how close to the ground a 90 kg man will come if he uses this cable to jump from the tower.

Solution


V T V mg1 1 10 0= = = (30)

Datum at k: V1

4

30 300 9 81

8 829 10

=

= ¥

( )( . )

.

m)( m

J

V Tg2 20 0, = =

V V V k

T V T V

k

g e22

1 1 2 2

4

01

25

0 8 829 10 0 12 5

= + = +

+ = +

+ ¥ = +

( )

. .

m

k = 7060 N/m b

(b) From (a), k = 7063 2. N/m

T

W g

V d

T

1

1

2

0

90 90 9 81

90 9 81 40

0

== ¥= ¥ ¥=

=-

.

. ( )

N

Datum: V V V d

V d

g e22

22

01

27063 2 40 25

3531 6 15

= + = +

=

( . )( )

. ( )

- -

-

294



d = distance from the ground T V T V1 1 2 2+ = +

0 90 9 81 40 0 3531 6 15

3531 6 104615 1 75929

2

2

+ = +

+

( )( . )( ) . ( )

. .

- -

-

d d

d d 44 0=

d = 16 903. m,12.7198 m

Discard 16 903. m (assumes cord acts in compression when rebound occurs).

d = 12 72. m b

295


Solution

Potential energy of the magnet.

F r dV dr

Vr

v T

V V V

V r

m

m g

m

= = -

=

= == +

=

300

300

0 0

300

2

1 1

1 1 1

1 1

/

( ) ( )

( )

/

/

-

-

r

r

Vm

12 2 2 2

1

1

100 12 200 200 100 42544

206 262

= + + ==

=

( ) [( ) ( ) ]

.

( )

- - mm

-- -300

206 2621 45446

112 0 061

..

( ) ( ) ( .

=

= =

N.mm

mm )(9.81)(112V mgg ))

N mm( ) . .Vg 1 65 9232=

V V V

V

g m1 1 1

1

65 9232 1 45446 64 46874

0 064469

= +

= ==

( ) ( )

. . .

. .

- N.mm

N m

TT mv v v

V V V

B B B

g m

22 2 2

2 2 2

1

2

1

20 06 0 03

0 06 9 81

= = ( ) =

= +

= ( )

. .

( ) ( )

. ( . )(112 mm) N mm

N.m

- 300 -

-12

17 9368

0 0179372

=

=

. .

.V

T V T V1 1 2 2+ = +

. . .

.

0 0 064469 0 03 0 017937

2

2

2

+ =

=

v

v

B

B

-

7746 2 2 m /s vB = 1 657. m/s b

PRoBlEM 13.192

A 60-g hollow steel sphere attached to a 300 mm cord can swing about Point O in a vertical plane. It is subjected to its own weight and to a force F exerted by a small magnet embedded in the ground. The magnitude of that force expressed in N is F r= 300 2/ , where r is the distance from the magnet to the sphere expressed in mm. Knowing that the sphere is released from rest at A, determine its speed as it passes through Point B.

296


PRoBlEM 13.193

A satellite describes an elliptic orbit about a planet of mass M. The minimum and maximum values of the distance r from the satellite to the center of the planet are, respectively, r0 and r1. Use the principles of conservation of energy and conservation of angular momentum to derive the relation

1 1 2

0 12r r

GM

h+ =

where h is the angular momentum per unit mass of the satellite and G is the constant of gravitation.

Solution

Angular momentum.

h r v r v

h r v r v

vh

rv

h

r

= == =

= =

0 0 1 1

0 0 1 1

00

11

,

(1)


T mv

VGMm

r

T mv

VGMm

r

T V T V

mvGMm

r

A

A

B

B

A A B B

=

=

=

=

+ = +

1

2

1

2

1

2

02

0

12

1

02

-

-

-00

12

1

02

12

0 1

1 0

1 0

1

2

21 1

2

=

=È

ÎÍ

˘

˚˙ =

È

ÎÍ

˘

˚˙

mvGMm

r

v v GMr r

GMr r

r r

-

- --

Substituting for v0 and v1 from Eq. (1)

hr r

GMr r

r r

hr r

r r

2

02

12

1 0

1 0

2 12

02

12

02

1 12-

-

-

È

ÎÍÍ

˘

˚˙˙

=È

ÎÍ

˘

˚˙

È

ÎÍÍ

˘

˚̊˙˙

= + =È

ÎÍ

˘

˚˙

+ÊËÁ

h

r rr r r r GM

r r

r r

hr r

2

12

02 1 0 1 0

1 0

1 0

2

0 1

2

1 1

( )( )--

ˆ̂¯̃

= +ÊËÁ

ˆ¯̃

=21 1 2

0 12

GMr r

GM

hQ.E.D.

297


Solution


T mv VGMm

r

T mv VGMm

r

GM gR

T V

B BB

A A AA

A A

= =

= =

=+ =

1

2

1

2

12 30

02

2

2

-

-

( . )Eq.

TT V

mvgR

rm mv

gR

rm

B B

BA

A

+

=1

2

1

202

22

2

- -

r

v vgR

r

r

r

r

A

AB

B

A

B

= + =

=ÊËÁ

ˆ¯̃

= + =

6370 400 6770

21

6370 60 64

202

2

km

- -

330

2 9 81 6 37 10

6 43 101

6430

67702

02

6 2

6

( . )( . )

( . )

km

v vA = ¥¥

ÊËÁ

ˆ¯̃

- -

Conservation of Angular Momentum. v vA2

02 66 21809 10= ¥- . (1)

r v r v

v v vA A B B i

A

== ∞ =

sin

/ sin

f ;

( ) .6430 6770 55 0 778010 0 (2)

Eqs. (2) and (1): [ ( . ) ] .1 0 77801 6 21809 10202 6- v = ¥ v0 3970= m/s b

PRoBlEM 13.194

A shuttle is to rendezvous with a space station, which is in a circular orbit at an altitude of 400 km above the surface of the earth. The shuttle has reached an altitude of 60 km when its engine is turned off at Point B. Knowing that at that time the velocity v0 of the shuttle forms an angle f0 55= ∞ with the vertical, determine the required magnitude of v0 if the trajectory of the shuttle is to be tangent at A to the orbit of the space station.

298


PRoBlEM 13.195

A 25-g steel-jacketed bullet is fired horizontally with a velocity of 600 m/s and ricochets off a steel plate along the path CD with a velocity of 400 m/s. Knowing that the bullet leaves a 10-mm scratch on the plate and assuming that its average speed is 500 m/s while it is in contact with the plate, determine the magnitude and direction of the average impulsive force exerted by the bullet on the plate.

Solution

Impulse and momentum.

Bullet alone:

m t mv F v1 2+ D =

t direction. + mv F t mvt1 215 20cos cos∞ D = ∞-

F

Ft

t

t

t

D = ∞ ∞D =

( . )[ ]

.

0 025 600 15 400 20

5 092

kg m/s cos m/s cos

-kkg m/s

m

m/ss

kg m/s)/(

-

-D = = = ¥

= ◊

tS

vBC

AV

t

0 010

50020 10

5 092

6.

( .F 22 s

kg m/s

kN

2

0 10

254 6 10

254 6

6

3

¥

= ¥ ◊=

- )

.

.Ft

n direction.

+ - ∞ + D = ∞mv F t mvn1 215 20sin sin

F

F

n

n

t

t

D = ∞ + ∞

D =

( . )[ ]

.

0 025 600 15 400 20

7 3025

kg m/s sin m/s sin

kkg m/s s◊ D = ¥t 20 10 6-

Fn = ◊ ¥

= ¥ ◊

( ( )

.

43025 20 10

365 1 10

6

3

kg m/s)

kg m/s2

-

Fn = 365 1. kN

299



Force on bullet.

F F Fn t= + = + =2 2 2 2365 1 254 1 445. . kN

tan..

.

q

q

= =

∞ = ∞

F

Fn

t

365 1254 1

15 40 1-

.q = ∞55 1

F = 445 kN 40 1. ∞

Force on plate. ¢ =F F- ¢ =F 445 kN 40 1. ∞ b

300


Solution

Velocity of the hammer at impact.


T1 0=

V mg

V

V

H

H

=

==

( . )

( . )( . )( . )

.

1 2

0 650 9 81 1 2

7 6521

m

kg m/s m

J

2

T m

V v v

V

H H

2

2 2 2

2

1

20 650

20 325

0

=

= =

=

..

T V T V

v

1 1 2 2

20 7 652 0 325

+ = +

+ =. .

v

v

2 23 54

4 852

==

.

.

m /s

m/s

2 2

Velocity of pile after impact.

Since the impact is plastic ( ),e = 0 the velocity of the pile and hammer are the same after impact.


The ground reaction and the weights are non-impulsive.

PRoBlEM 13.196

The 650-kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140-kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact ( ),e = 0 determine the average resistance of the ground to penetration.

301



Thus, m v m m v

vm v

m m

H H H p

H H

H p

= + ¢

¢ =+

=+

=

( )

( )

( )

( )( . .

650

650 1404 852 3 9 m/s) 992 m/s

Work and energy. d = 0 110. m

T U T2 2 3 3+ =-

T m m v

T

T

T

H H22

3

22

23

1

20

1

2650 140 3 992

6 295 10

= + ¢

=

= +

= ¥

( )( )

( )( . )

. J

U m m d F d

F

U

H p AV

AV

2 3

2 3

9

650 140 9 81 0 110 0 110-

-

-

-

= +

= +=

( )

( )( . )( . ) ( . )

8852 49 0 110. ( . )- FAV

T U T2 2 3 3+ =-

6 295 10 852 49 0 110 03. . ( . )¥ + =- FAV

FAV = = ¥( . ) ( . ) .7147 5 0 110 64 98 103/ N FAV = 65 0. kN b

302


PRoBlEM 13.197

A small sphere B of mass m is attached to an inextensible cord of length 2a, which passes around the fixed peg A and is attached to a fixed support at O. The sphere is held close to the support at O and released with no initial velocity. It drops freely to Point C, where the cord becomes taut, and swings in a vertical plane, first about A and then about O. Determine the vertical distance from line OD to the highest Point ¢¢C that the sphere will reach.

Solution

Velocity at Point C (before the cord is taut).

Conservation of energy from B to C.

T

V mg a mga

T mv V

T V T V

mga

B

B

C C C

B B C C

=

=Ê

ËÁˆ

¯̃=

= =

+ = +

+ =

0

22

22

1

20

0 21

2

( )

22

2 2

2mv C

v ga

C

C

+

=

Velocity at C (after the cord becomes taut).

Linear momentum perpendicular to the cord is conserved.

q = ∞45

+ - = ¢

¢ = ( )Ê

ËÁˆ

¯̃

¢ = =

mv mv

v ga

v ga ga

C C

C

C

sinq

2 22

2

2 214

Note: The weight of the sphere is a non-impulsive force.

Velocity at C':

C C to (conservation of energy)¢ : T m v V

T m v V

C C C

C C C

= ¢ =

= ¢ =¢ ¢ ¢

1

20

1

20

2

2

( )

( )

303



Datum: T V T V

m v m v

C C C C

C C

+ = +

¢ + = ¢ +

¢ ¢

1

20

1

202 2( ) ( )

¢ = ¢ ¢v vC C

¢ ¢¢C C to (conservation of energy):

T m v

T m ga

T mga

C C

C

C

¢ ¢

¢

¢

= ¢

= ( )=

1

21

22

2

2

2

1 4 2

( )

/

Datum: T V T VC C C C¢ ¢ ¢¢ ¢¢+ = +

V

T

V mgh

C

C

C

¢

¢¢

¢¢

===

0

0

2

20 0

2

2

mga mgh

h a

+ = +

= h = 0 707. a b

304


Solution

Disks A and B (total momentum conserved).

m m m m vA A B B A A B Bv v v+ = ¢ + ¢

Normal direction: m v m v m vA o A A n B B ncos ( ) ( )q + = ¢ + ¢0 (1)

Relative velocities

[ cos ( ) ] ( ) ( )

(cos ) ( ) ( )

v v e v v

v e v vA B n B n A n

B n A n

qq

- --

= ¢ ¢= ¢ ¢0 (2)

Multiplying Eq. (2) by mB and subtracting it from Eq. (1)

v m em m m v

v vm em

m m

A B A B A n

A nA B

A B

0

0

cos ( ) ( )( )

( ) ( cos )( )

( )

q

q

--

= + ¢

¢ =+

(3)

From Equation (3):

(a) m emA B> ( )¢vA n positive b

(b) m emA B< ( )¢vA n negative b

(c) m emA B= ( )¢ =vA n 0 b

PRoBlEM 13.198

Disks A and B of mass mA and mB, respectively, can slide freely on a frictionless horizontal surface. Disk B is at rest when it is hit by disk A, which is moving with a velocity v0 in a direction forming an angle q with the line of impact. Denoting by e the coefficient of restitution between the two disks, show that the n component of the velocity of A after impact is (a) positive if m emA B> , (b) negative if m emA B� , (c) zero if m emA B= .

305


PRoBlEM 13.199

Blocks A and B are connected by a cord which passes over pulleys and through a collar C. The system is released from rest when x = 1 7. m. As block A rises, it strikes collar C with perfectly plastic impact ( ).e = 0 After impact, the two blocks and the collar keep moving until they come to a stop and reverse their motion. As A and C move down, C hits the ledge and blocks A and B keep moving until they come to another stop. Determine (a) the velocity of the blocks and collar immediately after A hits C, (b) the distance the blocks and collar move after the impact before coming to a stop, (c) the value of x at the end of one compete cycle.

Solution

a) Velocity of A just before it hits C.

Conservations of energy.

Datum at j:

Position j:

( ) ( )v v

T

v

A B1 1

1

1

0

0

0

= ===

Position k: T m v m v

v v

A A B B

A B

22 21

2

1

2= +

=

( )

( )kinematics

T v v

V m g m g

g

V

A A

A B

22 2

2

2

1

25 6

11

21 7 1 7

5 6 1 7

= + =

===

( )

( . ) ( . )

( )( )( . )

--

-11 7. g

T V T V

v gA

1 1 2 2

20 011

21 7

+ = +

+ = - .

v

v

A

A

2 3 4

119 81

3 032

1 741

= ÊËÁ

ˆ¯̃

==

.( . )

.

.

m /s

m/s

2 2

306



Velocity of A and C after A hits C.

¢ = ¢v vA C ( )plastic impact

Impulse-momentum A and C.

+ m v T t m m vA A A C A+ D = + ¢( )

( )( . )5 1 741 8+ D = ¢T t vA (1)

v v v vB A B A= ¢ = ¢; ( )cord remains taut

B alone:

m v T t m v

T t vB A B A

A

- D- D

= ¢= ¢( )( . )6 1 741 6 (2)

Adding Equations (1) and (2), 11 1 741 14( . ) = ¢vA

¢ =vA 1 3679. m/s

¢ = ¢ = ¢ =v v vA B C 1 368. m/s b

(b) Distance A and C move before stopping.


Datum at k:

Position k:

T m m m v

T

T

V

A B C A22

22

2

2

1

214

21 3681

13 103

0

= + + ¢

= ÊËÁ

ˆ¯̃

==

( )( )

( . )

. J

307



Position l:

T3 0=

V m m gd m gd

V gd gdA C B3

3 8 6 2

= += =

( )

( )

--

T V T V2 2 3 3+ = +

13 103 0 0 2. + = + gd

d = =( . ) ( )( . ) .13 103 2 9 81 0 6679/ m d = 0 668. m b

(c) As the system returns to Position k after stopping in Position l, energy is conserved, and the velocities of A, B, and C before the collar at C is removed are the same as they were in Part (a) above with the directions reversed. Thus, ¢ = ¢ = ¢ =v v vA C B 1 3679. m/s. After the collar C is removed, the velocities of A and B remain the same since there is no impulsive force acting on either.

Conversation of energy.

Datum at k:

T m m v

T

T

A B A22

22

2

1

21

25 6 1 3679

10 291

= + ¢

= +

=

( )( )

( )( . )

. J

V2 0=

T V m gx m gx

V gxB A4 4

4

0

6 5

= ==

--( )

T V T V2 2 4 4+ = +

10 291 0 1 9 81. ( )( . )+ = x

x = 1 049. m b

308


Solution

Conservation of energy j k.

Datum at k:

Sphere A

Position j:

v

T

V mgl

A

A A

===

0

0

11

1 ( cos )- q

Position k:

T mv

V

A22

2

1

20

=

=

0 11

202+ = +mgl mvA A A( cos )- q

v glA A A2 2 1= ( cos )- q (1)

Conservation of momentum at k.

mv mv mv mvA B A B+ = ¢ + ¢

v v vA A B+ = ¢ + ¢0 (2)

Relative velocities at k.

( )v v e v v v e v vA B B A A B A- - -= ¢ ¢ = ¢ ¢ (3)

Adding Equations (2) and (3) and solving for ¢vB,

¢ = +v e vB A1

21( ) (4)

PRoBlEM 13.200

A small sphere A attached to a cord AC is released from rest in the position shown and hits an identical sphere B hanging from a vertical cord BD. If the maximum angle qB formed by cord BD with the vertical in the subsequent motion of sphere B is to be equal to the angle qA, determine the required value of the ratio l lB A/ of the lengths of the two cords in terms of the coefficient of restitution e between the two spheres.

309



Conservation of energy k l.

Sphere B

Position k: T m vB221

2= ¢( )

V

T2

3

0

0

==

V mglB B3 1= ( cos )- q

T V T V m v mglB B B2 2 3 321

20 0 1+ = + ¢ + = +( ) ( cos )- q

( ) ( cos )¢ =v glB B B2 2 1- q (5)

Substituting ¢vB from Eq. (4) into Eq. (5)

1

41 2 12 2( ) ( cos )+ =e v glA B A- q (6)

Dividing (1) into (6) and setting q qA B=

1

4

1 2 1

2 1

2 2

2

( ) ( cos )

( cos )

+=

e v

v

gl

glA

A

B B

A B

--

qq

l l eB A/ /= +( )1 42 b

310


PRoBlEM 13.201

A 2-kg block A is pushed up against a spring compressing it a distance x = 0 1. m. The block is then released from rest and slides down the 20º incline until it strikes a 1-kg sphere B, which is suspended from a 1 m inextensible rope. The spring constant k = 800 N/m, the coefficient of friction between A and the ground is 0.2, the distance A slides from the unstretched length of the spring d = 1 5. m and the coefficient of restitution between A and B is 0.8. When a = ∞40 , determine (a) the speed of B (b) the tension in the rope.

Solution

Data: m m k x dA B= = = = =2 1 800 0 1 1 5kg, kg N/m m, m, , . .

m q ak e l= = = ∞ = ∞ =0 2 0 8 20 40 1 0. , . , , , . m

Block slides down the incline.

+ Â =Fy 0:

N m gA- cosq = 0

N m g

F N

A

f k

== ∞== =

=

cos

( )( . )cos

.

( . )( . )

q

m

2 9 81 20

18 4368

0 2 18 4368

3

N

..6874 N

Use work and energy. Datum for Vg is the impact point near B.

T V k xe1 1 12 20

1

2

1

2800 0 1 4 00= = = =, ( ) ( )( . ) . J

( ) ( )sin ( )( . )( . )sin .V m gh m g x dg A A1 1 2 9 81 1 6 20 10 7367= = + = ∞ =q J

U F x df1 2 3 6874 1 6 5 8998Æ = + = =- - -( ) ( . )( . ) . J

T m v v v VA A A A22 2 2

21

2

1

21 1 000 0= = = =( )( ) .

T V U T V vA1 1 1 2 2 220 4 00 10 7367 5 8998 1 000 0+ + = + + + = +Æ : . . . .-

vA2 28 8369= . m /s2

vA = 2 9727. m/s 20∞

311



Impact: Conservation of momentum.

Both A and B, horizontal components

+

:

m v m v m vA A A A B Bcos cosq q+ = ¢ +0

(2)(2.9727) cos cos ( . )20 2 20 1 00∞ = ¢ +∞v vA B (1)

Relative velocities: ( ) ( ) [( ) ( ) ]¢ ¢ =v v e v vB n A n B n A n- -

¢ ¢ =¢ ∞ ¢ =v v e v

v vB A A

B A

cos [ ]

cos ( . )( . )

q - --

0

20 0 8 2 9727 (2)


¢ =¢ =

v

vA

B

1 0382

3 6356

.

.

m/s

m/s

Sphere B rises: Use conservation of energy.

T m v V

T m v V m gh m gl

B B

B B B

12

1

2 22

2 2

1

20

1

21

= ¢ =

= = =

( )

( cos )- a

T V T V m v m v m g

v v gl

B B B B

B

1 1 2 22

22

22 2

1

20

1

21

2

+ = + ¢ + = +

= ¢

: ( ) ( cos )

( )

-

-

a

(( cos )

( . ) ( )( . )( cos )

.

1

3 6356 2 9 81 1 40

8 6274

2

-

- -

a

= ∞

= m /s2 2

(a) Speed of B. v2 2 94= . m/s

(b) Tension in the rope. r

r

=

= = =

1 00

8 6274

1 008 62742

22

.

.

..

m

m/sav

n

+ Â =F m an B n :

T m g m a

T m a gB B n

B n

- cos

( cos )

( . )( . . cos )

aa

== += + ∞1 0 8 6274 9 81 40

T = 16 14. N

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