For Practice

FOC Questions:

1, 3, 7, 8, 10 and 15

Problems:

4, 17, 23, 25 & 47

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Note: The answers key of FOC is posted separately

Ch. 5 Uniform circular motion

• FOC 11. Two cars are traveling at the same constant speed v. As the drawing indicates, car A is

moving along a straight section of the road, while car B is rounding a circular turn.

Which statement is true about the accelerations of the cars? (a) The acceleration of both

cars is zero, since they are traveling at a constant speed. (b) Car A is accelerating, but car

B is not accelerating. (c) Car A is not accelerating, but car B is accelerating. (d) Both

cars are accelerating.

1. (c) The velocity of car A has a constant magnitude (speed) and direction. Since its velocity is constant, car A does not have an acceleration. The velocity of car B is continually changing direction during the turn. Therefore, even though car B has a constant speed, it has an acceleration (known as a centripetal acceleration).

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FOC 33. Two cars are driving at the same constant speed v around a racetrack. However, they

are traveling through turns that have different radii, as shown in the drawing. Which

statement is true about the magnitude of the centripetal acceleration of each car? (a) The

magnitude of the centripetal acceleration of each car is the same, since the cars are

moving at the same speed. (b) The magnitude of the centripetal acceleration of the car at

A is greater than that of the car at B, since the radius of the circular track is smaller at A.

(c) The magnitude of the centripetal acceleration of the car at A is greater than that of the

car at B, since the radius of the circular track is greater at A. (d) The magnitude of the

centripetal acceleration of the car at A is less than that of the car at B, since the radius of

the circular track is smaller at A.

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FOC 7Section 5.3 Centripetal Force

7. The drawing shows two identical stones attached to cords that are being whirled on a

tabletop at the same speed. The radius of the larger circle is twice that of the smaller

circle. How is the tension T1 in the longer cord related to the tension T2 in the shorter

cord? (a) T1 = T2 (b) T1 = 2T2 (c) T1 = 4T2 (d) 11 22

T T= (e) 11 24

T T=

𝐹 = 𝑚𝑎 = 𝑚𝑣2/r

d)

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FOC 88. Three particles have the following masses (in multiples of m0) and move on three

different circles with the following speeds (in multiples of v0) and radii (in multiples of r0):

Particle Mass Speed Radius

1 m0 2v0 r0

2 m0 3v0 3r0

3 2m0 2v0 4r0

Rank the particles according to the magnitude of the centripetal force that acts on them,

largest first. (a) 1, 2, 3 (b) 1, 3, 2 (c) 2, 1, 3 (d) 2, 3, 1 (e) 3, 2, 1

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FOC 10Section 5.4 Banked Curves

10. Two identical cars, one on the moon and one on the earth, have the same speed and are

rounding banked turns that have the same radius r. There are two forces acting on each car,

its weight mg and the normal force FN exerted by the road. Recall that the weight of an

object on the moon is about one-sixth of its weight on earth. How does the centripetal force

on the moon compare with that on the earth? (a) The centripetal forces are the same. (b)

The centripetal force on the moon is less than that on the earth. (c) The centripetal force on

the moon is greater than that on the earth.

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FOC 15Section 5.7 Vertical Circular Motion

15. The drawing shows an extreme skier at the bottom of a ski jump. At this point the track

is circular with a radius r. Two forces act on the skier, her weight mg and the normal force

FN. Which relation describes how the net force acting on her is related to her mass m and

speed v and to the radius r? Assume that “up” is the positive direction.

(a) 2

N

mvF mg

r+ = (b)

2

N

mvF mg

r− =

(c) 2

N

mvF

r= (d)

2mvmg

r− =

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Pr. 44. Speedboat A negotiates a curve whose radius is 120 m. Speedboat B negotiates a curve whose

radius is 240 m. Each boat experiences the same centripetal acceleration. What is the ratio vA/vB of the

speeds of the boats?

_______________________________________________________________________

4. REASONING In each case, the magnitude of the centripetal acceleration is given by 2

c/a v r= (Equation 5.2). Therefore, we will apply this expression to each boat and

set the centripetal accelerations equal. The resulting equation can be solved for the

desired ratio.

SOLUTION Using Equation 5.2 for the centripetal acceleration of each boat, we have

2 2A B

cA cBA B

and v v

a ar r

= =

Setting the two centripetal accelerations equal gives

2 2A B

A B

v v

r r=

Solving for the ratio of the speeds gives

A A

B B

120 m0.71

240 m

v r

v r= = =

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Pr. 1717. For background pertinent to this problem, review Conceptual Example 6. In Figure 5.7 the man hanging

upside down is holding a partner who weighs 475 N. Assume that the partner moves on a circle that has a

radius of 6.50 m. At a swinging speed of 4.00 m/s, what force must the man apply to his partner in the

straight-down position?

17. REASONING AND SOLUTION The force P supplied by the man will be

largest when the partner is at the lowest point in the swing. The diagram at

the right shows the forces acting on the partner in this situation. The

centripetal force necessary to keep the partner swinging along the arc of a

circle is provided by the resultant of the force supplied by the man and the

weight of the partner. From the diagram, we see that

P mgmv

r− =

2

Therefore,

Pmv

rmg= +

2

Since the weight of the partner, W, is equal to mg, it follows that m = (W/g) and

2 2 2( / ) [(475 N)/(9.80 m/s )] (4.00 m/s)(475 N) = 594 N

(6.50 m)

W g vP W

r= + = +

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Pr. 23*23. mmh A “swing” ride at a carnival consists of chairs that are swung in a circle by 15.0-m cables

attached to a vertical rotating pole, as the drawing shows. Suppose the total mass of a chair and its occupant

is 179 kg. (a) Determine the tension in the cable attached to the chair. (b) Find the speed of the chair.

23. REASONING

a. The free body diagram shows the swing ride and the two forces that act on a chair:

the tension T in the cable, and the weight mg of the chair and its occupant. We note

that the chair does not accelerate vertically, so the net force y

F in the vertical

direction must be zero, 0y

F = . The net force consists of the upward vertical

component of the tension and the downward weight of the chair. The fact that the net

force is zero will allow us to determine the magnitude of the tension.

b. According to Newton’s second law, the net force x

F in the horizontal direction

is equal to the mass m of the chair and its occupant times the centripetal acceleration

( )2

c/a v r= , so that

2

c/

xF ma mv r= = . There is only one force in the horizontal

direction, the horizontal component of the tension, so it is the net force. We will use

Newton’s second law to find the speed v of the chair.

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Pr. 23 SolutionSOLUTION

a. The vertical component of the tension is +T cos 60.0, and the weight is −mg, where

we have chosen “up” as the + direction. Since the chair and its occupant have no

vertical acceleration, we have that 0y

F = , so

+T cos60.0° - mg

Fyå

= 0 ()

Solving for the magnitude T of the tension gives

( )( )2179 kg 9.80 m/s3510 N

cos60.0 cos60.0

mgT = = =

b. The horizontal component of the tension is +T sin 60.0, where we have chosen

the direction to the left in the diagram as the + direction. Since the chair and its

occupant have a centripetal acceleration in this direction, we have

T sin60.0°

Fxå

= mac

= mv2

r

æ

èçö

ø÷ ()

From the drawing we see that the radius r of the circular path is r = (15.0 m) sin 60.0

= 13.0 m. Solving Equation (2) for the speed v gives

( )( )13.0 m 3510 N sin 60.0sin 60.014.9 m/s

179 kg

rTv

m

= = =

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Pr. 2525. Before attempting this problem, review Examples 7 and 8. Two curves on a highway have the

same radii. However, one is unbanked and the other is banked at an angle . A car can safely travel

along the unbanked curve at a maximum speed v0 under conditions when the coefficient of static friction

between the tires and the road is s = 0.81. The banked curve is frictionless, and the car can negotiate it

at the same maximum speed v0. Find the angle of the banked curve.

25. REASONING From the discussion on banked curves in Section 5.4, we know that a

car can safely round a banked curve without the aid of static friction if the angle of

the banked curve is given by ( )2

0tan /v rg = , where vo is the speed of the car and r

is the radius of the curve (see Equation 5.4). The maximum speed that a car can have

when rounding an unbanked curve is 0 s

v gr= (see Example 7). By combining

these two relations, we can find the angle .

SOLUTION The angle of the banked curve is ( )1 2

0tan /v rg − = . Substituting the

expression 0 s

v gr= into this equation gives

( ) ( )2

1 1 1 10 s

stan tan tan tan 0.81 39

v g r

rg rg

− − − −

= = = = =

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Pr. 47*47. A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a

circular arc with a radius of 45.0 m. Determine the maximum speed that the cycle can have while moving

over the crest without losing contact with the road.

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47. REASONING Because the crest of the hill is a circular arc, the motorcycle’s speed v

is related to the centripetal force Fc acting on the motorcycle: 2c

F mv r= (Equation

5.3), where m is the mass of the motorcycle and r is the radius of the circular crest.

Solving Equation 5.3 for the speed, we obtain 2c

v F r m= or c

v F r m= . The free-

body diagram shows that two vertical forces act on the motorcycle. One is the weight

mg of the motorcycle, which points downward. The other is the normal force FN

exerted by the road. The normal force points directly opposite the motorcycle’s

weight. Note that the motorcycle’s weight must be greater than the normal force. The

reason for this is that the centripetal force is the net force produced by mg and FN

and

must point toward the center of the circle, which lies below the motorcycle. Only if

the magnitude mg of the weight exceeds the magnitude FN of the normal force will

the centripetal force point downward. Therefore, we can express the magnitude of the

centripetal force as Fc = mg − FN. With this identity, the relation c

v F r m=

becomes

( )Nmg F r

vm

−= (1)

SOLUTION When the motorcycle rides over the crest sufficiently fast, it loses contact

with the road. At that point, the normal force FN is zero. In that case, Equation (1)

yields the motorcycle’s maximum speed:

( )0mg r mv

m

−= =

gr

m( )( )29.80 m/s 45.0 m 21.0 m/sgr= = =

Donerville Rd.

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