Kevin S. Huang

2010: Problem 24

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

Solution:

The moment of inertia of the original disk is

I_0=\frac{1}{2}MR^2

To calculate the moment of inertia of the removed disk, note that since $M \propto R^2$ , it has radius $r=R/2$ and mass $m=M/4$ . Its moment of inertia around the axis (through its edge) is

I_1=\frac{1}{2}mr^2+mr^2=\frac{3}{2}mr^2=\frac{3}{2}\frac{M}{4}\left(\frac{R}{2}\right)^2=\frac{3}{32}MR^2

where we used the parallel-axis theorem. Thus, the moment of inertia of the remaining object is

I=I_0-I_1=\frac{1}{2}MR^2-\frac{3}{32}MR^2=\frac{13}{32}MR^2

so the answer is B.

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Kevin S. Huang

2010: Problem 24

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2010: Problem 24

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