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Moment of inertia of a uniform horizontal solid cylinder of mass $M$ about an axis passing through its edge and perpendicular to the axis of the cylinder when its length is 6 times its radius $R$ is
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Verified Answer
The correct answer is:
$\frac{49 M R^2}{4}$
Moment of inertial of solid cylinder, about the axis passing through its centre of mass and perpendicular to its plane
$$
I_{X Y}=M\left(\frac{l^2}{12}+\frac{R^2}{4}\right)
$$
From theorem of parallel axis, moment of inertial of cylinder ahout an axis passing through its edge and perpendicular to it plane.
$$
\begin{aligned}
I_{A B} & =I_{X Y}+M\left(\frac{l}{2}\right)^2 \\
\therefore \quad I_{A B} & =M\left(\frac{l^2}{12}+\frac{R^2}{4}\right)+M \frac{l^2}{4} \\
& =M\left(\frac{l^2}{3}+\frac{R^2}{4}\right)
\end{aligned}
$$
But $\quad l=6 R$
$$
\begin{aligned}
\therefore \quad I_{A B} & =M\left[\frac{(6 R)^2}{3}+\frac{R^2}{4}\right] \\
& =M\left[12 R^2+\frac{R^2}{4}\right] \\
& =M\left(\frac{49 R^2}{4}\right)=\frac{49 M R^2}{4}
\end{aligned}
$$
$$
I_{X Y}=M\left(\frac{l^2}{12}+\frac{R^2}{4}\right)
$$
From theorem of parallel axis, moment of inertial of cylinder ahout an axis passing through its edge and perpendicular to it plane.
$$
\begin{aligned}
I_{A B} & =I_{X Y}+M\left(\frac{l}{2}\right)^2 \\
\therefore \quad I_{A B} & =M\left(\frac{l^2}{12}+\frac{R^2}{4}\right)+M \frac{l^2}{4} \\
& =M\left(\frac{l^2}{3}+\frac{R^2}{4}\right)
\end{aligned}
$$
But $\quad l=6 R$
$$
\begin{aligned}
\therefore \quad I_{A B} & =M\left[\frac{(6 R)^2}{3}+\frac{R^2}{4}\right] \\
& =M\left[12 R^2+\frac{R^2}{4}\right] \\
& =M\left(\frac{49 R^2}{4}\right)=\frac{49 M R^2}{4}
\end{aligned}
$$
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