Average velocity is the average rate at which position changes between two clock times. Since the velocity might be changing between two given clock times, we speak of the average rate.

The average rate at which any quantity changes in time is

average rate = (change in quantity) / (change in clock time).

In the present case the changing quantity is position so we have

average velocity = average rate of position change = (change in position) / (change in clock time),

or

vAve = Ds / Dt.

It follows that the displacement of an object is the product of its average velocity and the time interval over which it moves:

Ds = vAve * Dt

and that the time required to move through a given displacement at a given velocity is found by dividing displacement by velocity

Dt = Ds / vAve.

Examples:

If an automobile travels 30 meters in 5 seconds, it moves and average of 6 meters every second so its average velocity is 6 meters / second.

If an automobile travels at 50 miles / hour for 6 hours, it travels 50 miles six times, so it travels 300 miles.

If an automobile travels 200 miles at 50 miles / hour, it requires 200 miles / (50 miles / hour) = 4 hours. We can think of dividing the 200 mile trip into four 50-mile segments.

Average Acceleration

Average acceleration is the average rate at which velocity changes between two clock times. Since the acceleration might be changing between two given clock times, we speak of the average rate.

The average rate at which any quantity changes in time is

average rate = (change in quantity) / (change in clock time).

In the present case the changing quantity is velocity so we have

average acceleration = average rate of velocity change = (change in velocity) / (change in clock time),

or

aAve = Dv / Dt.

It follows that the velocity change of an object is the product of its average acceleration and the time interval over which it moves:

Dv = aAve * Dt

and that the time required for velocity to change by a given amount, at a given acceleration, is found by dividing velocity change by acceleration

Dt = Dv / aAve.

Examples:

If the velocity of an automobile changes by 10 mph in 5 seconds, then its velocity changes by an average of 2 mph every second and its average acceleration is 2 mph / sec.. We note that the speedometer will move through a change of 2 mph every second, on the average.

If the velocity of an automobile changes by 18 m/s in 6 seconds, then its velocity changes by an average of 3 m/s every second and its average acceleration is 3 m/s / sec..

If an automobile changes its velocity at the average rate of 4 m/s / sec (or 4 m/s^2), then in 5 seconds its velocity will change by 20 m/s.

If an automobile accelerates at an average rate of 2 m/s / sec, then to change its velocity by 8 m/s will require 4 seconds.

Average Velocity

If a quantity changes at a uniform rate from some initial value to a final value, then the average value of that quantity is the average of the initial and final values:

average value = (initial value + final value) / 2, IF RATE OF CHANGE IN UNIFORM!!!!

If the rate of change is not uniform, this average of the initial and final values might be a reasonable approximation of the average value, but will not usually equal the average value.

If velocity increases uniformly from initial velocity v0 to final velocity vf, then the average velocity is

vAve = (vf + v0) / 2, IF ACCELERATION IS UNIFORM!!!!

If the acceleration is not uniform then the average of vf and v0 might be a reasonable approximation of the average velocity but will not heavily equal the average velocity.

Examples:

If a car's velocity increases uniformly from 30 mph to 50 mph, its average velocity will be 40 mph. If the increase is not uniform, then the average velocity might be greater than 40 mph (e.g., the car accelerates quickly to 49 mph, then slowly to 50 mph, spending more time near 50 mph than near 30 mph) or less than 40 mph (spending more time near 30 mph).

Change in Velocity

Change in velocity is regarded in this course as a fundamental quantity for the purposes of thinking clearly about motion. It isn't fundamental in the standard equations of uniformly accelerated motion, which involve the variables vf, v0, a, Dt and Ds and therefore isn't important for an algebraic approach to physics. It is however a central concept without which it is difficult to think about accelerated motion. It also needs to be distinguished from average velocity.

The change in a quantity is equal to

change in quantity = later value of quantity - earlier value

so change in velocity is

Dv = vf - v0

We can obtain the final velocity of an object by adding its change in velocity to its original velocity:

vf = Dv + v0,

and we can obtain an object's original velocity by subtracting the velocity change from the final velocity:

v0 = vf - Dv.

Examples:

If an automobile increases its velocity from 30 mph to 50 mph, its change in velocity will be 20 mph.

If an automobile moving at 20 mph increases its velocity by 10 mph, its final velocity will be 30 mph.

If an automobile increases its velocity by 15 mph to 40 mph, its original velocity was 25 mph.

Newton's Second Law

If we apply a net force F_net to a mass m, the mass will accelerate more for a larger force, less for a larger mass. It turns out that the relationship is simple: a = F_net / m instead of something like a = `sqrt(F_net) / m^2 or something similarly inconvenient. So we have

a = F_net / m.

If we apply various net forces to a mass, we find that the acceleration is directly proportional to the force (e.g., twice the net force will give twice the acceleration). We express this proportionality as

F_net = m * a,

and this is the most common form in which Newton's Second Law is stated. This is simply a rearrangement of the a = F_net / m relationship of the preceding paragraph.

We can also think of what the net force on an object and its resulting acceleration tell us. Greater force and lesser acceleration will be associated with greater mass, which is consistent with the relationship

m = F_net / a.

It is important to understand that F in the above expressions of Newton's Second Law always stands for the net force, the total of all the forces acting on the object.

The unit of a net force which gives a 1 kg mass an acceleration of 1 m/s^2 is F_net = ma = (1 kg) (1 m/s^2) = 1 kg m/s^2, and is called 1 Newton.

The gravitational force on a mass m in a gravitational field of strength g is F_grav = m * g, in the 'downward' direction. This force is called the weight of the object.

Examples:

If a net force of 30 Newtons is applied to a 10 kg object its acceleration will be 30 Newtons / 10 kg = 3 m/s^2.

If we wish to give a 20 kg object an acceleration of 2 m/s^2, the net force required is (20 kg) * (2 m/s^2) = 40 Newtons.

If a net force of 200 Newtons gives a mass and acceleration of 4 m/s^2, the mass of the object is (200 Newtons) / (4 m/s^2) = 50 kg.

In the vicinity of the Earth an object under the influence of gravity, with no other forces, is observed to accelerate "downward" at 9.8 m/s^2. For a typical 75-kg male, this requires a force of 75 kg * 9.8 m/s^2 = 735 Newtons. This is the weight on Earth of any 75-kg object.

Vectors in the x-y Plane

A vector in the x-y plane has a magnitude and a direction. The direction is measured as an angle `theta, measured in the counterclockwise direction with respect to the positive x axis. The magnitude of the vector is usually designated by a letter like V or R. Here we will use R for the magnitude of the vector.

The vector quantity is completely equivalent in its effect to the combination of its two components Rx and Ry. Here we have

Rx = R cos(`theta) and Ry = R sin(`theta).

If we know the components Rx and Ry, then the magnitude R is easily found from the Pythagorean Theorem:

R = `sqrt(Rx^2 + Ry^2)

and the angle with respect to either the positive or negative x axis is ATN(Ry / Rx):

`theta = ATN(Ry / Rx) if Rx >= 0

`theta = 180 degrees + ATN(Ry / Rx) if Rx = `pi radians + ATN(Ry / Rx) if Rx < 0

The position, velocity or acceleration of an object moving in the x-y plane is a vector quantity with magnitude and components s, sx and sy (position); v, vx and vy (velocity); a, ax and ay (acceleration).

A force applied in the x-y plane is a vector quantity with magnitude F and components Fx and Fy.

Other quantities such as momentum, angular momentum, torque, electric and magnetic fields, temperature gradients and many others, are also vector quantities. The above methods apply to all vector quantities.

Examples:

A velocity with magnitude 30 m/s in the direction defined by angle `theta = 120 deg has components

vx = 30 m/s * cos(120 deg) = - 15 m/s

and

vy = 30 m/s * sin(120 deg) = 26 m/s (approx).

This object can be 'tracked' by two points, one moving at -15 m/s in the x direction, the other at 26 m/s in the y direction.

A force with components -30 Newtons in the x direction and -50 Newtons in the y direction will have magnitude

Fx = `sqrt( (-30 N) ^ 2 + (-50 N) ^ 2) ) = 58 N (approx),

and direction

`theta = 180 deg + ATN(-50 N / (-30 N) ) = 180 deg + 59 deg = 239 deg.

Note that this direction is in the third quadrant, consistent with the negative x and y components of the force.

A force of 58 N at 239 deg is completely equivalent to the combined effect of forces of -30 N in the x direction and -50 N in the y direction. A vector is always completely equivalent in its effect to be combination of its components.

Work

When a constant force is exerted parallel to the direction in which an object moves, then the work done is equal to the product of the force and the distance through which it is applied. This work is equal to the energy expended by the net force.

When a constant force is exerted in a direction other than parallel to the direction in which an object moves, then the work done is equal to the product of the force component in the direction of motion and the distance through which it is applied. A force F exerted in a direction making angle `theta with the direction of motion will have component

Fperpendicular = F * cos(`theta).

The work done is therefore

DW = Fparallel * Ds.

Examples:

If you exert a constant force of 100 Newtons while pushing your car 200 meters, with the force exerted in the direction of motion, you will do work (100 Newtons) * (200 meters) = 20,000 Newton meters = 20,000 Joules. You will expend this much energy in doing the net work. (You will actually expend an additional 100,000 Joules or so in producing thermal energy, which will probably make you sweat; but this is in addition to the energy expended by the force you exert).

If you exert a constant force of 100 Newtons while pushing your car 200 meters, with the force exerted at 60 degrees to the direction of motion, you will apply 100 Newtons * cos(60 deg) = 50 Newtons in the direction of motion, and 100 Newtons * sin(60 deg) perpendicular to the direction of motion. The force exerted perpendicular to the direction of motion does no work. You therefore do work DW = (50 Newtons) * (200 meters) = 10,000 Newton meters = 10,000 Joules. You will expend this much energy in doing the net work (plus whatever it takes to keep your metabolism going during the process).

If your net force does work equal to 30,000 Joules while traveling a distance of 200 meters, then the net force exerted in the direction of motion is (30,000 Joules) / (200 meters) = 150 Newtons.

If you exert a net force of 150 Newtons and expend 7500 Joules of energy as through this force, then you must have traveled (7500 Joules) / (150 Newtons) = 50 meters.

Momentum

The momentum of an object of mass m moving at velocity v is the product p = m * v and, in any collision or other interaction where no external forces are applied, the total momentum of all the objects involved remains constant. Among other things momentum can be regarded intuitively as 'collision effectiveness': a large mass and/or a large velocity gives an object greater influence in a collision.

If we can determine the momentum of an object by its behavior in a collision, then if we know its mass we can determine its velocity

v = p / m,

or if we can also measure its velocity we can find its mass

m = p / v.

Examples:

If the mass of an object moving at 10 m/s is 12 kg, its momentum is (10 m/s) * (12 kg) = 120 kg m/s. If it was to collide head-on with an object with momentum - 80 kg m/s (moving in the opposite direction, as indicated by the -), its momentum would prevail and the net momentum after collision would be positive.

An object of mass 20 kg whose momentum is determined to be 100 kg m/s must be moving at velocity (100 kg m/s) / (20 kg) = 5 m/s.

An object moving at 20 m/s whose momentum is observed to be 160 kg m/s has mass (160 kg m/s) / (20 m/s) = 8 kg.

Power

Average power is the average rate at which work is done. That is,

Pave = DW / Dt,

where DW is the work done during time interval Dt.

Examples:

If you do 20,000 Joules of work in 400 seconds, then you are working at an average rate of 50 Joules / sec, or 50 watts.

If you are working at the rate of 100 watts, then in 30 seconds you will do (100 watts) * (30 sec) = (100 Joules/sec) * (30 sec) = 3000 Joules of work.

To perform 40,000 Joules of work at the rate of 500 watts requires (40,000 Joules) / (500 watts) = (40,000 Joules) / (500 Joules / sec) = 80 seconds.

Impulse and Momentum

When an average net force Fave is applied to any object for time of duration Dt, the momentum of the object changes by Fave Dt, so we say that

Dp = Fave Dt.

This quantity is called the impulse delivered by the force Fave to the object.

If we know how long it takes to accomplish a given momentum change Dp, we can determine the average force as

Fave = Dp / Dt.

If we know the average force Fave applied to accomplish a given momentum change Dp, we can determine the duration of that force as

Dt = Dp / Fave.

In a collision of two objects, the objects exert equal and opposite forces on one another and therefore experience equal and opposite momentum changes.

Examples:

A net force of 2000 Newtons, acting for .02 seconds, will result in a momentum change of (2000 Newtons) * (.02 seconds) = 40 Newton seconds = 40 kg m/s.

A momentum change of 30 kg m/s can be accomplished by an average force of 120 Newtons in (30 kg m/s) / (120 Newtons) = .25 second.

A momentum change of 50 kg m/s will be accomplished in .01 seconds by an average force of (50 kg m/s) / (.01 sec) = 5000 kg m/s^2 = 5000 Newtons.

Kinetic Energy

When a net force F acts on an object of mass m through a distance Ds, the work done is F * Ds and the quantity .5 m v^2 increases by amount F * Ds. The quantity .5 m v^2 therefore represents the energy imparted to the object. Since the quantity .5 m v^2 represents an energy that depends only on the velocity of the object, it is called kinetic energy.

If we know the kinetic energy of an object and its mass, we can easily find its velocity. Since

KE = .5 m v^2,

it follows that

v = `sqrt(2 * KE / m).

If we know the kinetic energy of an object and its velocity, we can find its mass by solving the same equation for m. We obtain

m = 2 * KE / v^2.

Examples:

A 1500 kg automobile moving at 10 m/s has kinetic energy .5 m v^2 = .5 (1500 kg) (10 m/s) ^ 2 = 75,000 kg m^2 / s^2 = 75,000 Joules. If the velocity is doubled, v^2 will quadruple and the kinetic energy will be 300,000 Joules, 4 times as much.

If 2000 Joules of work is done by some net force on a 100 kg object initially at rest, then the KE of the object will become 2000 Joules (since it started at rest, with no KE), and its velocity will be v = `sqrt(2 * 2000 Joules / (100 kg) ) = `sqrt(40 Joules / kg) = `sqrt(40 kg m^2/s^2 / (100 kg) ) = 6.4 m/s (approx).

If 2000 Joules of work is done by some net force on a 100 kg object initially moving at 10 m/s, then the KE of the object will become 2000 Joules + .5(100 kg) (10 m/s) ^ 2 = 7000 Joules (since it started at rest, with no KE), and its velocity will be v = `sqrt(2 * 7000 Joules / (100 kg) ) = `sqrt(140 Joules / kg) = `sqrt(140 kg m^2/s^2 / (100 kg) ) = 11.8 m/s (approx). Note that the increase in velocity is less than in the preceding example.

If 5000 Joules of work increases the velocity of an object from 0 to 10 m/s, then the mass of the object must be m = 2 * 5000 Joules / (10 m/s) ^ 2 = 100 kg.

If a 40-kg object moving at 20 m/s is slowed to 10 m/s, then its KE decreases from 8000 Joules to 2000 Joules and the work done on the object by the net force is -6000 Joules. The object therefore does 6000 Joules of work on whatever entity is applying the net force.

Weight, mass and gravitational field strength g

At a given point in the vicinity of a gravitational mass such as a planet or a star, the gravitational field strength g is the acceleration toward the center of that mass experienced by any object which is permitted to fall freely in that field. At the surface of the Earth g = 9.8 m/s^2.

The weight of an object is the force exerted on it by a gravitational field. Since F = m * a and the acceleration a is just the gravitational field strength g, we have

wt. = m * g.

If we know the weight of an object (e.g., by measuring how much a calibrated spring stretches in order to suspend the object), we can easily determine its mass:

m = wt. / g.

If we were on an unknown planet we could determine its gravitational field strength by measuring the weight of a known mass (e.g., by using a calibrated spring as above):

g = wt. / m.

Examples:

The weight of a 50-kg mass at the surface of the Earth is (50 kg) * (9.8 m/s^2) = 490 Newtons.

If a spring balance tells us that a certain fish weighs 300 Newtons, then its mass must be

m = wt. / g = 300 Newtons / 9.8 m/s^2 = 30.6 kg.

If on a certain asteroid a 5-kg mass is determined to have weight .2 Newtons, then the acceleration of gravity on that asteroid is g = .2 Newtons / (5 kg) = .04 m/s^2 (be careful about jumping--you will go very high, and you might just escape the asteroid completely).

Gravitational Potential Energy

When an object of mass m is raised at constant velocity from ground level to an altitude h in a constant gravitational field of strength g, the force exerted by gravity has magnitude mg and the force required to lift the object must also have magnitude mg, and must be directed in the vertical direction, parallel to the gravitational force (since the object is not accelerating the net force must be 0, so the force required to lift the object must be equal and opposite to the gravitational force). The work done on the object by the force is therefore DW = Fparallel * Ds = mg * h, and the lifting force does work DW = m g * h.

If the object is released and permitted to fall freely back to the ground, gravity will do work DW = m g * h on the object and the object will therefore gain kinetic energy KE gain = PE loss = m g * h

The lifting force therefore imparted an energy m g * h, which was recovered in the form of KE as the object fell back to Earth. At altitude h the object therefore had, by virtue of its position, the potential to regain the energy given to it in the form of KE.

This energy is called the Gravitational Potential Energy of the object relative to ground level (we don't have to measure potential energy from ground level; it can be measured from any convenient fixed point):

PEgrav = m * g * h.

We might measure the gravitational potential energy of an object by using a metered energy supply of some type to lift the object. Then if we know the mass and gravitational field strength we can determine the altitude to which the object is lifted:

h = PEgrav / (m * g).

Or knowing the mass and altitude to which the object is lifted we can determine the gravitational field strength:

g = PEgrav / (m * h).

If we know the gravitational field strength and the altitude we could also determine the mass of the object:

m = PEgrav / (g * h).

Examples:

If an object of mass 50 kg is raised from a reference level to an altitude of 20 meters above this level, in the vicinity of the Earth's surface, then the weight of the object is wt. = (50 kg) * (9.8 m/s^2) = 490 Newtons and the work required is DW = 490 Newtons * 20 meters = 9800 Joules. If the object is dropped and falls freely gravity will do this much work on it as it falls and it will attain a KE of 9800 Joules at the instant it reaches the reference level. (We can determine this velocity by the definition KE = .5 m v^2, obtaining v = `sqrt(2 * KE / m) = 19.7 m/s (approx.)). Relative to the reference level we see that the object had potential energy 9800 Joules at the 20 meter altitude.

By slowly lifting a 50-kg mass hight into a foggy sky using an electric motor equipped with an electric meter, and with very nearly frictionless pulleys, we determine that we expended .04 kW-hrs. of energy, or 144,000 Joules. We conclude that the object attained a PEgrav of 144,000 Joules. Since its weight is 50 kg * 9.8 m/s^2 = 490 Newtons, the altitude of the object must be (144,000 Joules) / (490 Newtons) = 294 meters (approx.). We could also easily enough determine how fast the object would be traveling if it broke free and fell through the fog onto our heads (set KE gain = PE loss and solve for v).

Using an appropriate mechanism we determine that the potential energy gained by a 50-kg mass when lifted to an altitude of 40 meters relative to its starting point is 45,000 Joules. We can find the gravitational field strength to be g = PEgrav / (m * h) = 45,000 Joules / (50 kg * 40 meters) = 22.5 m/s^2 (either gravity is extra strong today or we are on another planet).

Using a water wheel and measuring the amount of water turning the wheel we estimate that the potential energy gained by a certain horse, when raised 4 meters, is 60,000 Joules. The mass of the horse is therefore m = PEgrav / (g * h) = 60,000 Joules / (9.8 m/s^2 * 4 meters) = 1530 kg (approx.). (Very big or very fat horse--that's over 3000 lbs).

Linear Restoring Force

When a light strong spring is displaced x units from its equilibrium position, a force very nearly of the form F = - k * x tends to pull back toward the equilibrium position. This force is linear in x, and is therefore called a linear restoring force. The actually restoring force for a spring is very nearly but not exactly linear.

A simple pendulum, when displaced x units from equilibrium, experiences a very nearly linear restoring force as long as x is small compare to the length the pendulum.

A rubber band experiences a force which is fairly close to, but significantly deviates from linear.

Whenever we have a situation which we wish to approximate by a linear restoring force, we use the form F = - k * x for the relationship between the force and the displacement from equilibrium. The proportionality constant k is called the restoring force constant.

Examples:

If a spring is at equilibrium, then extends to a 10 cm displacement from equilibrium when a mass of .5 kg is hung from it, the force exerted by the spring to support the weight is -(.5 kg) * 9.8 m/s^2) = -4.9 Newtons (the - sign occurs because the force is in the opposite direction to the displacement--if the displacement is down, the force is up, and vice versa). The restoring force constant is therefore k = - F / x = -(-4.9 Newtons) / (.1 meter) = 49 Newtons / meter.

If a pendulum has restoring force constant 30 Newtons / meter, then a displacement of 15 cm from equilibrium will result in a restoring force of - ( 30 Newtons / meter) * (.15 meters) = -4.5 Newtons.

Elastic Potential Energy

When an object subject to a linear restoring force F = - k * x is displaced from equilibrium to position x from equilibrium, the force which displaces the object changes from 0 to k * x (the displacing force will be in the direction opposite to the restoring force and will hence not have the - sign). The force varies linearly from 0 to k * x and hence averages (0 + k * x) / 2 = .5 k * x. This force is applied through distance x, so the work done is DW = Fave * Ds = (.5 k * x) * x = .5 k x^2. If the object is then released, it will be accelerated back to the equilibrium position by forces equal and opposite to those which displaced it and this work will be regained in the form of kinetic energy.

We thus say that, at position x relative to equilibrium, and object subject to the linear restoring force F = - k * x has potential energy PEelastic = .5 k * x^2.

Examples;

If the force constant of a spring is 100 N / m, then at a displacement of x = 20 cm from the equilibrium position an object attached to the spring will have elastic potential energy PEelastic = .5 (100 N/m) * (.2 m) ^ 2 = 2 Joules. If the object is then released, and if no energy is dissipated, it will have 2 Joules of KE when it reaches equilibrium.

The same object would have elastic potential energy PEelastic = 8 Joules at the x = 40 cm position. If released from that position then upon reaching the 20 cm position, where PEelastic = 2 Joules, the 6 Joule difference will manifest itself as kinetic energy.

If 40 Joules are required to displace a spring 20 cm from its equilibrium position, then since PEelastic = .5 k x^2 implies that k = 2 * PEelastic / x^2, we see that k = 2 * 40 Joules / (.2 m) ^ 2 = 200 N / m.

Energy Conservation

When an object moves from one position to another, its potential and kinetic energies can change, and it can do work.

If no work is done by the object, then an increase in kinetic energy requires an equal decrease in potential energy, and an increase in potential energy requires an equal decrease in kinetic energy (e.g., when a metal ball falls its potential energy decreases as its kinetic energy increases by the same amount; if it is tossed up into the air its potential energy increases as its kinetic energy decreases by the same amount).

If in addition the object does a certain amount of work, the sum total of the kinetic and potential energies must decrease by the same amount in order to supply the energy to perform the work.

This situation can be expressed as DKE + DPE + DW = 0, which implies that a difference in one of the quantities must be balanced by an equal and opposite difference in the total of the other two.

Examples:

If a 5 kg mass slides down a ramp, decreasing its altitude by 3 meters, and does no work in the process (assuming therefore a frictionless ramp and no air resistance), then its potential energy decreases, changing by DPE = (5 kg) * (9.8 m/s^2) * (-3 meters) = -147 Joules. This should have the effect of increasing the kinetic energy by 147 Joules. The conservation of energy equation ensurers this: Since DW is assumed to be 0, the only way DKE + DPE + DW = 0 is if DKE = + 147 Joules.

If the same mass slides down the same ramp but, more realistically, does 30 Joules of work against a frictional resistance, then we would expect a kinetic energy increase of only 117 Joules. The conservation of energy equation ensures this: DKE + DPE + DW = 0, so DKE + (-147 Joules) + 30 Joules = 0 and we easily solve to get DKE = 117 Joules.

If a pendulum is pulled back to a 20 cm displacement, giving it an elastic potential energy of 200 Joules, then after it is released it travels through water and does 80 Joules of work against the resistance of the water, it should clearly have 120 Joules of KE at its equilibrium position. The conservation of energy equation ensures this: DKE + DPE + DW = 0, so DKE + (-200 Joules) + 80 Joules = 0 and we easily solve to get DKE = 120 Joules.

Instantaneous Velocity

Suppose that a car is accelerating from rest. It goes from 0 m/s to 10 m/s with strictly increasing velocity. Then it seems obvious that the car will have a different position and a different velocity at every instant, and that it will take every velocity between 0 and 10 m/s (these things seem obvious, but at a deeper level these assumptions need to be carefully analyze and they may not be so).

If we know the position at every instant we can determine the average velocity between any two instants, since we will be able to determine the displacement Ds from the two positions and the time interval Dt between the two instants. Then we would obtain vAve = Ds / Dt.

If we wish to determine the velocity at the instant when the clock time is t, we can come as close as we wish by determining the average velocity between clock times t and t + Dt, choosing Dt to be very small so that the average velocity is determined over a very short time interval following clock time t. As Dt its smaller and smaller, the average velocity gets closer and closer to the velocity at the instant t.

For every Dt there is a corresponding Ds, and an average velocity vAve = Ds / Dt. As Dt shrinks toward 0, vAve approaches v, the velocity at instant t. So we say that v is the limit as Dt approaches 0 of the average velocity Ds / Dt. In the notation used here, we say that v = lim(Dt -> 0) [Ds / Dt] .

Centripetal Acceleration

When an object moves at speed v in a circular path of radius r, it must be constantly turned toward the center of the circle. This changes its velocity by changing the direction of its motion. Any change in velocity entails an acceleration. In this case the acceleration is a = v^2 / r, and is directed toward the center of the circle. This acceleration is called a centripetal (toward-the-center) acceleration.

Examples:

If we swing a 3-kg mass around in a circle of radius 2 meters at a speed of 10 m/s, we must accelerate the mass at a = v^2 / r = (10 m/s)^2 / (2 meters) = 50 m/s^2 toward the center of the circle. This requires a force of (3 kg) * (50 m/s^2) = 150 Newtons, which is equal to the force we must apply to a string or rope in order to keep the object moving in a circle.

If you stand at the outer rim of a merry-go-round of radius 7 meters, and if due to the motion of the merry-go-round you are traveling at 6 meters / second, then your centripetal acceleration is (6 m/s) ^ 2 / 7 meters = 5.14 m / s^2. If your mass is 50 kg, then the frictional force between your feet and the surface of the merry-go-round had better be at least 50 kg * 5.14 m/s^2 = 257 Newtons (compared to your weight, which is 50 kg * 9.8 m/s^2 = 490 Newtons). Otherwise you won't stay in a circular path and will depart the merry-go-round.

If when swinging a 2-kg ball on a string of length 2 meters, resulting in a circular path of radius 2 meters, you somehow measure a force component of 40 Newtons toward the axis of rotation, then this component is the centripetal force; the centripetal acceleration is thus (40 Newtons) / (2 kg) = 20 m/s^2. The velocity of the ball can therefore be found by solving a = v^2 / r for v, obtaining v = `sqrt(a * r), and substituting to obtain v = `sqrt(20 m/s^2 * 2 meters) = 6.4 m/s (approx.).

Newton's Law of Universal Gravitation

When the centers of mass of masses m1 and m2 are separated by distance r, a gravitational force of F = G * m1 * m2 / r^2, with G = 6.67 * 10 ^ -11 N m^2 / kg^2, acts on each object in the direction from its center of mass to that of the other. The objects mutually attract each other along the line connecting their centers of mass.

A model of this law assigns a 'total gravitational flux' of 4 `pi G m to any mass. This total flux is spread out uniformly over every sphere whose center coincides with the object's center of mass and which sphere contains the entire mass. The flux density is the total flux 4 `pi G m divided by the area 4 `pi r^2 of the sphere, and this density gives the gravitational field strength at the surface of the sphere. This gravitational field strength is the acceleration given to a freely falling object at that point; this acceleration is directed toward the center of mass of the object.

Examples:

The force on 1 kg at a distance of 4 * 10 ^ 8 meters from the center of the Earth, the approximate distance at which the Moon orbits the Earth, is approximately F = 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10 ^ 24 kg) * (1 kg) / (4 * 10 ^ 8 m) ^ 2 = .025 Newtons. Compare this with the 9.8 Newton force of gravity on a 1 kg mass at the surface of the Earth.

At the surface of the Earth all free objects accelerate toward the center at 9.8 m/s^2, so an object of mass m will experience force F = m * 9.8 m/s^2. The surface of the Earth is approximately 6.4 * 10 ^ 6 meters from its center, so if M represents the mass of the Earth we have F = G * M * m / r^2. Thus m * 9.8 m/s^2 = G * M * m / r^2, and Earth has mass M = 9.8 m/s^2 * r^2 / G = 9.8 m/s^2 * (6.4 * 10^6 m) ^ 2 / (6.67 * 10^-11 N m^2 / kg^2) = 6 * 10 ^ 24 kg, approximately. This determination of Earth's mass requires only that we know the values of G, the universal gravitational constant, and the radius r of Earth.

The inverse-square nature of Newtons Law of Universal Gravitation, arising from the r^2 in the denominator, implies that a doubling of the separation of two objects will result in an increase by a factor of 4 in the denominator, which will result in 1/4 of the force. The flux model demonstrates this clearly, since doubling the distance doubles the radius of the sphere and hence quadruples its area, spreading the gravitational influence over 4 times the area. More generally if the ratio of two separations of the same two objects is (r2 / r1), the ratio of the gravitational forces is (r1 / r2) ^ 2.

Circular Orbits

A satellite in a circular orbit around a much more massive planet experiences a centripetal force equal to the gravitational force exerted by the planet (what else is there to supply this force?). Thus the gravitational force Fgrav must be equal to the centripetal force m * (centripetal acceleration) = m * v^2 / r: Fgrav = m * v^2 / r.

This equality can be expanded as G * M * m / r^2 = m * v^2 / r; the mass m of the satellite cancels and we are left with G * M / r^2 = v^2 / r. This equation can be solved for M, for v or for r, yielding

M = v^2 * r / G (mass of planet can be found from velocity and radius of orbit),

v = `sqrt(G * M / r) (velocity of orbit found from planet mass and radius), or

r = G * M / v^2 (radius of orbit found from planet mass and orbital velocity).

Gravitational Field at Distance r from Mass M

The gravitational field at distance r from the center of mass of a mass M is F / m = (G * M * m / r^2) / m = G * M / r^2. We think of this acceleration as being a function g(r) of distance from the planet: g(r) = G * M / r^2.

If we know the gravitational acceleration at a given distance from the center of a planet we can determine the mass of the planet: solving g(r) = G * M / r^2 for M we have M = g(r) * r^2 / G.

If we know the gravitational acceleration and the mass of the planet we can determine our distance from the center of a planet: solving g(r) = G * M / r^2 for r we have r = `sqrt(G * M / g(r) ).

Examples:

At a distance of 10000 km (10^7 meters) from the center of Earth (mass 6 * 10^24 kg) the gravitational acceleration is g(10^7 meters) = 6.67 * 10^-11 N m^2 / C^2 * (6 * 10^24 kg) / (10^7 meters) ^ 2 = 1.1 m/s^2 (approx.).

If we can determine the gravitational acceleration at a known distance (say 3 m/s^2 at distance 10^7 meters) from a planet (found for example from the velocity and radius of an orbit), we can determine the mass of the planet: M = (3 m/s^2) * (10^7m) ^ 2 / (6.67 * 10^-11 N m^2 / kg^2) = 4.5 * 10^24 kg.

Potential and Kinetic Energies of a Satellite

The work required to move a satellite of mass m MODIFY THE WORKSHEET from distance r to infinite distance from a planet of mass M is G * M * m / r (found by integration of the force function). As a satellite 'falls' from near-infinite distance to distance r, it gains KE and loses PE. If we take the infinite point as the reference point for PE, the PE at distance r must therefore be negative. So from this standard reference point the PE at distance r is PE = - G * M * m / r.

The KE is easily found by first finding the orbital velocity at distance r: v = `sqrt(G * M / r). It follows that the KE is .5 m v^2 = .5 G * M * m / r--exactly half the magnitude of the PE.

The total energy is KE + PE = .5 G * M * m / r - G * M * m / r = - .5 G * M * m / r.

To change to a 'higher' orbit therefore involves gaining PE by going to a 'less negative' total energy. When a satellite fires its engines to climb higher, it ultimately slows down as a result of the PE gain.

Examples:

At a distance of 5000 km from a planet of mass 6 * 10^24 kg, the PE of a 1000 kg satellite is PE = -6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) * (1000 kg) / (5 * 10^6 m) = -8 * 10^10 Joules. Its KE is 4 * 10^10 Joules (half the magnitude of the PE) and its total energy is -4 * 10^10 Joules.

If the above satellite climbs to 7000 km, its PE increases to about -5.7 * 10^-8 Joules, (which is an increase because to get -5.7 we add 2.3 to 8).  The corresponding KE will decrease from 4 * 10^10 Joules to approximately 2.8 * 10^10 Joules.   The increase in PE is about 2.3 * 10^10 Joules and the decrease in KE is about 1.2 * 10^10 Joules (actually it is exactly half the PE increase), so the total energy increases by about 1.2 * 10^10 Joules.  A satellite would fire its engines to increase is total energy, and its PE would increase while its KE ultimatly decreased; the net energy change is positive since the PE increase is twice the KE decrease.

Angular Velocity and Acceleration

Angular velocity measures the rate at which the angular position of a rotating object changes.  Since different parts of a rigid rotating object move at different speeds, standard velocity will have different values at different points on the object.   At a given instant the angular velocity, however, is a property of the entire object.

The average angular velocity is designated by the Greek letter `omega, and is defined in terms of the angle `theta which measures the angular position of the object.   An angular displacement is a change in `theta:  angular displacement = `theta2 - `theta1.  Just as average velocity is the average rate at which position changes, displacement divided by time interval, average angular velocity is angular displacement divided by time interval:  `omegaAve = D`theta / Dt.

Angular acceleration is defined as the rate at which angular velocity changes:   `alphaAve = D`omega / Dt.

Examples:

If a merry-go-round rotates from angular position 30 degrees to angular position 45 degrees (as measured from some standard position) in .5 seconds, then the angular position is changing at an average rate of 15 degrees / .5 seconds = 30 degrees per second.  Using radians, which are much easier to work with for most angular applications, we see that the angular position change is `pi / 4 radians - `pi / 6 radians = `pi / 12 radians, so the average angular velocity is `pi / 12 radians / (.5 sec) = `pi / 6 radians / second.

If the angular velocity of a merry-go-round increases from `pi / 6 radians / second to 2 * `pi / 3 radians / second in 5 seconds, then the average rate at which angular velocity changes is `alpha = (2 * `pi / 3 radians / sec - `pi / 6 radians / second) / (5 seconds) = `pi / 10 radians / second.

If a force of 30 Newtons is applied perpendicular to the handle of a wrench, at a distance of 20 centimeters (.2 meters) from the axis of rotation of a stuck lid, then the torque applied to the lid is 30 Newtons * .2 meters = 6 meter Newtons.

If a force of 30 Newtons is applied at an angle of 60 degrees to the handle of a wrench, at a distance of 20 centimeters (.2 meters) from the axis of rotation of a stuck lid, then the component of the force perpendicular to the handle is 30 Newtons * sin(60 deg) = 26 Newtons. The torque applied to the lid is therefore 26 Newtons * .2 meters = 5.2 meter Newtons.

A person pushing a merry go round, exerting the force parallel to the rim at the point of application, will have the same effect as if that person was pushing at that point on a wrench plant onto the axle of the merry go round. From the geometry of the situation is clear that the force is perpendicular to the handle of the hypothetical wrench, and the wrench has a length which is equal to the radius of the merry go round. The torque is therefore a equal to the product of the force and the radius of the merry go round. If this torque is greater than the frictional resistance of the merry go round then that torque will be positive in the merry go round will accelerate.

Moment of Inertia

If a very small mass m is constrained to rotate at distance r from an axis, then a net force F at the position of the mass and directed always parallel to its direction of motion will result in an acceleration a = F / m, equivalent to and angular acceleration `alpha = a / r = F / (m r). The force F will produce a torque `tau = F * r, so F = `tau / r and the relationship between angular acceleration and torque can be written

• `alpha = `tau / (m r^2).

The quantity m r^2 is called the Moment of Inertia of the mass m, and if we let I = m r^2 we can write

• `alpha = F / I.

This relationship extends to any solid object constrained to move about a fixed axis, provided I is the sum of all the m r^2 contributions from every part of the object. Using 'Sigma to denote the capital Greek letter which connotes summation, we thus write

• I = `Sigma( m r^2).

For a thin hoop of mass M and radius R, constrained to rotate about an axis through its center, all the mass is effectively concentrated at distance R from the axis, so when all the m r^2 contributions from different parts of the hoop are added up we find that the moment of inertia is simply I = M R^2.

For a uniform disk of radius R and mass M, constrained to rotate about axis through its center, some parts of the disk are closer to the center than others, so r varies from 0 to R. Less mass is concentrated near the center than at further distances from the center. This makes it somewhat challenging to determine the sum of all the m r^2 contributions. The methods of calculus can be used to show that this sum must be I = .5 M R^2.

The methods of calculus also show that

• The moment of inertia of a uniform sphere of radius R and mass M, constrained to rotate about axis through its center, is I = .4 * M R^2.

• The moment of inertia of a thin uniform rod of length L and mass M, constrained to rotate about an axis perpendicular to the rod and through its center, is I = 1/12 * M L^2.

If the rod is constrained to rotate about an axis perpendicular to the rod and through one of its ends then the individual masses making up the rod lie on the average twice as far from the axis of rotation, so that r^2 is on the average four times as great as for an axis through the center and I is therefore four times as great.

• Thus for the axis through the end of the rod I = 1/3 M L^2.

Examples:

A hoop of mass 3 kg and radius1.3 meters has a moment of inertia I = M R^2 = 3 kg * (1.3 m)^2 = 5.07 kg m^2. If a net force of 6 Newtons is applied at the rim of the hoop, and parallel to the rim, then it exerts a torque of 6 N * 1.3 m = 7.8 meter Newtons. The angular acceleration is easily found by solving `tau = I * `alpha for `alpha. We obtain angular accel = `alpha = `tau / I = 7.8 m N / ( 5.07 kg m^2) = 1.5 rad / sec^2.

Two uniform discs are joined and mounted on a thin light axle which coincides with the central axis of each disc. The first disc has radius 10 cm and mass .8 kg, the second has radius 30 cm and mass 3 kg. If a net force of 20 N is applied to the first disc at its rim, directed perpendicular to the rim, then the angular acceleration is found by dividing the net torque by the moment of inertia of the system. The moment of inertia is this sum of the moments of inertia of the two individual discs. The moment of inertia of a disc is .5 M R^2, so the total moment of inertia is I = .5 * .8 kg * (.1 meter)^2 + .5 * 3 kg * (.3 meter)^2 = .139 kg m^2. The torque is `tau = .1 m * 20 N = 2.0 m N. Thus the angular acceleration is `alpha = `tau / I = 2.0 m N / (.139 kg m^2) = 14 rad / s^2, approx..

Torque:  `tau = I * `alpha

A force F applied to an object at a distance r from its axis of rotation, and at an angle `theta to the 'moment arm', results in a 'twisting force', or torque, equal to F * r * sin(`theta). Torque is often designated by the Greek letter `tau, so we write

• `tau = F * r * sin(`theta).

The 'moment arm' is a line from the axis to the point of application of the force, with the line running perpendicular to the axis of rotation. The length of the moment arm is therefore equal to the distance r from the axis of rotation to the point of application.

If the force F is applied in the direction perpendicular to the moment arm then it will have its maximum possible effect on the rotational motion of the object. If it is applied parallel to the moment arm it will have no effect on the rotational motion of the object. Common sense and experience tell us this much.

• This is consistent with the definition of torque as `tau = F * r * sin(`theta), because sin(`theta) takes its maximum value 1 when `theta = 90 degrees (force perpendicular to moment arm) and its minimum value 0 when `theta = 0 (force parallel to moment arm).

For an object constrained to rotate about a fixed axis, the angular acceleration `alpha, the moment of inertia I and the net torque `tau acting on the object have the relationship

• `tau = I * `alpha.

Examples:

A straight lever 2 meters long is being used to lift a heavy object. A force of 250 Newtons is applied at the end of the lever, at an angle of 60 degrees to the direction of the lever. This force results in a torque of `tau = F * r * sin(`theta) = 250 N * 2 m * sin(60 deg) = 435 m N.

If the lever is in the form of a uniform rod of mass 12 kg, and if the load on the lever is suddenly released so that the lever now pivots around the end opposite that to which the force is applied, the lever will undergo an angular acceleration due to the applied torque. The moment of inertia of the lever is I = 1/3 M L^2 = 1/3 * 12 kg * (2 m)^2 = 16 kg m^2. Its angular acceleration will thus be `alpha = `tau / I = 435 m N / (16 kg m^2) = 27 rad / s^2.

The distance traveled around its arc by a point at distance r from the axis of rotation, as the object rotates through angle `theta, is

• arc distance:  s = r * `theta (where `theta is measured in its default units of radians). 

The speed v of the point when angular velocity is `omega is

• speed:  v = `omega * r (`omega measured in standard units of radians / sec).

The acceleration of the point in the direction of the arc is

• acceleration:  a = `alpha * r (`alpha measured in standard units of radians / sec^2).

If the net torque on the rotating object with moment of inertia I is `tau, then the angular acceleration is related to these quantities by

• `tau = I * `alpha,

which is easily solved for `alpha:

• `alpha = I / `tau.

Example:

Suppose the wheel of radius 25 cm accelerates uniformly through 1/3 of a complete turn, starting from rest, in .4 seconds. The angular displacement of the wheel is therefore `d`theta = 1/3 * 2 `pi radians = 2 `pi / 3 radians. A point on its rim will move through an arc distance of `ds = r * `d`theta = 25 cm * 2 `pi / 3 rad = 50 `pi / 3 cm = 52.5 cm, approx..

The average angular velocity of the wheel is `omegaAve = `d`theta / `dt = 2 `pi / 3 rad / (.4 sec) = 5 `pi / 3 rad / sec..

Since acceleration is uniform and the wheel starts from rest its final angular velocity will be `omegaFinal = 2 * `omegaAve = 10 `pi / 3 rad / sec.. The final speed of a point on the rim will be vf = r * `omegaF = 25 cm * 10 `pi / 3 rad / sec = 250 `pi / 3 cm / sec = 260 cm / sec, approx..

The angular acceleration of the wheel will be `alpha = `d`omega / `dt = (10 `pi / 3 rad / sec) / (.4 sec) = 25 `pi / 3 rad / sec^2 = 26 rad / sec^2, approx.. The acceleration around the arc of a point on the rim will be a = r * `alpha = 25 cm * 26 rad/sec^2 = 650 cm / sec^2.

If the wheel has moment of inertia .08 kg m^2, then the net torque required to accelerate it at 26 rad/sec^2 is `tau = I * `alpha = .08 kg m^2 * 26 rad/sec^2 = 2.1 m N, approx..

Moment of Inertia from Net Torque and Angular Acceleration

If we know the net torque on an object and its angular acceleration we easily rearrange the relationship

`tau = I * `alpha

among net torque, moment of inertia and angular acceleration to get

I = `tau / `alpha.

Example:

If an object experiences an angular acceleration of 50 rad / sec^2 when subjected to a net torque of 2.4 m N, then its moment of inertia is I = `tau / `alpha = 2.4 m N / (50 rad/sec^2) = .048 kg m^2.

Example:

A simple pendulum of length L and mass m, when displaced a small distance x from its equilibrium position, will experience a net force very close to F = - (m * g / L) * x. This proportionality is easily obtained by considering the forces involved.

Thus a pendulum with amplitude of motion small compared to its length experiences a restoring force of the form F = - k * x with k = m g / L, and as a result it will undergo simple harmonic motion within angular frequency `omega = `sqrt( k / m) = `sqrt( (m g / L) / m ) = `sqrt ( g / L).

For example a simple pendulum of length 2 meters and mass 3 kg will experience a linear restoring force with constant k = m g / L = 3 kg * 9.8 m/s^2 / ( 2 meters) = 14.7 kg / m^2 = 14.7 N / m. Its angular frequency will be `omega = `sqrt( k / m) = `sqrt ( 14.7 N/m / (3 kg) ) = `sqrt( 4.9 s^-2) = 2.2 rad / sec., approx..

This will be the angular frequency of any simple pendulum of length 2 meters, since as seen above m divides out of the expression for `omega, leaving `omega = `sqrt(g / L) = `sqrt ( 9.8 m/s^2 / (2 m) ) = `sqrt(4.9 s^-2) = 2.2 rad/sec, approx..

Note that a complete cycle around the reference circle corresponds to an angular displacement of 2 `pi rad = 6.28 rad, approx.. At 2.2 rad/s the complete cycle therefore take 6.28 rad / (2.2 rad/s) = 2.7 seconds, approx..

As another example a mass of .5 kg attached to a light spring with force constant 100 N / m will undergo simple harmonic motion with angular frequency `omega = `sqrt( k / m) = `sqrt( 100 N/m / (.5 kg) ) = `sqrt( 200 s^-2) = 14 rad/sec, approx..

This implies that it will complete a cycle in period T = 2 `pi rad / (14 rad / sec) = .44 sec, approx.. Its frequency will be f = 14 rad / sec / (2 `pi rad / cycle) = 2.3 cycles / sec..

m = k / `omega^2 = 1200 N/m / (8.3 rad/s)^2 = 1200 N/m / (69 rad^2 / s^2) = 17 kg. approx..

This is equivalent to calculating KE = 1/2 k ( A^2 - x^2) = 1/2 ( 1000 N/m) ( (.3 m)^2 - (.2 m)^2 ) = 25 J.

amplitude = A = `sqrt( 2 * total energy / k ) = `sqrt ( 2 * 1.6 Joules / (1200 N/m) ) = `sqrt(.0028 m^2) = .053 meters.

Ddd