# Kinetic energy of disc pivoting about its edge

Suppose we have a disc of radius $R$ suspended vertically from a pivot on its edge. This disc is then given an initial angular displacement $θ$ and permitted to freely swing without friction.

A problem arises when I attempt to calculate the maximum velocity (at equilibrium, when $θ$ is zero).

$$Mgh=\frac{1}{2}Mv^2$$ $$Mg(R-Rcos\theta)=\frac{1}{2}Mv^2$$ $$v=\sqrt{2gR(1-cos\theta)}$$

Using angular quantities, however, we have

$$Mgh = \frac{1}{2}I\omega^2$$ $$Mg(R-Rcos\theta) = \frac{1}{2}(\frac{1}{2}MR^2 + MR^2)\omega^2$$ $$\omega = \sqrt{\frac{4g}{3R}(1-cos\theta)}$$

There is an obvious inconsistency between the results when we manipulate them using $\omega R = v$ to find the linear velocity from the angular result

$$v= \sqrt{\frac{4gR}{3}(1-cos\theta)}$$

My textbook derives the maximum velocity using angular quantities. My question is - is there not a translational component to the kinetic energy as well? (The centre of mass is certainly not stationary) Why is the first result incorrect?