PS 17A: Uniform Circular Motion
Name: ________________________
1. Can you go around a curve with zero acceleration? Explain.
NO. GOING AROUND A CURVE REQURES A CONSTANT CHANGE IN DIRECTION.
VELOCITY IS A VECTOR QUANTITIY THAT IS CONSTANTLY CHANGING.
ACCELERATION = Δv/Δt.
2. Can you go around a curve with constant acceleration? Explain.
NO. SINCE ACCELERATION IS A VECTOR QUANTITY AND THE DIRECTION OF
ACCELERATION IS ALWAYS CHANGING THE ACCELERATION IS ALWAYS
CHANGING.
BUT THE MAGNITUDE OF THE ACCELERATION IS CONSTANT.
3. In order to cause a moving object to follow a circular path, what type of force is needed?
A CENTRIPETAL FORCE.
4. What provides the centripetal force on a car rounding a curve on a level road?
FRICTION BETWEEN THE TIRES AND THE ROAD.
5. If the radius of the path of an object in uniform circular motion is doubled, how must the
centripetal force be adjusted to maintain the same tangential speed?
Since 𝐹𝑐 = 𝑚 ∙
𝑣2
𝑟
and mass and speed are kept constant, then r = 2x would mean that Fc = 1/2X
6. What centripetal force is needed to keep a 70 kg person going in a 5 meter radius circle at a
speed of 10. m/s?
7. What is constant in uniform circular motion?
THE SPEED OF THE OBJECT IS CONSTANT.
8. On a merry-go-round, where should you sit to experience the greatest tangential speed: On an
inside horse or an outside horse?
IF YOU SIT ON THE OUTSIDE HORSE YOU WILL EXPERIENCE THE GREATEST
SPEED.
9. Does an inward force or an outward force act on the clothes during the spin cycle of an automatic
washer?
INWARD FORCE
THE AGENT OF THE FORCE IS THE PUSHING FORCE (NORMAL FORCE) PROVIDED
BY THE WALLS OF THE WASHING MACHINE.
10. To obtain uniform circular motion, what is the relationship between the direction of net force and
the direction of the instantaneous velocity of the moving object?
THE DIRECTION OF THE NET FORCE (Fc) MUST BE PERPENDICULAR TO THE
INSTANTANEOUS VELOCITY OF THE OBJECT IN UNIFORM CIRCULAR MOTION.
11. If you whirl a yo-yo about your head in a horizontal circle, in what direction must a force act on
the yo-yo? What is the agent that exerts the centripetal force?
INWARD
THE AGENT IS THE STRING TENSION. Fc = Ftension
12. You swing one yo-yo around your head in a horizontal circle. Then you swing another one with
twice the mass, but you don’t change the length of the string or the period.
a. How do the tension forces of the strings differ?
IF YOU HAVE 2X THE MASS, YOU WILL HAVE 2X THE FORCE
b. How do the centripetal forces differ?
THE AGENT OF THE CENTRIPETAL FORCE IS THE STRING TENSION, SO THE
TENSION WILL BE 2X GREATER.
13. A runner moving at a speed of 8.8 m/s rounds a bend with a radius of 25 meters.
a. What is the centripetal acceleration of the runner?
3.1 m/s2
b. What agent causes the force on the runner?
THE FRICTIONAL FORCE OF THE TRACK ON THE RUNNER’S SHOES.
c. If the runner has a mass of 68 kg, calculate the centripetal force.
210.8 N
(210 N – sig figs)
14. Racing on a flat track, a car going 32 m/s rounds a curve 56 meters in radius.
a. What is the car’s centripetal acceleration?
18 m/s2
b. What minimum coefficient of static friction between the tires and road would be needed
for the car to round the curve without slipping?
15. An athlete whirls a 7.00-kg hammer tied to the end of a 1.3-m chain in a horizontal circle. The
hammer makes one revolution in 1.0 seconds.
a. Calculate the centripetal acceleration of the hammer.
51.3 m/s2
b. Calculate the tension in the chain. Ftension = Fc = 357 N
c. Calculate the centripetal force.
357 N
16. According to the Guinness Book of World Records (1990) the highest rotary speed ever attained
was 2010 m/s (4500 mph). The rotating rod was 15.3 cm (6 in.) long. Assume that the speed
quoted was that of the end of the rod.
a. What is the centripetal acceleration of the end of the rod? 2.64 x 107 m/s2
b. If you were to attach a 1.0-g object to the end of the rod, what force would be needed to
hold it on the rod? What might be the agent of this force? 26,400 N
17. Early skeptics of the idea of a rotating Earth said that the fast spin of Earth would throw people
at the equator into space. The radius of Earth is about 6400 km. Show why this objection is
wrong by calculating the following.
a. The speed of a 97-kg person at the equator. 465 m/s
b. The force needed to accelerate the person in the circle.
3.3 N
c. The weight of the person.
950.6 N
d. The normal force of Earth on the person. In other words, calculate the person’s apparent
weight.
953.3 N
18. The Round Up carnival ride has a 2.0-m radius and rotates once each 0.90 seconds.
a. Find the speed of a rider.
14 m/s
b. Find the centripetal acceleration of a rider. 98 m/s
c. What agent produces this acceleration?
FORCE OF WALLS OF RIDE (NORMAL
FORCE)
d. When the floor drops down, riders are held up by friction. Draw a free-body diagram of
the situation, showing all forces involved.
e. What coefficient of static friction is needed to keep the riders from slipping?
19. What is a centrifugal force? Is this a true force? Give an example.
CENTRIFUGAL FORCE = IS AN EFFECT OF ROTATION. IT IS NOT PART OF AN
INTERACTION AND THEREFORE IS NOT A TRUE FORCE.
EXAMPLE: FEELING THROWN AGAINST THE CAR DOOR AS YOU GO AROUND A
TURN.
20. Mr. Pelz is riding on The Demon at Great America. He experiences a downward acceleration of
15.6 m/s2 at the top of the loop and an upward acceleration of 26.3 m/s2 at the bottom of the
loop. Use Newton's second law to determine the normal force acting upon Mr. Pelz's 864 kg
roller coaster car.3
a. At the top of the loop.
5011 N downward
b. At the bottom of the loop.
31,190 N upward