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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Aicharya · Accepted Answer

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} $

Moment of inertia of a uniform horizontal solid cylinder of mass $M$ about an axis passing through its edge…

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