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.

Question

Aicharya · Accepted Answer

Step 1: Moment of Inertia of the Original Disc
The moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about its central axis (perpendicular to its plane) is given by:
$I_{\text {original }}=\frac{1}{2} M R^2$
Step 2: Moment of Inertia of the Removed Part
The removed part is a smaller disc of radius $R / 2$.
Since, the original disc has uniform mass distribution, the mass of the smaller disc (proportional to its area) is:
The moment of inertia of a smaller disc about its own center is:
$\begin{gathered}
M_{\text {removed }}=M \times \frac{\pi(R / 2)^2}{\pi R^2}=M \times \frac{1}{4}=\frac{M}{4} \\
I_{\text {removed,center }}=\frac{1}{2} M_{\text {removed }}\left(\frac{R}{2}\right)^2 \\
I_{\text {removed opeter }}=\frac{1}{2} \times \frac{M}{4} \times \frac{R^2}{4}=\frac{1}{32} M R^2
\end{gathered}$
$\begin{gathered}
I_{\text {removed }}=I_{\text {removed,ceater }}+M_{\text {removed }} d^2 \\
I_{\text {remored }}=\frac{1}{32} M R^2+\left(\frac{M}{4} \times \frac{R^2}{4}\right) \\
I_{\text {removed }}=\frac{1}{32} M R^2+\frac{1}{16} M R^2 \\
I_{\text {removed }}=\frac{1}{32} M R^2+\frac{2}{32} M R^2=\frac{3}{32} M R^2
\end{gathered}$
Step 3: Moment of Inertia of the Remaining Part
The moment of inertia of the remaining part is:
$\begin{gathered}
I_{\text {remaining }}=I_{\text {original }}-I_{\text {removed }} \\
I_{\text {remaining }}=\frac{1}{2} M R^2-\frac{3}{32} M R^2 \\
I_{\text {rvmaining }}=\frac{16}{32} M R^2-\frac{3}{32} M R^2 \\
I_{\text {remaining }}=\frac{13}{32} M R^2
\end{gathered}$

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

Solution