fe/eit dynamics sample problems

8/9/2019 FE/EIT Dynamics sample problems

1/37

inematics

a

acceleration

gravitational

acceleration

position

radius

distance

time

velocity

ft/sec

2

m/s2

ft/sec

2

m/s2

ft

rn

ft m

ft

m

sec

s

ft/sec m/s

9

r

v

S y m b o l s

0:

e

rad/sec''

rad

tad/see

rad/s

2

rad

rad/s

angular acceleration

angular position

angular velocity

deals only with relationships among position, velocity,

acceleration, and time.

A body in motion can be considered a particle if rota-

tion of the body is absent or insignificant. A particle

does not possess rotational kinetic energy. All parts of

a particle have the same instantaneous displacement,

velocity, and acceleration.

A

rigid body

does not deform when loaded and can be

considered a combination of two or more particles that

remain at a fixed, finite distance from each other. At

any given instant, the parts (particles) of a rigid body

can have different displacements, velocities, and acceler-

ations if the body has rotational as well as translational

motion.

If r is the position vector of a particle, the instantaneous

velocity and acceleration are

r

[position] 1 4 . 1

dr

[velocity]

1 4 . 2

=-

dt

dv d

2

r

[acceleration]

1 4 .3

=-=-

dt dt

2

R E C T I L IN E A R M O T I O N

A rectilinear system is one in which particles move only

in straight lines. (Another name is

linear system.)

The

relationships among position, velocity, and acceleration

for a linear system are given by Eqs. 14.4 through 14.6.

S u b s c r i p t s

s(t) = v(t)dt =

J

a(t)de

1 4 . 4

0

initial

f

final

v(t)

=

d:;t)

=

ja(t)dt

n

normal

1 4 . 5

r

radial

( ) _ dv(t) _ d

2

s(t)

t

tangential

a t - dt - dt

2

1 4 . 6

e

transverse

I N T R O O O C T I O N T o K I N E M A T I C S

Dynamics

isthe study ofmoving objects. The subject is

. divided into kinematics and kinetics.

Kinematics

is the

study of a body's motion independent of the forces on

the body. It is a study of the geometry of motion with-

out consideration of the causes of motion. Kinematics

R e c ta n g u la r C o o r d in a te s

The position of a particle is specified with reference to

a coordinate. system. Three coordinates are necessary

to identify the position in three-dimensional space; in

two dimensions, two coordinates are necessary. A coor-

dinate can represent a linear position, as in the rectan-

gular coordinate 'system, or it can represent an angular

position, as in the polar system.

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o f e s s i o n a l P ub lu ut io n s, ln e .


2/37

out of the integrals in Eqs. 14.4 and 14.5. The ini

distance from the origin is

so;

the initial velocity

constant, va; and a constant acceleration is denoted

Consider the particle shown in Fig. 14.1. Its position,

as well as its velocity and acceleration, can be speci-

fied in three primary forms: vector form, rectangular

coordinate form, and unit vector form.

F i g u r e

1 4 .1

R e c t a n g u l a r C o o r d i n a t e s

~ath of particle

y

(x,

y.

z

x

k

z

The vector form of the particle's position is

r,

where

the vector

r

has both magnitude and direction. The

rectangular coordinate form is (x, y, z). The unit vector

form

is

0\

r

=

xi

+

yj

+

zk

1 4 .7

The velocity and acceleration are the first two deriva-

tives of the position vector, as shown in Eqs. 14.8

and 14.9.

dr

v=-

dt

= xi + yj + zk

dv d

2

r

a=-=-

dt dt

2

=xi+jjj+ik

1 4 .9

1 4 .8

C o n s t a n t A c c e le r a ti on

Acceleration is a constant in many cases, such as a free-

.falling

body with constant acceleration g. If the accel-

eration is constant, the acceleration term can be taken

a(t) =

o

v(t)

=

ao

J

dt

=

va

+

aot

s(t) =o J J dt

2

1 4 . 1

] 4 .1 ]

aot2

=S o

+

vat

+

-2-

v

2

(t)

=

6

+

2ao(s - so)

1 4 .1

1 4 .1

C U R V I L I N E A R M O T I O N

Curvilinear motion describes the motion of a part

o

along a path that isnot a straight line. Special exam

of curvilinear motion include plane circular motion

projectile motion. For particles traveling along cu

linear paths, the position, velocity, and accelerat

may be specified in rectangular coordinates as they

w

for rectilinear motion, or it may be more convenien

express the kinematic variables in terms of other c

dinate systems (e.g., polar coordinates).

T r an s v e r se a n d R a d ia l C o m p o n en t s

In polar coordinates, the position of a particle is

scribed by a radius,

r,

and an angle, B . The posit

may also be expressed as a vector of magnitude r

direction specified by unit vector e

r

Since the velo

of a particle is not usually directed radially out from

center of the coordinate system, it can be divided

two components, called

radial

and

transverse,

wh

are parallel and perpendicular, respectively, to the

radial vector. Figure 14.2 illustrates the radial

transverse components of velocity ina polar coordin

system, and the unit radial and unit transverse vect

e

r

and

ee,

used in the vector forms of the motion eq

tions.

[position]

1 4 .1

v=;e;

+

Veee

= fer + reee

[velocity)

]4. ]

.. B 0 2 )

=

r-r e

r

+ r B +

2 rB )ee

[acceleration] / 4 . / 6


3/37

K in e m a t i ( s 14~3

F i g u r e i 4 . 2 R a d i a l

; ; d

T r ~ ~ 5 v e ~ 5 e C o o r d i ~ ~ t e ~ .

y

r

path of

particle

\ 8

eQ~ _ _e ~r _

~l _

x

T a n g e n t i a l a n d N o r m a l C o m p o n e n t s

Aparticle moving in a curvilinear path will have instan-

taneous linear velocity and linear acceleration. These

linear variables will be directed tangentially to the path,

and, therefore, are known as

tangential velocity, Vt,

and

tangential acceleration, at,

respectively. The force that

constrains the particle to the curved path will generally

be directed toward the center of rotation, and the par-

ticle will experience an inward acceleration perpendic-

ular to the tangential velocity and .acceleration, known

a s

the

normal acceleration, an.

The resultant accelera-

tion, a, is the vector sum of the tangential and normal

accelerations. Normal and tangential components of ac-

o celeration are illustrated in Fig. 14.3. The vectors en

and

et

are normal and tangential to the path, respec-

tively. p is the principal radius of curvature.

1 4 . 1 7

1 4 . 1 8

F i g u r e 1 4 . 3 T a n g e n t i a l a n d N o r m a l C o o r d i n a te s

instantaneous

center of rotation

y

x

P la n e C ir tu la r M o tio n

Plane circular motion (also known as rotational particle

motion, angular motion, or circular motion) is motion

of a particle around a fixed circular path. The behav-

ior of a rotating particle is defined by its angular po-

sition,

B ,

angular velocity, w, and angular acceleration,

a. These variables are analogous to the s, v, and a

variables for linear systems .. Angular variables can be

substituted one-for-one in place of linear variables in

most equations.

B

[angular position]

1 4 . 1 9

dB

[angular velocity]

1 4 . 2 0=-

dt

dJ.JJ

a=-

dt

d

2

e

[angular acceleration] 1 4 . 2 1

dt

2

R E L A T I O N S H IP S B E T W E E N f i N E A if A N D

R O T A T IO N A L V A R IA B L E S

..

- _

_ _ .

. . . . . . . . . . . . . . . . . .. . . . . . . . . . . . .

s

=

rB

1 4 . 2 2

1 4 .2 3

1 4 . 2 4

Vt =rw

dVt

at = ra = dt

v r 2

an=-=rw

r

1 4 . 2 5

P R O J E C T IL E M O T I O N

A projectile is placed into motion by an initial impulse.

(Kinematics deals only with dynamics during the flight.

The force acting on the projectile during the launch

phase is covered in kinetics.) Neglecting air drag, once

the projectile is in motion, it is acted upon only by

the downward gravitational acceleration (i.e., its own

weight). Thus, projectile motion is a special case of

motion under constant acceleration.

Consider a general projectile set into motion at an an-

gle of

B

from the horizontal plane, and initial velocity

Vo ,

as shown in Fig. 14.4. In the absence of air drag,

the following rules apply to the case of travel over a

horizontal plane.

The trajectory is parabolic.

The impact velocity is equal to initial velocity, Vo

The range is maximum when

B =

45 .

_The time for the projectile to travel from the

launch point to the apex is equal to the time to

travel from apex to impact point.

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o f e s s i o n a l P u b li c a t i o n s , I n c .


4/37

1 4 - 4 F E R e v i e w M a n u a l

1 -

The time for the projectile to travel from the apex

of its flight path to impact is the same time an ini-

tially stationary object would take to fall straight

down from that height.

F ig u r e 14 .4 P r o i e c t i l e M o t i o n

y

v{t)

path of projectile

~----

x

The following solutions to most common projectile

problems are derived from the laws of uniform accel-

eration and conservation of energy.

ax =0 14.26

a

y

=g 1 4 . 2 7

Vx= Vxo= Vocosf 14 .28

Vy=VyO-

gt

=Vosine -

gt 1 4 . 2 9

x

= vxot = votcose

1 4 . 3 0

1

2

1

2

Y

= Vyot - 29t = vot sin e - 29t

1 4 . 3 1

s A M p L E P R O B L E M s

Problems 1-3 refer to a particle whose curvilinear mo-

tion is represented by the equation s

= 2 0t + 4t2 - 3t

3

.

1. What is the particle's initial velocity?

(A)

2 0

mjs

(B) 2 5 mjs

(C) 30 mfs

(D) 32

mfs

76394

S o l u t io n :

ds

2

v =d =

20

+ 8t - 9t

t .

At

t=

,

v

=

20

+

(8) (0) - (9) (0)2

=

20

mfs

Answer is A.

2 .

What is the acceleration of the particle at tiuie

t = O ?

(A ) 2 mfs2

(B) 3 mf

s

2

(0) 5

mfs2

(D) 8

mf

s2

S o lu t i o n :

At t

=

0,

a

=8

mffp

Answer

is D.

#17 394 .

3. What is the maximum speed reached by the par-

ticle?

(A)

21.8

mfs

(B)

27.9

mjs

(0) 34.6 mfs

(D) 48.0

mfs

#78394

S o l u t io n :

The maximum of the velocity function is found

b y

equating the derivative of the velocity function to zero

and solving for t.

v=

20 +

8t - 9t

2

d v

- = 18t =0

dt

8

t=-

s

=

0.444 s

18

Vmax

=

0

+

(8)(0.444 s) - (9)(0.444 S)2

= 21.8 mfs

Answer is A.

4. Choose the equation that best represents a rigid

body or particle under constant acceleration.

(A) a =

9.81

mjs2 + voft

(B) v =o +

aot

0)

v

= vo +

l o t a(t)dt

D

a

=

vVr

.

P ro f es s io n al P u b li ca tio n s , I n c . - - -


5/37

14 6 FE R ev ie w M an u al _

S o l u t i o n :

gt

2

Y =

ot sin e 2

J .. .

t

2

- votsine

+

y

=

0

2

y = -1500 m since it is below the launch plane.

( 9

m )

~12

m

. 2 s

t

2

- (1000 -; )

t

sin 30 - 1500m=

m

2

m

4.905 -

t -

500 - t - 1500

m

=0

S2 S

-b ...jb

2 -

4ac

t

=. [quadratic formula]

2a

500

V

-500)2 - (4)(4.905)(-1500)

(2 ) (4.905)

= +104.85 s, -2.9166 s

x

=

votcos8

=

1000 :) (104.85 s) cos 30

=90803 m (90800 m)

Answer is D.

F E - S T Y L E E X A M P R O B L E M S

Problems 1 and 2 refer to a particle for which the posi-

tion is defined by

set )

=

2 sin

ti

+

4 cas tj [t in radians]

1. What is the magnitude of the particle's velocity at

t =4 rad?

(A) 2.61

(B) 2.75

(C) 3.30

(D) 4.12

8 2689

2. What is the magnitude of the particle's acceleration

at

t

= 7r?

(A) 2.00

(B) 2.56

(C) 3.14

(D ) 4.00

8 3689

3. For the reciprocating pump shown, the radius of

the crank 'is ' T ' = 0.3 m,and the rotational speed is

n

=

350

rpm. What is the tangential velocity of point

A

on the crank corresponding to an angle of

e

=35 from

the horizontal?

(A) 0 m/s

(B ) 1.1 mjs

(C) 10

m/s

(D) 11 m/s

4. A golfer on level ground attempts to drive a golf

ball across a 50 m wide pond, hitting the ball so that it

travels initially at 25

isi].

The ball travels at an initial

angle of 45 to the horizontal plane. How far will the

golf ball travel, and does it clear the pond?

(A) 32

m;

the ball does not clear the pond

(B) 45 m; the ball does not clear the pond

(C) 58 m; the ball clears the pond

(D) 64 m; the ball clears the pond

5. Rigid link AB is 12 m long. It rotates counterclock-

wise about point A at 12 rev/min. A thin disk with ra-

dius 1.75m is pinned at its center to the link at point B.

The disk rotates counterclockwise at 60 rev /min with

respect to point

B.

What is the maximum tangential

velocity seen by any point on the disk?

(A) 6 m/s

(B) 26 m/s

(C )

33

m/s

(D) 45 mjs

#84394

#85994

P r o f e s s i o n a l P u b l i c a t i o n s , I n c .

IIIIiIII .;.. __ .;.. -- -.~


6/37

K in em a ti c s 1 4 - 7

-

For the following pr-ob lerns use the NCEES Hand-

book as your only reference.

6. A particle has a tangential acceleration of at (rep-

resented by the equation given) when it moves around

a point in a curve with instantaneous radius of 1 m.

What is the instantaneous angular velocity (in rad/s)

of the particle?

at =2t - sin t + 3 cot t [inm/s2]

A t

2

+ cost + 3 In [csc r]

(B) t

2

-cost+3lnlcsctl

C t2 - cost

+

3 In [ s in t]

(D) t

2

+cost+3Inlsintl

3555794

7. A stone is dropped down a well. 2.47 s after the

stone is released, a splash is heard. If the velocity of

sound in air is 342

ti s] ,

find the distance to the surface

of the water in the well.

A 2.4

m

B.

7.2 m

C 28 m

(D) 30 m

2878687

Problems 8 and 9 refer to the following situation.

A motorist is traveling at 70 km/h when he sees a traffic

light in an intersection 250 m ahead turn red. The

light's red cycle is 15 s. The motorist wants to enter

the intersection without stopping his vehicle, just as

the light turns green.

8. What. uniform deceleration of the vehicle will just

put the motorist in the intersection when the light turns

green? .

(A) 0.18 m/s

2

(B) 0.25 m/s2

(C) 0.37m/s

2

(D )

1.3 m/s2

28 84687

9. If the vehicle decelerates at a constant rate of

0.5 m/s2, what will be its speed when the light turns

green?

(A) 43 km/h

(B) 52 km/h

(C)

59 km/h

(D) 63 km/h

2885687

Problems 10 and 11 refer to the following information.

The position (in radians) of a car traveling around a

curve is described by the following function of time (in

seconds).

8 (t)

= t

3

- 2 e - 4t + 10

10. What is the angular velocity at

t

= s?

(A) -16 rad/s

(B) -4 rad/s

(C) 11 rad/s

(D)

15

rad/s

2 886687

11. What is the angular acceleration at t = 5 s? .

(A )

4 rad/s

2

(B)

6 rad/s

2

(C) 26 rad/s

2

(D) 30 rad/s

2

. 2886 687

12. The rotor of a steam turbine is rotating at

7200 rev /min when the steam supply is suddenly cut

off. The rotor decelerates at a constant rate and comes

to rest after 5 min. What was the angular deceleration

of the rotor?

(A) 0.40 rad/s?

(B)

2.5 rad/s

2

(C ) 5.8 rad/s

2

(D) 16 rad/s

2

2887687

13. A flywheel rotates at 7200 rev /min when the power

is suddenly cut off. The flywheel decelerates at a con-

stant rate of 2.1

rad/s

and comes to rest

6

min later.

How many revolutions does the flywheel make before

coming to rest?

(A )

18000 rev

(B) 22000 rev

(C)

72000 rev

(D) 390000 rev

2888687

Problems 14-i6 refer to the following situation.

A projectile has an initial velocity of 110 m/s and a

launch angle of 20 from the horizontal. The surround-

in IT terrain is level and air friction is to be disregarded.

b ,

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o f e s s io n a l P u b lic a ti o n s ; I n c .


7/37


IIIIII I ___

14. What is the flight time of the projectile?

(A) 3.8 s

(B) 7.7 s

(C )

8.9

s

(D) 12

s

O OOOMM 298

15. What is the horizontal distance traveled by the

projectile?

(A) 80

m

(B) 400 m

(C ) 800 m

(D) 1200

m

OOO OMM 298

16. What is the maximum elevation achieved by the

projectile?

(A) 72 m

(B) 140 m

(C)

350

m

(D ) 620 m

OOOOMM 298

S O L U T I O N S T O F E - S T Y L E E X A M P R B L E M S

S o l u t i o n

1 :

()

. ds( t)

2. .

v t

=

--;It

= cos

zi -- 4sintJ

At

t=

4

rad,

v(4) =2cos (4 rad) - 4sin (4 rad)j

=

1.3li - (-3.03)j

Iv 4)1

=

V(-1.31)2

+

(3.03)2

=.30

Answer is C.

S o lu ti o n 2 :

From Prob.

1,

v(t) = 2cos ti - 4 sin tj

( )

dv(t) . .

at = -- = -2Sintl- 4costJ

dt

a(r r)

=

-2sin1l'i-4cos1l'j

= O i

+ 4.0j

\a(1l) \=V(O)2

+

(4.0)2

=.0

Answer is D.

S o l u t i o n 3 :

Use the relationship between the tangential and rota-

tional variables.

Vt =

w

w =

angular velocity of the crank

=350 :i:) (21l' ::~) (60 ~)

mm

=36.65

rad/s

(

rad)

Vt

=

0.3

m) 36.65 -s-

=11.0 m/s

This value is the same for any point on the crank at

r =

0.3 m.

Answer is D.

S o l u t i o n 4 :

The elevation of the ball above the ground is

gt

2

. gt

2

Y =Vyot - 2 =vat sin e - 2

When the ball hits the ground, y =0, and

gt

2

votsine

=

2

Solving for t, the time to impact is

2va sin ()

t= - ---

9

Substitute the time of impact into the expression for

x

to obtain an expression for the range.

(

2vosine)

x

=

vatcose =vc 9 cose

2v

2

=

_0sin e cos ()

9

(2 ) (25

m r

---'----;m~s~ sin 45 cos45

9.812'

s

=63.7 m

Answer is D.


8/37

S o l u t i o n 5 : S o lu t i o n 7 :

The elapsed time is the sum of the time for the stone to

fall and the time for the sound to return to the listener.

The distance, x, traveled by the stone under the influ-

ence of a constant gravitational acceleration is

The maximum tangential velocity of point B with re-

spect to point A is

Vt;BIA

=

rw

= r(27rf

(

rad) ( rev)

(12 m) 27r - 12 -.

rev nun

1 2

X =o +

v o t

+ 2 ,gt

60~

min

Xo and Vo are both zero.

=

15.08

m/s

1

x =_gt

2

2

The maximum tangential velocity of the periphery of

the

disk with respect to point B is

Solving for

t,

the time for the stone to drop is

Vt,disklB = rw =r(27r f

(1.75 m) (27r rad)

rev

(

60 re.v)

min

The time for the sound (traveling at velocity c) to return

to the listener is

60 _s_

min

=11.00 s s s ]

The total time taken is

he velocities combine when the two velocity vectors

coincide in direction. The maximum velocity of the

periphery of the disk with respect to point

A

is the sum

of the magnitudes of the two velocities.

h + t2

=

2.47 s

f

x x

-- + -

= 2.47 s

c

Vt,disklB

=

Vt,BIA + Vt,disklB

m m

=15.08 -

+

11.00 -

s s

Substitute values for 9 and c.

R

x

m + ------m =.47 s

9.81 - 342-

S2 s

=6.08 m/s

Answer is B.

By trial and error with the answer choices given (or by

solving the quadratic equation),

x

=28 m

S o lu t i o n 6 :

Answer is C.

dVt

at=: dt

Vt

= at dt

=J(2t~sint+3 cott) dt

= t

2

+ cos t + 3 In [sin tl [inm/s)

(70 ~) (1000 ~)

v o

=

3600 ~

=19.44

mls

S o lu t i o n 8 :

The initial speed of the vehicle is

Vt t2 + cost + 3 In [sin r]

=- =- - - - - - - - - - - - ~- - ~

rim

t

2

+ cas t + 3 In

[ s in

tl [in rad/s)

The distance traveled under a constant deceleration is

Answer is D.

1 2

x =Xo + vot - 2 , at

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o f e s s i o n a l P u b l i c a t i o n s , I n c .


9/37

1 4 - 1 0 F E R e v i e w M a n u a l

-

Letting

Xo =

0, the acceleration required to travel a

distance x in time

t

starting with velocity

v o

is

(2)(vot -

x)

a =~------ .,,-----

t

2

(2) [( 19.44 ~) (15

s ) -

250

m ]

(15 S )2

=

0.37 m/s2

Answer is C.

S o l u t i o n 9 :

v

=

va

+

at

krn ( -0.5 ~) (158) (3600 ~)

=70 _ + .:....:._

, -8

------=--_~-

h 1000 k~

=

43 km/h

Answer is A.

S o l u t i o n 1 0 :

d B 2

W ( t) = - = 3t - 4t - 4

. dt ..

w (3 ) =

(3)(3)2 - (4)(3) - 4

=

11 rad/s

Answer is C.

S o l u t i o n 1 1 :

a( t)

=

dw (t)

dt

6t - 4

a( 5) =

6)(5) - 4

=6 rad/s

2

Answer is C.

S o lu t i o n 1 2 :

W = W o - at

(

1200

_r~v) (271 _rad)

mm rev (s )

o = ---6-0-~s~------- - a( 5 min) 60 ~

min

e x

=

2.51 rad/s

2

Answer is B.

S o l u t i o n 1 3 :

1 2

B

=

B o

+

wo t - 2 at

=0 + (1200 : : i : )

(2 71) (6

ruin) - ~ (2.1 r : 2 d )

x

[ ( 6

min) (60 rr:n)]

2

=135.4

X

10

3

rad

e

=

135.4 x 10

3

rad

. 271

=1.5 X 10

3

rev

Alternative solution: The average rotational speed dur- .

ing deceleration is

rev .

7200 -. -0

mm

P r o f e s s i o n a l P u b l i c a t i o n s , I n c . =

2

=

600 rev /min

(

rev)

e =t =

3600 min (6 min)

=1,600 rev

Answer is B.

S o l u t i o n 1 4 :

The vertical component of velocity is zero at the apex.

Vy =

a sine -

gt

0= n o

7 )

sin 20 - (9.81 ~) t

t

=.84 s

The projectile takes an equal amount of time to return

to the ground from the apex. The total flight time is

T

= (2)(3.84 s) = 7.68 s (7.7 s)

Answer is B.


10/37

K in e m a t i c s 1 4 - 1 1

s o l u t i o n 1 5 :

Calculate the range from the horizontal component of

velocity.

S o l u t i o n 1 6 :

The elevation at time

t

is

x

=

vxt

=Vo

coset

= (110 7 ) cos200(7.68 s)

=794m

y =

vot sin e ~ ~

gt

2

=

(110 7 ) (3.84 s) sin

20

- (~) (9.81 ~) (3.848)2

=72 m

Answer is C.

Answer is A.

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o fe s s io n a l P u b li ca t io n s , In c


11/37

inetics

S u b s c r ip t s

0

initial

f

friction

k

dynamic

n

normal

r

radial

R resultant

s

static

t tangential

e

transverse

I N T R O D U C T I O N T O K I N E T I C S

Kinetics is the study of motion and the forces that cause

N o m e n d a t u r e

motion. Kinetics includes an analysis of the relationship

acceleration

ft/sec

2

m/s2

between the force and mass for translational motion and

a

between torque and moment of inertia for rotational

f

linear frequency

Hz

Hz

motion. Newtons laws form the basis of the governing

F force

lbf

N

9

gravitational

theory in the subject of kinetics.

acceleration

ft/sec

2

m/s2

.............................................

9c

gravitational

M O M E N T U M

constant (32.2)

lbm- ft /lbf-sec

2

k

spring constant

lbf/ft

N/m

The vector linear momentum (momentum) is defined

m mass lbm kg by Eq. 15.1. It has the same direction as the velocity

N

normal force

lbf

N

vector. Momentum has. units of force x time (e.g., lbf-

p

linear momentum

lbf-sec

N-s

sec or Ns).

r

position

ft

m

r

radius

ft

m

p=mv

[ 8 1 ]

1 5 . 1 0

R

resultant force

lbf

N

mv

[U .S .]

1 5 .1 b

distance

ft

m

p=-

t

time

sec

gc

s

T

period

sec

s

Momentum is conserved when no external forces act on

v

velocity

ft/sec

m/s

W

weight

lbf

N

a particle. If no forces act on the particle, the velocity

and direction of the particle are unchanged. The

law o

conservation of momentum states that the linear mo-

S y m b o l s

mentum is unchanged if no unbalanced forces act on the

particle. This does not prohibit the mass and velocity

Q

angular

from changing, however. Only the product of mass and

acceleration rad/sec?

rad/s

2

velocity is constant.

S

deflection

ft

m

)

angular position

rad

rad

j . L

coefficient of

N E W r O N / S F i R S T A N D S E C O N D i A W S O F

M O T I O ~ . . . . . . .

friction

.r:

Newtons first law of motion

states that a particle will

ngle deg

deg

w

natural frequency

rad/sec

rad/s

remain in a state of rest or will continue to move with

P r of e ss io n a l p u b li ca t i o n s , I n c


12/37

15 2 FER ev iew M an u a l _

constant velocity unless an unbalanced external force

acts on it.

This law can also be stated in terms of conservation of

momentum: If the resultant external force acting on a

particle is zero, then the linear momentum of the par-

ticle is constant.

Newtons second law of motion

states that the acceler-

ation of a particle is directly proportional to the force

acting on it and is inversely proportional to the particle

mass. The direction of acceleration is the same as the

direction of force.

This law can be stated in terms of the force vector re-

quired to cause a change in momentum: The resultant

force is equal to the rate of change of linear momentum.

F= dp

dt

1 5 . 2

For a fixed mass,

F _ dp _ d(mv)

- dt

;It

dv

=m-

dt

=ma [S I]

[US.]

1 5 . 3 a

F= ma

gc

1 5 . 3 b

s

W E I G H T

The

weight,

W, of an object is the force the object exerts

due to its position in a gravitational field, g.

W=mg

W=

mg

gc

1 5 . 4 a

S I]

[US.]

1 5 . 4 b

gc

is the gravitational constant, approximately 32.2 lbm-

ft/lbf-sec

2

.

Friction is a force that always resists motion or impend-

ing motion. It always acts parallel to the contacting

surfaces. If the body is moving, the friction is known

as

dynamic friction.

If the body is stationary, friction

is known as

static friction.

The magnitude of the frictional force depends on the

normal force,

N,

and the

coefficient of friction, IL,

be-

tween the body and the contacting surface.

1 5 . 5

The static coefficient of friction is usually denoted with

the subscript

s,

while the dynamic coefficient of friction

is denoted with the subscript

k. ILk

is often assumed

to be 75 percent of the value oflLs. These coefficients

are complex functions of surface properties. Experi-

mentally determined values for various contacting con-

ditions can be found in handbooks.

For a body resting on a horizontal surface, the normal

force

is the weight of the body. If the body rests on an

inclined surface, the normal force is calculated as the

component of weight normal to that surface, as illus-

trated in Fig. 15.1.

N=mgcos

N

=

mgcos

9 c

1 5 . 6 a

1 5 . 6 b

1 5 .1

[S I]

[US.]

F i g u r e 1 5 . / F r i c t i o n a l a n d N o r m a l F o r c e s

impending motion

P r o fe s s io n a l P u b li c a ti o n s, I n c. - - - -

The frictional force acts only in response to a disturbing

force, and it increases as the disturbing force increases.

The motion of a stationary body is impending when the

disturbing force reaches the maximum frictional force

ILsN.

Figure 15.1 shows the condition of impending

motion for a block on a plane. Just before motion starts,

the resultant,

R,

of the frictional force and normal force

equals the weight of the block. The angle at which

motion is just impending can be calculated from the

coefficient of static friction.

=an

-lJ.Ls

Once motion begins, the coefficient of friction droPS

slightly, and a lower frictional force opposes movement.

This is illustrated in Fig. 15.2.


13/37

K in e t i c s 1 5 - 3

F i g u r e 1 5 . 2 F r i c t i o n a l F o r c e ~ e r s u s D i s t u r b i n g F o r c e . . . . .

. I impending motion

no rnotron ~ I

~e~~~b~~~~~~e~ I

~motion

4~

~sN

disturbing force

K i N E T I C S O F A P A R T i C L E

Newton's second law can be applied separately to any

direction in which forces are resolved into components.

The law can be expressed in rectangular coordinate

form (i.e., in terms of z- and y-component forces), in

polar coordinate form (i.e., in tangential and normal

components), or in radial and transverse component

form.

R e c ta n g u la r C o o r d in a t es

Equation 15.8 is Newton's second law in rectangular

coordinate form and refers to motion in the x-direction.

Similar equations can be written for the y-direction or

any other coordinate direction.

[ 8 1 ]

1 5 . 8

In general,

Fx

may be a function of time, displacement,

and/ or velocity. If

Fx

is a function of time only, then

the motion equations are

J

(F x(t))

vx(t) =xo + - - - : ; : n - dt

x( t) = o + vxot + J vx( t)dt

1 5 . 1 0

[ 8 1 ]

1 5 .9 '

If Fx is constant (i.e., is independent of time, displace-

ment, or velocity), then the motion equations become

[ 8 1 ]

1 5 . 1 1

vx (t)

=

v-o

+ (~) t

1 5 . 1 2

F. t

2

x(t) =Xo + vxot

+

2~

a

x

t

2

=Xo + vxo t + --

2

1 5 .1 3

T a n g e n t i a l a n d N o rm a l C o m p o n e n t s

For a particle moving along a circular path, the tangen-

tial and normal components of force, acceleration, and

velocity are related.

LFn = man=

m ( v I )

[ 8 1 ]

1 5 . 1 4

LF

t

=

m a t

=m

d ~ t )

[ 8 1 ]

1 5 . 1 5

R a d i a l a n d T r a n sv er se C o m p o n e n t s

For a particle moving along a circular path, the radial

and transverse components of force are

[ 8 1 ] 1 5 . 1 6

LFe =mae

[ 8 1 ]

1 5 . 1 7

F R E E V I B R A T I O N

.

: : ,

.

Vibration is an oscillatory motion about an equilibrium

point. If the motion is the result of a disturbing force

that is applied once and then removed, the motion is

known as

natural

(or free )

vibration.

If a continuous

force or single impulse is applied repeatedly to a system,

the motion is known as

forced vibration.

A simple application of free vibration is a mass sus-

pended from a vertical spring, as shown in Fig. 15.3.

After the mass is displaced and released, it will oscillate

up and down. If there is no friction (i.e., the vibration

is undamped), the oscillations will continue forever.

F i g u r e

1 5 . 3

S i m p l e M a s s -S p r i n g S y s t e m

The system shown in Fig. 15.3 is initially at rest. The

mass is hanging on the spring, and the equilibrium po-

sition is the static deflection,

8

st

.

This is the deflection

due to the gravitational force alone.

mg =k 8

st

[ 8 1 ]

1 5 . 1 8 0

[u.s.] IS . 1 8 b

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o f e s s i o n a l P u b l i c a t i o n s I n c


14/37

1 5 - 4 F E R e v i e w M a n u a l _

The system is then disturbed by a downward force (i.e.,

the mass is pulled downward from its static deflection

and released). After the initial disturbing force is re-

moved, the mass will be acted upon by the restoring

force (-

kx )

and the inertial force (

-mg) . .Both

of these

forces are proportional to the displacement from the .

equilibrium point, and they are opposite in sign from

the displacement. The equation of motion is

F=ma

[ 8 1 J

1 5 .1 9

mg - k(x + Ost ) = m/i:

[ 8 1 ]

1 5 . 2 0

kO st - k(x + Ost ) = mii:

[ 8 1 J

1 5 .2 1

mx+kx = 0

[ 8 1 ]

1 5 .2 2

The solution to this second-order differential equation is

x(t)

=

C

1

coswt

+

C

2sinwt

1 5 .2 3

C

1

and C

2

are constants of integration that depend on

the initial displacement and velocity of the mass. w is

known as the natural frequency of vibration or angular

frequency. It has units of radians per second. It is not

the same as the linear frequency,

i

which has units of

hertz. The period of oscillation, T, is the reciprocal of

the linear frequency.

w = {

[ 8 1 ]

1 5 . 2 4 0

k

gc

[U .8 .]

1 5 . 2 4 b

= -

m

f= ~ = ~

1 5 .2 5

2 1 1

T

T

= ~ = 211

1 5 .2 6

f

w

For the general case where the initial displacement is

Xo

and the initial velocity is va, the solution of the equation

of motion is

x(t)

=

ocoswt + ~) sinwt 1 5 . 2 7

For the special case where the initial displacement is

Xo

and the initial velocity is zero, the solution of the

equation of motion is

x t) =xocos w t

1 5 .2 8

s A M p f p R O B I E M S .

....................

P ro fe s si on a l P u b li ca t i o n s , In c . - - - .

1.

For which of the following situations is the net force

acting on a particle necessarily equal to zero?

(A) The particle is traveling a~ constant velocity

around a circle. .

(B)

The particle has constant linear momentum.

C The particle has constant kinetic energy.

(D) The particle has constant angular momentum.

#86691

S o l u t i o n :

This is a restatement of Newton's first law of motion

which says that if the resultant external force

acting

on a particle is zero, then the linear momentum of the

particle is constant.

Answer is B.

2.

One newton is the force required to

(A) give a 1

g

mass an acceleration of 1m/s2.

(B)

accelerate a

10

kg mass at a rate of

0.10

mjs2.

(C) accelerate a 1 kg mass at a rate of 1.00 em/s2.

(D) accelerate a 1 kg mass at a rate of 9.81 m/s2.

#87689

S o l u t i o n :

Newton's second law can be expressed in the form of

F =ma. The unit of force in S1 units is the newton,

which has fundamental units of kgm/s

2

. A newton is

the force required to accelerate a 1kg mass at a rate of

1

m/s2 or a

10

kg mass at a rate of

0.10

m/s2.

Answer is B.

3. A 550 kg mass initially at rest is acted upon by a

force of 50e

t

N. What are the acceleration, speed, and

displacement of the mass at

t =4

s?

(A) 4.96 m/s2, 4.87 mis, 19.5 m

(B )

4.96 m/s2, 4.96 mis, 19.8 m

(C) 4.96

m/s2,

135.5

mis,

1466

m

D

4.96

m/s

2

,

271

us],

3900

in

#88691

S o l u t i o n :

F 50e

4

N

a = - = - - -

550 kg

=4.96

rn/s

2


15/37

K i n e t i c s 1 5 - 5

r

4

50e

t

N e

t

4 1

v = )0 550 kg dt = 1110 = 4.96 - 11

= 4.87 m/s

i

t i t 50e

t

N

i t

(t) = --k-

de

=

(4.87 m

dt

a a 550 gas

s =

~87tl~

=

(4.87

7 )

(4 s ) - 0

=19.48 rn

Answer is A.


A 5 kg block begins from rest and slides down an

inclined plane .

After 4 s, the block has a velocity of 6 m/s.

4. If the angle of inclination is 45, how far has the

block traveled after 4 s?

(A ) 1.5 m

(B) 3

m

(C) 6

m

(D ) 12 m

89689

S o l u t i o n :

v(t) =a

+

aot

m

v(t) -va 6 -;-0

ao= =

_o? _

t 4

s

=.5 m/s2

1 2

s C t =o + vat + 2 ao t

=0+0+ (~) (1.5 ~) (4S)2

=

12

m

Answer is D.

5. What is the coefficient of friction between the plane

- and the block?

(A ) 0.15

(B) 0.22

(C)

0.78

(D ) 0.85

90 689

S o l u t i o n :

Choose a coordinate system so that the x-direction is

. parallel to the inclined plane.

LFx

= ma

x

= mg

x - Ff

ma

x

=mgsin45 - p,mgcos45

mgsin45 -:-ma

x

p,=

mgcos45

gsin45 - ax

9 cos 45

(9.81 ~) s in 45 -1.5 ~

=

(9.81 : Z cos 45

=

0.78

Answer is C.

6. A constant force of 750 N is applied through a pulley

system to lift a mass of 50 kg as shown. Neglecting

the mass and friction of the pulley system, what is the

acceleration of the 50 kg mass?

F=

750 N

(A ) 5.20 m/s2

(B )

8.72 m/s2

(C) 16.2 m/s2

(D) 20.2 m/s2

91 694

S o l u t i o n :

Apply Newtons second law to the mass and to the

two frictionless, massless pulleys. Refer to the following

free-body diagrams.

T,

1 .

F

T,

mass A

pulley B

pulley C


16/37

1 5 6 F E R e v i e w M a n u a l - - - - - - - - - - - - - - - - - - - - - - - _

mass A: Tl - mg =ma

pulley

B: 2T2 - T , =0

pulley C: T2 = F = 750 N

Tl -

mg

2T2 -

mg

2F -

mg

a=

m m

m

(2) (750 N) - (50

kg)

(9.81 ~)

50

kg

=

20.2

m/s2

Answer is

7. A mass of 10 kg is suspended from a vertical spring

with a spring constant of 10 N/m. What is the period

of vibration?

A 0.30 s

(B) 0.60 s

(C )

0.90

s

(D) 6.3

s

9 2394

S o l u t i o n :

m ~

T =

2 7 rV

k =

27 r ~

10 ~

=6.3 s

Answer is

F E -S T Y L E E X A M P R O B L E M S

1.

If the sum of the forces on a particle is not equal to

zero, the particle is

(A)

moving with constant velocity in the direction

of the resultant force.

(B)

accelerating

in

a direction opposite to the re-

sultant force.

(C)

accelerating in the same direction as the resul-

tant force.

(D)

moving with a constant velocity opposite to

the direction of the resultant force.

93689

2.

A varying force acts on a 40 kg weight as shown in

the following force versus time diagram. What is the

object's velocity at

t =

4 s if the object starts from

rest?

F N

3

2

2

t (5)

(A) 0 m/s

B

0.075

m/s

(C ) 0.15 m/s

(D) 0.30 ta]

9 4691


The 52 kg block shown starts from rest at posi-

tion A and slides down the inclined plane to posi-

tion

B.

When the block reaches position B, a 383 N hori-

zontal force is applied.

The block comes to a complete stop at position

C.

The coefficient of friction between the block and

the plane is f .- L =0.15.

c

3.

Find the velocity at position B.

(A) 2.41 m/s

(B) 4.12

m/s

(C) 6.95

m/s

(D)

9.83

m/s

. 9568 1

4. Find the distance between positions Band C.

(A) 3.23 m

(B) 4.78 rn

(C )

7.78 m

(D) 10.1

m

966 81

P r o f e s s i o na l P u b l ic a t io n s , I n c . .- - -


17/37

problems 5 and 6 refer to the following pulley system.

In standard gravity, block

A

exerts a force of 10000 N

and block

B

exerts a force of 7500 N. Both blocks are

initially held stationary. There is no friction and the

pulleys have no mass. ,~

10000 N

5.

Find the acceleration of block

A

after the blocks are

released.

(A) 0 m/s2

(B ) 1.4 m/s2

(C ) 2.5 m/s2

(D) 5.6

m/s2

9799 4

6 .

Find the velocity of block

A

2.5 s after the blocks

are released.

\

(A )

B

(C )

(D)

o m/s

3.5 m/s

4.4 m/s

4.9 m/s

9 8994

For the following problems use the NCEES Hand-


7. What is the period of a pendulum that passes the

center point 20 times a minute?

(A) 0.2 s

(B) 0.3 s

(C ) 3 s

(D) 6 s

2142689

8. A variable force of (40 N)cos

e

is attached to the end

of a spring whose spring constant is 50 N

[ux.

There is

no deflection when e =90. At what angle, e , will the

spring deflect 20 em from its equilibrium position?

(A) -14

(B) 25

(C ) 64

(D) 76

39331294

K ineti s 15 7

9. A spring has a constant of 50 N/m. The spring is

hung vertically, and a mass is attached to its end. The

spring end displaces 30 em from its equilibrium position.

The same mass is removed from the first spring and

attached to the end of a second (different) spring, and

the displacement is 25 em. What is the spring constant

of the second spring?

(A )

(B )

C

(D)

46

N/m

56 N/m

60 N/m

6 3 N /m

39351294

10. A cannonball of mass 10 kg is fired from a cannon

of mass 250 kg. The initial velocity of the cannonball

is 1000

km/h.

All of the cannon's recoil is absorbed by

a spring with a spring constant of 520 NIcm. What is

the maximum recoil distance of the cannon?

A 0.35 m

B

0.59

m

(C ) 0.77 m

(D )

0.92 m

3 942295

11. A child keeps a 1 kg toy airplane flying horizontally

in a circle by holding onto a 1.5 m long string attached

to its wing tip. The string is always in the plane of the

circular flight path.

If

the plane flies at 10

us],

what

is the tension in the string?

(A) 7 N

(B ) 15 N

(C )

28

N

(D ) 67 N

3 471 193

12. A car with a mass of 1530 kg tows a trailer (mass

of 200 kg) at 100 km/h. What is the total momentum

of the car-trailer combination?

(A) 4600 N-s

(B) 22000 N-s

(C) 37000Ns

(D) 48 000 Ns

1584 689

13. A car is pulling a trailer at 100 km/h. A 5 kg cat

riding on the roof of the car jumps from the car to the

trailer. What is the change in the eat's momentum?

(A) -25 N-s(loss)

(B) a N-s

(C) 25 Ns (gain)

{D) 1300 N-s (gain)

#4.169 795


18/37

1 5 - 8 F E R e v i e w M a n u a l 1 _

-

14. A 3500 kg car accelerates from rest. The constant

forward tractive force of the car is 1000 N, and the con-:

stant drag force is 150 N. What distance will the car

travel in 3 s?

A 0.19

m

(B )

1.1

m

C

1.3

m

(D )

15 m

2108 689

S O L U T I O N S T O F E - S T Y L E E X A M P R O B L E M S

S o lu ti o n 1 :

Newton's second law, F =ma, can be applied sepa-

rately to any direction in which forces are resolved into

components, including the resultant direction.

Since force and acceleration are both vectors, and mass

is a scalar, the direction of acceleration is the same as

the resultant force.

Answer is C.

S o l u t i o n 2 :

Use the impulse-momentum principle. The impulse is

the area under the F-t curve. There are two right tri-

angles.

F= dp

d t

FD.. t = m :::.v

FD.. t

(2) [(~) (3N)(2S)]

D ..v

= -- =

--=-~--:,------=-

tri

40 kg

=

0.15

m/s

Answer is C.

S o l u t i o n 3 :

Choose a coordinate system parallel and perpendicular

to the plane, as shown.

L F x =max

Wx - flN =ma

x

mg sin e - flmg cas e = ma;

ax =

9

sin e -

fl9

cas e

=9.81 ~) [1

5

3 - (0.15) G ~ ]

=

2.415

m/s2

v

2

=v~

+

2ao( s - so )

v a

=

o =

v

2

= 2 aos

= (2) (2.415 ~) (20

m)

=

96.6

m2/s2

v

= J

96.6 ~2

=

9.83 iis]

Answer is D.

S o l u t i o n 4 :

L F x =ma

mg

sin

e - pcose - fl mgcose + Psine)

=

ma

a = (~) [mgsine - pcose - fl mgcose + P sin e)]

P

g sin e - flg cos e - - (cose

+

fl sin e)

m

=9.81 ~) [ 1

5

3 - (0.15TG~)]

_ (383

N ) [12 + (0.15) (~)]

52 kg 13 13

=

4.809

m/s2

P r o f e s s io na l P ub li c a t i o n s , I n c . - - -


19/37

K inetin 1 5 9

2 2

v =vo+2aO(S-30)

vo =

9.83 m/s

[from Prob. 3]

v = 30 = 0

2 _ (9.83 m ) 2

-vo S

3 =

2ao

= (2 ) (-4.809 ;)

=

10.05

m :

Answer is D.

S o lu t io n 5 :

Refer to the following free-body diagrams.

T

T

Apply Newton's second law to the free body of mass A.

\

But

m=Wig,

so

1 0 0 0 0 : )

(aA)

=

10000

N -

TA

9.812

s

Apply Newton's second law to the free body of

mass B.

Combine the equations and solve for

aA,

setting

TA =

TB, and aA =-aB'

W

A

- mAaA = WB - mBaB = WE + mBaA

WA - WB g(WA - WB)

aA= =

mB+mA WB+WA

m (10000 N -7500 N)

= 9.81 s2 7500 N + 10000 N _

= 1 .4 m /s2

A l t e r n a t e S o lu ti o n :

But F

=

ma, so

Since the tension in the rope is the same everywhere,

10000 N - T T - 7500 N

10000 N 7500 N

T

=

8571 N

From block A,

=

g

g(WA -

T)

A

(9.81 ~) (10000 N - 8571 N)

10000 N

=1.4

m/s2

Answer is B.

S o lu ti on 6 :

VA

=

Vo

+ aAt

=

+

1 . ~ ; (2.5 s)

=3.5 ui]

Answer is B.

S o lu t io n 7 :

A pendulum will pass the center point two times during

each complete cycle. Therefore,

10

cycles are completed

in

60 s.

elapsed time

T = - ------,--

no. of cycles

60

s

- 10

= 6

s

Answer is D.

___________________________________ . ~h~oo~~~~m~ ~


20/37

15 10 FE R ev iew M anua l _

S o lu t io n 8 :

From Hookes law, the relationship between force, F,

and deflection,

x,

or a linear spring is

F=kx

(50 :) (20 cm)

(40 N )c o se

=

cm

100 -

m

c as e =

0.25

e

=

cos-1(0.25)

-;; 75.5 (76)

Answer is D.

S o lu ti on 9 :

The gravitational force on the mass is the same for both

springs. From Hooke ~ law, .

F =

k1Xl

=

kZX2

kz =k1Xl

X2

(50 ~) (30 cm)

25 cm

=0 N/m

Answer is C.

Solution 1 0 :

Use the conservation of momentum equation to deter-

mine the velocity of the cannon after the ball is fired.

Initially, the cannon and cannonball are both at rest.

Since the cannon recoils, its velocity direction will be

opposite (i.e., negative) to the direction of the cannon-

ball.

(10 kg) (0)+(250 kg) (0)

=

10 kg) (1000 ~) + (250 kg) (v.)

Vc

=

40 km/h

Use the conservation of energy principle to determine

the compression of the spring. The kinetic energy of

the cannon will be equal to the elastic potential energy

stored in the spring.

KE=PE

(6.5)(250 kg) [(40~) (l~OO ~) r

3600

h J

= (0.5) (520 c:) (100 :)

x2

:r;

=

0.77 m

Answer is C.

So lu tio n 1 1:

The normal acceleration (perpendicular to the path of

the airplane) is

v

Z

an

=

.l.

T

(10

~ 2

1.5 m

=66.7 m/s2

The tension in the string is equal to the centripetal

force.

=

1 kg) (66.7 ~)

= 66.7 N

Answer is D.

Solution 12 :

(100 k:) (1000 :)

v = ------- ---.,. ----

3600 ~

= 27.78

sii]

P=mv

= (1530 kg + 200 kg) (27.78 7 )

=

8060

N-s

Answer is D.

S o lu ti on 1 3 :

The law of conservation of momentum states that the

linear momentum is unchanged if no unbalanced forces

act on an object. This does not prohibit the mass and

velocity from changing; only the product .of mass and

P ro f e s s io n a l P u b li ca l io n s, In c .


21/37

locity is constant. In this case, both the total mass

d the velocity are constant. Thus, there is no change.

nswer is

B

l u ti o n 1 4 :

F=1000 N - 150 N

=

850 N

F

a

m

850 N

3500 kg

=

0.243

m/s2

1

2

S

= v o t

+ zat

=+ ~

(0.243 ~) (3 s)2

=

1.09

m

nswer is B

K i n e t i c s 1 5 - 1 1


22/37

Kinetics of

Rotational Motion

S u b s c r i p t s

o initial

c centroidal

f friction

n normal or natural

o origin or center

S static

t tangential or torsional

N o m e n c l a tu r e

a

acceleration

ft/sec

2

m/s2

A

area

ft2 m2

M A S S M O M E NT O F I N E R T I A

d

distance

ft

m

F

force lbf N

The

mass moment of inertia

measures a solid object's

9

gravitational

acceleration

ft/sec

2

m/s2

resistance to changes in rotational speed about a specific

axis.

Ix , Iy,

and

Iz

are the mass moments of inertia with

. g e

gravitational

respect to the

x-, y-,

and z-axes, respectively. They are

constant

(32.2) lbm-ft.Zlbf-sec

not components of a resultant value.

G

shear modulus

lbf/ft

2

Pa

h

angular momentum

ft-lbf-sec

Nms

I mass moment

i: = J

(y2 + z2)dm

1 6 . 1

of inertia

Ibm-ft2

kg.m

2

J

area polar moment

of inertia

ft4

m4

t;

=

J

(x

2

+ z2)dm

1 6 .2

k

t

torsional spring

constant

ft-lbf/rad

N-m/rad

t=

J (x

2

+

y2)dm

length

ft

rn

1 6 .3

m

mass

lbm kg

M

moment

ft-lbf

N-m

r

radius

ft

rn

The

centroidal mass moment of inertia, Ie ,

is obtained

r

radius of gyration

ft

m

when the origin of the axes coincides with the object's

R

moment arm

ft

m

center of gravity. Once the centroidal mass moment of

t

time

sec s

inertia is known, the

parallel axis theorem

is used to find

v

velocity

ft/sec

m/s

the mass moment of inertia about any parallel axis. In

W

weight

lbf N

Eq.

16.4,

d is the distance from the center of mass to

the parallel

ax is .

S y m b o l s

. 2

1 6 .4

Q

angular acceleration

rad/sec''

rad/s

2

Iny parallel axis = Ie +md

o

angular position

rad

rad

)

superelevation angle

deg deg

For a composite object, the parallel axis theorem must

f J ,

coefficient of friction

be applied for each of the constituent objects.

p

density

Ibm/ft

3

kg/m

3

w

angular velocity

rad/sec

radj's

to

natural frequency

rad/sec rad/s

-2 . 2

1 6 . 5

I = Ie ,l +m]:d

1

+ Ic ,2 +m2

d

2 + ...

P r o f e s s i o l o l P ub li c ati o n s, I n c


23/37

1 6 - 2 F E R e v i e w M a n u a l _

The radius of gyration, r , of a solid object represents the

distance from the rotational axis at which the objects

entire mass could be located without changing the mass

moment of inertia.

For a rigid body rotating about an axis passing through

its center of gravity located at point 0, the scalar value

of angular momentum is given by Eq. 16.9.

ho =Iw

[S I]

] 6 . 9 0

r=ff

] 6 .6

Iw

[U .S .]

] 6 . 9 b

o=-

gc

1= r2m

] 6 .7

Table 16.1 (at the end of this chapter) lists the mass

moments of inertia and radii of gyration for some stan-

dard shapes.

P L A t ~ f M o t i o N O F

A

R I G ID B O D Y

General rigid body plane motion, such as rolling wheels,

gear sets, and linkages, can be represented in two dimen-

sions (i.e., the plane of motion). Plane motion can be

considered as the sum of a translational component and

a rotation about a fixed axis, as illustrated in Fig. 16.1.

F i g u r e ]

6.]

C o m p o n e n t s

of

P l a n e M o ti o n

p

plane motion

p

translation

rotation

R o t a t i o n A b o u t a F i x e d A x i s

Rotation about a fixed axis describes a motion in which

all particles within the body move in concentric circles

about the axis of rotation.

The angular momentum taken about a point is the

moment of the linear momentum vector. Angular mo-

mentum has units of distance x force x time (e.g.,

ft-lbf-sec or Ncn-s). It has the same direction as the

rotation vector and can be determined from the vectors

by use of the right-hand rule (cross product).

ho

=

r X mv

mv

ho=rx-

. 9c

]6 . 8 0

1 6 . 8 b

[S I]

[ U .S .]

Although Newton's laws do not specifically deal with

rotation, there is an analogous relationship between ap-

plied moment and change in angular momentum. For

a rotating body, the moment (torque),

M,

required to

change the angular momentum is

dho

M=-

dt

] 6 . ]0

The rotation of a rigid body will be about the center

of gravity unless the body is constrained otherwise. If

the moment of inertia is constant, the scalar form of

Eq.

16.10

is

[S I] ] 6 . 1 1 0

[U .S .] ] 6 . 1 1 b

Velocity and position in terms of rotational variables

can be determined by integrating the expression for ac-

celeration.

] 6 .1 2

1 6 .] 3

] 6 .1 4

I n s t a n t a n e o u s C e n t e r o f R o t a t io n

Analysis of the rotational component of a rigid body's

plane motion can sometimes be simplified if the loca-

tion of the body's

instantaneous center

is known. Using

the instantaneous center reduces many relative motion

problems to simple geometry. The instantaneous center

(also known as the

instant center

and

Ie)

is a point at

which the body could be fixed (pinned) without chang-

ing the instantaneous angular velocities of any point on

the body. Thus, for angular velocities, the body seems

to rotate about a fixed instantaneous center.

The instantaneous center is located by finding two

points for which the absolute velocity directions are

known. Lines drawn

per pen dicu lar to these two velo c -

ities will intersect at the instantaneous center. (This

P r o f e s s i o n a l P u b l ic a t i o n s , In c . --


24/37

_ . _ K in et ic s o f R o t at io n al M o tion 163

crraphic procedure is slightly different if the two veloci-

o

ties are parallel, as Fig. 16.2 shows.) For a rolling wheel,

the instantaneous center is the point of contact with the

supporting surface.

.. ,... .

F i g u r e 1 6 . 2 . G r a p h i c M e th o d o f F in d i n g t h e I n s t a n ta n e o u s C e n t e r

IC

IC

The absolute velocity of any point, P, on a wheel rolling

(Fig. 16.3) with translational velocity, vo, can be found

by geometry. Assume that the. wheel is pinned at point

C and rotates with its actual angular velocity,

e

=

w =

voir. The direction of the point's velocity will be per-

pendicular to the line of length

I

between the instanta-

neous center and the point.

lvo

v=lw= -

r

1 6 . 1 5

F i g u r e

1 6 . 3

I n s t a n t a n e o u s C e nt e r o f a R o ll i n g W h e e l

C E N t R I F U G A C F O R C f

t

Newton's second law states that there is a force for every

acceleration that a body experiences. For a body mov-

ing around a curved path, the total acceleration can be

separated into tangential and normal

components,

By

Newton's second law, there are corresponding forces in

the tangential and normal directions. The force associ-

ated with the normal acceleration is known as the

cen-

tripetal force. The centripetal force is a real force on the

body toward the center of .rotation. The so-called cen-

trifugal force

is an apparent force on the body directed

away from the center of rotation. The centripetal and

centrifugal forces are equal in magnitude but opposite

in sign.

Equation 16.16 gives the centrifugal force on a body of

mass m with distance r from the center of rotation to

the center of mass.

mvi 2

F e = man = -- = mrw

r

2 2

F

_ man _ mv

t _

mrw

g e g e

r

ge

[S I] 1 6 . 1 6 0

[u.s.]

1 6 .1 6 b

B A N K IN G O F C U R V E S .

If a vehicle travels in a circular path on a fiat plane with

instantaneous radius r and tangential velocity

Vt,

it will

experience an apparent centrifugal force. The centrifu-

gal force is resisted by acombination of roadway bank-

ing (superelevation) and sideways friction. The vehicle

weight, W, corresponds to the normal force. For small

banking angles, the maximum frictional force is

Ff

=J.lsN

=

.lsW

1 6 . 1 7

For large banking angles, the centrifugal force contributes

to the normal force. If the roadway is banked so that

friction is not required to resist the centrifugal force, the

superelevation angle,

e ,

can be calculated from Eq, 16.18.

v

2

tan 8

= -.i.

gr

1 6 . 1 8

T O R S I O N A L F R E E V I B R A T IO N

The torsional pendulum in Fig. 16.4 can be analyzed

in a manner similar to the spring-mass combination.

Disregarding the mass and moment of inertia of the

shaft, the differential equation is

1 6 . 1 9

F i g u r e

1 6 . 4

T o r s i o n a l P e n d u l u m

L

G

)9


25/37

,

1 6 - 4 F E R e v i e w M a n u a l _

For the torsional pendulum, the torsional spring con-

stant

k

t

can be written

1 6 . 2 0

The solution to Eq.

16.20

is directly analogous to the

solution for the spring-mass system.

B (t) =ocosw n t + (::) sinwnt

1 6 .2 1

S A M p L E P R O B L E M s

1.

Why does a spinning ice skater's angular velocity

increase as she brings her arms in toward her body?

(A) Her mass moment of inertia is reduced.

(B) Her angular momentum is constant.

(e) Her radius of gyration is reduced.

(D )

all of the above

99689

S o l u t i o n :

As the skater brings her arms in, her radius of gyra-

tion and mass moment of inertia decrease. However, in

the absence of friction, her angular momentum,

h,

is

constant. From Eq. 16.9,

h

w

I

Since angular velocity, w, is inversely proportional to

the mass moment of inertia, the angular velocity in-

creases when the mass moment of inertia decreases.

Answer is D.

2. Link AB of the linkage mechanism shown in the

illustration rotates with an instantaneous counterclock-

wise angular velocity of 10 rad/s, What is the instan-

taneous angular velocity of link Be when link AB is

horizontal and link CD is vertical?

,W A B

= 10 rad/s counterclockwise)

~/~5m

--5-m--I ~

I

4 C . -

5m

D._

~

(A) 2.25 rad/s (clockwise)

(B) 3.25 rad/s (counterclockwise)

(e) 5.50 rad/s (clockwise)

(D) 12.5 rad/s

(clockwise)

1 00694

S o l u t i o n :

Find the instantaneous center of rotation. The absolute

velocity directions at points Band e are known. The

instantaneous center is located

by

drawing perpendicu-

lars to these velocities as shown. The angular velocity

of any point on rigid body link Be is the same at this

instant.

VB = 50 m/s

t

B

I/IC

4m

II

- - - - - - - - - - - - 1 -

1

I

I

13m

I

I

I

4 I

vc-

C

VB =ABwAB

(

rad)

= (5 m) 10 -s- = 50m/s

50 m

VB

wBC = OB = 4 ~ = 12.5 rad/s

[clockwise]

Answer is D.

3. Two 2 kg blocks are linked as shown. Assuming

that the surfaces are frictionless, what is the velocity of

block B if block A is moving at a speed of 3

m/s?

P r of e ss io n a l P u b li ca l io n s , I n c. - -


26/37

16 6 FE R ev iew M an u a l _

1. A 2 kg mass swings in a vertical plane at the end

of a 2 m cord. When e = 30, the magnitude of the

tangential velocity of the mass is

1

ui]. What is the

tension in the cord at this position?

(A ) 18.0 N

(B) 19.6 N

(C) 24.5 N

(D) 29.4 N

104 694

2. A 2 kg mass swings in the horizontal plane of a circle

of radius 1.5 m and is held by a taut cord. The tension

in the cord is 100 N. What is the angular momentum

of the mass?

(A) 5.77 Nms

(B) 26.0

N-m s

(C) 113Nun-s

(D)

150 N-ms

105691

3. A disk rolls along a fiat surface at a constant speed

of 10 m/s. Its diameter is 0.5 m. At a particular in-

stant, point P on the edge of the disk is 45 from the

horizontal. What is the velocity of point P at that in-

s tan t?

o

f------------j

10

m/s ~

(A) 10.0 m/s

(B ) 1 5 .0 m/s

(C) 16.2

m/s

(D) 18.5 m/s

106193

4. A car travels around an unbanked 50 m radius curve

without skidding. The coefficient of friction between

the tires and road is 0.3. What is the car's maximum

speed?

(A)

14

km/h

(B) 25 km/h

(C) 44 km/h

(D) 54 km/h

107394

5. Traffic travels at 100 km/h around a banked high-

way curve with a radius of

1000

m. What banking angle

is necessary such that friction will not be required to re-

sist the centrifugal force?

(A )

1.4

(B) 2.8

(C) 4.5

(D) 46

2107689

Forthe following problems use the NCEES Hand-


6. The center of gravity of a roller coaster car is 0.5 ill

above the rails. The rails are 1m apart. What is the

maximum speed that the car can travel around an un-

banked curve of radius 15 m without the inner wheel

losing contact with the top of the rail?

(A) 8.58

m/s

B

12.1

m/s

(C) 17.2

m/s

(D) 24.2 m/s

2133689

7. A 50 kg cylinder has a height of 3 m and a radius of

50 em. The cylinder sits on the z-axis and is oriented

with its major axis parallel to the y-ax is . W hat is the

mass moment of inertia about the z-axis?

P r o fe s s io n a l P u b l k c fi o n s, I n c . -- -


27/37

A

4.1 kgm

2

(B )

16 kgm

2

(C ) 41 kgm

2

(D )

150 kgm

2

3553794

8. A

uniform thin disk has a radius of

30 em

and a mass

of

2

kg.

A

constant force of

10 N

is applied

tangentially

at a varying, but unknown, distance from the center

of the disk. The disk accelerates about its axis at 3t

rad/s

2

. What is the distance from the center of the

disk at which the force is applied at t

= 12

s?

Cl W

~

R

= 30 em lever

(A )

32.4 em

(B) 3.6.0 em

(C )

54.0 em

(D ) 108

em

3975395

9. A torsional pendulum consists of a 5 kg uniform disk

with a diameter of

50

em attached at its center to a rod

1.5

m in length. The torsional spring constant is

0.625

Nrri/rad. Disregarding the mass of the rod, what is the

natural frequency of the torsional pendulum?

(A) 1.0 rad/s

(B) 1.2 radys

(C ) 1.4 rad/s

(D) 2.0 rad/s

4170 795

10. A 3 kg disk with a diameter of

0.6

m is rigidly at-

tached at point B to a 1 kg rod 1 m in length. The

rod-disk combination rotates around point A. What is

the mass moment of inertia about point A for the com-

bination?

0.6 m

(A) 0.47 kgm

2

(B) 0.56 kgm

2

(C )

0.87

kgrrr

(D) 3.7 kgm

2

2095689

11. A 1kg uniform rod 1 m long is suspended from

the ceiling by a frictionless hinge. The rod is free to

pivot. What is the product of inertia of the rod about

the pivot point?

(A) 0

kgm

2

(B) 0.045 kg-rrr

(C)

0.13 kg.m

2

(D) 0.33 kgm

2

2096689

12.

A wheel with a radius of 0.75

m

starts from rest

and accelerates clockwise. The angular acceleration (in

rad/s'') of the wheel is defined by a=6t - 4. What is

the resultant linear acceleration of a point on the wheel

rim at

t

=

2

s?

(A ) 6

m/s2

(B ) 12 m/s2

(C ) 13

m/s2

(D ) 18m/s2

2 893 687

13. A uniform rod (AB) of length L and weight W

is pinned at point C and restrained by cable OA. The

cable is suddenly cut. The rod starts to rotate about

point C, with point A moving down and point B moving

up. What is the instantaneous linear acceleration of

point B?

~

a

I__

Cl ~

L

- .. ; . P ro f e s sio n a l P u b li ca t io n s, I n c


28/37

16 8 FE R ev iew M an ua l 1 _

(A)

g

16

B

g

4

(C )

g

7

D

g

4

2911 687

14. A

uniform rod

(AB)

of length

L

and weight

W

is pinned at point

C.

The rod is accelerating with an

instantaneous angular acceleration (in

rad/s'')

of

0:

=

12g

/7

L. What is the instantaneous reaction at point

C?

A

1 .

4

B

L

A)

W

-

4

(B)

W

-

3

(C )

4W

-

7

(D)

7W

12

2912687

15. A 1530 kg car is towing a 300 kg trailer. The coef-

ficient of friction between all tires and the road is

0.80.

How fast can the car and trailer travel around an un-

banked curve of radius

200

m without either the car or

trailer skidding?

(A) 40.0 km/h

(B) 75.2

km/h

(C )

108.1

km/h

(D) 143

km/h

1579689

16. A 1530 kg car is towing a 300 kg trailer. The coef-

ficient of friction between all tires and the road is

0.80.

The car and trailer are traveling at

100

km/h around

a banked curve of radius

200

m. What is the neces-

sary banking angle such that tire friction will not be

n eces s ary to p rev en t

skidding?

(A) 8

(B) 21

(C) 36

(D) 78

158 2689

17. A wheel with a

0.75

m radius has a mass of

200

kg.

The wheel is pinned at its center and has a radius of

gyration of 0.25 m. A rope is wrapped around the wheel

and supports a hanging

100

kg block. When the wheel

is released, the rope begins to unwind. What is the

angular acceleration of the wheel?

A

5.9

rad/s

2

(B) 6.5

rad/s

2

(C ) 11 rad/s

2

(D) 14 rad/s

2

2907687

S O L U T I O N S T O F E -S T Y L E E X A M P R O B L E M S

So lu ti on 1:

Use tangential and normal components.

P r of es s io n a l P u b li ca t io n s , In c . - - -

v2 1 7 ) 2

an=

. . . 1 .

=---- -.:....-

r

2 m

=.5

m/s2

Sum forces in the normal direction.

LPn

=

man

=

T - Wsin60

T=man

+

mg sin 60

=

(2

kg)

(0.5 ~)

+

(2

kg)

(9.81 ~)

(sin

60)

=8.0 N

Answer is A.

So lu tio n 2 :

tension

=

centripetal force


29/37

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - lK in e t i c s o f R o t a t i o n a l M o t io n 1 6 - 9

mv

2

T= __

t

r

r

mv

2

-- =f1mg

r

v = vf1gr

= V

(0.3) (9.81 ~) (50 m )

=

12.13

m / s

(12.13 ~) (3600 ~)

v=

1000

r:

=

43.67

km/h

Answer is C.

S o lu t i o n 5 :

Since there is no friction force, the superelevation angle,

B ,

can be determined directly.

v

2

tanB = ~

gr

e

= tan

( ; )

(

(100 ~) (1000 ~))

2

3600 ~

=tan-l~--~--~~------~~

(9.81 ~) (1000 m)

100 N

(2

kg)(1.5 m)

=5.77

rad/s

ho

=

rmv

=

r

2

mw =

(1.5 m)2 (2 kg) (5.77 r:d)

=26.0 Nms

Answer is B.

S o l u t io n 3 :

Use the instantaneous center of rotation to solve this

problem. Assume the wheel is pinned at point A.

Z 2

=

(2)(0.25

m)2 -

(2)(0.25

m)2(cos

135)

[lawof cosines]

= 0.2134 m 2

I =

VO.2134m 2

=

0.462 m

Zvo (0.462 m) (10 ~)

vp = ---;;:- 0.25 m

=

18.5

m/s

The velocity of point P is perpendicular to the line AP.

Answer

is D.

S o lu t i o n 4 :

The car uses friction to resist the centrifugal force.

=

4.50

Answer

is

C.

S o lu t i o n 6 :

The wheel will lose contact with the top of the rail when

the reaction on the rail is zero. Refer to the following

illustration. Wheel A is the inner wheel.

YCG

=0.5 m

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ P r o fe s s io n a l P u b l i c a ti o n s , I n c .


30/37

1 6 - 1 0 F E R e v i e w M a n u a l

1 _

The forces acting on the car are the centrifugal force, F e ,

its weight, and the reaction at the outer wheel. (The

reaction at the inner wheel is zero.) Take moments

about rail B.

LMB = 0 =

WXCG - F eYCG

mv;YcG

=mgxCG-

_.. .::.. : :.... .:....:c .

r

S o l u t io n 9 :

The radius of the disk is

R= .~D

2

50 em

- - 2 (100 ::)

=0.25 m

v =g XCGr

YCG

(9.81 ~) (~) (15 m)

0.5 m

The mass moment of inertia of the disk is

I = ~MR2

2

=(~) (5 kg)(0.25 m)2

=0.15625 kg-rrr

Natural frequency can be determined from the equation

for the torsional spring constant.

k

t

=

w

2

I

0.15625 kgm

2

= rad/s

= 12.1 m/s

Answer is B.

S o l u t i o n 7 :

Find the formula for

Ix

in the Mass Moments of Inertia

table.

m 3R2 + 4h2)

Ix

=----------

12

(50 kg)[(3)(0.5 m)2 + (4)(3 m)2]

12

=153.1 kg-rrr (150 kgrn )

w = /

~----,--,--

0.625 Nm

rad

Answer is D.

Answer is D.

S o l u t io n 8 :

The centroidal moment of inertia is

I

=

~mR2

2

=0.5)(2 kg)(0.3 m)2

= 0.09 kgm

2

The acceleration at t=12 s is

a =

3t

= ( 3 r:d) (12 s)

=

6 rad/s

2

Mo=F r=Ia

Ia

r

F

(0.09 kgm

2

) (36 ~)

10 N

=.324 m (32.4 em)

So lution 1 0 :

The mass moment of inertia of the rod about its end is

ML2

Irod,A

=3

(1 kg)(1 m)2

3

=0.33 kg.m

2

The mass moment of inertia of the disk about its own

center is

MR2

Idisk,c

=

-2-

(3 kg) Tr

2

=

0.135 kgm

2

Answer is A.

The distance AC is

AC =JAB2

+

BC

2

1 m)2 + (O.~ ill

r

=

1.04

rn

.

- - - - - - - - _ _ - - - - - - - - - - - - - - - _ . .

_

.

ro f e s si on a l P u b li ca t i o n s , I n c .


31/37

Using the parallel axis theorem, the mass moment of

inertia of the disk about point A is

Idisk,A =Idisk,e +mdisk

AC2

=0.135

kgm

2

+

(3

kg)

(1.04

m)2

=3.38 kgrrr

The total moment of inertia of the rod and disk is

IA =

Irod,A

+

Idisk,A

=0.33kgm

2

+

3.38

kgm

=

3.71

kg.m

2

Answer is D.


The product of inertia for the rod is zero because the

pivot point lies on an axis of symmetry.

Answer is A.


The angular acceleration at t

=

2 s is

a(2 s)

=

(6 )(2 s) - 4 =8 rad/s

2

The equation for the angular velocity is

w

=

a(t)dt

J(6t - 4)dt

3t

2

- 4t +wo

However, the wheel starts from rest, so W o =

O .

At t = 2 s, the angular velocity is

w (2

s ) = (3)(2 S)2 - (4)(2 s )

= rad/s

The tangential acceleration of the point is

(

rad)

=

(0.75

m ) 8 ~

=6 m/s2

The normal acceleration (directed toward the center of

the wheel) is

(

rad)2

=(0.75 m) 4S

=2 m/s2

The resultant acceleration is

a

=

V a ; + a ~

( 6 ~ )

2

+ ( 1 2 ~ )

2

=13.4

m/s2

Answer is

C.


Point C is L/4 from the center of gravity of the rod.

The mass moment of inertia about point C is

Ie =eG + md

2

=

1

1

2 )

m L

2

+ m ( ~

r

=

m L 2

C ~

+

1

1

6 )

: 8 ) m L

2

The sum of moments on the rod is

W L m g L

4 4

The angular acceleration is

L M e

a

Ie

m g L

4

: 8 ) m L 2

12 g

7L

The tangential acceleration of point B is

= (~) ~1

g

7

Answer is C.


32/37

1 6 - 1 2 F E R e v i e w M a n u a l - - - - - - - - - - - - - - - - - - - - - - - - - - - - - _

S o l u ti o n 1 4 :

The mass moment of inertia of the rod about its center

of gravity is

Take moments about the center of gravity of the rod.

All moments due to gravitational forces will cancel. The

only unbalanced force action on the rod will be the re-

action force, Rc, at point C.

LMCG =Rc (~)

LMCG =CGacG

Rc= 4W

7

Answer is C

S o lu ti o n 1 5 :

To keep the vehicle and trailer from skidding, the cen-

tripetal force must be less than or equal to the frictional

force. At the limit,

Fe =F

f

p,N

a

n

m

p,mg

= -- =p,g

m

=

0.8) (9.81 ~)

=

7.848

m/s2

The normal acceleration can be calculated from the tan-

gential velocity.

Vt =

Janr

=

J

(7.848 ~) (200

m )

=

39.6 m/s

(39.6 ~) (3600 ~)

1000 :

=143

km/h

Answer is D

S o lu tio n 1 6 :

The velocity is

(100 k:) (1000 :)

v - -- --------;;----

- 3600 ~

=

27.78

m/s

The necessary superelevation angle is

Answer is B

S o l u ti o n 1 7 :

The mass moment of inertia of the wheel is

The unbalanced torque (moment) on the wheel is

M=

F R

=

mg - ma)R = mRg - a)

= mblockRg - Ra)

The acceleration is given by

M=Ia

mblockRg - Ra)

= =

mwheelr2a

mblockRg

_ (100 kg)(0.75

m )

(9.81 ~)

- (200 kg)(0.25 m )2 + (100 kg)(0.75 m )2

=

0.7

rad/s

2

Answer is C

P r o f e s s i o n a l P u b l i c a ti o n s , I n c . - - - -


33/37

nergy and Work

N o m e n c l a t u r e

a

acceleration

e

coefficient of

restitution

E

energy

F

force

g gravitational

acceleration

\

gravitational

gc

constant

(32.2)

h

height above

datum

I

mass moment

of inertia

Imp

impulse

k

spring constant

m

mass

p

linear momentum

r

distance

t

time

v

velocity

W

work

X

displacement

S y m b o l s

w

angular velocity

E N E R G Y A N D W O R I C

ft-lbf

lbf

J

N

Ibm-ftj'lbf-sec-'

ft

m

lbm-ft2

Ibf-sec

lbf/ft

lbm

lbm-ft/sec

ft

kgm

2

Ns

N/ m

kg

kg-m.'s

m

s

m/s

J

sec

ft/sec

ft-lbf

ft

m

rad/sec

rad/s

The

energy

of a mass represents the capacity of the mass

to do work. Such energy can be stored and released.

There are many forms that the stored energy can take,

including mechanical, thermal, electrical, and magnetic

energies. Energy is a positive, scalar quantity, although

the change in energy can be either positive or negative .

Wark, W,

is the act of.changing the energy of a mass.

Work is a signed, scalar quantity. Work is positive when

a fo rce acts in the d irec tio n o f m o tio n an d m ov es a m as s

from one location to another. Work is negative when

a force acts to oppose motion. (Friction, for example,

always opposes the direction of motion and can only do

negative work.) The net work done on a mass by more

than one force can be found by superposition.

The work performed by a force is calculated as a dot

product of the force acting through a displacement.

w=

l=

1 7 . 1

K i n e t i c E n e r g y

Kinetic energy

is a form of mechanical energy associ-

ated with a moving or rotating body. The

linear kinetic

energy of a body moving with instantaneous linear ve-

locity v is

1

KE =

Zmv

2

[S I]

1 7 . 2 0

2

KE=

mv

[U .S . ] 1 7 . 2 b

2g

c

The

rotational kinetic energy

of a body moving with

instantaneous angular velocity

w

is

KE = ~Iw2

[S 1]

1 7 . 3 0

KE = Iw2

[U .S .]

/ 7 . 3 b

2gc

For general plane motion in which there are transla-

tional and rotational components, the kinetic energy is

the sum of the translational and rotational forms.

The change in kinetic energy is calculated from thedif-

ference of squares ofvelocity, not from the square of the

velocity difference.

1 7 . 4 0

1 7 . 4 b

P r o f e s s i o n a l P u b l i c a t i o n s , l i n e .


34/37


1 _

P o te n t ia l E n e rg y

Potential energy (also known as gravitational potential

energy)

is a form of mechanical energy possessed by

a mass due to its relative position in a gravitational

field. Potential energy is lost when the elevation of a

mass decreases. The lost potential energy usually is

fe/eit dynamics sample problems

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