Question 437: Centre of Mass and Rotational Motion

Question

A uniform solid cylinder with radius R and length L has moment of inertia I$$_1$$, about the axis of the cylinder. A concentric solid cylinder of radius $$R'=\frac{R}{2}$$ and length $$L'=\frac{L}{2}$$ is carved out of the original cylinder. If I$$_2$$ is the moment of inertia of the carved out portion of the cylinder then $$\frac{I_1}{I_2}=$$ __________.
(Both I$$_1$$ and I$$_2$$ are about the axis of the cylinder)

Accepted Answer

$I_{1}=\frac{\left(ho \pi R^{2} Light) R^{2}}{2} \quad$ ( $ho$ : density of cylinder)
$$
\begin{aligned}
& I_{2}=\frac{\left[ho \pi\left(\frac{R}{2}ight)^{2} \frac{L}{2}ight]\left(\frac{R}{2}ight)^{2}}{2} \\
& 	herefore \quad \frac{I_{1}}{I_{2}}=\frac{32}{1}
\end{aligned}
$$

Centre of Mass and Rotational Motion

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