Download Circular motion review packet

Advanced Placement
PHYSICS B
Circular Motion
Student
2013-2014
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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involved in the production of this material.
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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
Tension
FT , T
Applied
FSubscript
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.
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
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Circular Motion, Gravity, & Orbits
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
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
Fc = ma = mac = m ⎢ ⎥ = m ⎢ ⎝
= 2
⎢ 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.
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Circular Motion, Gravity, & Orbits
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
(a) On the diagram below, draw all of the forces acting on the sphere when it is at position 4.
T
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.
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Circular Motion, Gravity, & Orbits
T = 2W = 2mg =
mv 2
R
v 2 = 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
T
W
W
(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.
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Circular Motion, Gravity, & Orbits
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.
time in the air depends upon vertical displacement
t=
2y
2(2 R )
R
=
=2
g
g
g
horizontal distance depends on horizontal velocity
⎛ R⎞
gR 2
gR ⎜ 2
= 2R
⎟=2
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
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1999
PHYSICS B
SECTION II
Time—90 minutes
7 Questions
Directions: Answer all seven questions, which are weighted according to the points indicated. The suggested time
is about 15 minutes for answering each of questions 1-4, and about 10 minutes for answering each of questions 5-7.
The parts within a question may not have equal weight. Show all your work in this booklet in the spaces provided after
each part, NOT in the green insert.
1. (15 points)
The Sojourner rover vehicle shown in the sketch above was used to explore the surface of Mars as part of the
Pathfinder mission in 1997. Use the data in the tables below to answer the questions that follow.
Mars Data
Radius: 0.53 ´ Earth’s radius
Mass: 0.11 ´ Earth’s mass
Sojourner Data
Mass of Sojourner vehicle:
Wheel diameter:
Stored energy available:
Power required for driving
under average conditions:
Land speed:
11.5 kg
0.13 m
5.4 ´ 10 5 J
10 W
6.7 × 10 −3 m/s
(a) Determine the acceleration due to gravity at the surface of Mars in terms of g, the acceleration due to
gravity at the surface of Earth.
(b) Calculate Sojourner’s weight on the surface of Mars.
(c) Assume that when leaving the Pathfinder spacecraft Sojourner rolls down a ramp inclined at 20 ° to the
horizontal. The ramp must be lightweight but strong enough to support Sojourner. Calculate the minimum
normal force that must be supplied by the ramp.
(d) What is the net force on Sojourner as it travels across the Martian surface at constant velocity? Justify
your answer.
(e) Determine the maximum distance that Sojourner can travel on a horizontal Martian surface using its stored
energy.
(f) Suppose that 0.010% of the power for driving is expended against atmospheric drag as Sojourner travels
on the Martian surface. Calculate the magnitude of the drag force.
Page 8
1999 PHYSICS B
. ( 10
t.
27
´
-
C
Disk
0.14 m
5. (10 points)
A coin C of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center, as shown
above. The disk rotates at a constant rate in a counterclockwise direction as seen from above. The coin does
not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.
(a) The figure below shows the disk and coin as viewed from above. Draw and label vectors on the figure
below to show the instantaneous acceleration and linear velocity vectors for the coin when it is at the
position shown.
(b) Determine the linear speed of the coin.
(c) The rate of rotation of the disk is gradually increased. The coefficient of static friction between the coin
and the disk is 0.50. Determine the linear speed of the coin when it just begins to slip.
(d) If the experiment in part (c) were repeated with a second, identical coin glued to the top of the first coin,
how would this affect the answer to part (c) ? Explain your reasoning.
Page 9
2001 AP® PHYSICS B FREE-RESPONSE QUESTIONS
PHYSICS B
SECTION II
Time—90 minutes
7 Questions
Directions: Answer all seven questions, which are weighted according to the points indicated. The suggested time
is about 15 minutes for answering each of questions 1-4, and about 10 minutes for answering each of questions 5-7.
The parts within a question may not have equal weight. Show all your work in the pink booklet in the spaces
provided after each part, NOT in this green insert.
1. (15 points)
A ball of mass M is attached to a string of length R and negligible mass. The ball moves clockwise in a vertical
circle, as shown above. When the ball is at point P, the string is horizontal. Point Q is at the bottom of the circle
and point Z is at the top of the circle. Air resistance is negligible. Express all algebraic answers in terms of the
given quantities and fundamental constants.
(a) On the figures below, draw and label all the forces exerted on the ball when it is at points P and Q,
respectively.
(b) Derive an expression for u min , the minimum speed the ball can have at point Z without leaving the circular
path.
(c) The maximum tension the string can have without breaking is Tmax . Derive an expression for u max , the
maximum speed the ball can have at point Q without breaking the string.
(d) Suppose that the string breaks at the instant the ball is at point P. Describe the motion of the ball
immediately after the string breaks.
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Page 10
2004 AP® PHYSICS B FREE-RESPONSE QUESTIONS (Form B)
PHYSICS B
SECTION II
Time—90 minutes
6 Questions
Directions: Answer all six questions, which are weighted according to the points indicated. The suggested time is
about 17 minutes for answering each of questions 1-4, and about 11 minutes for answering each of questions 5-6.
The parts within a question may not have equal weight. Show all your work in the goldenrod booklet in the spaces
provided after each part, NOT in this lavender insert.
1. (15 points)
A designer is working on a new roller coaster, and she begins by making a scale model. On this model, a car of
total mass 0.50 kg moves with negligible friction along the track shown in the figure above. The car is given an
initial speed u0 = 1.5 m s at the top of the first hill of height 2.0 m. Point A is located at a height of 1.9 m at
the top of the second hill, the upper part of which is a circular arc of radius 0.95 m.
(a) Calculate the speed of the car at point A.
(b) On the figure of the car below, draw and label vectors to represent the forces on the car at point A.
(c) Calculate the magnitude of the force of the track on the car at point A.
(d) In order to stop the car at point A, some friction must be introduced. Calculate the work that must be done
by the friction force in order to stop the car at point A.
(e) Explain how to modify the track design to cause the car to lose contact with the track at point A before
descending down the track. Justify your answer.
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