Question

A uniform thin circular disc of mass \(M\) and radius \(R\) is shown in the figure. The moment of inertia of the shaded region about the diameter AB of the disc is:

• A. \(\frac{7MR^2}{16}\)
• B. \(\frac{7MR^2}{64}\)
• C. \(\frac{7MR^2}{32}\)
• D. \(\frac{MR^2}{16}\)

Hint:

Whenever a portion is removed from a body, \[ I_{\text{remaining}} = I_{\text{original}} - I_{\text{removed}}. \] Also remember: \[ I_{\text{diameter of disc}} = \frac{MR^2}{4}. \]

Answer 1

The Correct Option is B

Concept: The shaded portion consists of \[ \text{(Left semicircle of radius }R\text{)} - \text{(two semicircular holes of radius }R/2\text{)}. \] The required axis is the diameter \(AB\). Hence, \[ I_{\text{shaded}} = I_{\text{left semicircle}} - 2I_{\text{small semicircle}}. \]

Step 1: Moment of inertia of the left semicircle about \(AB\).
For a complete disc, \[ I_{AB}=\frac{MR^2}{4}. \] Since the axis \(AB\) divides the disc into two equal halves and the integrand \(x^2\) is symmetric, \[ I_{\text{left semicircle}} = \frac12\left(\frac{MR^2}{4}\right). \] Therefore, \[ I_{\text{left semicircle}} = \frac{MR^2}{8}. \]

Step 2: Mass of each removed semicircle.
Radius of each small semicircle: \[ r=\frac{R}{2}. \] Mass is proportional to area. The area of one semicircle is \[ \frac12\pi\left(\frac{R}{2}\right)^2 = \frac{\pi R^2}{8}. \] Since the area of the full disc is \[ \pi R^2, \] the mass of one small semicircle is \[ m = M\left(\frac{\pi R^2/8}{\pi R^2}\right) = \frac{M}{8}. \]

Step 3: Moment of inertia of one small semicircle about \(AB\).
For a full circle of radius \(r\), \[ I_{\text{diameter}} = \frac{m_{\text{full}}r^2}{4}. \] A semicircle contributes exactly half of this value about the same diameter. Hence \[ I_{\text{small}} = \frac{mr^2}{4}. \] Substituting \[ m=\frac{M}{8}, \qquad r=\frac{R}{2}, \] we obtain \[ I_{\text{small}} = \frac{1}{4} \left(\frac{M}{8}\right) \left(\frac{R}{2}\right)^2. \] \[ = \frac{MR^2}{128}. \]

Step 4: Subtract the two removed semicircles.
\[ I_{\text{shaded}} = \frac{MR^2}{8} - 2\left(\frac{MR^2}{128}\right). \] \[ = \frac{16MR^2}{128} - \frac{2MR^2}{128}. \] \[ = \frac{14MR^2}{128}. \] \[ = \frac{7MR^2}{64}. \]

Step 5: Final answer.
\[ \boxed{ I_{AB} = \frac{7MR^2}{64} } \]