Question

A solid sphere of mass \( M \) and radius \( R \) has moment of inertia \( I \) about its diameter. It is recast into a disc of thickness \( t \) whose moment of inertia about an axis passing through its edge and perpendicular to its plane, remains \( I \). Radius of the disc will be

• A. \( \frac{R}{\sqrt{19}} \)
• B. \( \frac{R}{\sqrt{15}} \)
• C. \( \frac{2R}{\sqrt{15}} \)
• D. \( \frac{2R}{\sqrt{19}} \)

Hint:

When converting the shape of an object (like from a sphere to a disc), remember to equate the moments of inertia to ensure the correct dimensions are maintained. The radius of the new shape is related to the original shape through this relation.

Answer 1

The Correct Option is C

Step 1: Moment of inertia of a solid sphere.
The moment of inertia of a solid sphere about its diameter is: \[ I_{\text{sphere}} = \frac{2}{5} M R^2 \] Step 2: Moment of inertia of a disc.
The moment of inertia of a disc about an axis passing through its edge and perpendicular to its plane is: \[ I_{\text{disc}} = \frac{1}{2} M R_{\text{disc}}^2 + M R_{\text{disc}}^2 = \frac{3}{2} M R_{\text{disc}}^2 \] where \( R_{\text{disc}} \) is the radius of the disc. Step 3: Equating the moments of inertia.
Since the moment of inertia of the recast disc is equal to that of the sphere, we set the two equal: \[ \frac{2}{5} M R^2 = \frac{3}{2} M R_{\text{disc}}^2 \] Solving for \( R_{\text{disc}} \): \[ R_{\text{disc}} = \frac{2R}{\sqrt{15}} \] Step 4: Conclusion.
Thus, the radius of the disc is \( \frac{2R}{\sqrt{15}} \), which is option (C).