Question

From a uniform circular thin disc of mass \(9M\) and radius \(R\), a small disc of radius \(\frac{R}{3}\) is removed. The centre of the small disc is at a distance \(\frac{2R}{3}\) from the centre of the original disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the centre of the original disc is

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

Hint:

For removed parts, subtract their moment of inertia using the parallel axis theorem when required.

Answer 1

The Correct Option is A

Step 1: Moment of inertia of full disc.
Moment of inertia of a uniform disc about its centre is \[ I_{\text{full}} = \frac{1}{2}(9M)R^2 = \frac{9}{2}MR^2 \]

Step 2: Mass of removed small disc.
Mass is proportional to area, \[ m = 9M \times \left(\frac{R/3}{R}\right)^2 = M \]

Step 3: Moment of inertia of removed disc.
About its own centre, \[ I_c = \frac{1}{2}MR^2\left(\frac{1}{3}\right)^2 = \frac{MR^2}{18} \] Using parallel axis theorem, \[ I = I_c + m\left(\frac{2R}{3}\right)^2 = \frac{MR^2}{18} + \frac{4MR^2}{9} = \frac{9MR^2}{18} = \frac{MR^2}{2} \]

Step 4: Moment of inertia of remaining disc.
\[ I_{\text{remaining}} = I_{\text{full}} - I_{\text{removed}} = \frac{9}{2}MR^2 - \frac{1}{2}MR^2 = 4MR^2 \]