Class Notes, Fall 2001

Week of	Aug 22	Sep 5	Oct 1	Nov 5	Phy 205 page
	Aug 27	Sep 10	Oct 8	Nov 12
		Sep 17	Oct 15	Nov 19
		Sep 24	Oct 22	Nov 26
			Oct 29		Notes for Fall, 2000

Phy 205, Week 12

7 Nov Angle, angular velocity, angular acceleration:

These behave just like the variables used to describe motion in one dimension back in chapter 2, x, dx/dt, dv_x/dt. Here though you are using the one coordinate θ to describe rotation about a single axis instead of the one coordinate x to describe motion along a line.

Start with the relation between arc length along a circle, s = R  θ, differentiate with respect to time to get

relating the speed along the circumference of the circle to the angular speed.

Kinetic energy of a rotating object. Instead of doing the general case, I'll start with the first simple case: two point masses. I specified two masses (m₁ and m₂) at respective distance from the axis of ration (r₁ and r₂). The total kinetic energy of this system is of course

The masses are rotating at a rate ω, so I can figure out their speeds from the preceeding equation, v = R ω, and substitute.

and rewrite this as

The combination in the parentheses is called the moment of inertia (for two point masses.) If you have more masses you have more terms. If you have a continuous distribution of mass, it will translate into an integral.

I showed how to get the moment of inertia for the simple case of a thin wheel for which the mass is concentrated almost all at a constant distance from the axis. ( MR² )

For other cases you have to do some integrals, and I postponed doing them. I applied this to the question of rolling something downhill. Use conservation of energy to do so and show why one object will roll faster than another.

9 Nov Show how to compute a moment of inertia: a disk. Start by asking how to find the area of a circle. The standard answer, πR² may be well-known, but where does it come from?

Divide the circle using concentric circles. The region between the two closely-spaced circle of radii r and r + Δr is pretty close to a rectangle. A long, thin rectangle. One side is Δr and the other side is 2 π r. This gives the area as the base times the height, 2 π r Δr. Of course this is valid only for small enough θr, but you eventually take a limit as this approached zero. The total area is a sum over all of these.

N å i=1  2 π ri Δ ri

The limit of these as all the Δr's approach zero is the definition of the integral

ó õ R 0  2 π  r dr

and this is πR².

Apply the same method to finding the moment of inertia of a disk. Use the same picture to divide the disk into tiny rings. For each ring the moment of inertia is easy. All the mass of the ring as at essentially a constant distance from the origin. The piece of the moment of inertia from this ring is then

to find the Δm, assume that the mass is uniformly distributed throughout the disk. The amount of mass in one ring is then proportional to the area of that ring.

where A is the area of the whole disk, πR,², and ΔA = 2 π r Δr.

The contribution to the total moment of inertia is then

The moment of inertia of the disk is the integral of

from zero to R. The result is M R²/2.

Show how to use energy to solve the problem of a mass hung over a pulley and attached to another mass on a horizontal track. Start everything from rest and choose the zero point of potential energy at the position of the mass (m₂) on the track. The hanging mass is a distance d down from the height of the track.

Later the energy comes from the kinetic energy and the changed potential energy of the masses. When the upper mass m₂ moves a distance x, the hanging mass moves the same amount.

If there is no friction these two energies are equal. Use the fact that v = R ω and that the moment of inertia of the disk-shaped pulley was just calculated to re-write this as

The value of "d" cancels out. This is then

You can actually find the acceleration from this if you differentiate the equation with respect to time. The left side is no problem; that's just v=dx/dt. How do you differentiate v² with respect to time?

There are only a few ways that you have to do derivatives.
1. You've memorized it.
2. You go back to the definition of the derivative.
3. You use the chain rule.
4. You use the product rule.

This time it's the chain rule.
All of these v's are really v_x's and the a's are a_x's, but in the middle of the calculation I'll drop the subscripts and then restore them at the end.

d v2 dt =   dv2 dv   dv dt = 2 v a

Differentiate the energy equation above and put these into it.

Rewrite this, canceling various factors.

or

ax =   m1 g m1 + m2 + M/2