Question

Consider a disc of radius \( R \) and mass \( M \). A hole of radius \( \frac{R}{3} \) is created in the disk such that the center of the hole is \( \frac{R}{3} \) away from the center of the disc. The moment of inertia of the remaining system about the axis perpendicular to the disc and passing through the original center is:

• A. \( \frac{MR^2}{2} \)
• B. \( \frac{13}{27} MR^2 \)
• C. \( \frac{1}{3} MR^2 \)
• D. \( 4MR^2 \)

Hint:

Use parallel axis theorem when subtracting moments for composite bodies with removed mass.

Answer 1

The Correct Option is B

Moment of inertia of complete disc about its center is \( I = \frac{1}{2}MR^2 \).
Hole radius \( r = \frac{R}{3} \Rightarrow \text{mass removed} = M' = M \cdot \left(\frac{r^2}{R^2}\right) = M \cdot \frac{1}{9} \).
Moment of inertia of the removed part about center of disc: \[ I_{\text{hole}} = \frac{1}{2}M'\left(\frac{R^2}{9}\right) + M'\left(\frac{R}{3}\right)^2 = \frac{1}{2} \cdot \frac{M}{9} \cdot \frac{R^2}{9} + \frac{M}{9} \cdot \frac{R^2}{9} = \frac{M R^2}{162} + \frac{M R^2}{81} = \frac{M R^2}{54} \] Final moment of inertia = \( \frac{1}{2}MR^2 - \frac{MR^2}{54} = \frac{27 - 1}{54}MR^2 = \frac{13}{27}MR^2 \)