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)
Answer (integer)
32
Solution
$I_{1}=\frac{\left(\rho \pi R^{2} L\right) R^{2}}{2} \quad$ ( $\rho$ : density of cylinder)
<br/><br/>
$$
\begin{aligned}
& I_{2}=\frac{\left[\rho \pi\left(\frac{R}{2}\right)^{2} \frac{L}{2}\right]\left(\frac{R}{2}\right)^{2}}{2} \\\\
& \therefore \quad \frac{I_{1}}{I_{2}}=\frac{32}{1}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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