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AP PHYSICS B
Circular Motion, Gravity, & Orbits
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Circular Motion, Gravity, & Orbits
What I Absolutely Have to Know to Survive the AP* Exam
Force is any push or pull. It is a vector. According to Newton’s Second Law, bodies in motion continue to move in a
straight line unless acted upon by a force. Forces that change the direction of a body’s velocity vector (versus
changing the magnitude of the body’s velocity vector) are called center seeking forces or centripetal forces. A
centripetal force is NOT a new kind of force, but is any naturally occurring force in a system that changes the
direction of a body’s motion, like the normal, frictional, weight, tension, and gravitational force. Centripetal forces
produce centripetal accelerations. The centripetal acceleration a body experiences is related to the radius of the
body’s path and the magnitude of the body’s velocity. When a body is being pushed or pulled out of straight line
motion, there exists a single force, the component of a single force, or the combination of forces and components that
act centripetally. The NET centripetal force is always directed toward the center of the circular motion. If a body is
following a curved path, there must be a naturally occurring force like tension, component of the normal force, or
gravitation that is directed toward the center of the circular path. Centripetal acceleration is always perpendicular to
the velocity vector or line of motion.
Key Formulas and Relationships
ac =
v2
r
where v =velocity or speed of an object in UCM or orbiting satellite
ac = centripetal acceleration
v=
2π r
=2π rf
T
r =radius of circular motion or radius of orbit
T =period of revolution
f =frequency
Fc = mac =
mv 2
r
Fc = centripetal force
2
⎛ 2π r ⎞
m⎜
⎟
4π 2mr
T ⎠
=
Fc = ⎝
r
T2
GM
v=
r
m =mass
Nm 2
=universal gravitational constant
kg 2
M =mass of object being orbited
G =6.67x10-11
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Circular Motion, Gravity, & Orbits
Centripetal Force is NOT a new force, but one or more of these normally occurring forces or their components
Weight
Fg , W
Gravitational
FG
Normal
FN , N
Fg = mg
Always directed toward the Earth’s center.
Force on a free falling body, if we neglect air friction.
mm
FG = G 1 2 2
r
A force of attraction between any two massive objects.
When the Earth is one of the two bodies involved, then the force felt by
the second body while positioned on the Earth’s surface will always be
directed toward the Earth’s center.
A force of support, provided to an object by a surface in which the object
is in contact.
Always directed perpendicular to and away from the surface providing
the support.
Friction is produced by the atomic interaction between two bodies as
they either slide over one another (kinetic friction) or sit motionless in
contact with one another (static friction).
F f s max ≤ μs FN (static)
F f k = μk FN (kinetic)
Friction
Ff , f
WARNING! The two quantities f s and f k may look the same, but they
tell us different things. Kinetic friction is typically less than static friction
for the same two surfaces in contact. Note that the normal force
sometimes equals the weight but not always. When you draw a free body
diagram of forces acting on an object or system of objects, be sure to
include the frictional force as opposing the relative motion (or potential
for relative motion) of the two surfaces in contact.
Tension
FT , T
Applied
FSubscript
FT is a force that is applied to a body by a rope, string, or cable.
FT is applied along the line of the string and away from the body in
question.
Push me, pull you force that does not fall into one of the above
categories, for example, a friend shoves you. The magnitude of the force
is characterized by an F, with an appropriate subscript.
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Circular Motion, Gravity, & Orbits
Strategy on Centripetal Force Problems
1. Take one body in the system and draw a Free Body Diagram (FBD) for it. Remember that the
centripetal force is not a NEW force, so do NOT include it on your FBD. Your FBD will have
your old friends like the weight force, normal force, frictional force, tension in a string, the
gravitational force, etc. providing the center seeking force.
2. Choose x and y axes and place them beside your FBD. One axis must be in the direction of the
centripetal acceleration you are trying to find. Remember that the centripetal acceleration and
centripetal force always point toward the center of the circular path. The velocity vector is
perpendicular to both and tangential to the path of the object.
3. If there are forces on the FBD that are not along the x and y directions, find their respective x
and y components. Remember the x or y component might be the centripetal or center seeking
force.
4. Using Newton’s 2nd Law, sum the forces in the direction of the centripetal force and set them
mv 2
equal to
.
r
Important Concepts
An object that moves in a circle at constant speed, v, is said to experience uniform circular motion (UCM).
The magnitude of the velocity remains constant, but the direction of the velocity is continuously changing,
as shown in the diagram below. Notice that the velocity vectors are not pointing in the same direction. The
velocity vector is always tangent to the circular path.
So far, the only accelerations we have dealt with have been those that change the magnitude of an object’s
velocity, i.e. that make an object speed up or slow down. This type of acceleration is called tangential
acceleration, in which the object speeds up if the acceleration and the velocity are in the same direction.
The object slows down if the acceleration and the velocity are in opposite directions. There is another way
that an acceleration can change an object’s motion: it can change the direction of the object’s velocity
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Circular Motion, Gravity, & Orbits
vector. An object revolving in a circle is continuously accelerating, even when the speed remains constant.
The acceleration vector is always pointing toward the center of the circular path and it is always
perpendicular to the velocity vector, as shown below.
Centripetal (center-seeking) acceleration, ac, is defined as:
ac =
v2
r
If we measure the period, T, the time needed for the object to make a complete revolution, we can calculate
the speed of the object in UCM. During this time, it travels a distance equal to the circumference of the
2π r
circle, 2πr, where r is the radius of the circular path. The object’s speed, then, is represented by v =
.
T
According to Newton’s Second Law (ΣF = ma), an object that is accelerating must have a net force acting
upon it and we call this force, the centripetal force. This force acts in the same direction as the acceleration
which is toward the center of the circle. The equation for centripetal force can be expressed as
⎡ ⎛ 2π r ⎞ 2 ⎤
⎢⎜ 2 ⎟ ⎥
2
⎡ v2 ⎤
T ⎠ ⎥ 4π r
= 2
Fc = ma = mac = m ⎢ ⎥ = m ⎢ ⎝
⎢ r
⎥
T
⎣r ⎦
⎢
⎥
⎣⎢
⎦⎥
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Circular Motion, Gravity, & Orbits
Centripetal force is not a new kind of force, but the net force which points toward the center of the circle. It
may be a single force, a component of a force, or a combination of forces that are center seeking. We are
not talking about a new kind of force, but a new kind of situation in which the same old forces such as, the
gravitational force, frictional force, normal force, the tension in a string, or a combination thereof constrain
the object to follow a curved path. The centripetal force does NO work on an object, since work is defined
as W = Fr cos θ since the angle between the centripetal force and r is 90º and the cosine of 90º=0. In
addition, since there is no change in the speed of the object, by the work kinetic energy theorem, the
energy of the system remains constant unless work is done on the system and hence the work done is zero.
Centripetal force is a term used to identify forces that motivate an object out of straight line motion and
into curved motion. These center seeking forces must exist naturally within a system. Do not confuse the
centripetal force with the centrifugal force which is a fictitious force. Centrifugal force is a term used to
identify forces that must be assumed to exist if one is to use Newton’s Second Law in analyzing problems
from a non-inertial frame of reference. Do NOT use the term, centrifugal force on the AP Exam since it is
a force that does not exist.
Example UCM
1. A sphere on the end of a string is swung in a vertical circle. The mass of the sphere is m and the radius
of the circle is R. The position of the ball is marked every quarter of a revolution in the diagram below.
1
4
R
2
3
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Circular Motion, Gravity, & Orbits
(a) On the diagram below, draw all of the forces acting on the sphere when it is at position 4.
T
W
The only forces acting on the sphere in position 4 are the weight force and the Tension in the string. The
tension provides the centripetal force. Do not include the centripetal force on your FBD since in this case,
the tension is the centripetal force.
(b) When the sphere is at position 4, the tension force in the string is twice as great as the weight of the
sphere. Determine the speed of the sphere at position 4 in terms of the given quantities and any
fundamental constants.
mv 2
T = 2W = 2mg =
R
2
v = 2 gR
v = 2 gR
(c) On the diagrams below, draw and label the forces acting on the sphere
i. at the top of the circle (position 1) and
ii. at the bottom of the circle (position 3).
Position 1
Position 3
T
T
W
W
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Circular Motion, Gravity, & Orbits
(d) Write NSL for the sphere when in position 1. Assume speed is v at position 1.
Note that the centripetal force is the combination of the sum of the tension and the weight of the sphere
which both point toward the center.
mv 2
Fc = ΣF = mac =
r
2
mv
T +W =
R
(e) What is the minimum speed of the sphere at position 1 such that the sphere follows a circular path at
the top.
vmin occurs when T → 0
2
mvmin
r
mv 2
0 + mg = min
r
vmin = gR
T + mg =
(f) Write NSL for the sphere when in position 3. Assume the speed of the sphere is v at the bottom.
Note that the centripetal force is the combination of the difference between the tension and the weight of
the sphere since they point in opposite directions.
Fc = ΣF = mac =
T − mg =
mv 2
r
mv 2
R
(e) Suppose the string breaks just as the sphere is at position 1.
i. Describe the subsequent motion of the sphere.
The sphere will travel tangent to the path and follow a parabolic trajectory.
ii. Assuming the speed at position 1 is vmin , the minimum speed necessary to maintain the circular
path, find the horizontal distance traveled by the sphere using the ground level as being located at
position 1.
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Circular Motion, Gravity, & Orbits
time in the air depends upon vertical displacement
t=
R
2y
2(2 R )
=
=2
g
g
g
horizontal distance depends on horizontal velocity
⎛ R⎞
gR 2
2
gR ⎜ 2
=
= 2R
⎟
g
⎝ g⎠
If the force causing the object to move in a circle magically disappears as it did here, then remember the
problem becomes a 2 D projectile problem with the initial velocity being the tangential velocity of the
object which was undergoing.
x = vxt =
(
)
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Circular Motion, Gravity, & Orbits
Free Response
+y
θ
θ
1. (15 points)
A student swings a stopper of mass 0.40 kg in a horizontal circle (parallel to the floor), as
shown in the figure above. The string has a length L = 0.50 m, and is swung so that it
remains at an angle of θ = 60° from the vertical. A constant tension in the string is
maintained as the stopper is swung in a circular path at this angle.
A. On the diagram above, draw all of the forces acting on the stopper as it swings in
a circular path at this angle.
B. Determine the centripetal force on the stopper which causes it to travel in a
circular path.
C. Determine the linear speed v necessary to keep the stopper revolving in this
circular path.
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Circular Motion, Gravity, & Orbits
D. Determine the period (time for one complete revolution) of the stopper as it is
swung in this circular path.
E. Is it possible to keep the stopper whirling in a perfectly horizontal circle such that
the stopper does not droop and θ=90º from the vertical as shown in the diagram
below? Check Yes or No and explain your answer.
θ
Yes ____
No ____
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Circular Motion, Gravity, & Orbits
FREE RESPONSE
Question 2 (10 pts)
m
v
M
r
A satellite of mass m is moving in a circular orbit with linear speed v, around a planet of
mass M, orbiting at a particular distance r from the center of the planet.
A. Determine the radius of revolution, r of the satellite, in terms of the given
quantities and any fundamental constants?
B. What is the relationship between the radius of revolution and the mass of the
satellite which is undergoing uniform circular motion? Justify your response.
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Circular Motion, Gravity, & Orbits
C. Assume the acceleration due to gravity g at a distance r from the center of the planet
of mass M is 9 m/s2. In terms of the radius of revolution r, what would the speed of the
satellite have to be to remain in a circular orbit around this planet at this distance?
Multiple Choice
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Circular Motion, Gravity, & Orbits
I
II
III
Questions 1 – 4 A ball on the end of a string is being swung in a vertical circle, rotating
clockwise as shown above.
1. Which of the following vectors represents the centripetal force acting on the ball at position I?
A)
B)
C)
D)
E)
2. If the string were suddenly cut when the ball is at position II shown in the figure above, the
subsequent motion of the ball would be
A)
B)
C)
D)
E)
to move to the right.
to move to the left.
to move to the top of the page.
to move down and to the right
to move up and to the left.
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Circular Motion, Gravity, & Orbits
3. The tension in the string is
I. greater at position I than at position III.
II. greatest at position II.
III. greater at position III than at position I.
A)
B)
C)
D)
E)
I only
II only
III only
II and III only
I and II only
4. The ball has a mass m and a speed v as it moves around the vertical circle of radius r.
Which of the following expressions can be used to find the minimum speed of the ball at
position I such that the circular path is maintained?
(A) v = gr
(B) v = gr
(C) v =
(D) v =
(E) v =
(T − mg ) r
m
(T + mg ) r
m
(T − mg ) r
m
5. A 30 kg child sits on the edge of a merry go round at a radius of 6 m. The tangential speed of
the child is 12 m/s. The work done by the centripetal force during one complete revolution is
A)
B)
C)
D)
E)
0J
1440 J
5760 J
18086 J
36173 J
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Circular Motion, Gravity, & Orbits
6. An object revolves in uniform circular motion. Which of the graphs below best represents the
centripetal force on the vertical axis vs. the speed of the object on the horizontal axis?
(A)
(B)
(C)
(D)
(E)
7. The larger the orbital radius of a satellite
I. the greater the gravitational force acting on it.
II. the larger its acceleration.
III. the longer its orbital period.
A)
B)
C)
D)
E)
I only
II only
III only
II and III only
I, II, and III
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Circular Motion, Gravity, & Orbits
Questions 8 - 10. A planet orbits the sun and is moving from point A to point E as shown in the
figure below.
A
E
B
D
C
8. The speed of the planet is greatest at point
A)
B)
C)
D)
E)
A
B
C
D
E
9. The centripetal force exerted on the planet is at its lowest value at point
A)
B)
C)
D)
E)
A
B
C
D
E
10. The centripetal acceleration of the planet is greatest at point
A)
B)
C)
D)
E)
A
B
C
D
E
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Circular Motion, Gravity, & Orbits
11. A coin of mass m is placed on a vinyl stereo record of radius R and moves at a constant
tangential speed v. If the frictional force between the coin and the vinyl record is at its maximum
value, then which of the following expressions can be used to find the value of the coefficient of
friction between the coin and the vinyl record?
C)
gR
v
gv
μ=
R
μ = vgR
D)
μ=
A)
B)
E)
μ=
v2
gR
mv 2
μ=
R
12. A student holds one end of a 1.5 m string and swings a 1.0 kg ball on the other end of the
string in a horizontal circular path with a constant speed and a centripetal acceleration of 1.5
m/s2. The centripetal force is most nearly
A)
B)
C)
D)
E)
0N
0.7 N
1.5 N
3.0 N
4.5 N
13. A carnival ride has a 2.0 m radius and rotates once each 0.90 s. The centripetal
acceleration of the rider is most nearly
A)
B)
C)
D)
E)
1.8 m/s2
2.5 m/s2
14 m/s2
56 m/s2
98 m/s2
14. A line drawn from the sun to a planet sweeps out equal areas in equal
A)
B)
C)
D)
E)
distances
displacements
speeds
accelerations
times
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