# Equations of motion of a disk oscillating inside a cylinder A disk of radius r and mass M is oscillating inside a cylinder with a bigger radius R, without slipping. The goal is to find the dependency on $$\omega$$, the angular velocity of the disk, and $$\frac{d\theta}{dt}$$.

I've seen this problem solved using the relation of the length traveled by the center of mass $$s=(R-r) \theta$$ and it's velocity $$s= v_{cm}t$$. So:

$$v_{cm} t= (R-r) \theta\\ \Rightarrow \frac{\omega r}{(R-r)} = \frac{d\theta}{dt}$$

But this solution doesn't consider aceleration of the center of mass. As the normal force isn't even constant, either should be angular acceleration. So my solution would have started from:

$$v_{cm} t + \frac{1}{2} a_{cm} t^2 = (R-r) \theta$$

and gets impossible complicated. I wonder if there is anything that I'm not aware about that makes the first (and simpler) solution ok rather than mine, or if it's just a simpler aproximation.

Solve by Conservation of Energy:

$$K+U=\text{constant}\tag{1}$$

The kinetic energy has two components, a translational one and a rotational one:

$$K=\frac12 Mv^2+\frac12 I\dot{\theta}^2$$

For slipping without sliding, $$v=r\dot{\theta}$$

The potential energy (gravity):

$$U=MgR(1-\cos\theta)$$

Insert everything into $$(1)$$ and take the time derivative:

$$\frac{\text{d}K}{\text{d}t}+\frac{\text{d}U}{\text{d}t}=0$$

This the Newtonian equation of motion of the system. When solved it yields $$\theta(t)$$.