physics 1 brief synopsis of ideas

Physics I

Brief Synopsis of Central Ideas

** indicates a detail required of only University Physics students (phy 231 and phy 241)

* indicates a detail required of only College and University Physics students (not required of phy 121 students)

Idea 1: Units and Rates	#Idea 13: Gravitation is an Inverse-Square Force
Idea 2: Velocity and Acceleration as Rates	#Idea 14: For objects in circular orbits, Centripetal Force = Gravitational Force
Idea 3: Applying the Ideas of Rates to Position, Velocity and Acceleration	#Idea 15: Change in gravitational PE = Average gravitational force * Displacement
Idea 4: Uniform Acceleration	#Idea 16: Between two circular orbits, `dKE = -1/2 * `dPE
Idea 5: Sequential Observations and Graphical Representation of Positions and Velocities	#Idea 17: Angular Motion is completely analogous to Linear Motion
Idea 6: Newton's First Law and situations involving Uniform Acceleration	#Idea 18: Newton's Second Law can be formulated in terms of Angular Motion and Moment of Inertia
Idea 7: Reasoning Out Uniformly Accelerated Motion	#Idea 19: The definition of Work can be reformulated in terms of Angular Quantites
Idea 8: Newton's Laws of Motion	#Idea 20: Angular Impulse-Momentum gives rise to a new conservation law.
Idea 9: Energy (net force applied through displacement) and the Work-Energy Theorem	#Idea 21: Simple Harmonic Motion results from a Linear Restoring Force
Idea 10: Impulse (force applied over time interval) and Momentum; Conservation of Momentum	#Idea 22: Simple Harmonic Motion can be modeled by the projection of a point moving about a Reference Circle
Idea 11: The Work-Energy Theorem can be rephrased in terms of conservative and nonconservative forces	#Idea 23: Velocity and Acceleration in SHM follow the Reference Circle Model
Idea 12: Vector Quantites	#Idea 24: Energy Relationships in SHM are consistent with the Reference Circle Model

Part I: Main Ideas Through Major Quiz

Idea 1: Units and Rates are Keys to Understanding

When we do calculations we always write down and think about the meaning of the units of the quantities involved. This is a very time-effective strategy for catching errors and for developing the ability to think through and solve problems.

We also do the algebra involved in determining the units of our results.

Simple rates are easy and natural if they concern money and the rate at which it is earned, and fairly easy in the context of velocity as rate of change of position. While thinking in the context of money or motion, simple rate problems can be solved using common sense.

Definition of an average rate: The average rate of change of A with respect to B is

average rate = change in A / change in B

** Rate problems develop into the idea of the derivative in calculus, and into differential equations. These concepts are much easier if we think in the context of simple rates.

An average rate of change of a quantity with respect to clock time is represented as the slope between two points of a graph of that quantity vs. clock time.

The change of a quantity during a short time interval is approximated by the area under the trapezoid formed by the graph of the rate of change of the quantity with respect to clock time vs. time clock time, for that time interval.

Idea 2: Velocity and Acceleration as Rates

Velocity and acceleration are defined as rates:

The average velocity of an object between two clock times is the average rate at which its position changes between those clock times. The average rate at which position changes is ave rate = change in position / change in clock time = ave velocity.

The average velocity of an object moving along a straight line is therefore represented by the slope of a graph of position vs. clock time.

** The velocity function v(t) is the derivative of the position function s(t) for and object.

The average acceleration of an object between two clock times is the average rate at which its velocity changes between those clock times. The average rate at which velocity changes is ave rate = change in velocity / change in clock time = ave acceleration.

The average acceleration of an object moving along a straight line is therefore represented by the slope of a graph of velocity vs. clock time.

** The acceleration function a(t) is the derivative of the velocity function v(t) for and object.

Idea 3: Applying the Ideas of Rates to Position, Velocity and Acceleration

All the concepts of rates apply to position, velocity and acceleration and their relationships.

Applying the concepts of rates we see that changes in position and velocity can be obtained from knowledge of velocities and accelerations, respectively:

The change in the position of an object over a given time interval is equal to the product of its average velocity and the duration of the time interval.

The change in position of an object moving along a straight line is therefore approximated as the area under the appropriate trapezoid on a graph of velocity vs. clock time.

The precise change in position is equal to the exact area under the graph of velocity vs. clock time, between the two given clock times.

** The change in position `ds between two clock times is the integral of the velocity function v(t) between the two corresponding clock times.

The change in the velocity of an object over a given time interval is equal to the product of its average acceleration and the duration of the time interval.

The change in velocity of an object moving along a straight line is therefore approximated as the area under the appropriate trapezoid on a graph of acceleration vs. clock time.

The precise change in velocity is equal to the exact area under the graph of acceleration vs. clock time, between the two given clock times.

** The change in velocity `dv between two clock times is the integral of the acceleration function a(t) between the two corresponding clock times.

The velocity of an object may change at a constant, an increasing, or a decreasing rate; the graph of velocity vs. clock time will reveal which.

Idea 4: Uniform Acceleration

If the velocity of an object changes at a constant rate, i.e., if the object accelerates uniformly, then

a graph of velocity vs. clock time forms a straight line and, as a result
the average velocity of the object over a time interval is the average of its initial and final velocities for that time interval.

If velocity does not change at a constant rate this is usually not the case.

** For uniform acceleration, a(t) = a = constant, so

v(t) is an antiderivative v(t) = a t + c of the acceleration function
the arbitrary constant c is equal to v(0) so we can write v(t) = a t + v0

the position function is the antiderivative x(t) = .5 a t^2 + v0 t + c of the acceleration function
the artitrary constant c is equal to x(0) so we write x(t) = .5 a t^2 + v0 t + x0
the change in position `ds between clock times 0 and `dt is `ds = x(`dt) - x(0) = .5 a `dt^2 + v0 `dt + x0 - x0, so

`ds = v0 `dt + .5 a `dt^2.

If an object accelerates uniformly over a time interval then the average velocity is halfway between the initial and final velocities. It follows that the change from the initial to the average velocity is the same as the change from the average velocity to the final velocity. Initial, average and final velocities will be 'equally spaced'.

As a special case of the above, when a uniformly accelerating object is observed over a time interval in which it starts from rest, and only when it starts from rest, its average velocity will be half of its final velocity on that time interval; the final velocity will thus, under these special conditions, be double the average velocity.

Idea 5: Sequential Observations and Graphical Representation of Positions and Velocities

If the position of an object is observed a series of clock times, its average velocity `ds/ `dt over each time interval can be easily determined. From the resulting approximate average velocity vs. clock time information we can then estimate the approximate average rate `dv /` dt at which velocities change between midpoint times, giving us an approximate graph of acceleration vs. clock time.

If the velocity of an object is observed at a series of clock times, its approximate average velocity over each time interval is easily determined. The approximate distance the object moves over each time interval is then easily found. From the distances moved over the successive time intervals, the total change in the position of the object from the first clock time to any other clock time is then easily calculated.

Given a graph of velocity vs. clock time over a range of clock times, we can partition the graph by a series of time intervals.

For each time interval we can determine the approximate average velocity from the velocities at the beginning and end of the time interval. The approximate average velocity over an interval is the average 'altitude' of the corresponding trapezoid.
From the average velocities and the durations of the various time intervals, we can calculate the approximate displacement corresponding to each interval. Adding displacements to initial position we obtain position vs. clock time information and can use this information to construct an approximate position vs. time graph.
The displacement corresponding to each interval is represented by the area under the v vs. t graph which corresponds to that interval.
The displacement corresponding to an interval is approximated by the area of the corresponding trapezoid. The accumulated area through each trapezoid approximates the displacement from the initial clock time through the clock time at the end of that trapezoid.

Given a graph of position vs. clock time over a range of clock times, we can partition the graph by a series of time intervals.

The average slope between any two clock times represents the approximate velocity at the midpoint of those two clock times. If we make a table of average slope vs. midpoint clock time and construct the corresponding graph we obtain an approximate graph of velocity vs. clock time.

The velocity vs. time graph will represent the slopes of the resulting position vs. time graph; the position vs. time graph will represent the accumulated areas under the velocity vs. time graph. ** This observation is equivalent to the Fundamental Theorem of Calculus.

The position vs. clock time, velocity vs. clock time, and rate of velocity change vs.clock time (i.e., acceleration vs. clock time) chronicles can be represented by graphs, with the 'y' coordinate of each graph representing the rates of change, or slopes, of the preceding graph. (i.e., the velocity graph represents as 'y' coordinates the slopes of the position graph, the acceleration graph represents at 'y' coordinates the slopes of the velocity graph).

Idea 6: Newton's First Law and situations involving Uniform Acceleration

The rate at which the velocity of an object changes is called its acceleration. Acceleration is rate of change of velocity.

If an object is accelerating in the direction of its motion, it is speeding up. If it is accelerating in the direction opposite to that of its motion, it is slowing down.

For example, an automobile coasting freely on a south-facing slope accelerates to the south, whether it is traveling north or south. If it is traveling south, it travels with the slope and speeds up. If it is traveling north it travels against the slope and slows down. If it is standing still, it will in the next instant be traveling south and speeding up.

If an object accelerates perpendicular to its direction of motion, with zero acceleration in its direction of motion, then its direction of motion will change but its speed will not.

Newton's First Law observes that, in the absence of a net force, an object will not accelerate. An object which does not accelerate will change neither its speed nor its direction of motion.

In the vicinity of the surface of the Earth any freely falling object is observed to accelerate at very nearly the same constant rate, independent of where on Earth it is.

A falling object, or an object on a uniform incline, which accelerates freely without resistance accelerates uniformly. On a uniform incline whose slope as measured from the horizontal direction is small this acceleration is very nearly equal to the product of the acceleration of gravity and the slope.

An object accelerating freely, except for the influence of friction, on a uniform incline with small slope will have greater acceleration for greater slope. The change in the acceleration from one slope to another will be very nearly equal to the product of the acceleration of gravity and the difference in the ramp slopes. Thus the slope of a graph of acceleration vs. ramp slope will be very nearly equal to the acceleration of gravity.

When an object is in free fall near the surface of the Earth, with no external forces other than gravity acting on it, the net force on the object is vertical, with no horizontal component.

Its vertical motion is therefore characterized by a uniform acceleration equal to that of gravity, while its horizontal motion is characterized by zero acceleration, resulting in constant horizontal velocity.

Idea 7: Reasoning Out and Formulating Uniformly Accelerated Motion

We can organize our thinking about a problem by using simple 'flow diagrams' showing the 'flow' of our reasoning. These diagrams can be extremely useful in 'mapping out' our solution strategies on complex situations.

For uniformly accelerated motion in one direction, we can reason out the motion using the quantities `ds, `dt, v0, vf, a, `dv, and vAve, and in terms of the units of these quantities.

In order to think clearly about what is going on, we need to think in terms of all seven of these quantities.
We can, however, formulate uniformly accelerated motion in terms of any five of these variables, provided the five are independent.
The consensus formulation is in terms of `ds, `dt, v0, vf, and a.

To formulate uniformly accelerated motion in terms of `ds, `dt, v0, vf, and a, we begin by formulating the definitions of average acceleration and average velocity in terms of these five variables, obtaining the two equations `ds = (vf + v0) / 2 * `dt and vf = v0 + a * `dt.

If we eliminate vf from the two equations we obtain a third equation `ds = v0 `dt + .5 a `dt^2, and if we eliminate `dt from the two equations we obtain a fourth equation vf^2 = v0^2 + 2 a `ds.
Using these equations, given the values of three of the five variables we can solve for the remaining two without having to think much.
It is OK not to think too much about a situation provided we already understand situation thoroughly, as we do if we can reason out the basic situations without resorting to formulas, and if we can derive the formulas from the definitions.

Part II: Major Quiz through Test #1

Idea 8: Newton's Laws of Motion

Newton's First Law tells us that whenever the net force on an object is 0, the magnitude and direction of the velocity of the object will not change.

We therefore say that any object either at rest or in motion at constant velocity with respect to some fixed direction is in equilibrium with respect to that direction.

From Newton's First Law we conclude some useful things:

When an object is either at rest or moving along a constant incline, then since it is not moving in the direction perpendicular to the incline it must be in equilibrium with respect to that perpendicular direction.
Hence the sum of all forces perpendicular to the incline must be 0.

When an object is suspended at rest or suspended and moving at constant velocity by a number of ropes, chains, rubber bands, cables, or similar devices, the sum of all the forces exerted on the object in any direction must be zero.

If all the suspending devices lie in a plane, it is possible to express all the forces on the object as the sum of their components in the vertical and horizontal (y and x) directions.

Newton's Third Law tells us that for every force exerted on an object by some source the object exerts an equal and opposite force on the source.

From Newton's Third Law we conclude some useful things:

If we push on an object while standing on a frictionless skateboard, the object will exert an equal and opposite force on us, and our mass plus that of the skateboard will have the acceleration determined by the magnitude of this force and our mass, and the direction of this force.

If a string exerts a tension force on a suspended mass, then the mass exerts an equal and opposite force on the string.

If a frictional force resists the motion of an object as it slides across some surface, then the object exerts a force equal and opposite to the frictional force, and therefore in the direction of the object's motion.

As the tires on your car attempt to spin along the road surface in some direction, they exert a frictional force on the road and the road exerts an equal and opposite force on the tires. It is this frictional force exerted on the tires that accelerates the automobile.

Using Newton's Third Law and similar principles we can sketch and analyze the forces acting on any system we care to define.

Newton's Second Law tells us that the net force Fnet required to accelerate a mass m at rate a is Fnet = m * a.

Standard situations include:

Atwood machine, where Fnet = m1 g - m2 g and mass = m1 + m2. String tension is a force internal to this system so does not contribute to the net force.
Person standing on a scale in an accelerating elevator, where Fnet = Fscale - m g and Fnet = m a so Fscale - m g = m a.
Mass m1 on level frictionless table attached by cord to mass m2 hanging over light frictionless pulley, where Fnet = m2 g acts on system with mass is m = m1 + m2. String tension is a force internal to this system so does not contribute to the net force.
Pendulum with mass m, length L pulled back distance x from equilibrium and released, where Fnet = m g / L * x directed toward equilibrium position.
Object of mass m moving freely on frictionless incline at angle theta, where Fnet = m g * cos(270 deg + theta) = m g * sin(theta) is the component of the weight parallel to the incline, directed down the incline. The normal force is exerted perpendicular to the incline and balances the perpendicular component of the weight.

Idea 9: Work (net force applied through displacement) and the Work-Energy Theorem

We compare the effect of applying a net force F to a mass m

while the mass moves through a known displacement `ds with the effect of a
pplying the same net force to the same mass for a time interval `dt.

If we apply a net force Fnet to a mass m while that mass moves through the displacement `ds in the direction of the force, the object's velocity in that direction changes at rate a = F / m.

The effect of this force on the velocity of the object is found by substituting a = F / m into the equation vf^2 = v0^2 + 2 a `ds.

We obtain the equation

Fnet * `ds = .5 m vf^2 - .5 m v0^2.

This tells us that the product Fnet * `ds gives us the change in the quantity .5 m v^2.

We define kinetic energy, or energy of motion, as the quantity

KE = .5 m v^2.

With this definition the difference .5 m vf^2 - .5 m v0^2 is expressed as KEf - KE0 = `dKE, and we have

Fnet * `ds = `dKE.

The quantity F * `ds, where `ds is displacement in the direction of the force F, is called the work done by the force F.

So our statement that Fnet * `ds = `dKE can be restated as

work by net force on system = change in kinetic energy.

The above statement is the most rudimentary form of the Work-Energy Theorem:

Work-Energy Theorem: work by net force on system = change in kinetic energy.

We can observe that by Newton's Third Law the work done by the net force on a system must be equal and opposite to the work done by the system against that net force, so that

work done by system = -(work done on system) = -`dKE,

and that therefore

work done by system against net force on system + `dKE = 0, or
`dWby + `dKE = 0.

If it is understood that `dW stands for the work done BY the system and `dKE for the KE change of the system we have

Work-Energy Theorem (first form): `dW + `dKE = 0 (note that `dW is work done BY the system)

** Using calculus we can generalize the Work-Energy Theorem to say that the integral of (net force on syste) * (displacement in the direction of that force) is equal to the change in the kinetic energy of the object which the force is applied:

** int( F ds, s, a, b) = `dKE,

** where the force is applied from displacement s = a to s = b and `dKE is the change in the kinetic energy of the object between these two displacements.

** This integral corresponds to the area under the force vs. displacement curve.

In all cases it is understood that the displacement is the displacement in the direction of the force.

Idea 10: Impulse (force applied over time interval) and Momentum; Conservation of Momentum

In contrast to the situation where we apply a net force through a give displcacement, here we apply the net force over a given time interval.

If we apply a net force Fnet to a mass m for a time interval `dt, the effect of the force is found by substituting a = F / m into the equation vf = v0 + a * `dt.

We obtain the equation

Fnet * `dt = m vf - m v0.

We call the quantity mv the momentum of the object and the quantity Fnet * `dt the impulse of the force.

The statement that Fnet * `dt = m vf - m v0 can therefore be restated as

impulse = change in momentum.

This is called the Impulse-Momentum Theorem.

** The Impulse-Momentum Theorem is easily generalized using calculus to include situations in which either or both the mass and velocity of the object changes while the force is applied.
** In this form the Impulse-Momentum Theorem reads

** F dt = dp,

** where p = mv stands for momentum.
** A simple rearrangement gives us

** F = dp / dt: force is the instantaneous rate of change of momentum.

By Newton's Third Law the forces exerted by two interacting objects one one another are equal and opposite. It follows that the impulses applied by the object to one another are equal and opposite. Hence the net change in momentum is equal and opposite. This idea is easily extended to any closed system of objects--i.e., any system in which the only forces exerted on the objects in the system are exerted by other objects (and in fact to any system where the vector sum of all the forces exerted on the system is zero). Impulses in such a system are applied in the form of equal and opposite forces, resulting in equal and opposite momentum changes. The result is that the net momentum change of the system must always be zero, so that the total momentum of the system remains constant.

Conservation of momentum: The total momentum of any closed system is unchanging.

Idea 11: The Work-Energy Theorem can be rephrased in terms of conservative and nonconservative forces

The Work-Energy Theorem can be restated in terms of conservative and nonconservative forces, introducing the idea of potential energy.

A conservative force is one that stores energy in such a way that we can recover the work done against the force.

For example a gravitational force is conservative since the kinetic energy we obtain from dropping an object from a certain height, in the ideal situation in which air resistance and other nongravitational forces are eliminated, is equal to the work that must be done against gravity to raise the object to that height.
This occurs because the gravitational force acts as the object is raised, requiring an equal and opposite force to raise the object at constant velocity, and then acts as the object falls through the same distances.

The elastic forces exerted by a spring or a rubber band as it is stretched are in the ideal case conservative, since they are experienced again in the opposite direction upon release (this is not quite the case for real rubber band and springs, which have some internal friction both when being stretched and released so that in fact more work must be done in stretching than is done in the 'snap back'; however we take it to be so for 'ideal' springs and rubber bands, in which the frictional forces are considered to be negligible) so that the work done against the elastic force can be recovered.

The work done against a conservative force is therefore considered to be 'stored' in the system until it is released; upon release it can perform work on another system and/or increase the kinetic energy of the system.

We therefore say that the work done by a system against a conservative force increases the 'potential energy' of the system, and that upon release potential energy is converted into some other form of work and/or energy.

A nonconservative force is one from which we can get back none of the work done against this force.

For example work done against friction typically increases the random kinetic energy (thermal energy) of objects which are being moved across one another.
This energy is random and diffuses away from its points of origin.
While theoretically possible it is statistically impossible for such energy to reorganize itself and do useful work.

Recall the Work-Energy Theorem:

Work-Energy Theorem (first form): `dW + `dKE = 0 (note that `dW is work done BY the system AGAINST the net force acting on the system)

If we denote by `dWnoncons only the work done against nonconservative forces in a system and by `dPE the work done by the system against conservative forces, then `dW = `dWnoncons + `dWcons and the Work-Energy Theorem can be restated as follows:

Work-Energy Theorem (second form): `dWnoncons + `dPE + `dKE = 0

The system can do positive work `dWnoncons against nonconservative forces only by decreasing the sum of its potential and kinetic energy.
If no external forces act on it, then `dW = 0 so that the system can increase its potential energy only at the expense of an equal amount of its kinetic energy (remember that `dW cannot be negative), and can increase its kinetic energy only at the expense of an equal amount of its potential energy.

Idea 12: Vector Quantites

A vector quantity is one that has magnitude and direction. Examples include velocity, acceleration, force and momentum.

A vector in two dimensions can be represented on a two-dimensional coordinate system with an x and a y axis.
If we measure the angle theta of the vector in the counterclockwise direction from the positive x axis then the action of the vector is completely equivalent to the combined action of its x and y components V cos(theta) and V sin(theta), respectively, where V is the magnitude of the vector.
To add two or more vectors we can proceed as follows:

Add the x components of all the vectors to get the x component Rx of the resultant vector.

Add the y components of all the vectors to get the y component Ry of the resultant vector.

The magnitude of the resultant vector is R = sqrt(Rx^2 + Ry^2) and the angle is theta = arctan(Ry / Rx), provided Rx is positive, or arctan(Ry / Rx) + 180 degrees if Rx is negative.

Vector quantities have magnitude and direction and include, among many others, displacement, velocity, acceleration, force, momentum and gravitational field.

The following, among many others, have no direction and therefore not vector quantities: distance, time, mass, volume, density, energy.

Important applications of vectors to forces include the following:

The weight w of an object on a incline which makes angle theta with horizontal acts in the vertical downward direction; if the xy coordinate system is oriented so the x axis is directed parallel to the incline with the y axis perpendicular, then the weight is at angle 270 deg - theta as measured counterclockwise from the positive x axis, so it has x and y components wParallel = w cos(270 deg - theta) parallel to the incline (i.e., in the x direction), and wPerpendicular = w sin(270deg - theta) perpendicular to the incline (i.e., in the y direction). [Since | cos(270 deg - theta) | = | sin(theta) | and | sin(270 deg - theta) | = | cos(theta) |, these formulas are consistent with those obtained using right-triangle trigonometry. Your text likely uses right-triangle trigonometry rather than the circular-model definitions used in most of this course. The two approaches give the same results, but the pictures are slightly different.]
The tension T in a pendulum is, for displacements from equilibrium which are much less than the length of the pendulum, very nearly equal to the weight of the pendulum. If the angle of the pendulum from vertical is theta, then tension vector makes angle 90 deg + theta as measured counterclockwise from the positive x axis. This vector therefore has component w cos(90 deg + theta) in the x direction, pointing back toward the equilibrium position. Since the displacement from equilibrium is x = L cos(90 deg + theta), it follows that cos(90 deg + theta) = x / L so that the x component of the tension is Tx = x * w/ L. That is, the tension component tending to restore the pendulum to equilibrium is proportional to the displacement from equilibrium. Note also that the horizontal tension is in the same proportion to pendulum weight as displacement x is to pendulum length. [Note that right-angle trigonometry can also be used to analyze these forces, with the result that the horizontal component of tension has magnitude w sin( | theta | ). Since | cos(90 deg + theta) | = | sin(theta) | this is equivalent to the result obtained using the circular model. ]

Part III: Test #1 through Test #2

Idea 13: Gravitation is an Inverse-Square Force

The gravitational field has units of acceleration, and in the vicinity of a spherically symmetric planet is inversely proportional to the square of the distance from the center of the planet.

It follows that the gravitational field in the vicinity of such a planet is given by a function g(r) = k / r^2, where k is a proportionality constant and r the distance from the center of the planet.
If we know the value of g at some r beyond the surface of the planet, then we can evaluated the proportionality constant k and obtain a general expression for the field g(r).

The gravitational force between two point masses m1 and m2 is F = G m1 m2 / r^2, where r is the distance between the masses.

Idea 14: For objects in circular orbits, Centripetal Force = Gravitational Force

Centripetal acceleration is given by Fcent = v^2 / r, where Fcent is the centripetal force holding the object in its circular path, v the velocity of the object and r the radius of the circle.

For a satellite in a circular orbit the centripetal force is the net force, which is equal to the gravitational force. For a satellite with mass m in its orbit around a planet of mass M, with M >> m and r = radius of orbit we therefore have m v^2 / r = G M m / r^2, expressing the equality between centripetal force and gravitational force.

This equation is easily solved to obtain an expression for v in terms of orbital radius r.

Idea 15: Change in gravitational PE = Average gravitational force * Displacement

The gravitational potential energy change between two points is equal to the work required to move the object at constant velocity against the gravitational field, from the first point to the second.

The work is equal to the product of the force we must exert against gravity with the displacment parallel to this force.
Since in moving parallel to the gravitational force we change our distance r from the center, we multiply the average gravitation force by the change in r t obtain `dPE = Fave * `dr, where Fave is the average of the (nonlinear) gravitational force on an object and `dr the change in the distance of the object from the center of a planet.
We can approximate Fave by averaging the force at the first distance with the force at the second. However because of the upward concavity of the F vs. r graph this estimate will overestimate the average force.

** To get the accurate change in potential integral we must integrate force with respect to distance from the center. This gives us

** `dPE = integral( G M m / r^2 with respect to r, from r1 to r2), where M is the mass of a planet, m the mass being moved from distance r1 to distance r2 with respect to the center of the planet.

The precise change in potential energy from one distance r = r1 to another distance r = r2 from center is

`dPE = G M m / r1 - G M m / r2

The gravitational potential at distance r from the center of the planet is

gravitational potential = - G M / r.

Gravitational potential is in units of Joules / kg.

The change in potential energy between two points is equal to the change in gravitational potential from the first point to the second, multiplied by the mass of the object being moved.

Idea 16: Between two circular orbits, `dKE = -1/2 * `dPE

For a small mass m in circular orbit around a planet with large mass M, gravitational PE is - G M m / r, while gravitational KE is 1/2 G M m / r.

It follows that to move from a circular orbit at radius r1 to a circular orbit at radius r2 the gravitational PE must increase from - G M m / r1 to -G M m / r2 while KE decreases from 1/2 G M m / r1 to 1/2 G M m / r2.

Gravitational potential - G M / r is negative for all values of r, approaching zero as r approaches infinity. Thus as we move away from a planet we begin with a large negative gravitational potential and as we move further and further away our gravitational potential increases, approaching but never reaching zero.

Gravitational potential energy - G M m / r is also negative for all r, approaching zero as r approaches infinity. If at distance r = R we give the mass m a large enough kinetic energy, so that KE - G M m / r > 0, and if the object moves through empty space so that no energy is dissipated, the work-energy theorem tells us that the increase in PE will never be enough to decrease the KE to zero. That is, the object will never stop.

'escape velocity' is the velocity at which KE - G M m / r = 0
this velocity occurs when KE = G M m / r, so that .5 m v^2 = G M m / r
escape velocity is found by solving this equation for v, obtaining v = sqrt( 2 G M m / r).

Idea 17: Angular Motion is completely analogous to Linear Motion

A one-radian angle any central angle for which the arc distance along a circle is equal to the radius of the circle.

The reasoning for angular motion is identical to that for linear motion, with meters of displacement replaced by radians of angular displacement.

The arc distance corresponding to angular displacement theta is `dsArc = r * theta, where r is the radius of the circle.

Lowercase omega is the Greek letter corresponding to angular velocity, which is the rate of change of angular position and is measured in units of radians per second.

Lowercase alpha is the Greek letter corresponding to angular acceleration, which is the rate of change of angular velocity and is measured in units of radians per second.

The speed of the motion of a point moving around the arc of a circle is equal to the product of the angular velocity of the corresponding radial line and the radius of the circle: v = omega * r.

The acceleration component in the direction of motion of a point moving around the arc of a circle is equal to the product of the angular acceleration of the corresponding radial line and the radius of the circle: vParallel = alpha * r, where vParallel is velocity in (i.e., parallel to) the direction of motion.

Idea 18: Newton's Second Law can be formulated in terms of Angular Motion and Moment of Inertia

The torque exerted by a force is tau = F * r * sin(theta), where tau is the torque exerted by force F applied at distance r from axis of rotation, F making angle theta with moment arm.

In the special case where the force is exerted perpendicular to the moment arm, we have tau = F * r.
Torque in angular dynamics is analogous to force in linear dynamics.

The effect of a net torque on an object depends on the torque and on the difficulty of achieving angular acceleration of the object, in a way analogous to the relationship between mass and net force, where mass is among other things a measure of the difficulty of accelerting an object.

For angular motion the resistance to angular acceleration is measured by the quantity I, called moment of inertia.
For a particle of mass m at distance r from axis of rotation we have I = m r^2.
Once more, moment of inertia is analogous to mass in linear dynamics; just as greater mass requires greater force for a given acceleration, the greater the moment of inertia the greater the torque required to achieve a given angular acceleration.

In general for a real object, I = sum( m r^2), which is the moment of inertia I of a collection of particles of various masses at various distances from axis of rotation.

In general the summation of the m r^2 contributions requires calculus. Among the results we obtain are the following:

For a homogenous circular cylinder with mass M and radius R with axis of rotation through the axis of the cylinder we have I = 1/2 M R^2
For a uniform homogeneous rod with mass M and length L rotating about an axis through the center of the rod and perpendicular to the rod we have I = 1/12 M L^2

Just as Fnet = m a for linear motion, tauNet = I * alpha, where tauNet is the net torque required to achieve angular acceleration alpha on an object with moment of inertia I.

Idea 19: The definitions of Work and KE can be reformulated in terms of Angular Quantites

When we apply torque through an angular displacement we are in fact applying a force through a distance and doing work. Analogous to `dW = F * `ds for linear motion we have `dW = tau * `d`theta, where `dW is the work done by a torque tau acting through angle `d`theta.

Just as work W is equal to the integral of force in direction of motion multiplied by displacement (W = integral ( Fparallel * `ds) we have W = integral ( tau with respect to theta, from theta1 to theta2)

Every particle making up a rotating object has a mass and a velocity and hence a kinetic energy. The kinetic energy of the entire rotating object is the sum of the kinetic energies of all particles making up the object. Particles closer to the axis of rotation have less velocity and hence contribute less to the kinetic energy of the object than particles of the same mass which lie further from the axis of rotation. The moment of inertia I makes it easy to calculate this complicated quantity. We have

KE = .5 * I * omega^2, where KE is the kinetic energy of an object with moment of inertia I rotating about its axis with angular velocity omega. This is analogous to KE = .5 m v^2 for linear motion.

Idea 20: Angular Impulse-Momentum gives rise to a new conservation law.

For linear motion we have impulse `dp = F * `dt, which in a closed system is conserved since objects act on one another with equal and opposite forces. Since rotating objects in contact with one another exert equal and opposite torques on one another they exert equal and opposite angular impulses on one another, where

angular impulse = tau * `dt, where tau is torque and `dt is time interval.

Just as impulse is equal to change in momentum, angular impulse is equal to change in angular momentum, where

angular momentum = I * omega, with I the moment of inertia and omega the angular velocity.

Just as linear momentum is conserved in a closed system, angular momentum is conserved in a closed system.

Angular momentum is not the same as linear momentum. Energy, momentum and angular momentum are all different quantities, each conserved in any closed system.

Part IV: Test #2 through Final

Idea 21: Simple Harmonic Motion results from a Linear Restoring Force

When the net force Fnet acting on a mass m is proportional to displacement of the object from some equilibrium position x we can write Fnet = - k x. The proportionality constant k is called the force constant.

Since Fnet = m a we have m a = - k x, or a = -k/m * x.

** Since a(t) = x '' (t) it follows that x(t) = B sin( omega * t) + C cos(omega * t) = A cos(omega * t + theta0), where omega = sqrt(k/m) and theta0 can be any real number.

When this proportionality exists between Fnet and x, an object displaced from equilibrium and released will oscillate about the equilibrium position with angular frequency omega = sqrt( k / m ).
The motion of such an object is called Simple Harmonic Motion.
The detailed motion can be described by the equation x(t) = A cos( omega * t ), where A is the maximum displacement from equilibrium. A is called the amplitude of the motion. This model assumes that the object reaches its maximum positive displacement from equilibrium at clock time t = 0.
If the model x(t) = A cos( omega * t ) describes the motion then the velocity is described by the equation v(t) = -omega * A * sin(omega * t).
** the velocity function v(t) is the derivative v(t) = x ' (t) = dx/dt of the position function x(t)
If the model x(t) = A cos( omega * t ) describes the motion then the acceleration is described by the equation a(t) = -omega^2 * A * cos( omega * t ).

Note that this acceleration is not uniform. Simple harmonic motion cannot be analyzed using the equations or the reasoning processes we use with uniformly accelerated motion.

** the acceleration function a(t) is the derivative a(t) = v ' (t) = da/dt of the velocity function v(t), the second derivative a(t) = x '' (t) of the position function

* If the position is not at the maximum positive displacement when t = 0 then the model x(t) = A cos( omega * t ) is not appropriate. However a slight modification can give the position function corresponding to any specific initial condition:

* If the initial position is not at the maximum positive displacement, choosing an appropriate theta0 for the general equation of motion x(t) = A cos( omega * t + theta0 ).

* Another possible form of the general equation is x(t) = A sin( omega * t + theta0 ) .

*Using either form theta0 must be chosen to satisfy specified conditions on the motion.

* The position function is x(t) = A sin( omega * t + theta0) gives us velocity function v(t) = -omega * A * sin(omega * t + theta0).

** the velocity function is the derivative of the position function

* The position function is x(t) = A sin( omega * t + theta0) gives us velocity function a(t) = -omega^2 * A * cos( omega * t + theta0 ).

** the acceleration function is the derivative of the velocity function, the second derivative of the position function

* 21.10. Other possible position functions include x(t) = A sin(omega * t) or y(t) = A sin(omega * t) or x(t) = A sin(omega * t + theta0) or y(t) = A sin(omega * t + theta0).

Idea 22: Simple Harmonic Motion can be modeled by the projection of a point moving about a Reference Circle

Simple harmonic motion can be modeled as the projection onto the x axis (or onto the y axis, or indeed onto any axis thru the center of the circle) of motion at constant angular velocity around a reference circle whose radius is equal to the amplitude of the motion.

The forms x(t) = A cos(omega * t), x(t) = A cos(omega * t + theta0) and y(t) = A sin(omega * t + theta0), as well as others, can be obtained from the circular model:

If motion around the reference circle at angular velocity omega starts at the positive x axis when t = 0 then x(t) = A cos(omega * t) is the x coordinate of that point at clock time t.
If motion around the reference circle at angular velocity omega starts at angular position theta0 when t = 0 then x(t) = A cos(omega * t + theta0) is the x coordinate that point at clock time t.
If motion around the reference circle at angular velocity omega starts at the positive x axis when t = 0 then y(t) = A sin(omega * t) is the x coordinate of that point at clock time t.

Idea 23: Velocity and Acceleration in SHM follow the Reference Circle Model

The formulas given earlier for v(t) and a(t) can be found directly from the circular model:

The reference-circle point, which moves on a circle of radius A with angular velocity omega has speed v = omega * A and is directed tangent to the circle, so that the velocity of the reference-circle point is v = omega * A at a right angle to the radian line (i.e., the line from the center of the circle to the reference point).
The centripetal acceleration of the reference-circle point is a = v^2 / A = omega^2 * A and is directed toward the center of the circle.
If the reference-circle point starts at the positive x axis when t = 0 then at clock time t the radial line makes angle omega * t with the positive x axis, so that at clock time t the velocity vector is at a right angle to omega * t. It follows that the x component of the velocity of the reference point is vx(t) = -omega * A * sin(omega * t).
If the reference-circle point which starts at the positive x axis when t = 0 then the direction of the centripetal acceleration is opposite to the direction omega * t of the radial line. It follows that the x component of the centripetal acceleration is ax(t) = -omega^2 * A * cos(omega * t).

Note that reference-circle velocity is parallel to the x axis when the reference circle point lies on the y axis. At these points the reference-circle angle omega * t is at a right angle with the x axis so that | sin(omega * t) | = 1 and | vx(t) | = omega * A.

Thus when the reference circle angle crosses the y axis, the x velocity matches the speed of the point on the reference circle.

Note also that reference-circle centripetal acceleration is parallel to the x axis when the reference circle point lies on the x axis. At these points the reference-circle angle omega * t is parallel to the x axis so that | cos(omega * t) | = 1 and | ax(t) | = omega^2 * A.

Thus when the reference circle angle crosses the x axis, the x acceleration matches the centripetal acceleration of the point on the reference circle.

Idea 24: Energy Relationships in SHM are consistent with the Reference Circle Model

The PE relative to the equilibrium position of a simple harmonic oscillator at displacement x from equilibrium is the work done against conservative forces in order to displace the oscillator from equilibrium to that position.

The restoring force Fnet = - k x is conservative. The work done against this conservative force will be the negative of the work done by the conservative force.
The work done by Fnet between equilibrium and position x is equal to Fave * `ds, where Fave is the average value of Fnet between the two positions and `ds is the displacement.
Since Fnet changes linearly from zero at equiibrium to -k x at position x its average value is Fave = (0 + (-kx) ) / 2 = -k x / 2. This simple average works only because Fnet is linear with respect to x.
Displacement from equilibrium to position x is `ds = x.
Thus the work done by the restoring force is Fave * `ds = -kx / 2 * x = - k x^2 / 2.
The work done against the restoring force is `dPE = - (-k x^2 / 2) = k x^2 / 2.

Thus the PE of the oscillator at position x, relative to equilibrium, is PE = k x^2 / 2.

The PE of the oscillator is therefore maximized at the extreme positions x = A and x = -A.
At maximum amplitude we have PE = .5 k A^2.
At maximum amplitude the velocity of the oscillator is zero so its KE is zero.
The total mechanical energy of the oscillator at maximum amplitude is therefore just its PE
Since the net force Fnet = - k x on the oscillator is conservative there are no dissipative forces and conservation of energy tells us that `dPE + `dKE = 0, so that PE + KE must be constant. It follows that for a simple harmonic oscillator
total mechanical energy = PE + KE = 1/2 k A^2.

Since at position x we have PE = 1/2 k x^2, conservation of energy tells us that

1/2 k x^2 + KE = 1/2 k A^2 so that
KE = 1/2 k A^2 - 1/2 k x^2.

At position x = 0 we have PE = 0 so that velocity is maximized at this point and

KE at equilibrium = 1/2 k A^2
1/2 m vMax^2 = 1/2 k A^2 so that
vMax = sqrt( k A^2 / m). Since omega = sqrt(k/m) we have
vMax = omega * A.

Recall that from the circular model we have already concluded that vMax = omega * A. Thus the energy picture is consistent with the circular model, which itself is consistent with the model obtained using Newton's Second Law and calculus.