Laying the foundation-Circular motion€¦ · AP* PHYSICS B Circular Motion, Gravity, & Orbits...

AP* PHYSICS B

Circular Motion, Gravity, & Orbits

Teacher Packet

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Objective To review the student on the concepts, processes and problem solving strategies necessary to successfully answer questions on circular motion, gravity, and orbits. Standards Circular motion, gravity, and orbits are addressed in the topic outline of the College Board AP* Physics Course Description Guide as described below. AP Physics Exam Connections Topics relating to circular motion, gravity, and orbits are tested every year on the multiple choice and in most years on the free response portion of the exam. The list below identifies free response questions that have been previously asked over circular motion, gravity, and orbits. These questions are available from the College Board and can be downloaded free of charge from AP Central. http://apcentral.collegeboard.com. Free Response Questions

2004 Question 1 2004 Form B Question 1 2001 Question 1 2002 Form B Question 2 1999 Question 5 1999 Question 1 (a)

I. Newtonian Mechanics E. Circular motion and rotation

1. Uniform circular motion 2. Torque and rotational statics

F. Oscillations and Gravitation 4. Newton’s law of gravity 5. Orbits of planets and satellites (circular)

http://apcentral.collegeboard.com/




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

2

where =velocity or speed of an object in UCM or orbiting satellite

centripetal acceleration 2 =2 r

c

c

va vr

arv

Tπ π

=

=

= f =radius of circular motion or radius of orbit

=period of revolution

r

T

2

2

2

c 2

2-11

2

=frequency

centripetal force

24F =mass

Nm =6.67x10 =kg

c c c

fmvF ma F

rrm

mrT mr T

GMv Gr

ππ

= = =

⎛ ⎞⎜ ⎟⎝ ⎠= =

= universal gravitational constant

=mass of object being orbited M



Centripetal Force is NOT a new force, but one or more of these normally occurring forces or their components

gF mg= Always directed toward the Earth’s center. Force on a free falling body, if we neglect air friction.

Weight gF , W

1 22G

m mF Gr

=

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.

GF Gravitational

NF , N

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.

Normal

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).

maxf s sF NFμ≤ (static)

(kinetic)f k k NF Fμ=

WARNING! The two quantities sf and kf 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.

F , f Friction f

TF is a force that is applied to a body by a rope, string, or cable.

TF is applied along the line of the string and away from the body in question.

, T Tension TF

SubscriptF 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.

Applied



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

equal to 2mv

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



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:

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

circle, 2πr, where r is the radius of the circular path. The object’s speed, then, is represented by 2 rvTπ

= .

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

2

cvar

=

2

22

2

24

c c

rv rTF ma ma m mr r

π

Tπ

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎡ ⎤ ⎝ ⎠⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

2



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 cosW Fr 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.

2

1

4

3

R



(a) On the diagram below, draw all of the forces acting on the sphere when it is at position 4.

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.

2

2 2 mvT W mgR

= = =

2 2

2

v gR

v g

=

= R

(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

W

T

W T

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.

2

2

c cmvF F ma

rmvT W

R

= Σ = =

+ =

(e) What is the minimum speed of the sphere at position 1 such that the sphere follows a circular path at the top.

min2min

2min

min

occurs when 0

0

v T

mvT mgr

mvmgr

v gR

→

+ =

+ =

=

(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.

2

2

c cmvF F ma

rmvT mg

R

= Σ = =

− =

(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 , 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.

minv



( )2

time in the air depends upon vertical displacement

2 2(2 ) 2

horizontal distance depends on horizontal velocity

2 2 2x

y R Rtg g g

R gRx v t gR Rg 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.



Free Response Question 1 (15 pts)

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 below, draw all of the forces acting on the stopper as it swings in a circular path at this angle. Teaching Note: students should not draw components on their FBD as the latest rubrics count the components as extraneous vectors.

2 points One point for each correctly drawn and labeled force vector (tension and weight).

1 point for no extraneous forces, such as the centripetal force or even the x and y components of the tension.

B. Determine the centripetal force on the stopper which causes it to travel in a circular path.

( ) 2

0

m0.040 kg 10 0.40 Ns

y

y WT F mg

=

⎛ ⎞= = = =⎜ ⎟⎝ ⎠

∑F

( ) ( )tan 0.40 N tan 60 0.69 Nx

x y

m

T T θ

=

= = ° =∑F a

1 point For indicating that the sum of the vertical components of the forces is zero 1 point For indicating that the sum of the horizontal components of the forces is not zero 1 point For determining that the centripetal force acting on the stopper is the x component of the tension 1 point For the correct answer with units

+y θ

T

FW



C. Determine the linear speed v necessary to keep the stopper revolving in this circular path.

1 point For setting the x component of the tension in the string equal to the centripetal force

( ) ( )( ) ( )

2

2

,

where the radius sin0.50 m sin 60 0.43 m

0.69 N 0.43 m m2.7 0.040 kg s

x

x

x

mvmR

mvTR

R LR

T Rvm

θ

= =

=

=

= ° =

= = =

∑F a

1 point For attempting to find R and not using the length of the string 1 point For substituting the mass of the stopper in the equation for centripetal force 1 point For the correct answer with units consistent with the earlier answer for the centripetal force

D. Determine the period (time for one complete revolution) of the stopper as it is swung in this circular path.

1 point For a correct equation for the period of revolution

( )

2 =

2 0.43m1.0 sm2.7

s

rTv

T

π

π= =

1 point For the correct answer with units consistent with the earlier answer for speed and radius

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 __x__ It is not possible to keep the mass revolving in a horizontal circle with θ=90º, since the force of gravitational attraction (FW) must pull the stopper down to an equilibrium angle. The stopper could only revolve in a horizontal circle if it had no mass and no weight to pull the string down away from the horizontal. As θ approaches 90º the string would break as the tension in the string approaches infinity.

1 point For checking NO 1 point For a correct explanation involving the weight force and the tension in the string



Question 2 (10 pts)

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.

v

r

M

A. Determine the radius of revolution, r of the satellite, in terms of the given quantities and any fundamental constants?

1 point For indicating that the gravitational force plays the role of the centripetal force

2

2

2

2

c GF F

mv GmMr r

mrv GmMGMrv

=

=

=

=

1 point For a correct application of Newton’s law of Universal Gravitation 1 point For a correct application of

2satellitem v

r

1 point For a correct expression for the radius of revolution in terms of the given quantities and the fundamental constant G



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.

1 point For indicating that there is no relationship between the radius and the mass of the satellite

Radius of revolution is independent of the mass of the satellite.

1 point For a proper justification, i.e. the mass of the satellite divides out of both sides of the equation

Radius of revolution depends upon the mass of the planet the satellite is orbiting, not the mass of the satellite.

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?

1 point For indicating that the weight force plays the role of the centripetal force

w cF F= 2

2

mvmgr

gr v

=

=

1 point For indicating that the mass of the satellite divides out of the equation 9 3v gr r= = = r

1 point For substituting 9 m/s2 for g 1 point For the correct expression in terms of the radius of revolution, r



Multiple Choice

I

III

II

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)

D Tension provides the centripetal force that causes the ball to follow a circular path.

The velocity vector is always tangent to the path and in the direction of motion, and the centripetal force and acceleration are toward the center of the circle.

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) to move to the right. B) to move to the left. C) to move to the top of the page. D) to move down and to the right E) to move up and to the left.



C Uniform Circular Motion At the instant the ball is in the position shown,

it is moving tangent to the path, so up toward the top of the page

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) I only B) II only C) III only D) II and III only E) I and II only

C Tension provides the centripetal force that causes the ball to follow a circular path.

At position I, the string goes slack due to the weight of the ball pulling downward. At position II, the weight of the ball pulls downward, but the tension in the string pulls toward the center. At position III, the weight is directed downward, but the tension is directed upward, causing the string to be tighter than it is at I or II.

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 g= r (B) v g r=

( )T mg rv

m−

=(C)

( )T mg rv

m+

=(D)

( )T mg r

vm−

=(E)



A At top of circle the tension approaches zero at minimum velocity

2

2

0 at minimum velocity at top

mvT mgr

Tmvmg

rv gr

+ =

→

=

=

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) 0 J B) 1440 J C) 5760 J D) 18086 J E) 36173 J

cosW FdW KE

θ== Δ

A Centripetal forces do no work

Angle between F and d is 0 degrees. Energy remains constant unless there is work done on the system.



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)

C Centripetal force is directly proportional to the square of

the velocity

2

cF mvr

=



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) I only B) II only C) III only D) II and III only E) I, II, and III

C Kepler’s third law According to Kepler’s third law, the period of revolution squared is proportional to the average orbital radius cubed, causing a larger orbit to have a longer period.

Questions 8 - 10. A planet orbits the sun and is moving from point A to point E as shown in the figure below.

A

B

C D

E



8. The speed of the planet is greatest at point A) A B) B C) C D) D E) E Speed of a satellite E The speed is greatest when the planet is

closest to the sun.

9. The centripetal force exerted on the planet is at its lowest value at point A) A B) B C) C D) D E) E

A The speed is least at point A, r is the greatest, so the centripetal force exerted on the planet is at its lowest value at point A.

2

cF mvr

=

10. The centripetal acceleration of the planet is greatest at point A) A B) B C) C D) D E) E

E The acceleration of the planet is greatest at the point where the gravitational force between the sun and the planet, point E.

2

cvar

=



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?

vgR

=μ A)

Rgv

=μB)

vgR=μ C)

gRv 2

=μD)

Rmv2

=μE)

D Frictional force plays the role

of the centripetal force RmvFf N

2

== μ mgFN =, and , so

Rmvmg

2

=μgRv 2

=μ, and

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) 0 N B) 0.7 N C) 1.5 N D) 3.0 N E) 4.5 N

C

( )

2

21.0 1.5 1.5 N

c c

c

mvF mar

mF kgs

= =

⎛ ⎞= =⎜ ⎟⎝ ⎠

2

cF mvr

=


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) 1.8 m/s2 B) 2.5 m/s2 C) 14 m/s2 D) 56 m/s2 E) 98 m/s2

0.90

2 2 (2.0 m) m140.90 s s

T sd rvt T

π π=

= = = =

E Period to find speed, then find centripetal acceleration

2

2

2

m14ms 98

2.0 m scvar

⎛ ⎞⎜ ⎟⎝ ⎠= = =

14. A line drawn from the sun to a planet sweeps out equal areas in equal A) distances B) displacements C) speeds D) accelerations E) times

E Kepler’s second law This is a statement of Kepler’s second law of planetary motion and implies that a planet travels at a higher speed when it is closer to the sun.


Laying the foundation-Circular motion€¦ · AP* PHYSICS B Circular Motion, Gravity, & Orbits...

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Transcript of Laying the foundation-Circular motion€¦ · AP* PHYSICS B Circular Motion, Gravity, & Orbits...