Centripetal Force and Universal Gravitation Review Sheet Answers
Physics
Review Questions:
1. When you turn around a corner in your car, you lean toward the outside side of your car.
a) What forces are acting on you as you round the corner?
Gravity pulls you down and the normal force from the seat pushes you up, of course. In addition, the
friction between the seat and your rump (and a little bit on your back where you lean against the chair)
pushes you toward the inside of the turn.
b) Why do you lean to the outside of the turn?
You don’t actually lean outward. Instead, your body tries to go in a straight line (inertia – Newton’s 1st
Law) and the force of friction has to make your body turn. Since the friction is primarily between your
rump and the seat, it is immediately affected, but your upper body keeps going straight until there is an
angle to your spine, which means that your spine can pull your upper body toward the center of the circle.
So it would be more correct to say that your rear end slides out from under your upper body than to say
that your upper body leans outward.
c) Why do motorcycle and bicycle riders have to lean into turn? What would happen if they tried to take the turn
upright?
A cyclist of any kind must lean into a turn for several reasons. In this context the most important is that
when they lean the seat is pushing on the rider at an angle – partly upward, supporting them against
gravity, and partly inward, pushing them toward the center of the turn and making them change direction.
If they tried to take the turn upright, the friction on the ground would begin to make the motorcycle turn,
but there would be very little force making the rider turn and they would be in danger of coming off the
bike.
2. Why can it be dangerous to hit your brakes while taking a corner in a car? (Hint: What force or forces slow
you down while you’re braking? What force or forces make you turn? What do you remember about that force
or those forces?)
When braking the force of friction between the tires and the road is what makes you stop. When turning
the force of friction between the tires and the wheels is what makes you turn. If you hit the brakes while
turning, the friction force has to do two things at once. That is okay, but it takes a larger amount of
friction to do both than to do either one separately. Since the tire’s friction is limited by the coefficient of
friction between the wheel and ground, it is possible that braking and turning could require more force
than friction can supply and the car will begin to skid.
3. In a circus act, a motorcycle rider drives into a ball made of metallic mesh and drives around in vertical
circles. How is it possible for the motorcyclist to ride upside down?
A motorcyclist can only ride upside down in particular situations. A motorcyclist cannot ride upside
down on a flat ceiling without special equipment. A motorcyclist with an everyday street bike can ride
upside down inside a ball because they are constantly being turned by the walls of the ball. As the cycle
goes over the top of the ball, the floor and gravity are both pushing down on it toward the center of the
ball and together provide the centripetal force that makes the bike turn so that it comes around in a loop.
As long as the bike is going fast enough that it needs at least as much force as gravity provides to make
the turn, the bike will stay on the ball. The gravity will not move the ball toward the floor if it is all
needed just to make the turn along the ceiling.
4. If you are riding on a merry-go-round and let go of the railing, you will come off the
merry-go-round. Which direction will you be going as you come off?
According to Newton’s 1st Law, when you let go off the railing and the merry-goround can’t exert any force to change your direction, you will keep going in the same
direction that you were already going. In this case, you are moving sideways around
the outside of the circle, so you will keep moving sideways in a straight line that is
tangent to the circle of the merry-go-round [as shown in the diagram]
5. A 1000 kg car traveling at 20 m/s makes a circular turn with a radius of 50 m on a flat road.
a) What is the car’s centripetal acceleration as it goes around the turn?
v2 (20 m/s)2
aC = r = 50 m = 8 m/s/s
b) How big a centripetal force is required to bring them around the corner?
FC = m·aC = (1000 kg)(8 m/s/s) = 8000 N
c) What force provides the centripetal force?
The force of friction is what pushes the car toward the center of the turn and makes it change
direction.
d) What is the minimum coefficient of friction required for the car to make it around the turn?
On flat ground, FN = Fg = mg = 1000 kg · 9.8 m/s/s = 9800 N
Ff = FN
Ff 8000 N
 = F = 9800 N = 0.816
N
e) (extreme challenge) What if the corner was banked toward the center of the turn at 35°? How
strong would friction have to be then?(Hint: First find coefficient of friction?)
If the corner is banked then the normal force has to push at an angle and
one component of it will act up, resisting gravity, while the other pushes in
as a part of the centripetal force. In addition, the force of friction will act
at an angle and one component will push down, helping pull the car down,
while the other pushes in as part of centripetal force. If we look at a force
diagram drawn from the back of the car as it rounds the corner, we will see
how the forces combine.
In the vertical direction, the gravitational force and one component of the
frictional force act downward against a component of the normal force.
In the horizontal direction, components of the normal force and
frictional force act together toward the center of the circle to make
up the centripetal force. The amount of centripetal force hasn’t
changed just because the road is banked - there still has to be 8000
N - the only difference is where it’s coming from.
Vertical Forces must cancel because car is not moving up or down:
FNsin55° - Ffcos55° - Fg = 0
but the Ff = FN
FNsin55° - FNcos55° - Fg = 0
FN(sin55° - cos55°) = Fg
Horizontal forces act together to provide the necessary centripetal force to make the turn:
v2
(20 m/s)2
FNcos55° + Ffsin55° = FC = m r = 1000 kg 50 m = 8000 N
FNcos55° + FNsin55° = 8000 N
FN(cos55° + sin55°) = 8000 N
Now we have two equations with two unknowns. We will solve one equation for one variable
and then substitute in the other equation:
FN(cos55° + sin55°) = 8000 N
FN =
8000 N
cos55° + sin55°
8000 N
(sin55° - cos55°) = (1000 kg)(9.8 m/s/s)
cos55° + sin55°
8000 N (sin55° - cos55°) = 9800 N (cos55° + sin55°)
(8000 N)sin55° - (8000 N)cos55° = (9800 N)cos55° + (9800 N)sin55°
-(9800 N)sin55° - (8000 N)cos55° = (9800 N)cos55° - (8000 N)sin55°
- [(9800 N)sin55° + (8000 N)cos55°] = (9800 N)cos55° - (8000 N)sin55°
- [12616.3 N] = -932 N
- = -0.0739
 = 0.0739
Now we can calculate how big the friction has to be. First we have to find out how big the
normal force is:
FN(cos55° + sin55°) = 8000 N
FN(0.634) = 8000 N
FN = 12616.3 N
Then we can calculate the force of friction.
Ff = FN = (0.0739)(12616.3 N) = 932 N
6. A 0.5 kg tetherball is hit so that it goes in a horizontal circle around the pole with a radius of 2 m.
a) If the string can exert a maximum of 900 N of tension, what is the fastest speed the ball could have so
that the string wouldn’t break?
v2
T = FC = m r
T·r 900 N · 2 m
v2 = m = 0.5 kg = 3600 Nm/kg
v = 3600 m2/s2 = 60 m/s
b) Assume the ball is hit at 10 m/s and keeps going that same speed. What will happen to the
centripetal force on the ball as the string wraps around the pole?
As the ball wraps around the pole, the string gets shorter and shorter, so the ball moves in a
tighter circle. With a smaller radius, the amount of centripetal force needs to be larger (because
the formula for centripetal force involves dividing by the radius and dividing by a smaller
number gives a larger result). Since the string’s tension is the centripetal force, a larger
centripetal force means more tension in the string.
7. Arielle, 55 kg, rides her skateboard up and around a complete loop inside a 3m diameter sewer pipe.
She enters at the bottom of the pipe going 10 m/s. Assume there is no friction or air resistance.
a) How much normal force does the pipe exert on her as she starts turning to make the loop?
When at the bottom of the loop, the normal force from the bottom of the pipe pushes upward on
the skateboard (and Arielle). This force points to the center of the circle and it is what makes
Arielle start turning around the loop of pipe. However, the force of gravity pulls downward on
Arielle and some of the normal force is needed to cancel
gravity, so only part of the normal force makes her turn. The
part of the normal force that makes the turn is what is left when
gravity is canceled out. In an equation we write this:
v2
FN – Fg = FC = m r
v2
FN – mg = m r
(10 m/s)2
FN – 55kg·9.8 m/s/s = 55 kg
1.5 m
FN – 539 N = 3667 N
FN = 4206 N
b) If she reaches the top going 6.42 m/s, how much normal force does the top of the pipe have to exert
on her as she goes around the loop?
At the top, the gravity and normal force both point downward – toward the center of the circle –
so they are both working together to make Arielle keep turning. So together they provide the
necessary centripetal force:
v2
FN + Fg = FC = m r
v2
FN + mg = FC = m
r
(6.42 m/s)2
FN + 55kg·9.8 m/s/s = 55 kg 1.5 m
FN + 539 N = 1511 N
[Since it is going slower, it easier to turn & needs less force]
FN = 962 N
[Because gravity is helping, the pipe has to exert a lot less force]
c) What is the slowest speed she could be traveling at the top and still make it around the loop?
Look at the equation that describes how the forces affect her motion when
v2
she is at the top of the loop: FN + Fg = FC = m r . Clearly, the faster she
goes, the more force is necessary to make her turn. Gravity is always
present, but the normal force can chip in if more force is needed. The
converse is also true: the slower she travels, the less force needs to be
applied to make her turn. However, gravity never changes, so the smallest
force that can act on her is gravity. If she goes slow enough, that is too
much force to carry her around the turn and she ends up making a tighter turn and coming off the
pipe [as shown by the dotted line in the picture to the right]
v2
Fg = FC = m r
v2
539 N = 55 kg 1.5 m
14.7 m2/s2 = v2
3.8 m/s = v
8. Eric likes to drive his 1200 kg car
really fast over the hills in the
countryside around his house.
a) He goes over one hill, which curves
with a 15 m radius. What is the fastest
that he can go over the hill without
losing contact with the ground?
At the top of the circle, the
force of gravity pulls down
toward the center of the circle that the car is following and the normal force pushes up on the car
away from the center of the circle. Since the centripetal force is the total force toward the center,
it will be whatever part of gravity is left over once the normal force cancels out as much as it
can:
v2
Fg – FN = FC = m
r
This means that as Eric goes faster, he needs more centripetal force to keep him on the road. To
get more centripetal force, the normal force has to drop because gravity never changes. Once the
normal force drops to 0, that is the biggest force that could possibly act toward the center of the
circle is just the force of gravity.
v2
Fg = FC = m r
v2
(1200 kg)(9.8 m/s/s) = (1200 kg)15 m
147 m2/s2 = v2
12.1 m/s = v
b) After that hill, he comes down and goes through a valley which curves up with a radius of 25 m. The
springs on the car can take a maximum force of 94,000 N before the car bottoms out. How fast can he
drive through the valley without bottoming out his car?
At the bottom of the valley, the force of gravity pulls down on the car while the normal force
pulls up. The normal force is toward the center of the circle and would be the centripetal force
except that the gravity cancels out part of the normal force. This means that the actual
centripetal force is the part of the normal force that is not canceled out. The limit on the speed
comes because the springs can only exert a limited force in reaction to the normal force, this
means that the maximum the normal force can be is 94,000 N.
FN – Fg = FC = m
v2
r
v2
94,000 N – (1200 kg)(9.8 m/s/s) = (1200 kg)25 m
v2
94,000 N – (1200 kg)(9.8 m/s/s) = (1200 kg)25 m
v2
82240 N = (1200 kg)25 m
1713.3 = v2
41.4 m/s = v
9. In an attempt to build the world’s largest pendulum (and get their
names into the Guinness book of world records), Steve & Jake hang a
5000 kg wrecking ball on the end of a 300 m long cable and hang it from
the bottom of a bridge over a deep gorge. They then hoist the ball up to
the edge of the canyon (so it is even with the suspension point) and let it
go from rest. (Clue: a, c and e can be skipped since the answer is given
to you)
a) Ignore air resistance and use the law of conservation of energy to
prove that the speed of the wrecking ball when it reaches the bottom of
its swing is 76.7 m/s.
The law of conservation of energy states that the total energy at
the end is equal to the total energy at the beginning. At the top,
it is elevated 300 m above the bottom (it can only fall 300 m since that’s how long the rope is) so it has
gravitational potential energy. There is no other significant type of energy present at the start. At the end,
it’s at the end of its rope so it can’t fall any further, which means there is no gravitational potential
energy, but it’s moving so there is kinetic energy.
Ef = Ei
KEf = GPEi
1
2
2 mv f = mghi
v2f = 2ghi
2
v f = 2(9.8 m/s/s)(300 m) = 5880 m2/s2
vf = 5880 m2/s2 = 76.7 m/s
b) How big is the tension in the cable when the pendulum is at the bottom of it’s swing?
The tension is pulling up on the wrecking ball but gravity is pulling down. To keep it turning in a circle,
there must be a net force to the center (the centripetal force). This means that the tension must be larger
than gravity and by just the right amount to keep it moving in a 300 m radius circle.
v2
T – Fg = FC = m r
v2
T – mg = m
r
(76.7 m/s)2
T – (5000 kg)(9.8 m/s/s) = (5000 kg)
300 m
T – 49,000 N = 98,048 N
T = 147,048 N
c) Unfortunately, neither of them noticed the pipeline running across the gorge 200 m beneath the bridge while
they were setting up. As the pendulum swings, the cable runs into the pipe and is bent by it so that the pendulum
begins to wrap itself around the pipeline. Use the law of conservation of energy to prove that the speed of the
pendulum when it comes even to the pipeline is 62.6 m/s.
The law of conservation of energy states that the total energy at the end is equal to the total energy at the
beginning. At the beginning, it is at the lowest point so there is no gravitational potential energy but it is
moving so there is kinetic energy. As it wraps around the pipe it will rise up and gain gravitational
potential energy. When it comes even with the pipe, it will be 100 m higher than at the start.
1
1
2
2
mv
+
mgh
f=
f
2
2 mv i
1
1
2
2
2(5000 kg)v f + (5000 kg)(9.8 m/s/s)(100 m/s) = 2 (5000 kg)(76.7 m/s)
(2500 kg) v2f + 4900000 J = 14707225 J
(2500 kg) v2f = 9807225 J
v2f = 3922.9 m2/s2
vf = 3922.9 m2/s2 = 62.6 m/s
Ef = Ei
or
KEf + GPEf = KEi
or
d) How big is the tension in the cable when the pendulum is even with the pipe?
The ball is now moving in a circle around the pipe; since there were 100 m of rope sticking out
beyond the pipe, the radius of the circle is 100 m. The only force acting toward or away from the
center of the circle is the tension which pulls the wrecking ball toward the pipe. That tension is
the centripetal force in this case:
v2
T = FC = m r
v2
(62.6 m/s)2
T = m r = (5000 kg) 100 m = 195,938 N
e) Use the law of conservation of energy to prove that the speed of the pendulum when it reaches the
point over the pipeline is 44.3 m/s.
The law of conservation of energy states that the total energy at the end is equal to the total
energy at the beginning. At the beginning, it is at the lowest point so there is no gravitational
potential energy but it is moving at 76.7 m/s so there is kinetic energy. As it wraps around the
pipe it will rise up and gain gravitational potential energy. When it reaches the point above the
pipe, it will be 200 m higher than at the start.
Ef = Ei
KEf + GPEf = KEi
1
1
2
2
2 mv f + mghf = 2 mv i
1
1
2
2
2(5000 kg)v f + (5000 kg)(9.8 m/s/s)(200 m/s) = 2 (5000 kg)(76.7 m/s)
(2500 kg) v2f + 9800000 J = 14707225 J
(2500 kg) v2f = 4907225 J
v2f = 1962.9 m2/s2
vf = 1962.9 m2/s2 = 44.3 m/s
f) How big is the tension in the cable when the pendulum is over the pipe?
It is still moving in a 100 m radius circle when it reaches the top. This time, the tension is
pulling down toward the pipe and gravity is also pulling down. Since they are both acting in the
same direction, they work together to make the centripetal force.
v2
T + Fg = FC = m r
v2
T + mg = m r
T + (5000 kg)(9.8 m/s/s) = (5000 kg)
(44.3 m/s)2
100 m
T + 49,000 N = 98,124.5 N
T = 49124.5 N
10. How much gravitational force does an 80 kg person exert on an 8 kg bowling ball that is sitting on
a shelf 0.4 meters away from the center of their body?
Fg =
GmM (6.67 x 10-11 Nm2/kg2)(80 kg)(8 kg)
= 2.67 x 10-7 N = 0.000000267 N
r2 =
(0.4 m)2
11. Most asteroids are not big enough to be spherical. Ceres, the largest known asteroid is very close,
with an average radius of 950 km and a mass of 8.7 x 1020 kg. Let’s imagine that Rachel (65 kg), pays a
visit to Ceres.
a) As she stands on the surface of Ceres, how much would she weigh? Compare that to her weigh on
Earth – what percentage of her Earth weight would she feel on Ceres?
On Earth, Rachel weighs Fg = mg = (65 kg)(9.8 m/s/s) = 637 N.
On Ceres, Rachel weighs Fg =
GmM (6.67 x 10-11 Nm2/kg2)(65 kg)(8.7 x 1020 kg)
= 4.14 N
r2 =
(950,000 m)2
4.14 N
Rachel’s weight on Ceres is 637 N = 0.0065 = 0.65% of her weight on Earth.
b) How fast would Rachel have to run to orbit Ceres just above it’s surface?
If Rachel is in orbit, that means that the force of gravity on her is exactly the right amount to
provide all of her centripetal force.
v2
Fg = FC = m r
v2
4.14 N = (65 kg) 950000 m
60507.7 = v2
246 m/s = v
c) How long would it take her to go around Ceres once at that speed?
Right at the surface, Ceres circumference will be 2r = 2(950,000 m) = 5969026 m
d 5969026 m
since d = vt, if we solve for time we will get t = v = 246 m/s = 24264 s = 6.74 hours
d) How fast would Rachel have to move to orbit Ceres at a height of 1500 km above the surface?
At a height of 1500 km above the surface, Rachel would be orbiting with a radius of 1500 km +
950 km = 2450 km = 2450000 m
At that height, Fg =
GmM (6.67 x 10-11 Nm2/kg2)(65 kg)(8.7 x 1020 kg)
= 0.622 N
r2 =
(2,450,000 m)2
Since the gravity is the centripetal force
v2
Fg = FC = m r
v2
0.622 N = (65 kg) 2450000 m
23444.6 = v2
153.1 m/s = v
e) How long would it take her to go around Ceres once at that speed?
In that orbit, the circumference will be 2r = 2(2,450,000 m) = 15393804 m
d 15393804 m
since d = vt, if we solve for time we will get t = v = 153.1 m/s = 100,547 s = 1.16 days
12. The 1.8 x 1020 kg Death Star is placed into orbit around the forest moon of Endor. It completes one
300,000 km-radius orbit in 20 days.
a) How fast is the Death Star moving in its orbit?
d 2r 2·300000000m 1884955592m
v= t = t =
= 1728000 s = 1090.8 m/s
20 d
b) How much force is required to turn the Death Star in its orbit? (What specific force is responsible?)
The force that makes the Death Star turn is the centripetal force for that orbit:
v2
(1090.8 m/s)2
FC = m r = 1.8 x 1020 kg 300000000 m = 7.139 x 1017 N
c) How much mass must Endor have to keep the Death Star in its orbit?
Because it is the gravitational force from Endor that keeps the Death Star in its orbit, the
gravitational force must equal the amount of centripetal force that the Death Star needs applied
to stay in orbit:
Fg = FC = 7.139 x 1017 N
If we plug that into Newton’s Law of Universal Gravitation, we can solve for Endor’s mass:
Fg =
GmM
r2
7.139 x 1017 N =
(6.67 x 10-11 Nm2/kg2)(1.8 x 1020 kg)(M)
(300,000,000 m)2
5.35 x 1024 kg = M