Question

A circular portion of diameter $R$ is cut out from a uniform circular disc of mass $M$ and radius $R$ as shown in figure. The moment of inertia of the remaining (shaded) portion of the disc about an axis passing through the centre O of the disc and perpendicular to its plane is:

• A. $\frac{15}{32}M{R}^2$
• B. $\frac{7}{16}M{R}^2$
• C. $\frac{13}{32}M{R}^2$
• D. $\frac{3}{8}M{R}^2$

Answer 1

The correct option is C $\frac{13}{32}M{R}^2$

Let $I$ be the Moment of inertia of the disc about O without removal of the circular portion.
Moment of inertia of remaining portion $I_{remainder}=I-I_{cut-off}$
$I=\frac{1}{2}M{R}^2$ (using the formula for MI of a solid disc about the centre and perpendicular to the plane of the disc).
Radius of the cut-off disc is $R_{cut-off}=\frac{R}{2}$
By using parallel axis theorem, $I_{cut-off}=I_O^{\prime }+\left(M_{cut-off}\right)d^2$
where $I_O^{\prime }$ is the Moment of inertia of cut-off disc about the axis passing through O' (which is cm) and $d$ is the distance between the centres O and O'.
Mass of cut-off portion $M_{cut-off}=\pi \rho (R^2-(\frac{R}{2}{)}^2)=\pi \times \rho \times \frac{3}{4}{R}^2$
$\therefore$ Moment of inertia of cut-off disc $I_{cut-off}=\frac{1}{2}{M}_{cut-off}\times (\frac{R}{2}{)}^2=\frac{1}{2}\pi \rho \times \frac{3}{4}{R}^2\times (\frac{R}{2}{)}^2=\frac{3\rho \pi }{32}{R}^4$
$\therefore I_{remainder}=\frac{1}{2}\pi \rho R^2\times R^2-\frac{3\pi \rho }{32}{R}^4=\frac{13}{32}\pi \rho R^2\times R^2=\frac{13}{32}\times \left(\pi \rho R^2\right)\times R^2=\frac{13}{32}M{R}^2$