A uniform circular disc of radius ' R ' and mass ' M ' is rotating...

A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius

\mathrm{R} / 2

is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above. A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing

A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing

Options

A $\frac{7}{32} \mathrm{MR}^2$
B $\frac{9}{32} \mathrm{MR}^2$
C $\frac{17}{32} \mathrm{MR}^2$
D $\frac{13}{32} \mathrm{MR}^2$

Correct answer

D. $\frac{13}{32} \mathrm{MR}^2$

Step-by-step solution

\begin{aligned} & I_{\text{complete }}=\frac{M R^2}{2} \\ & \begin{aligned} I_{\text{removed }} & =\left(\frac{M}{4}\right)\left(\frac{R}{2}\right)^2 \frac{1}{2}+\left(\frac{M}{4}\right)\left(\frac{R}{2}\right)^2 \\ & =\frac{3 M R^2}{32} \\ I=\frac{M R^2}{2} & -\frac{3 M R^2}{32}=\frac{13}{32} M R^2\end{aligned}

\end{aligned}$

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