Question

A circular disc of radius R and thickness (R)/(6) has moment of inertia I about an axis passing through its centre perpendicular to its plane. It is melted and recast into a solid sphere. The moment of inertia of the sphere about its diameter is:

• A. \(I\)
• B. \(\dfrac{2I}{8}\)
• C. \(\dfrac{I}{5}\)
• D. (I)/(10)

Hint:

Always conserve mass (or volume) when objects are reshaped.

Answer 1

The Correct Option is D

Step 1: Volume conservation:
\( \pi R^2 \left(\dfrac{R}{6}\right) = \dfrac{4}{3}\pi r^3 \)
\( \Rightarrow r = \dfrac{R}{2} \) 

Step 2: Disc moment of inertia:
\( I = \dfrac{1}{2}MR^2 \) 

Step 3: Sphere moment of inertia:
\( I_s = \dfrac{2}{5}Mr^2 = \dfrac{2}{5}M\left(\dfrac{R}{2}\right)^2 = \dfrac{MR^2}{10} \) 

Step 4: Hence:
\( I_s = \dfrac{I}{5} \)