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2.2.2 In Class or Homework Exercise
1. Sue whirls a yo-yo in a horizontal circle. The yo-yo has a mass of 0.20 kg and
is attached to a string 0.80 m long.
a. If the yo-yo makes 1.0 complete revolution each second, what force
does the string exert on it?
m  0.20kg
r  0.80m
f  1.0 Hz
FT  ?
If we draw a free body diagram, we see that the string cannot be
perfectly horizontal since there is no force other than the string to
balance the force of gravity:
However, for this and other problems involving horizontal circles we will
assume (unless told otherwise) that the string is horizontal and only
exerts a horizontal force.
1
f
1

1.0
 1.0 s
T
2 r
v
T
2 (0.80)

1.0
 5.03m / s
v2
r
(5.03) 2

0.80
 31.6m / s 2
ac 
Fc  mac
FT  mac
 (0.20)(31.6)
 6.3 N
b. If Sue increases the speed of the yo-yo to 2.0 revolutions per second,
what force does the string now exert?
Since she is doubling the frequency, she is cutting the period in half.
This will double the speed and will quadruple the acceleration (since
ac  v 2 ). This will require a force 4 times as large
FT  4(6.3)
 25 N
UNIT 2 2D Motion
RRHS PHYSICS
Page 59 of 136
2. A car rounds a curve on a flat road of radius 50.0 m at a speed of 50.0 km/h.
Will the car make the turn if (a) the pavement is dry and the coefficient of
static friction is 0.60, (b) the pavement is icy and   0.20 ?
There are a number of ways in which this problem may be approached. We
will first calculate the required coefficient of friction to make the turn at the
given speed.
r  50.0m
v  50.0km / h  13.9m / s
 ?
The free body diagram looks like this:
So
Fc  mac
F f  mac
 FN  mac
 mg  m
v2
r
v2

gr
(13.9) 2

(9.80)(50.0)
 0.39
This tells us the required (or minimum) coefficient of friction.
a. Since the dry coefficient of friction is 0.60, there will be enough friction
to make the turn.
b. Since the wet coefficient of friction is only 0.20, there will not be
enough friction to make the turn
UNIT 2 2D Motion
RRHS PHYSICS
Page 60 of 136
3. A gravitron circus ride consists of a cylindrical room with vertical walls (as
shown below) that rotates about its axis
People are positioned against the circular wall. When the room is rotating at a
certain speed, the floor drops out and the people in the room remain
suspended against the outer wall. Consider a gravitron circus ride that has an
8.0 m radius and rotates 0.36 times per second.
a. Draw a free body diagram indicating all of the forces involved.
Notice that the normal force is horizontal (perpendicular to the wall) and
friction is up (opposing the person trying to slide down the wall). There
is no outward force on the person.
b. What coefficient of friction is necessary to prevent the people from
sliding down the wall?
UNIT 2 2D Motion
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r  8.0m
f  0.36 Hz
 ?
T
1
f
1
0.36
 2.78s

2 r
T
2 (8.0)

2.78
 18.1m / s
v
In order for people to not slide down the wall, the force of friction must
balance the force of gravity:
Vertical
Ff  Fg
 FN  mg
Horizontal
The centripetal acceleration (and therefore the centripetal force) in the
diagram is to the left. It must therefore be provided by the normal force
exerted by the wall:
Fc  mac
FN  mac
If we combine the horizontal and vertical equations, we get
 FN  mg
 ( mac )  mg
v2
 g
r
gr
 2
v
(9.80)(8.0)

(18.1) 2
 0.24
c. People often describe the ride by saying that they were being pressed
against the wall. Is this true?
Not really. There body has a natural tendency to continue travelling in a
straight line (Newton’s First Law), but the wall is in the way. The wall is
exerting an inward force to make them travel in a circular path and this
is what they are feeling.
UNIT 2 2D Motion
RRHS PHYSICS
Page 62 of 136
4. What minimum speed must a roller coaster be travelling when upside down at
the top of a circle if the passengers are not to fall out. Assume a radius of
curvature of 8.00 m.
In general for the people in the roller coaster, there would be two forces
exerted downward at the top of the path, the force of gravity and the normal
force of the roller coaster cart; however, the slower the cart is travelling the
lower the magnitude of the normal force. At the minimum speed, the normal
force is zero, so the only force acting on the person is the force of gravity. This
means that the person has a downward acceleration of 9.80 m/s2. Since there
is no upward force, the person is falling at 9.80m/s 2 at this point. For the
passengers to no fall out of the car, it must also have a downward
acceleration of 9.80m/s2.
v2
ac 
r
v2
9.80 
8.0
v  8.85m / s
If the cart goes slower than 8.85 m/s, its downward centripetal acceleration is
less than the 9.80 m/s2 that the person is accelerating at and the person will
fall out of the cart.
5. A coin is placed 18.0 cm from the axis of a rotating turntable of variable
speed. When the speed of the turntable is slowly increased, the coin remains
fixed on the turntable until a rate of 58 rpm is reached. What is the coefficient
of static friction between the coin and the turntable?
r  18.0cm  0.180m
f  58rpm  0.97 Hz
 ?
T
1
f
1
0.97
 1.03s

UNIT 2 2D Motion
2 r
T
2 (0.180)

1.03
 1.10m / s
v
RRHS PHYSICS
v2
r
(1.10) 2

0.180
 6.72m / s 2
ac 
Page 63 of 136
Fc  mac
F f  mac
 FN  mac
 mg  mac
 (9.80)  6.72
  0.69
6. A ball on a string is revolving at a uniform rate in a vertical circle of radius 96.5
cm. If its speed is 3.15 m/s and its mass is 0.335 kg, calculate the tension in
the string
a. at the top of its path
It should be noted that although we are assuming a constant speed
throughout the vertical circle in this problem, this is not a realistic
situation. The ball’s speed must change in order to satisfy conservation
of energy (this will be addressed in the next problem).
r  96.5cm  0.965m
v  3.15m / s
m  0.335kg
FT  ?
Fg  mg
 (0.335)(9.80)
 3.28 N
v2
r
(3.15) 2

0.965
 10.3m / s 2
ac 
The acceleration is downward. Using down as positive,
UNIT 2 2D Motion
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Page 64 of 136
Fc  mac
mac   F
mac  FT  Fg
mac  FT  Fg
(0.335)(10.3)  FT  3.28
FT  0.17 N
b. at the bottom of its path
m  0.335kg
ac  10.3m / s 2
Fg  3.28 N
FT  ?
Since we are now at the bottom of the path, the acceleration is upward.
We will use up as positive.
Fc  mac
mac   F
mac  FT  Fg
mac  FT  Fg
(0.335)(10.3)  FT  3.28
FT  6.73 N
c. at the middle of its path (halfway between top and bottom)
m  0.335kg
ac  10.3m / s 2
Fg  3.28 N
FT  ?
UNIT 2 2D Motion
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Page 65 of 136
Since the centripetal acceleration is horizontal at this point, only the
tension in the string contributes to (and therefore provides) the
centripetal force
Fc  mac
FT  mac
 (0.335)(10.3)
 3.45 N
d. Assuming a constant speed throughout the entire vertical circle, where
would the string be most likely to break and in what direction would the
ball travel immediately after the string broke?
Since the tension is greatest at the bottom of the path, the string would
be most likely to break at the bottom. Since the ball is travelling
horizontally at the bottom of the path, it would continue to move
horizontally (either left or right) immediately after the string breaks.
7. A 5.0 kg mass is being swung in a vertical circle on a 3.0 m rope.
a. What is the critical speed (i.e. the minimum speed at which the ball will
maintain a circular path) for this mass?
If the ball is to fail to maintain a circular path, this will occur at the top of
the path. Consider a mass on a string being swung in a vertical circle;
as the mass is swung faster, the string must exert more force to
maintain a circular path. As the mass is swung more slowly, the string
exerts less force. If you slow the mass down too much, it falls out of the
circle at the top. The critical speed occurs at the exact moment that the
tension goes to zero so that the entire centripetal force at this time is
being provided by the force of gravity. The mass must therefore have
an acceleration equal to the acceleration due to gravity:
v2
ac 
r
v2
g
r
v  gr
 (9.80)(3.0)
 5.4m / s
.
UNIT 2 2D Motion
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Page 66 of 136
b. Calculate the tension in the rope at the ball's lowest point, assuming
that the ball is travelling at its critical speed at the top of the circle.
Assume no change in energy for the system.
Fg  mg
m  5.0kg
r  3.0m
FT  ?
 (5.0)(9.80)
 49 N
Since we are assuming no change in energy for the system, the ball
must be travelling faster at the bottom since it is at a lower height and
will have less potential energy. If we use the lowest point in the circle
as the reference level,
Ei  E f
E pi  Eki  Ekf
mghi  12 mvi2  12 mv 2f
(9.80)(6.0)  12 (5.4) 2  12 v 2f
v f  12.1m / s
v2
ac 
r
(12.1) 2

3.0
 48.8m / s 2
Since we are now at the bottom of the path, the acceleration is upward.
We will use up as positive.
UNIT 2 2D Motion
RRHS PHYSICS
Page 67 of 136
Fc  mac
mac   F
mac  FT  Fg
mac  FT  Fg
(5.0)(48.8)  FT  49
FT  290 N
8. Tarzan plans to cross a gorge by swinging in an arc from a hanging vine. If his
arms are capable of exerting a force of 1500. N on the vine, what is the
maximum speed he can tolerate at the lowest point of his swing? His mass is
85.0 kg; the vine is 4.0 m long.
Fa  1500.N
Fg  mg
m  85.0kg
 (85.0)(9.80)
r  4.0m
 833 N
v?
Since Tarzan is at the bottom of the path, the acceleration is upward. We will
use up as positive.
Fc  mac
v2
ac 
mac   F
r
mac  Fa  Fg
v2
7.85 
4.0
mac  Fa  Fg
v  5.6m / s
(85.0)a  1500.  833
c
ac  7.85m / s 2
9. How many revolutions per minute would a 15 m diameter Ferris wheel need to
make for the passengers to feel weightless at the top?
UNIT 2 2D Motion
RRHS PHYSICS
Page 68 of 136
If the passengers are to feel weightless, the must be in free fall which means
that their downward acceleration must be 9.80m/s2. This must be the
centripetal acceleration.
ac  9.80m / s 2
r  7.5m
f ?
v2
r
v2
9.80 
7.5
v  8.6m / s
ac 
2 r
T
2 (7.5)
8.6 
T
T  5.5s
v
1
T
1

5.5
 0.18r / s  60 s / min
f 
 11r / min
10. A person is swinging a 0.150 kg ball on a string in a horizontal circle of radius
0.600 m. The ball makes 2.00 revolutions per second.
a. Although the circle is horizontal, the string is not. Explain why not.
If we draw a free body diagram, we see that the string cannot be
perfectly horizontal since there is no force other than the string to
balance the force of gravity:
A horizontal string cannot exert a vertical force to balance the
downward force of gravity; therefore, the string must have a vertical
component.
b. Ignoring the fact that the string is not perfectly horizontal (due to the
weight of the ball), calculate the tension in the string.
m  0.150kg
r  0.600m
f  2.00 Hz
FT  ?
T
1
f
1
2.00
 0.500 s

UNIT 2 2D Motion
2 r
v
T
2 (0.600)

0.500
 7.54m / s
RRHS PHYSICS
v2
ac 
r
(7.54) 2

0.600
 94.8m / s 2
Page 69 of 136
Since we are assuming that the string is horizontal, the tension in the
string will be providing the centripetal force:
Fc  mac
FT  mac
 (0.150)(94.8)
 14.2 N
c. Find the tension in the string (and the angle it makes with the
horizontal) if we do not ignore the effect of the weight of the ball. 5.91,
14.3N
Vertical
ma   F
0  Fg  FTy
0  Fg  FTy
FTy  mg
 (0.150)(9.80)
 1.47 N
Horizontal
mac   F
mac  FTx
(0.150)(94.8)  FTx
FTx  14.2 N
tan  
FT  FTx2  FTy2
FTy
FTx
1.47
14.2
 5.91

 (14.2)2  (1.47) 2
 14.3N
As can be seen the tension is less than 1% different when accounting
for the angle of the string.
UNIT 2 2D Motion
RRHS PHYSICS
Page 70 of 136
11. A projected space station consists of a circular tube which is set rotating about
its center (like a tubular bicycle tire). The circle formed by the tube has a
diameter of 1.600 km.
a. On which part of the inside of the tube will people be able to walk?
The people will be able to walk on the inside outer edge of the tube. It
is similar to placing a marble on a turntable – as the turntable spins, the
marble will roll and will continue in a straight line toward the outer edge
of the turntable. In the space station, the people will move in a straight
line toward the inside outer edge. From their point of view, it will be as if
there is a force pushing them toward this part of the space station.
b. What must be the rotation speed (revolutions per day) if an effect equal
to gravity at the surface of the earth (1 g ) is to be felt?
r  800.m
ac  9.80m / s 2
f ?
2 r
T
2 (800.)
88.5 
T
T  56.8s
v2
r
v2
9.80 
800.
v  88.5m / s
ac 
f 
v
1
T
1
56.8
 0.0176r / s  86400 s / day

 1520r / day
UNIT 2 2D Motion
RRHS PHYSICS
Page 71 of 136
12. When you drive rapidly on a hilly road or ride in a roller coaster, you feel
lighter as you go over the top of a hill and heavier when you go through a
valley.
a. Sketch each situation, including the relevant forces, and explain these
sensations.
When the car is going over the top of the hill, the center of curvature for
the hill is below the car, so there is a downward centripetal
acceleration. This means that the downward force of gravity is greater
than the upward normal force on the person.
Since the normal force is smaller than if the car were at rest, the person
will feel lighter.
When the car is going through a valley, the center of curvature for the
valley is above the car, so there is an upward centripetal acceleration.
This means that the upward normal force is greater than the downward
force of gravity on the person.
Since the normal force is larger than if the car were at rest, the person
will feel heavier.
b. Consider a car going over the top of a hill (radius = 95 m). How fast
would the car have to go so that the normal force acting on the driver is
zero?
Since the normal force is zero, the only force acting on the person will
be the force of gravity so the acceleration must be 9.80 m/s 2. The
person will feel weightless in this situation.
UNIT 2 2D Motion
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Page 72 of 136
r  95m
ac  9.80m / s 2
v?
v2
r
v2
9.80 
95
v  30.5m / s
ac 
 110km / h
13. A bucket of water can be whirled in a vertical circle without the water spilling
out, even at the top of the circle when the bucket is upside down. Explain.
This is similar to the upside down roller coaster problem. If the water were to
fall out of the bucket, it will fall with an acceleration of 9.80 m/s 2. As long as
the bucket is being swung fast enough so that its downward centripetal
acceleration at the top of the path is at least 9.80 m/s2, it will be pushing down
on the water and the water will remain with the bucket. If the bucket slows
down so that its centripetal acceleration is less than 9.80 m/s 2, the water will
now have a greater downward acceleration and will leave the bucket.
14. For a car travelling with speed v around a curve of radius r , determine a
formula for the angle at which a road should be banked so that no friction is
required.
In this case, the centripetal acceleration is horizontal toward the inside of the
curve. We will use a horizontal coordinate system and break the normal force
up into horizontal and vertical components. The angle  in the free body
diagram will be the same angle as the angle that the road makes with the
horizontal.
UNIT 2 2D Motion
RRHS PHYSICS
Page 73 of 136
Since there is no friction, the only force with any horizontal components is the
normal force exerted by the road. The horizontal component of this normal
force must provide the required centripetal force.
Fc  mac
FNx  mac
v2
r
v2
mg tan   m
r
2
v
tan  
gr
FNy tan   m
 v2 
  tan  
 gr 
1
UNIT 2 2D Motion
RRHS PHYSICS
Page 74 of 136