and now we substitute the rolling constraint:

You can do this! It isn’t really that di cult (or that di erent from what you’ve done before).

Note well that the pulley behaves like an extra mass βM in the system – all of this mass has to be accelerated by the actual force di erence between the two masses. If β = 1 – a ring of mass

– then all of the mass of the pulley ends up moving at v and all of its mass counts. However, for a disk or ball or actual pulley, β < 1 because some of the rotating pulley’s mass is moving more slowly than v and has less kinetic energy when the pulley is rolling.

Also note well that the strings do no net work in the system. They are internal forces, with T2 doing negative work on m2 but equal and opposite positive work on M , with T1 doing negative work on M , but doing equal and opposite positive work on m1. Ultimately, the tensions in the string serve only to transfer energy between the masses and the pulley so that the change in potential energy is correctly shared by all of the masses when the string rolls without slipping.

Example 5.7.2: Unrolling Spool

Figure 74: A spool of fishing line is tied to a pole and released from rest to fall a height H, unrolling as it falls.

In figure 74 a spool of fishing line that has a total mass M and a radius R and is e ectively a disk is tied to a pole and released from rest to fall a height H. Let’s find everything: the acceleration of the spool, the tension T in the fishing line, the speed with which it reaches H.

We start by writing Newton’s Second Law for both the translational and rotational motion. We’ll make down y-positive. Why not! First the force:

We use the rolling constraint (as shown) to rewrite the second equation, and divide both sides by R. Writing the first and second equation together:

Next, we tackle the energy conservation problem. I’ll do it really fast and easy:

Example 5.7.3: A Rolling Ball Loops-the-Loop

m,r

H

R

Figure 75: A ball of mass m and radius r rolls without slipping to loop the loop on the circular track of radius R.

Let’s redo the “Loop-the-Loop” problem, but this time let’s consider a solid ball of mass m and radius r going around the track of radius R. This is a tricky problem to do precisely because as the normal force decreases (as the ball goes around the track) at some point the static frictional force required to “keep the ball rolling” on the track may well become greater than µsN , at which point the ball will slip. Slipping dissipates energy, so one would have to raise the ball slightly at the