Question

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre ?

• A. $13 MR^2/32 $
• B. $11 MR^2/32$
• C. $9 MR^2/32$
• D. $15 MR^2/32$

Answer 1

The Correct Option is A

Moment of Inertia Calculation for a Disc

The moment of inertia of a full disc is given by:

\(I_{\text{Total disc}} = \frac{MR^2}{2}\)

When a portion of the disc is removed, the mass is reduced. The mass removed is proportional to the area removed. Since the area of the removed portion is one-quarter of the total area, the mass removed is one-quarter of the total mass:

\(M_{\text{Removed}} = \frac{M}{4}  Mass\ is\ proportional\ to\ are\)

The moment of inertia of the removed portion can be calculated considering the perpendicular axis theorem. The formula for the moment of inertia of the removed portion is the sum of two parts:

\(I_{\text{removed}} = \frac{M}{4} \frac{(R/2)^2}{2} + \frac{M}{4} \left( \frac{R}{2} \right)^2\)

Simplifying, we get:

\(I_{\text{removed}} = \frac{3MR^2}{32}\)

The moment of inertia of the remaining disc is the total moment of inertia minus the moment of inertia of the removed portion:

\(I_{\text{Remaining disc}} = I_{\text{Total}} - I_{\text{Removed}}\)

Substituting the values:

\(I_{\text{Remaining disc}} = \frac{MR^2}{2} - \frac{3MR^2}{32} = \frac{13MR^2}{32}\)

Conclusion:

The moment of inertia of the remaining disc after the removal of a portion is \( \frac{13MR^2}{32} \).