Question 20 of 280
M and R be the mass and radius of a disc. A small disc of radius is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis AB passing through the centre O and perpendicular to the plane of disc is . The value of is ________.
Show answer and solution
Answer: 9
Without cavity
$\begin{aligned}
\text { Mass of removed disc }= & \frac{\mathrm{M}}{\pi \mathrm{R}^2} \times\left(\frac{\mathrm{R}}{3}\right)^2 \pi \\
& =\left(\frac{\mathrm{M}}{9}\right)
\end{aligned}$
M.I. of removed disc
$\begin{aligned}
& =\frac{\mathrm{MR}^2}{18} \\
& \mathrm{I}=\mathrm{I}_1-\mathrm{I}_2=\frac{\mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{18}=\frac{4 \mathrm{MR}^2}{9} \\
& (\mathrm{n}=9)
\end{aligned}$
$\begin{aligned}
\text { Mass of removed disc }= & \frac{\mathrm{M}}{\pi \mathrm{R}^2} \times\left(\frac{\mathrm{R}}{3}\right)^2 \pi \\
& =\left(\frac{\mathrm{M}}{9}\right)
\end{aligned}$
M.I. of removed disc
$\begin{aligned}
& =\frac{\mathrm{MR}^2}{18} \\
& \mathrm{I}=\mathrm{I}_1-\mathrm{I}_2=\frac{\mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{18}=\frac{4 \mathrm{MR}^2}{9} \\
& (\mathrm{n}=9)
\end{aligned}$
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