Question

A solid sphere of radius R has its outer half removed to become radius \( \frac{R}{2} \). What is the moment of inertia about its diameter?

• A. \( \frac{1}{2} \) of its initial volume
• B. unchanged
• C. \( \frac{1}{16} \) of initial volume
• D. \( \frac{1}{32} \) of initial volume

Hint:

Moment of inertia scales as \( I \propto MR^2 \); adjust both mass and radius if object is resized.

Answer 1

The Correct Option is D

Initial moment of inertia of sphere: \[ I = \frac{2}{5} MR^2 \] New mass \( M' = M \cdot \left(\frac{1}{8}\right) \) (volume is \( \frac{1}{8} \)) New radius \( R' = \frac{R}{2} \) \[ I' = \frac{2}{5} \cdot \frac{M}{8} \cdot \left(\frac{R^2}{4}\right) = \frac{2}{5} \cdot \frac{M R^2}{32} = \frac{1}{32} \text{ of original } I \]