AP Physics C

Handbook I

Introduction to Mechanics

Chapter Topic Pages

1. Mechanics Review from year one 2 - 11

2. Momentum in two dimensions 12 - 17

3. Torque 18 - 24

4. Rotational Motion

5. Moment of inertia 29 - 31

6. Dynamics with rolling objects 32 - 39

7. Center of Mass 41 - 45

8. Finding moments of inertia using calculus 46 - 50

9. Parallel Axis Theorem 51 - 52

10. Gravitational attraction 53 - 56

11. Energy and rotation 57 – 62

12. Angular Momentum 63 - 69

13. Gravitational PE on a planetary level 70 - 72

14. Elliptical Orbits 73 – 74

15. Cumulative review problems 75 - 82

16. Equations sheets 83 - 84

Springs and Simple Harmonic Motion will be covered inBook II Electricity and Magnetism

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Chapter 1 – General Understandings from first year physics

Basic terms, concepts, and problem solving techniques that you should have learned include:

Inertial mass (m) is an object’s resistance to change in motion, or conversely, its desire to continue in its current motion. Inertia is a scalar. It has magnitude but not direction. In the MKS system it is measured in Kilograms

The basic law of inertia as stated by Newton: Objects in uniform motion want to stay in uniform motion forever; unless acted upon by a net unbalanced force.

Gravitational mass (m) an object’s ability to attract other matter and its ability to curve space. The gravitational attraction created by mass is experienced as gravitational force. This force is equal on both masses but opposite in direction. G is a constant 6.67 E-11 in the MKS system

Fg = G m1 m2 / r2

Note: It has never been proved that inertial mass and gravitational mass are equivalent. On the positive side, there has never been a case where the two were not equivalent.

Force (F) The amount of push or pull exerted on matter. It is important to note that force is a vector. It has both magnitude and direction.

The basic law of force as stated by Newton: Force causes objects to accelerate

Fnet = maFfrict = Fnormal

Fslope = m g sin Fg = mg

Fg = G m1 m2 / r2

2

Kinematics: the study of how things move.

Acceleration (a) is a measure of how fast an object speeds up, slows down, or changes direction. It is measured in meters/sec2. In general it is given by

a = v / t

For circular motion it is called acceleration centripetal and is given by the equations

ac = v2 / r

ac = 4 2 R / T2

ac = 4 2 R f2

Basic Kinematics equations (note that both Vickie and Dickie require constant acceleration)

Vickie vf2 = vo

2 + 2 a (d)

Dickie df = ½ a t2 + vo t + do

Arthur a = v / t

Newtonian reference frame: For Newton motion in the universe is all relative. There is no prime location in the universe. Each frame of reference works as well as all others. The only thing that really mattered is that for Newton’s equations to work out you needed to be in a non-accelerated frame of reference.

dobserved = din reference frame + dof reference frame

vobserved = vin reference frame + vof reference frame

Notice: if you are in an inertial (non-accelerated) frame of reference, all of the kinematics and dynamics equations will work and give working answers in that frame and in any other observing inertial frame of reference.

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Momentum and Force: First Universal method of studying the world.

Momentum () which is an inertial measure of an object’s current motion = mv is the basic equation for the determination of momentum. Unlike inertia, momentum is a vector. When working with momentum you must include direction. In a closed system momentum is always conserved. Total momentum in a closed system remains constant.

Momentum before = Momentum after1o + 2o + . . . = 1f + 2f + . . .

= mv so

m1v1o + m2v2o + . . . = m1v1f + m2v2f + . . .

Impulse (J) is the measure of the strength of a mechanical interaction between objects. It is seen in the momentum change in objects that the interaction produces J = (mv) It is also expressed in terms of the force of the interaction J = F t

Impulse = Impulse

(mv) = F t

It is important to remember that momentum and impulse are vector quantities. You have to keep track of positive an negative directions.

4

Energy and Force: the second major way of studying the world

Energy (E) is an alternative way of measuring the mechanical world. Energy comes in many forms and it is measured by its ability to do workwith the assumption being that in a closed system the total energy will be constant. The forms of energy that we used were

Kinetic energy KE = ½ m v2 Gravitational Potential energy PEg = mgh

Conservation of Energy

Energy Before = Energy After

Work (W) is a measure of how much energy has changed form or has entered or left a system. Mechanically, Work is found in terms of force W = F d. Most commonly when energy leaves a system it is called Wout and it is usually in terms of heat generated by friction.

W = F d

Power (P) is the rate at which energy changes form, enters, or leaves a systemIt is measured in Watts.

P = E / t

In situations where velocity is constant we can use

P = F v

The English units for power are Horsepower. One horsepower is equal to 745 Watts.

5

Sample review problems. Show your work

A 240 kg block initially stationary is acted upon by a 400 Newton force find the following after 6 seconds has passed:

1. the acceleration of the block.

2. the displacement of the block.

3. the velocity of the block.

4. the KE of the block.

5. the momentum of the block.

6. the impulse on the block.

6

A 24 kg block moving to the right with a speed of 12 m/s collides with and sticks to a 16 kg block moving to the left with a speed of 30 m/sec.

7. What is the total kinetic energy of the objects in the system before the collision?

8. What is the net momentum of the blocks in the system before the collision?

9. What is the final speed of the 24 kg after the collision?

10. What is the total kinetic energy of the objects in the system after the collision?

11. What is the net momentum of the objects in the system after the collision?

12. What is the impulse on the 24 kg object due to the collision?

13. What is the impulse on the 16 kg object due to the collision?

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A 1200 kg car is driving up a 4 degree slope at a speed of 30 meters/second. Answer the following:

14. What is the minimum coefficient of friction needed to maintain this speed?

15. How much power must the engine produce to maintain this speed?

16. What average force must the engine be producing to maintain this speed?

A 2 kg rock is thrown straight upwards from the top of a 50 meter cliff with an initial speed of 20 m/sec. The rock ends up at the ground at the bottom of the cliff, answer the following:

17. The rock’s initial KE is.

18. The rock’s maximum elevation above the bottom of the cliff is.

19. The rock’s impact velocity at the bottom of the cliff.

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20. The rock was in the air for how many seconds.

21. When the rock hits the ground, find the magnitude of the impulse it experiences.

22. If the final collision lasts 0.05 seconds, what is the magnitude of the average force of impact with the ground?

A 70 kg coasting bicycle enters a loop the loop with an initial speed of 14 m/sec. The loop the loop has a radius of 3 meters. Find the following:

23. The initial kinetic energy of the bicycle

24. The initial centripetal acceleration of the bicyclist.

25. The true weight of the bicycle

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26. The apparent weight of the bicycle at the bottom of the loop

27. The speed of the bicycle at the top of the loop

28. The centripetal acceleration of the bicycle at the top of the loop

29. The apparent weight of the bicycle at the top of the loop.

A 5 kg block is resting on a horizontal surface. The coefficient of friction between the block and the surface is 0.7

30. Find the maximum force of friction the surface can exert on the block

31. If you push down on the block with a force of 35 Newtons, now what is the maximum force the surface can exert?

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Simple concept problems

1. You are fired upwards with an initial speed of 10 m/sec, what will your acceleration be at the highest point of your projectile motion?

2. You are fired at an angle of 60 degrees with respect to the horizontal with an initial speed of 10 m/sec. What will your velocity be at the highest point of your projectile motion?

3. Which exerts a greater force on the other. The Earth’s gravity on the Moon or the Moon’s gravity on the Earth?

4. A bee flies between two bicyclists as shown below. How many Kilometers did the

bee travel before the two bicyclists collide?

5. In which case above, will block A move with a greater acceleration?

6. In the picture above, the blocks have a weight of 100 Newtons each. What is the

tension in the rope connecting them together?

7. In the problem above, Block A is initially moving downward and Block B is initially moving upwards with a speed of 5 m/sec. How does this effect the tension in the rope?

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Chapter 2 - Momentum and collisions.

We will be looking at collision from a graphical point of view. Below is a picture taken stroboscopically of a 0.0457 kg golf ball colliding with and sticking to a clay blob.

If the pictures are taken every 60th of a second answer the following using a cm rular.1. What was the initial speed of the golf ball?

2. What was the initial momentum of the golf ball?

3. What is the final speed of the ball blob combination?

4. What is the mass of the Blob?

5. What is the impulse on the blob by the golf ball?

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Expanding collisions into two dimensions is really quite easy. It is essentially the same as we have done in previous one dimensional situations. Only this time we:

1. Break the motions up into perpendicular components. (usually vertical and horizontal)

2. Realize that momentum in each direction must be conserved. This means that you will do Before = After conservation of momentum equations in each direction.

3. Recombine the final perpendicular components to determine the resultant speed and direction for each object.

Let us take the example given in the picture below. Assume that the strobe takes 60 frames per second. The silver ball has a mass of 0.2 kg. Using a cm stick to make

measurements.

1. What is the initial vertical velocity of the silver ball?

2. What is the final angular speed of the silver ball?

3. What is the final angle (relative to horizontal) of the silver ball’s motion?

4. What is the horizontal magnitude of the silver ball’s velocity?

5. What is the vertical magnitude of the silver ball’s velocity?

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6. What is the final angular speed of the black ball?

7. What is the angle of the black ball’s resultant motion?

8. What is the horizontal magnitude of the black ball’s velocity?

9. What is the vertical magnitude of the black ball’s velocity?

10. Using conservation of momentum in the vertical direction find the mass of the black ball?

11. Using conservation of momentum in the horizontal direction, find the mass of the black ball?

12. Compare your answers. To what extent are they in agreement?

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Here is another stroboscopic picture of collisions between two objects.. Assume the flash rate is 60 frames per second and that the bigger ball has a mass of 0.2 kg.

What are the following:Object Net speed off vertical Vert.

VelocityHoriz.Velocity

Big ball before

Small ball before

Big ball after

Small ball after

Using vertical components only, find the mass of the small ball.

Using horizontal components only find the mass of the small ball.

The results from which direction should you trust more, mass determined by vertical speed or horiz speed?

Here is one last collision. It is between two frictionless pucks that have spring loaded bumpers. Lets see how well you understand the pictures and

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the physics behind them. Both objects (x) and (o) are the same mass; 0.4 kilograms each. The flashes are at 0.5 second intervals.

1. Can you tell if this collision is elastic or not? Give proof.

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2. What is the approximate duration of the collision? Explain.

3. What is the magnitude of the average collision force acting between the two pucks?

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Chapter 3: Torque

Torques are twisting forces. Let us take the example of a door. If you push across the door at a point close to the hinge it will start to twist (that is assuming that the door is not closed). Push in the same direction but further out on the door it is easier to move the door. The farther out you apply the force the more easily it turns.

The distance from the hinge that the force is applied isn’t the only thing that has a bearing of the amount of twisting force. The angle at which the force is applied is also important.

Torque is given by the following equation:

τ = r x F which is the same as τ = r F sin (θ)

Applying torques to structures

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Basic assumption for structures. Successful structures remain stationary (static). To remain stationary their acceleration must be zero. If it didn’t the only natural acceleration they would experience would be downwards. That is unacceptable. To have no acceleration means that the sum of all forces working on structures must add up to zero.

Fnet = m afor successful structures

Fnet = 0

It turns out that Newton’s second law of dynamics applies to twisting forces as well. Force is replaced by torque. Mass is replaced by moment of inertia. And linear acceleration (a) is replaced by angular acceleration ()

net = I is the angular acceleration measured in radians/sec2

I is the moment of inertia (rotational inertia) measured in Kg m2

For successful structures the sum of all twisting forces must be zero as well.

τnet = 0

τ1 + τ2 + τ3 + . . . = 0

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Sample problem 1: The pivot point for a load support arm

You have a 100 Kg ball hanging from a 5 meter long horizontal beam that is projecting from the side of a building and we want to know how much tension there will be in the support rope. We will start by assuming that the beam’s weight is negligible. That being the case, then there are two torques that are acting upon the beam that are causing it to want to rotate around its pivot point. The ball is a applying a clockwise acting torque. The rope is reacting to it and applying a counter-clockwise torque. For the beam to remain stationary the net torque must be zero.

τnet = τrope + τball

0 = r x Frope + r x Fg

0 = ( ) (Frope) sin ( ) + ( ) ( ) sin ( )

Frope = _________ (app. 2000 N)

On the back side of this page find the tension on the ropes by finding the torques acting on the following beams. Show all of your work.

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Sample Problem 2: A triangular bridge face

Let us revisit the simple triangular truss made out of ultra light carbon fiber rods. There is a 98 N cat on top of the truss. You wanted to find the supporting force that side A must supply to keep the cat there. To do this we will select our pivot point (hinge) to be at the bottom of rod B. There are two forces acting on rod B that would cause it to rotate around the pivot point.. One of these torques is supplied by side A the other by the cat. Rod A’s torque is clockwise in direction (+) and the torque due to the cat’s weight is counter-clockwise (-). You will notice that it is not necessary to know the exact length of rod A or B in order to solve this problem.

τnet = τA + τcat

0 = r x FA + r x Fcat

0 = ( ) (FA) sin ( ) + ( ) ( ) sin ( )

FA = _________ (app. 60 N)

On the back side of this page find the supporting forces for structures A and B in the picture below. Show all of your work.

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Sample Problem 3: Including the mass of the beam

In real life beams have mass. This mass acts as a torque on the beam causing it to want to rotate. To simplify the problem we assume that all of the beam’s mass is at its midpoint (we will prove this later when we do center of mass.) Lets revisit a problem from the last page. This time we take into account that the beam has a mass of 40 kg.

(We do not need to know the beam’s length. Just assume it has a length “r”)

τnet = τball + τbar + τrope

0 = r x Fball + 0.5 r x Fbar + r x Frope

0 = r sin( ) ( ) + 0.5 r sin( ) ( ) + r sin ( ) F rope

Frope = ________________

Sample Problem 4: A ladder resting against a wall

We have a 4 meter long, 12 kg ladder resting against a wall at an angle of 70 degrees. First, find how hard it is pushing against the wall.

Label pivot point and torques Show your work here

_

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Sample Problem 5: A ladder resting against a wall. Find minimum to support it.

We have the same 4 meter long, 12 kg ladder resting against a wall at an angle of 70 degrees. Now we want to find the minimum coefficient of friction needed to support the ladder. This time we will move the pivot point to the top of the ladder.

Label pivot point and torques Show your work here

Sample Problem 6: Now a 100 kg person is 3 meters up the ladder.

Does adding a person to the ladder change the minimum needed coefficient of friction? Repeat the previous problem with one additional torque due to the person.

Label pivot point and torques Show your work here.

Sample Problem 7: A small crane.23

A 30 kg 10 meter long beam pivots around a point that is 2 meters from one end. A 30 kg blob hangs from one end of the beam. You have a choice of supporting the beam by a single rope at either point A, B, or C. At which rope must be strongest? By what percent more strong must it be there compared to the other locations.

Fa = ____________________ Fb = _______________________ Fc = ______________________

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Chapter 4: Rotational Motion

Rotational motion is just an extension of what we have learned when doing linear motions. Just as there are linear displacements, velocities and accelerations, we have rotational displacements, velocities and accelerations. New for many students is the radian circle. We will be measuring angles in terms of the radius of the circle that we will be rotating.

Linear Motion vs Rotational Motion

d (meters) Displacement θ (Radians)

v (meters/sec) Velocity ω (Radians/sec)

a (meters/sec2) Acceleration α (Radians/sec2)

360 degrees = 2 π radians or One radian = 180/π degrees

Kinematics equations are surprisingly similar for rotational motion

Linear motion Rotational motion

df = 1/2 a t2 + vo t + do θf = 1/2 α t2 + ωo t + θo

vf 2 = vo2 + 2 a Δ d ωf 2 = ωo

2 + 2 α Δ θ

a = Δ v / Δ t α = Δ ω / Δ t

There is a huge advantage if you are using radians for angle measure. Converting from rotational units to linear units is quick and easy.

d = r v = r a = r

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As a matter of review, let us do some simple linear motion problems first. Imagine that a ball falls from rest with an acceleration of 9.8 m/sec2.

1. How far does it fall in 10 seconds?

Equation _____________________________________ Answer ______________

2. After traveling 100 meters, how fast is it going?

Equation _____________________________________ Answer ______________

3. How many seconds does it take to reach 100 m/sec?

Equation _____________________________________ Answer ______________

Let us do some simple rotational motion problems. Imagine that a motorized gyroscope speeds up from rest with an acceleration of 2 radians/sec2.

4. How many radians does it turn in 10 seconds?

Equation _____________________________________ Answer ______________

5. After turning 100 radians, how fast is it going?

Equation _____________________________________ Answer ______________

6. How many seconds does it take to reach 100 radians/sec?

Equation _____________________________________ Answer ______________

If the gyroscope has a radius of 0.2 meters we can find the linear distances, and speeds of the outside edge

7. How far has the edge traveled in 10 seconds?

Equation _____________________________________ Answer ______________

8. After turning 100 radians, how fast is it going in m/sec?

Equation _____________________________________ Answer ______________

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It is important to have a rough idea of what angular velocity in radians per second actually represents. Let us find the angular speed in radians per second for several real life situations.

1. A jet turbine spins at 20,000 rpm. ______________________

2. A second hand on a clock turns once a minute ______________________

3. The moon goes around the Earth once every 26.6 days. ______________________

4. The Earth spins on its axis once every 24 hours. ______________________

5. The Earth goes around the Sun once every 365.25 days. ______________________

6. A bicycle has a 25 inch dia. wheels (63.5 cm). If the bicycle is traveling at 20 mph (10 m/sec) how fast is the wheel spinning?

______________________7. A car has a 15 inch dia. wheels (38 cm) if the car is traveling

down the highway at 30 m/sec how fast is the wheel spinning? ______________________

Lets look at the linear edge speed of some of the subjects in the previous problems.

1. The turbine has a radius of 20 cm. ______________________

2. The second had of the clock has a radius of 20 cm. ______________________

3. The Earth Moon distance is 3.75E8 meters. ______________________

4. The radius of the earth is 6.38E6 meters at the equator. ______________________

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Now lets look at situations where there is constant angular acceleration. Remember the kinematics equations for rotation.

θf = 1/2 tα 2 + ωo t + θo ωf2 = ωo

2 + 2 α Δ θ = / tα Δω Δ

1. A car goes from 0 to 30 m/sec in 5 seconds. If its wheels have a diameter of 38 cm, What is the angular acceleration of the wheels?

Equation

Substitutions Answer ______________________

2. A computer hard drive has a running speed of 5200 rpm and a diameter of 5 cm. If it can reach this speed in 3 seconds find the following:

Its operational angular speed,

Equation

Substitutions Answer _____________________

Its operational linear speed at the outside edge.

Equation

Substitutions Answer _____________________

Its angular acceleration while speeding up.

Equation

Substitutions Answer _____________________

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Chapter 5: Moment of Inertia Linear motion Rotational motion

Newton’s second law Fnet = m a net= I

There is a substantial difference between linear and rotational inertia. Linear inertia is just a measure of how much stuff you have. It does not change if the stuff is stretched of bent.

Rotational inertia is determined by both how much stuff you have and its shape. Interestingly, it turns out that the shape is actually much more important than how much stuff you have.

For a simple explanation let us look at a very simple case where a mass is welded on to the end of a very long (r) massless bar. The other end of the massless bar is attached to a frictionless bearing. If we apply a constant force (torque) to the mass that is always perpendicular to the bar, the system will spin faster and faster.

From a distance the mass moves in circular motion. And must obey

= r x F and

= I Because r is very large when viewed from up close to the mass, the mass appears to be moving in almost a straight line and must obey

F = m a

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Newton’s 2nd law for rotational motion

Substituting via a = r a

Substituting = r

Substituting from F = ma

Notice that the object’s location is

vastly more important than its

mass. A mass at the rotational

point has no moment of inertia. A

mass located one meter away from

the rotational point has a certain amount of rotational inertia but the same mass that is two meters from the rotational point has four times

as much rotational inertia.

We will be using this relationship for particles (I = m r2) when we develop the equations for the moments of inertia for common shapes such as hoops, spheres, disks, bars, etc.

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The moment of inertia for any body can be found by adding up the moments of all of its mass components.

I = mi ri2

In general, this can be a laborious task. However, we will be working with several shapes for which the moments of inertia are easily determined and well known.

a thin loop a disk a solid sphere

I = M R2 I = 1/2 M R2 I = 2/5 M R2

a bar rotating a bar rotating around its center around its end

I = 1/12 M R2 I = 1/3 M R2

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Chapter 6: Rotational Dynamics

1. Pulling on a Disk

Suppose we have a metal disk (M = 5 kg) with a thin ribbon wrapped around it. We will pull the disk forward by pulling on the ribbon (Fribbon = 10 N).

The disk will roll and not skid. This requires enough Ffriction . We do not know for certain the direction of the force of friction so we pick a direction.

It is important that neither the laws of linear motion nor the laws governing rotational motion be violated. We will keep track of what is happening in each case. We also want to make sure that positive motion in the linear direction is consistent with positive for rotation. Note how we have set both directions so that this happens.

Linear Motion Rotational Motion

Fnet = m a τnet = I α

Answer: ( Acc = 2.67 m/sec2 , Ffrict = -3.33 N) Notice that the value for Ffrict is a negative. This lets us know that we picked the wrong original direction for Ffrict. In this case, Friction goes in the same direction as the Fribbion. Before you go on, ask yourself, if this makes sense.2. A Falling Disk

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(with ribbon wrapped around it)

Suppose we have a metal disk (M) that is hanging from the ceiling with a thin ribbon wrapped around it. Gravity is pulling on the disk through its center of mass. The disk has a radius (R). We want to find two things

A. the disk’s linear acc. B. the tension in the ribbon

Assumption: In this problem we will assume that the disk will fall downward and that the ribbon will be pulling upwards against this.

Directions: Positive for linear motion will be downward and for linear motion in the counter-clockwise direction. (Notice that positive spin will give positive linear acceleration.)

Linear Motion Rotational Motion

Fnet = m a τnet = I α

Answer: Fribbon = 1/3 Fg and acc. = 6.54 m/sec2

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4. Disk Rolling Down a Hill

Suppose we have a metal disk (M) that is rolling down a slope (θ). Gravity is pulling on the disk through its center of mass. The disk has a radius (R). We want to find three things.

A. the disk’s linear acc. B. the force of friction acting on the diskC. the minimum needed to keep the disk from sliding

Linear MotionRotational Motion

Fnet = m a τnet = I α

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5. Spooling Around

You have a spool of mass (M) andRadius (R) is being pulled alongBy a thread. If the coefficientOf friction with the ground is0.5. Find the following:

1 The maximum tension in the rope.8. The maximum linear acceleration

of the spool.

Linear Motion Rotational Motion

Fnet = m a net = I

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6. More Spooling Around

You have a spool of mass (M) and Radius (R) is being pulled along by a thread. If the coefficient of friction with the ground is 0.5. Find the following:

A. The maximum tension in the rope.B. The maximum linear acceleration of

the spool.

Linear Motion Rotational Motion

Fnet = m a τnet = I α

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7. Spinning disk with a hanging weight

Imagine that we have a pulley of mass X and that there is a weight of mass A hanging from a string wrapped around the pulley. This is different than previous problems because the part of the system that rotates does not move linearly and the part of the system that moves linearly does not rotate.

What we will do is treat our big system as two systems. One system is purely rotational and other purely linear.

In our picture only one object is experiencing linear acceleration (the hanging mass). And only one object is experiencing rotational motion (the pulley).

Assumption:The rope (and its tension) is the unifying factor for both systems. It allows us to go between rotational and linear systems.

Linear Motion (Mass A) Rotational Motion (Pulley)

Fnet = m a τnet = I α

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Pulley Problem: Without Rotational Inertia

Assumptions:

1. At any given time all objects in the pulley system will move at the same speed.

2. All objects will experience the same acceleration.3. The ropes do not stretch while the objects move.

Procedure:First find the acceleration of the system.

Establish fence around the whole system.

Then look at tensions in the system.

Establish fence around portion of systemso that rope you are looking at goes outof the fence.

There is a reason for the fence. It is to establish theboundaries of the system. With this we can now useNewton’s laws of dynamics. Specifically, outside forcescause inside masses to accelerate.

1. In the picture to the right, let us assume that A is 50 kg, B is 10 kg and C is 35 kg. Find the acceleration of the system.

2. On the back of this page redraw the system and find the tension in rope A by drawing a fence through which rope A passes.

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Pulley Problem: Including Rotational Inertia

Assume that the pulley problem now must account for the moment of inertia of the disk type pulley. Let us assume that the mass of the disk is 10 kg and its radius is 0.4 meters. Block A is 50 kg, B is 10 kg and C is 35 kg. (these masses are the same as those in a previous problem)

Hint: Break the system into three separate fenced off areas; one rotational and two linear ones. The tension in the rope going from mass A will not be the same as the tension in the rope going up from mass B. (You should be able to explain why we can make this assumption.) This will give use three equations and three unknowns (acc, FropeA , and FropeB) Solve for all three.

Linear block A

Substitutions

Linear blocks B & C.

Rotational pulley

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More pulley problems

Problem 1 (no rotational inertia)Assume A = 50 Kg B = 30 Kg

Pulley = 0 kg R = 0.5 m

Find Acc ______________

Ten A ______________

Ten B ______________

Problem 2Assume A = 50 Kg B = 30 Kg

Pulley = 30 kg R = 0.5 m

Find Acc ______________

Ten A ______________

Ten B ______________

Problem 3Assume A = 50 Kg B = 30 Kg

Pulley = 30 kg R = 0.5 mCoefficient of friction between

A and the table is 0.5

Find Acc ______________

Ten A ______________

Ten B ______________

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Chapter 7: Finding an Object’s Center of Mass

Newton believed that even the most complex objects could be treated as though they had all of their mass located at a single point. This point was called the object’s center of mass. It is the point around which the object would want to spin. And, when thrown, it among all the other locations on the object is the only location that will follow a parabolic trajectory when the object is thrown. The following equation will allow you to find the CofM (center of mass) for any shape object.

For regularly shaped objects of uniform material density the center of mass is easy to find because the CofM is at the geometric center of the object.

For irregularly shaped objects we can break them down into regularly shaped parts and find the overall CofM by using the following equation

Mtotal Rcm = m1 r1 + m2 r2 + . . . + mn rn

Mtotal Rcm = ∑ mi ri

We will start with simple objects first. You have a two dimensional shape made out of 1 x 1 meter sheets of copper (2 kg per square). Find the center of mass of the shape.

X dimensionMtotal Rcm = m1 r1 + m2 r2 + m3 r3 + m4 r4

Y dimensionMtotal Rcm = m1 r1 + m2 r2 + m3 r3 + m4 r4

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Calculate the CofM for the following situations:

1. Two unequal masses, one of which is at the origin.

2. Two unequal masses, neither of which is at the origin

3. Three masses not at center.

4. The mass of the Earth is 6.0E24 kg. The mass of the Moon is 7.3E22 kg. From their centers they are 3.8E8 meters apart. How far from the center of the Earth is the CofM for the Earth-Moon system?

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Density In this class we will be looking at mass and charge distributions based on their densities. The way we do this is based on how many dimensions the material occupies

Line density (one dimension)

= material/length for masses is in kg/m

Surface density (two dimensions)

= material /area for masses is in kg/m2

Volumetric density (three dimensions)

= material /volume for masses is in kg/m3

1. Typical barbed wire has a density of 0.34 lg/m. What is the mass of a 300 meter length if it?

2. Quarter inch thick iron sheet metal has a surface density of 50 kg/m2. What is the mass of a 2m by 3m iron sheet?

3. Silver has a volume density of 1.05E4kg/m3 You have an ingot that is 12cm X 12cm X 40cm (This is roughly 4.74in x 4.75in x 16in) How massive is the ingot?

4. Silver has a volume density of 1.93E4kg/m3 You have an ingot that is 12cm X 12cm X 40cm (This is roughly 4.74in x 4.75in x 16in) How massive is the ingot?

5. Silver has a volume density of 2.15E4kg/m3 You have an ingot that is 12cm X 12cm X 40cm (This is roughly 4.74in x 4.75in x 16in) How massive is the ingot?

6. A neutron star has a volume density of about 1.05E17kg/m3. A volume of that star that is 12cm X 12cm X 40cm (This is roughly 4.74in x 4.75in x 16in) has about how much mass?

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Applying density when finding the Center of Mass

1. We have a rectangular sheet (2m by 4 m) of metal which is made out of material with a density of 2 kg per square meter. There is a hole cut out of the rectangle as shown below. Find the center of mass of the system.

2. We have the same rectangular sheet but This time we insert in the open hole a metal that has a density of 5 kg per square meter. Now find the center of mass.

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3. And now for a challenge. You have a disk with a radius of 1.5 m. Into one side of it is drilled. A hole with radius 1 m as is shown. If the material of the disk had a density of σ = 12 kg/m2

Where is the center of mass of the system?

4. If the void is filled with a material of density σ = 20 kg/m2 where is the cm now?

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Chapter 8: Calculus and Moment of Inertia

Remember that the moment of inertia for a particle is given by

I = m r2

For several particles it is

I = mi ri2

For continuous distribution of particles we integrate

I = ∫ m r2

To integrate we must break the integral into miniscule bits of mass

I = ∫ r2 dm

Finding the moment of inertia for a bar rotating around its end.

Approach: 1. We will be integrating from – r to + r (- r = - L/2 and r = L/2).

2. Find by λ dividing the total mass by the total length.λ = M/L

3. λ has the same value if we look at miniscule bits of mass. λ = dm /dL

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The general equation I = ∫ r2 dm

In this case dm = λ dr = M/L dr

Substituting we get I = ∫ r2 M/L dr

I = M/L ∫ r2 dr

I = M/L [ 1/3 r3 ] evaluated from –L/2 to L/2

I = M/L [ 1/3 ( (L/2)3 - (-L/2)3 ) ]

I = M/L [ 1/3 (L3/8 + L3/8) ]

I = 1/12 M L2 ______________________________________________________________

On the backside of this page determine the moment of inertia for a bar rotating around its end.

Approach: 1. We will be integrating from 0 to r = L

2. The mass per unit length λ = M/L

3. We will break the bar into short lengths with dm = λ dx

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Calculus and Moment of a Disk

We will find the Moment if Inertia of a disk with a radius of R and a mass of M.

Approach:

1. disk material has a density σ = M/A = M/π R2

2. we will divide the disk into a series of concentric hoops, each with a width of dr and mass dm = σ da (area)

3. Note that the area of each hoop

da = length x width

da = (2 r) dr

4. We will integrate from r = 0 to r = R

The general equation I = ∫ r2 dm

In this case dm = ________ dr = __________________ dr

Substituting we get I = ∫ dr

On the back side of this page calculate the moment of inertia of a cylinder with inner radius = Ra and outer radius Rb.

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Moment of Inertia of a Sphere with a radius R and a mass M.

Approach:

1. The sphere’s density ρ = M/(4/3 π R3 )

2. The volume of a disk is dv = r2 dz

3. And ρ = dm/(π r2 dr )

4. We will divide the sphere into a series of concentric very thin disks. Remember that I for a disk is I = ½ m r2

The general equation I = ∫ r2 dm

In this case dm = ________ dr

Substituting we get I = ∫ dr

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Here are a series of situations where the moment of inertia can be derived using calculus.

1. A bar of mass M = 2.5 kg and length L = 1.5 is swung at the end of a string that is 2 meters long.

2. A bar of mass M= 2.5 kg and length L = 4 meters is rotated about a point 1 meter from the end.

3. A brass disk or radius R = 1.5 meters and density 16 kg/m2 has a concentric hole drilled into it that has a radius of 0.75 meters.

4. If the hole in the above problem is filled by silver that has an area density of 28 kg/m2, what is the moment of inertia now?

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Chapter 9: Parallel Axis TheoremAn alternative way of finding moments of inertia that does not require calculus.

So far we have been dealing with objects that spin around their centers. What if the spin is off center? There is a way to modify our simple equations so that we can do this. It is by using the parallel axis theorem. The theorem is very elegant. We treat the object as actually two objects.

1. First, we pretend it is a simple object that rotates around its own center of mass.

2. Next, we treat the object as a point mass (all located at its center of mass) that is moving around the true pivot point.

Inet = Icm + M R2

R is the distance between the center of mass and the pivot point.

Lets look at an example where we already know the answer. Lets say we know that a bar rotating around it center of mass has a moment of inertia Icm= 1/12 M L2 and we want to rotate this bar around its end. Plugging into the equation we get.

Inet = Icm + M R2

Inet = (1/12) M L2 + M (L/2)2

Inet = (1/12) M L2 + (1/4) M L2

Inet = (4/12) M L2

Inet = (1/3) M L2

.

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More Parallel Axis Theorem

Lets take a look at what happens as we move the pivot point further and further away from the center of rotation further and further away from the center of mass. In the problems below let us assume that we have a 10 kg ball with a radius of 10 cm.

Final Equation Inertia in Kg m2

___________________ __________ Ball rotating around itself.

___________________ __________ Ball rotating 5 cm from center.

___________________ __________ Ball rotating at edge (10cm)

___________________ __________ Ball rotating 20 cm from center.

___________________ __________ Ball rotating 1 m from center.

___________________ __________ Ball rotating 10 m from center.

___________________ __________ Point mass 10 kg ten meters from center

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Compound Objects

Find the moment of inertia of this system of objects. Axis of rotation is the center of disk A:

Ball A Disk A is 10 kg, Bar B is 4 kg, Disk C is 5 kg

R = 0.5 meters

I = ___________

τ = ___________

Where is the center of mass of this system (measured from the pivot point)?

What is the net torque acting on the system?

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Chapter 10: Gravitational attraction.

Back when we first started to study classical gravity we ran into Newton’s universal law of gravitation

Newton’s assumption was that gravitational force required two objects to interact. Mass m1 will exert a force on a nearby “victim” mass m2. He also assumed that it didn’t matter which object was more massive. Between a person standing on the surface of the planet Earth both the Earth and the person felt the same amount of attraction toward the other.

There are three things that you must be aware of in the above equation.

A. (often referred to as r hat) is a unit vector. It has a length of one unit. Therefore, it does not effect the magnitude of the force in any way. But, it has a direction. Its direction is away from the acting mass and toward the victim mass.

B. Why is there a negative sign in front of the constant G? Well, we assume that all commonly found masses have a positive value. We also assume that all masses attract each other. Since the unit vector points away from the acting mass toward the victim mass. The force of attraction will be in the opposite direction (negative direction)

C. G is a constant of proportionality. In the MKS system it is equal to 6.67 x 10-11 N m2/kg2

Review problem: Sun and a closely orbiting planet. msun = 2 x 1030kg. mplanet = 2 x 1020kg. The distance between the two is 3.5 x 109 meters (center of sun to center of planet) Find the velocity of the planet and its period. Begin with Fnet = ma

Find the velocity Find the period

Fnet = m a Fnet = m a

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Most stars are binary star systems. That is, they have two stars orbiting each other. Usually one of the stars is larger than the other. It would be as if our sun Mass 2 x 1030kg and Radius 7 x 108 m were to have another star of mass 1.2 x 1030 kg and Radius 3 x 108 m running around it at a distance of 3.5 x 109 meters (center of one star to center of the other). Find the period of their orbit around each other. Hint: 1. Systems orbit around their center of mass. 2. We are assuming each has a circular orbit.

Fnet = m a

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Some orbit problems with Center of Mass issues

1.The Earth has a radius of 6.37E6 meters and a mass of 6.0E24 kg. The moon has a radius of lskdfj meters and a mass of lsdfjs kg. If he distance between the two (center to center) is lksdjf meters, find the CofM for this system.

2.In the problem above, what is the period of the moon’s orbit?

3.In the problem above, what is the period of the Earth’s orbit?

4. What is the radius of the Earth’s orbit around the Earth-Moon grouping?

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Chapter 11: Energy and Rotational SystemsIn the past, after we worked and worked problems using forces, we then discovered that most of these problems could be solved more easily using Energy. Let us see if that is so.

KE linear = 1/2 m v2 KE rotational = 1/2 I 2 .PE linear = m g h PE rotational does not exist

W mechanical = F d W mechanical = A. Let us revisit a 34 kg disk of radius 0.8 m that rolls down a slope. We can find its

final speed at the bottom of the slope by using conservation of Energy.

Forces and Torques WayF = m a = I

Energy Way Before = After

( P E g ) = ( ) + ( )

( m g h) = ( ) + ( 1/2 I 2 )

( m g h) = ( ) + ( )

v = ( )

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B. Then there is the disk with a mass hanging from a thin rope wrapped around it. How fast will the block be moving after it has dropped 5 meters from rest?Forces and Torques

F = m a = I

Energy WayBefore = After

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C. A disk speeds up as a mass fallsAssume A = 50 Kg Disk = 30 Kg

Find the time it takes for Mass A to fall 4 meters from restOld Way

Energy Method

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D. Cool Pool Table ProblemA cue ball is initially at rest when it is hit by a cue stick and is given an initial speed of 5 m/sec but no initial rotational motion. As the ball skids along the surface of the table it gains angular velocity and loses linear velocity until it stops skidding and begins to roll without skidding. How far has the ball skidded before this happens? ( sliding is 0.4 and static is 0.7)

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E. Here are several more questions related to the pool ball problem on the previous page?

a. what was the linear impulse that the ball received?

b. What was the rotational impulse that the ball received?

c. What was the net impulse on the ball?

d. How much linear energy did the ball have in the beginning?

e. How much total energy did the ball have at the end of its sliding?

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Chapter 12: Angular Momentum

We have studied rotational motion from the point of view of torques (τ = I α and τ = r x F) and energy (KErotational = 1/2 I ω2. One of the most powerful ways of studying rotational motion is by looking at objects’ angular momentum (L) . For particles momentum and angular momentum can be found using the following equations

Linear Momentum ρ = m v

Rotational Momentum L = r x ρ

L = r ( m v ) sin ( θ )

L = r m r ω 1 Convert linear to rotational vel.

L = m r2 ω

L = I ω Where I = moment of inertia

Spinning Counter clockwise (from top) Spinning Clockwise (from top)

Notice that angular momentum is the result of a cross product between an object’s radius and its angular velocity. To determine the direction of the momentum vector we use the same right hand rule (the hitch hiker’s rule) that was used to determine the direction of a magnetic field created by a flowing current.

Right Hand Rule Fingers curl in direction of the angular velocity Thumb points in direction of angular momentum

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Remember how we learned Newton’s second law as both

F Δ t = mv And as F = m a

In rotational settings Newton’s second law becomes

τ Δ t = Δ ( I ω ) And as τ = I α

Figure skating and angular momentum

You have undoubtedly seen skaters spinning on the ice. As they bring their arms in from an outstretched position they speed up. Then, as they spread their arms outward again they slow down. They are employing the rule of conservation of momentum as part of their routine.

Two Spinning Masses

Imagine two masses (6 kg each) that are spinning around their center of mass (r = 4 meters). Initially the system is spinning at a speed of 1 cycle per second.

1. What is the initial angular speed of the masses in radians/second?

Equation

Substitutions

Answer

2. What is the initial linear speed of the masses?

Equation

Substitutions

Answer

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Two spinning masses continued

3. Use conservation of momentum to find the angular speed of the system when the radius is cut to 2 meters.

Equation

Substitutions

Answer

4. What is the linear speed of the masses?

Equation

Substitutions

Answer

5. What is the frequency of rotation?

Equation

Substitutions

Answer

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Two spinning masses continued

6. Use conservation of momentum to find the angular speed of the system when the radius is cut to 1 meter.

Equation

Substitutions

Answer

7. What is the linear speed of the masses now?

Equation

Substitutions

Answer

8. What is the frequency of rotation?

Equation

Substitutions

Answer

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Two spinning masses continued

9. What was the KE of the system when the radius was 4 meters?

Equation

Substitutions

Answer After is 16 times the energy

10.What was the KE of the system when the radius was 1 meter?

Equation

Substitutions

Answer After is 16 times the energy

11.Compare the original rotational KE of the system to that of the system when r = 1 meter.

Equation

Substitutions

Answer After is 16 times the energy

12. Explain how this has happened?

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The Sun Dies

Eventually the sun which now has a period of rotation about its axis of 10 hours and a mean radius of 7 x 108 m will collapse when its fuel runs out. It will shrink to about the size of the earth, 6.37 x 106 meters.

1. Find the sun’s present angular momentum ( L ).

Equation

Substitutions

Answer

2. Find the sun’s present Rotational KE.

Equation

Substitutions

Answer

3. Find the dead sun’s final Angular momentum ( L ).

Equation

Substitutions

Answer

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4. Find the sun’s final Rotational KE.

Equation

Substitutions

Answer

5. What caused this increase in KE rotational.

The angular vs. linear momentum dilemma

Is it possible for two persons who are stationary with respect to each other to see the same object where one of the observers sees that the object has angular momentum and the other person sees the object as having only linear momentum? The answer is yes. Here is how. Imagine that there is an object ( mass m )moving through space in a straight line and at a high speed (v). It is aimed directly at observer “A”. Observer “B” is at a distance from “A” so that the velocity vector of our moving object is perpendicular to observer B’s line of sight at a distance (r).

From A’s point of view the object has a momentum of

ρ = m v

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Observer B has to turn his head to keep a bead on the object. He perceives the object as moving in an angular manner. At the point shown above, the object’s angular velocity is perpendicular to the observer’s line of sight. So, for an instant at least, observer B perceives that the object is moving in a circle. This being the case, he measures the object’s angular momentum as

L = I w

L = (m r2) w

L = m r2 v / r

L = m v r

The doubters in the crowd might say that this is only an instantaneously perceived angular momentum and that at times before and after this observer B will not measure this. To them we say, “Not true!” Imagine that observer B is looking at the moving object from some angle φ.

At this position the object has an apparent angular velocity of v (cos

)φ . If it has an apparent angular velocity then it has an apparent angular momentum.

The general equation for angular momentum is L = m vperp radius

In this case L = (m) (v cos φ ) ( R )

but cos φ = r /R so R = r / cos φthis gives L = ( ) ( ) ( )

Which is the same momentum as in the first case L = m v r

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Chapter 13: Gravitational Potential Energy (on a planetary level)

In the past when we have dealt with gravitational potential energy we looked at small objects being lifted up a small distance from the surface of the Earth. The PE gained was just equal to the work done to lift the object. The Work done was positive because the force applied and the change in displacement were in the same direction. We could use the equation for work.

W = ∫F dr

What happens when the distances are much greater? The force of gravity is no longer constant but is governed by Newton’s Universal law of gravitation.

Fg = G m1 m2 / r2.

We now must integrate to find the total work done using W = ∫ Fapp dx. Notice that the Force we must apply is a restraining force against gravity.

Fapplied = - Fg = - G m1 m2 / r2

This force is in the opposite direction to the change in displacement (dx). The angle between the two is 180 degrees .

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Definition of work W = ∫

Fapp in termsof Fg W = ∫

Fg and Dot product W = ∫ ( ) ( ) cos

Cos 180 = -1 W = ∫

Constants move out W = ( ) ∫

Integrated W = ( ) [ ]

Evaluating W = ( ) [ - ]

The final answer PEg = WORK DONE = - G m1 m2 / rf

We have assumed that at some infinite distance from a source of gravity there is no PEg and no KE if the object is at rest.

Moving the object closer to the gravitational source takes energy out of the system (Wout) Where does this energy come from? We cannot have negative KE. The only other possibility is that we create negative potential energy.

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3. Suppose a 10 kg ball of krypton falls to Earth from infinitely far away. How fast is it going upon impact? The Earth has a mass of 6.0 x 1024 kg and a mean radius of 6.38 x 106

meters. Let us use Conservation of Energy to solve this problem.

Before = After

0 = ( K E ) + ( )

0 = ( ) + ( )

vimpact = ____________

It makes sense that PEg must go negative if the KE of the krypton is to go positive.

4. Now let us imagine placing this krypton in a very low (read at the surface) circular orbit. Find the speed of the ball using Fnet and centripetal acceleration

Fnet = m ac

Raw Eq ( ) = ( ) ( )

Substitutions ( ) = ( ) ( )

Answer v = ( )

What is the net energy for this low flying satellite?

Total Energy = ( PEg ) + ( K E )

Raw equation = ( ) + ( )

Substitutions = ( ) ( )

Answer Energy net = ( )

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Chapter 14: Elliptical Orbits

Johannes Kepler studied the orbit of Mars and determined three things.

1. Planets move in elliptical orbits with the sun at one of the two focii. (a circular orbit is one extreme of an ellipse)

2. There is a relationship between planet radii (r3) and period of orbit (T2)3. Planets sweep out equal areas in equal times

We have been able to work out orbital problems where the orbit is a perfect circle. With knowledge about both angular momentum and Potential Energy based on the Work principle we can extend our problem solving abilities to include many aspects of the elliptical orbit.

1. Review: The mass of the Sun is 2E30 kg. The mass of the Earth is 6E24 kg. The distance between the two is 1.5E11 meters. Find the orbital velocity of the Earth.

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2. We know that the earth with a velocity of approximately 30,000 m/sec orbits the sun in an almost perfectly circular path. We decide to fire a satellite so that it moves relative to the sun with an initial speed of 15,000 m/sec. How close will the satellite come to the sun? To do this problem we will use both conservation of momentum and conservation of Energy

Conservation of Angular Momentum (L)Lat Rmax = L at rmin

( ) = ( )

Solve for v at rmin v =

Conservation of Energy

E at Rmax = E at rmin

Energy types ( ) + ( ) = ( ) + ( )

( ) ( ) = ( ) ( )

Substituting from Angular Momentum we get rmin = _________________________

9. If we want to sent an exploratory satellite out to Mars (r = 2.28 x 1011 m) How fast should it be moving as it leaves the Earth?

Picture Equations

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Chapter 15: Some cumulative Review problems

Colliding Masses

1. Perfectly inelastic and no rotation

Imagine that a 2 kg blob is moving at 5 m/sec. It hits the center of a stationary 4 kg 1 meter long bar dead center and the two stick.

A. If the initial distance between the blob and the bar is 5 meters. Where is the system’s center of mass?

B. What is the speed of the center of mass of the system?

Explain your reasoning

Find the resultant velocity of the 2 kg blob

Equation

Substitutions

Answer 1.667 m/s Problem continues on next page

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Find the impulse on the 2 kg blobEquation

Substitutions

Answer -

Find the percent conservation of Kinetic Energy in the system

2. Perfectly inelastic collision with rotation

Imagine that a 2 kg blob is moving at 5 m/sec. It hits the end of a stationary 4 kg 1 meter long stick as shown below. The two stick together and move on.

.

A. What is the velocity of the System’s Center of mass before the collision? And after?

B. How far below the blob is the combined vertical center of mass of the system?

Problem continues on next page

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C. What is the moment of inertia of the combined system? Assume that it rotates around the system’s vertical center of mass.

Equation

Substitutions

Answer

D. Find the resultant linear speed of the system

Equation

Substitutions

Answer

E. Find the resultant angular speed of the system

Equation

Substitutions

Answer

F. Find the rotational impulse on the 2 kg blob

Equation

Substitutions

Answer

G. Find the percent conservation of Energy in the system

2.

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2.3. Placing telescopes in space eliminates several of the thorniest problems we

encounter here on earth; fog, clouds, dust, and city lights. There is one problem however, how do you turn the telescope so that you can move from one star to another. Let us assume for one moment that we have mounted a telescope between a massive pair of metal disks (Radius 5 meters and total mass 1000 kg) Let us also assume for a moment that we have managed to get the telescope stationary in the first place. We want to turn the telescope 0.5 radians in the counter clockwise direction. To do so we have mounted a small motor with a small disk (radius 0.5 meters and mass 20 kg) attached to it as shown below.

Which way should we spin the motor disk to do this?

A. If we can spin this disk with a speed of 300 rpm how fast will the telescope be turning?

B. How many revolutions of the small disk must we make for the telescope to line up with the star?

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C. But the telescope is moving and we cannot push off anything outside because we are in space. How do we get the telescope to not rotate forever?

4. You have a 5000 kg SUV that is initially at rest on flat ground. Answer the following questions:

a. Your vehicle accelerates from 0 to 30 m/sec in 7 seconds. What average acceleration is this?

b. How much average power did your engine produce?

c. What total KE will the vehicle have at this speed?

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d. What minimum coefficient of friction was necessary to do this?

e. At 30 m/sec the car experiences about 445 N of drag force. What horsepower will it need to continue at this speed?

f. You design a regenerative braking system using a 200 Kg 0.5 meter radius disk. That speeds up as the car slows down. Assuming 100 percent efficiency, how fast will the disk be spinning when the car comes to a complete stop?

g. How fast will the outside edge of the disk be spinning?

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5. The following hammer is allowed to swing from the pivot point (a) Answer the following questions:

a. Where is the center of mass of the system?

b. What is the moment of inertia of the system?

c. Gravity acts on the hammer, what is the initial angular acceleration of the hammer

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d. When the hammer gets down to the vertical orientation, how much KErotational will it have?

e. What rotational velocity will the hammer have at the bottom point?

f. What linear velocity will the bottom tip of the hammer have at this point?

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Equations sheets

84

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