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?


1 Answer 1


The translational component is already taken into account by your use of the parallel axes theorem to derive the moment of inertia for a disk rotating about its edge. But if you just use translation you are not accounting for the rotation of the disk... At the lowest point in the arc the points are not all moving in the same direction: some move faster, some move slower, some have vertical components. None of that is taken into account if you assume only translation- but it is when you treat the disk as rotating (which it is, after all).


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