Kevin S. Huang

2008: Problem 12

Topic: Rigid Bodies
Concepts: Kinetic energy, Moments of inertia, Parallel-axis theorem

Solution:

For a disk rotating around its center, we have

E=\frac{1}{2}I_{\text{disk}}\omega^2

For a disk rotating around its edge, we have

E’=\frac{1}{2}I_{\text{edge}}\omega^2

Recall the moment of inertia of a disk is given by

I_{\text{disk}}=\frac{1}{2}MR^2

By the parallel axis theorem, a disk rotating around its edge has moment of inertia

I_{\text{edge}}=I_{\text{disk}}+MR^2=\frac{3}{2}MR^2

Thus,

\frac{E’}{E}=\frac{I_{\text{edge}}}{I_{\text{disk}}}=3

so the answer is D.