AP Uniform Circular motion Review
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2002 Free Response Question
1. A ball is attached to a string of length l swings in a horizontal circle, as shown above, with a constant
speed. The string makes an angle  with the vertical, and T is the magnitude of the tension in the
string. Express your answers to the following in terms of the given quantities and fundamental
constants.
a) On the figure below, draw and label vectors to represent all the forces acting on the ball when it is at
the position shown in the diagram. The lengths of the vectors should be consistent with the relative
magnitudes of the forces.

l
FT
Fg
b) Determine the mass of the ball.
 Fy = 0
c) Determine the speed of the ball.  Fx = mac
l sin  FTsinv =
m
Fg = FTy
FTx = mac
mg = FTcos
FTsin= mv2
r
m=FTcos
g
r= l sin 
lsin  FTsinl sin gsing l sin  tan
FTcos 
cos
g
d) Determine the how many revolutions it makes per second. (Determine the frequency of revolution
of the ball) v= 2  r (rev)
v=2lsin(rev) =
v = rev
rev = g l sin  tan
t
t
2lsint
t
2 lsin
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e) Suppose the string breaks as the ball swings in its circular path. Qualitatively describe the trajectory of
the ball after the string breaks but before it hits the ground.
The ball will act like a horizontally fired projectile. It’s initial horizontal velocity will be tangent to it’s
circular path. Vertically the ball will accelerate down at 9.8m/s/s. It’s path will be parabolic.
AP Chapt 5 Pre-Test
1984 Centripetal Motion
1. When a person stands on a rotating merry-go-round, the frictional force exerted
on the person by the merry-go-round is
a) greater in magnitude than the frictional force exerted on the person by the merry-go-round
b) opposite in direction to the frictional force exerted on the merry-go-round by the person
c) directed away from the center of the merry-go-round
d) zero if the rate of rotation is constant
e) independent of the person's mass
Newton’s Third Law
2. Each of five satellites makes a circular orbit about an object that is much more massive than any of
the satellites. The mass and orbital radius of each satellite are given below. Which satellite has
the greatest speed?
Mass
Radius
1
Fg=Fc
A. m
R
2
B. m
C. m
D. m
E. 2m
1
R
Gmmo =mv2
R2
R
R
2R
R
Gmo
R
2
= v Therefore v is inversely related to the
orbital radius.
3. A ball attached to a string is whirled around in a horizontal circle having a radius r. If the radius
of the circle is changed to 4r and the same centripetal force is applied by the string, the new speed
of the ball is which of the following?
A. One-quarter the original speed
Fc=mv2 = m(2v)2
B. One-half the original speed
r
(4r)
C. The same as the original speed
D. Twice the original speed
E. Four times the original speed
4. A racing car is moving around the circular track of radius 300 meters shown below. At the instant
when the car's velocity is directed due east, its acceleration is directed due south and has a
magnitude of 3 meters per second squared. When viewed from above, the car is moving
A. clockwise at 30 m/s
a= v2 = (30m/s)2= 30m/s
B. clockwise at 10 m/s
r
(300m)
C. counterclockwise at 30 m/s
clockwise
D. counterclockwise at 10 m/s
E. with constant velocity
F. 1988
5. The horizontal turntable shown above rotates at a constant rate. As viewed from above, a coin on
the turntable moves counterclockwise in a circle as shown. Which of the following vectors best
represents the direction of the frictional force exerted on the coin by the turntable when the coin is
in the position shown?
The frictional force allows for the coin to turn there it
must be center seeking
6. An object weighing 4 Newtons swings on the end of a string as a simple pendulum. At the bottom
of the swing, the tension in the string is 6 Newtons. What is the magnitude of the centripetal
acceleration of the object at the bottom of the swing?
a)
b)
c)
d)
e)
0
1/2g
g
3/2 g
5/2 g
Fc = FT - Fg
mac = 6 N - 4 N
ac = 2N
m
= 2N
(4 N/g )
ac=1/2g
7. A satellite of mass M moves in a circular orbit of radius R at a constant speed v. Which of the
following must be true?
mv 2
I. The net force on the satellite is equal to
and is directed toward the center of the orbit.
R
II. The net work done on the satellite by gravity in one revolution is zero.
III. The angular momentum of the satellite is a constant. We will discuss this in the next unit!
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
8. A car initially travels north and then turns to the left along a circular curve. This causes a package
on the seat of the car to slide toward the right side of the car. Which of the following is true of the
net force on the package while it is sliding?
a)
b)
c)
d)
e)
The force is directed away from the center of the circle.
The force is directed north.
There is not enough force directed north to keep the package from sliding.
There is not enough force tangential to the car's path to keep the package from sliding.
There is not enough force directed toward the center of the circle to keep the package from
sliding. The frictional force is not enough to keep the package in uniform circular motion.
9. A roller coaster is on a track that forms a circular loop in the vertical plane. If the car is to just
maintain contact with the track at the top of the loop, what is the minimum value for its centripetal
acceleration at that point?
a) g downward b) .5 g downward c) g upward d) 2 g upward
10. What force allows each of the following to travel in a curved path? 3pts
a) A car rounding a bend in the road.
Friction
b) A person in a rotor or gravitron at an amusement park.
Normal
c) A person in orbit on the space shuttle
Gravity
11. Which car is most likely able to negotiate its curve? 2pts
a) A car traveling around a 25 m curve at 20 m/s
b) A car traveling around a 50 m curve at 20 m/s
c) A car traveling around a 75 m curve at 20 m/s
C
B
A
Why? 2pts Fc is inversely related to the radius of the path Fc = mv2
r
12.
A .025 kg ball is swung in a .75 meter radius vertical loop .
a) Sketch what happens to a ball swung in vertical loop if it is let go at “ 3 O Clock.
Vertical
Loops
It is let go
here.
b) Label the force vector due to gravity and the tension vector in each of the following
Vertical
Loops
A
B
c)Where is the tension in the string the least and why? A B C 2pts
Least at A because Fc = Fg + FT therefore Fc-Fg = Fc
d)Where is the tension in the string the most and why? A B C 2pts
Most at C because Fc = FT - Fg therefore Fc+Fg = FT
C
The rest of this page must be answered on a separate piece of paper
13 e) What is the tension in the string in diagram B when the velocity of the .025 kg ball rotates
at 8.00 m/s in the .75 meter loop? 5pts F=mac
FT=Fc
FT=mv2
r
14 f)What is the maximum velocity the ball can be rotated at before a string will break at position B if the
string can withstand 2.00 N ? 5pts
F=mac
mv2 + mg = FT
r
15 g) What is the minimum velocity needed to keep the ball in a uniform circle in postion A?
F=mac
Fc = Fg + FT and FT = 0 then Fc = Fg mv2 = mg
r
Fc = FT - Fg
16. a) What is the weight of a 60 kg astronaut on the earth calculated by using big G.
Fg = G mm
r2
b) What is the velocity of the space shuttle in a 3.19 x 106 m orbit above the surface of the
earth?
re = 6.38 x 106 m
G = 6.67 x 10-11 N m2 / kg2
me= 5.98 x 1024 kg
F=mac
Gmme = mv2
Gme = v2
r2
r
r
c) Why does an astronaut in a 3.19 x 106 m orbit feel weightless? 2pts
The astronaut is In constant freefall towards the earth and therefore the only force keeping the astronaut in
uniform circular motion is gravity. The astronaut does not have a normal force on him and therefore feels
weightless.
Fg = Fc
17. What is the maximum radius turn can a car traveling at 30 m/s can negotiate if the coefficient of
friction between the tires and the road is .55?
F=mac
Ffr = mac
FN = mv2
r
r= mv2
FN
Fy =0
Fg=FN
r= mv2
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