chapter 4 motion in two and three dimensions

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Chapter 4Motion in Two and Three Dimensions

Position vector in three dimensions

.kzjyixr

,kzjyixr

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Instantaneous velocity

The direction of the instantaneous velocity of a particle is tangent to the path at the particle’s position.

The components of are :

tdrdv

v

tdzdvand,

tdydv,

tdxdv zyx

v

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Instantaneous acceleration

The components of are :

tdvda

a

tdvdaand,

tdvd

a,tdvda z

zy

yx

x

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Projectile Motion In projectile motion, the horizontal motion and the vertical

motion are independent of each other; that is, neither motion affects the other. A projectile with an initial velocity can be written as (see figure 4-10) :-

The horizontal motion has zero acceleration, and the vertical motion has a constant downward acceleration of - g.

jvivv y0x00

0v

00000x0 sinvvandcosvv y

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The range R is the horizontal distance the projectile has traveled when it returns to its launch height.

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Examples of projectile motion

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Horizontal motion

No acceleration

tvxx x00

t)cosv(xx 000

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Vertical motion (Equations of Motion ):-

)yy(g2)sinv(v

tgsinvv

tg21t)sinv(

tg21tvyy

02

002

y

00y

,200

2y00

1 )

2 )

3 )

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The equation of the path

In this equation, x0 = 0 and y0 = 0. The path, or trajectory, is a parabola. The angle is between and the + direction.

)trajectory()cosv(2

xgx)(tany 200

2

0

0 0v x

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The horizontal range

This equation for R is only good if the final height equals the launch height. We have used the relationsin 2 = 2 sin cos .

The range is a maximum when = 45o

t)cosv(R 00

To find t = time of flight, y - y0 = 0 means that :

0

20

00

20

200

2singv

R

cossingv2

R

tg21t)sinv(0

0 0 0

0

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Sample Problem In the figure shown, a

rescue plane flies at 198 km/h (= 55.0 m/s) and a constant elevation of 500 m toward a point directly over a boating accident victim struggling in the water. The pilot wants to release a rescue capsule so that it hits the water very close to the victim.

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(a)  What should be the angle of the pilot's line of sight to the victim when the release is made?

Solving for t, we find t = ± 10.1 s (take the positive root).

Solutionhxtan 1

t)cosv(xx 000

2000 tg

21t)sinv(yy

)s1.10()0(cos)s/m0.55(0x m5.555x

48m500m5.555tan 1

22 t)s/m8.9(21t)0(sin)s/m0.55(m500

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(b)  As the capsule reaches the water, what is its velocity in unit-vector notation and as a magnitude and an angle?

When the capsule reaches the water,

s/m0.55)0(cos)s/m0.55(cosvv 00x

s/m0.99)s1.10()s/m8.9()0(sin)s/m0.55(

tgsinvv2

00y

j)s/m0.99(i)s/m0.55(v

61ands/m113v

v

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Sample ProblemThe figure below shows a pirate ship 560 m from a fort defending the harbor entrance of an island. A defense cannon, located at sea level, fires balls at initial speed v0 = 82 m/s.

(a)  At what angle from the horizontal must a ball be fired to hit the ship?

0

15

SOLUTION: 

Which gives

There are two solutions

0

20 2singv

R

816.0sin

)s/m82()m560)(s/m8.9(sin

vRgsin2

1

2

21

20

10

63)3.125(21

27)7.54(21

0

0

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(b)  How far should the pirate ship be from the cannon if it is to be beyond the maximum range of the cannonballs?

SOLUTION:  Maximum range is :-

The maximum range is 690m. Beyond that distance, the ship is safe from the cannon.

.m690m686

)45x2(sins/m8.9)s/m82(2sin

gv

R2

0

20

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Sample ProblemThe figure below illustrates the flight of Emanuel Zacchini over three Ferris wheels, located as shown and each 18 m high. Zacchini is launched with speed v0 = 26.5 m/s, at an angle = 53° up from the horizontal and with an initial height of 3.0 m above the ground. The net in which he is to land is at the same height.

(a)  Does he clear the first Ferris wheel?

0

18

SOLUTION The equation of trajectory when x0 = 0 and y0 = 0 is given

by :

Solving for y when x = 23m gives

Since he begins 3m off the ground, he clears the Ferris wheel by (23.3 – 18) = 5.3 m

200

2

0 )cosv(2xgx)tan(y

m3.20)53(cos)s/m5.26(2)m23()s/m8.9()m23()53tan( 22

22

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(b)  If he reaches his maximum height when he is over the middle Ferris wheel, what is his clearance above it?

SOLUTION: At maximum height, vy is 0. Therefore,

and he clears the middle Ferris wheel by

(22.9 + 3.0 -18) m =7.9 m

0gy2)sinv(v 200

2y

m9.22)s/m8.9()2(

)53(sin)s/m5.26(g2

)sinv(y 2

22200

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(c)  How far from the cannon should the center of the net be positioned?

SOLUTION: 

m69

)53(2sins/m8.9)s/m5.26(2sin

gv

R 2

2

0

20

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Avoid rounding errors One way to avoid rounding errors and other

numerical errors is to solve problems algebraically, substituting numbers only in the final step.

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Uniform Circular Motion A particle is in uniform circular motion if it travels

around a circle at uniform speed. Although the speed is uniform, the particle is accelerating.

The acceleration is called a centripetal (center seeking) acceleration.

T is called the period of revolution.

).period(vr2T

),onacceleratilcentripeta(rva

2

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sinvtdxd

,cosvtdyd pp

.jtdxd

rvi

tdyd

rv

tdvda

.jrxv

iryv

v

.j)cosv(i)sinv(jvivv

pp

pp

yx

.jsinrvicos

rva

22

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Thus, , which means that is directed along the radius r, pointing towards the circle’s center.

tancosr

v

sinrv

aa

tan

,rv)(sin)(cos

rvaaa

2

2

x

y

222

22

y2

x

a

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Sample Problem

“Top gun” pilots have long worried about taking a turn too tightly. As a pilot's body undergoes centripetal acceleration, with the head toward the center of curvature, the blood pressure in the brain decreases, leading to loss of brain function.

There are several warning signs to signal a pilot to ease up: when the centripetal acceleration is 2g or 3g, the pilot feels heavy. At about 4g, the pilot's vision switches to black and white and narrows to “tunnel vision.” If that acceleration is sustained or increased, vision ceases and, soon after, the pilot is unconscious—a condition known as g-LOC for “g-induced loss of consciousness.”

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What is the centripetal acceleration, in g units, of a pilot flying an F-22 at speed v = 2500 km/h (694 m/s) through a circular arc with radius of curvature r = 5.80 km?

SOLUTION: 

If a pilot caught in a dogfight puts the aircraft into such a tight turn, the pilot goes into g-LOC almost immediately, with no warning signs to signal the danger.

g5.8s/m0.83m5800)s/m694(

rva 2

22

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Relative Motion in One Dimension

The coordinate xPA of P as measured by A is equal to the coordinate xPB of P as measured by B plus the coordinate xBA of B as measured by A. Note that x is a vector in one dimension.

BAPBPA xxx

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The velocity vPA of P as measured by A is equal to the velocity vPB of P as measured by B plus the velocity vBA of B as measured by A. Note that v is a one dimensional vector. We have deleted the arrow on top.

BAPBPA

BAPBPA

vvv

),x(tdd)x(

tdd)x(

tdd

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Observers on different frames of reference (that move at constant velocity relative to each other) will measure the same acceleration for a moving particle. Note that the acceleration is a one dimensional vector.

Because VBA is constant, the last term is zero.

)v(tdd)v(

tdd)v(

tdd

BAPBPA

PBPA aa

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Sample ProblemBarbara's velocity relative to Alex is a constant vBA = 52 km/h and car P is moving in the negative direction of the x axis.

(a)  If Alex measures a constant velocity vPA = -78 km/h for car P, what velocity vPB will Barbara measure?

SOLUTION:                           

h/km130vh/km52vh/km78

vvv

PB

PB

BAPBPA

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(b)  If car P brakes to a stop relative to Alex (and thus the ground) in time t = 10 s at constant acceleration, what is its acceleration aPA relative to Alex?

2

0PA

s/m2.2

h/km6.3s/m1

s10)h/km78(0

tvv

a

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(c)  What is the acceleration aPB of car P relative to Barbara during the braking?

SOLUTION:  To calculate the acceleration of car P relative to Barbara,

we must use the car's velocities relative to Barbara. The initial velocity of P relative to Barbara is vPB = -130 km/h. The final velocity of P relative to Barbara is -52 km/h (this is the velocity of the stopped car relative to the moving Barbara).

This result is reasonable because Alex and Barbara have a constant relative velocity, they must measure the same acceleration.

2

0PB

s/m2.2

h/km6.3s/m1

s10)h/km130(h/km52

tvv

a

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Relative Motion in Two Dimension

PBPA

BAPBPA

BAPBPA

aavvv

rrr

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Homework (due Feb 1)Review Questions 72-91Pages 126-126