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Text Assignments:
General College Physics: Chapter 8, Problems 7, 11, 17, 20, 23, 24, 29, 33, 38
University Physics: Chapter 9, Problems 58, 59, 64, 67, 71, 76, 79, 82
Bare-bones nutshell summary.
Velocity and Acceleration
Velocity is the rate of change of position with respect to clock time. Acceleration is the rate of change of velocity with respect to clock time.
Velocity can be represented by the slope of a position vs. clock time graph. Acceleration can be represented by the slope of a velocity vs. clock time graph.
Change in position can be represented as an area beneath a velocity vs. clock time graph. Change in velocity can be represented as an area beneath an acceleration vs. clock time graph.
Velocity is the derivative of position. Acceleration is the derivative of velocity.
Change in position is an integral of the velocity function. Change in velocity is an integral of the acceleration function.
If motion is uniformly accelerated then the velocity vs. clock time graph is linear and the average velocity over a given time interval is equal to the average of initial and final velocities. The equations
`ds = (v0 + vf ) / 2 * `dt,
vf = v0 + a `dt,
`ds = v0 `dt + .5 a `dt^2 and
vf^2 = v0^2 + 2 a `ds
all apply to any time interval `dt over which acceleration is uniform.
Newton's Three Laws of Motion
An object in motion remains in motion with no change in velocity unless acted upon by a nonzero net force.
If a nonzero net force F_net acts on an object of mass m the result is a nonzero acceleration. Net force, acceleration and mass are related by F_net = m a.
For every action there is an equal and opposite reaction. That is, if one object exerts a force on another, the other exerts an equal and opposite force on it.
Work/Energy
If net force F_net acts on an object through displacement `ds in the direction of the force then the KE of the object changes by amount F_net * `ds. This product is called the work done by the force, so we have `dW_net = `dKE, where `dW_net is the work done by the net force. This statement is called the work-energy theorem.
If force F_net acts on an object through displacement `ds in the direction of the force, and if F_net can be expressed as the sum F_cons + F_noncons of a a conservative force F_cons and a nonconservative force F_noncons, then
F_net * `ds = F_cons * `ds + F_noncons * `ds. F_cons * `ds is the work done by conservative force and is equal to the negative of the change `dPE in potential energy.
F_noncons * `ds is the work `dW_noncons done by the nonconservative force.
We therefore have `dW_noncons = - `dPE + `dKE, which is rearranged into the full statement of the work-energy theorem
`dW_noncons + `dPE = `dKE.
The total energy E = KE + PE of any isolated system remains constant.
Momentum
If net force F_net acts on an object of mass m for time interval `dt the product F_net * `dt is equal to the change in the quantity m v.
The quantity m v is called the momentum of the object, and
F_net * `dt is called the impulse delivered to the object. The fact that F_net * `dt = `d ( m v ) is called the impulse-momentum theorem.
Provided that the only forces present are the forces involved in the collision, the momentum of two colliding objects immediately after collision is the same as their total momentum before collision.
The total momentum of a close system remains constant.
The work done by a force F between two positions on a straight line is represented as the area beneath the force vs. position graph between those positions.
Gravitation
Newton's Law of Universal Gravitation is expressed as F_grav = G M m / r^2. This is equivalent to stating that the total gravitational flux of a massive object, over any surface containing the object, is 4 pi G M.
Setting this expression equal to the expression m v^2 / r for centripetal acceleration gives us the following relationships for circular orbits: v = sqrt( G M / r), KE = .5 G M m / r.
Dividing the circumference 2 pi r by the velocity v gives the period T of the orbit.
The work done against gravity in separating masses M and m from distance r to infinite separation is G M m / r, which is the area to the right of r beneath the graph of G M m / r^2 vs. r, i.e., the integral from r to infinity of G M m / r.
The gravitational potential function relative to infinite separation is - G M / r. The change in this function between two points is the change in gravitational PE per unit of mass between those points.
Angular Dynamics
Change in position, velocity and acceleration along the arc of a circle are related to angular position, angular velocity and angular acceleration by `ds = r * `dtheta, v = r * omega and a = r * alpha. The equations of uniformly accelerated motion all translate directly into the equations of uniformly accelerated angular motion.
Torque is tau = r * F_perpendicular and moment of inertia is I = sum(m r^2), which for continuous mass distributions becomes an integral. Newton's Second Law and the work-energy theorem also translate directly substituting moment of inertia for mass and torque for force, giving us
tau_net = I * alpha and
tau_net * `dTheta = `dW_net = `dKE.
Angular impulse is tau_net * `dt, which for a constant moment of inertia is equal to the product (I * `dOmega).
The quantity I * omega is analogous to momentum, but unlike the work done by a net torque (which is still expressed in Joules and is interchangeable with any other form of work-energy) angular momentum has different units and is a completely different quantity.
Angular momentum is conserved in collisions, and the total angular momentum of a closed system remains constant.
SHM
SHM results when F_net = - k x and results in motion with angular frequency omega = sqrt(k/m). The resulting position, velocity and acceleration functions can be expressed as
x = A cos(omega t + phi),
v = -omega A sin(omega t + phi) and
a = -omega^2 A cos(omega t + phi).
Energy relationships for SHM are as follows:
Total energy is E = k A^2 / 2,
potential energy at position x is k x^2 / 2 so KE at position x is
KE = E - PE = k A^2 / 2 - k x^2 / 2 = .5 k ( A^2 - x^2).
All aspects of SHM can be modeled in terms of a projection on an approporiate straight line through the origin of a point in motion with angular velocity omega = sqrt(k/m) about a circle of radius A centered at the origin.
Continuing from preceding class:
What is the angular KE of the 1-cm section of the half-meter stick between the 34.5 cm mark and the 35.5 cm mark, if the meter stick is rotating at 3 rad / sec?
Recall that the linear mass density of the meter stick is 1.2 g / cm.
The KE of this 1.2 cm section of the meter stick is .5 m v^2. At this position (which is on the average about 10 cm from the axis of rotation) we have
So the KE is
Alternatively, the moment of inertia of the 1-cm segment is approximately equal to m r^2 = .0012 kg * (.10 m)^2 = .000012 kg m^2, so its angular KE is
What is the angular KE of the 1-cm section between the 39.5 cm mark and the 40.5 cm mark, at 3 rad / sec?
This can be calculated in exactly the same way as before, using r = 15 cm.
v will be 1.5 times as great so .5 m v^2 will be 1.5^2 = 2.25 times as great as the .000054 Joules at the 10-cm distance.
r is 1.5 times as great so moment of inertia I = m r^2 is 1.5^2 = 2.25 times as great, so that .5 I omega^2 will be 2.25 times as great as the previous .000012 kg m^2, or .000027 kg m^2. At the same angular velocity omega, the produce .5 I omega^2 will be 2.25 times as great.
So either way KE is 1.5^2 = 2.25 times as great as before. We end up with KE = .000122 Joules, to 3 significant figures.
What is the angular KE of the 1-cm section between the 44.5 cm mark and the 45.5 cm mark, at 3 rad / sec?
Here r is twice as great as in the first case, which by reasoning similar to that above gives us 4 times the moment of inertia and 4 times the KE at the same angular velocity. The moment of inertia will be .000028 kg m^2/s^2 and the KE will be .000216 Joules.
Does the angular KE of a 1-cm section increase linearly with distance from the axis of rotation?
r increases from 10 cm to 15 cm to 20 cm, and KE increases from 1 * .0000540 J to 2.25 * .0000540 J to 4 * .0000540 J . So as r increases we see that KE / cm is increasing at an increasing rate.
Estimate the moment of inertia of 5-cm segments of the half-meter stick, from 0 to 5 cm relative to the axis of rotation, from 5 cm to 10 cm, etc..
From 0 to 5 cm the mass is 1.2 g/cm * 5 cm = 6 grams.
The midpoint of the segment is at 2.5 cm. Using this as an approximate average for r^2 (and remember that r^2 is nonlinear and doesn't take its average value at the midpoint so this is only an approximation) we get I = m r^2 = .006 kg * (.025 m)^2 = .00000375 kg m^2.
Similar approximations for other segments give us .000032 kg m^2, .00009375 kg m^2, .00015 kg m^2 and .000304 kg m^2.
Add these up to .000585 kg m^2. We have identical segments on the other side of the axis of rotation, so our total moment of inertia is approximately 2 * .00585 kg m^2 = .00117 kg m^2.
What is the actual moment of inertia of the stick?
The formula for a uniform rod of mass M and length L, rotating about its center, is 1/12 M L^2 = 1/12 * .060 kg * (.50 m)^2 = .00125 kg m^2. Our approximation based on 5-cm intervals was pretty good.
A graph of r^2 vs. r is concave up, and a midpoint approximation to any quantity which is proportional to r^2 is therefore an underestimate. We see that our approximation was indeed a about 6% lower than the actual value.
If a force of .3 Newtons is exerted at a distance of 25 cm from the axis of rotation and perpendicular to that axis, what is the torque?
Torque = force perpendicular to moment arm * length of moment arm = .3 N * .25 m = .075 m N.
What therefore is the angular acceleration of the meter stick?
For linear acceleration F = m a so a = F / m.
In rotation torque is represented by tau and we have tau = I * alpha, so alpha = tau / I and we have
What would be the resulting angular velocity of the meter stick?
2 cm is 2 / 25 rad = .04 rad.
Acceleration 240 rad/s^2 through .04 rad, from rest, results in final angular velocity
omega_f = sqrt( omega_0^2 + 2 alpha `ds) = ... = 2.2 rad/s.
How much work will be done on the meter stick if the force acts through a distance of 2 cm?
.3 N * 2 cm = .006 Joules.
What will be the resulting KE of the meter stick?
Oughta be .006 Joules.
Note that .5 I omega^2 for I = .00125 kg m^2 and omega = 2.2 rad/s^2 should be the same, up to roundoff error.
Check these figures to see that they work out.
What was the velocity with which the pearl pendulum struck the meter stick?
What was the angular momentum of the pearl before collision, relative to the axis of rotation?
What was the angular momentum of the meter stick immediately after collision?
What was the angular momentum of the pearl immediately after collision?
Was angular momentum conserved?