A uniform solid cylinder with radius R and le | vedclass

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(D) The moment of inertia of a solid cylinder of mass $M$ and radius $R$ about its central axis is given by $I = \frac{1}{2}MR^2$. Let the density of

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(D) The moment of inertia of a solid cylinder of mass $M$ and radius $R$ about its central axis is given by $I = \frac{1}{2}MR^2$. Let the density of the material be $\rho$. The mass of the original cylinder is $M_1 = \rho \cdot \pi R^2 L$. Thus, $I_1 = \frac{1}{2} M_1 R^2 = \frac{1}{2} (\rho \pi R^2 L) R^2 = \frac{1}{2} \rho \pi R^4 L$. The carved-out cylinder has radius $r = R/2$ and length $l = L/2$. The mass of the carved-out portion is $M_2 = \rho \cdot \pi (R/2)^2 (L/2) = \rho \cdot \pi (R^2/4) (L/2) = \frac{1}{8} \rho \pi R^2 L = \frac{M_1}{8}$. The moment of inertia of the carved-out portion about the same axis is $I_2 = \frac{1}{2} M_2 r^2 = \frac{1}{2} (\frac{M_1}{8}) (R/2)^2 = \frac{1}{2} \cdot \frac{M_1}{8} \cdot \frac{R^2}{4} = \frac{M_1 R^2}{64}$. Since $I_1 = \frac{1}{2} M_1 R^2$, we have $M_1 R^2 = 2 I_1$. Substituting this into $I_2$, we get $I_2 = \frac{2 I_1}{64} = \frac{I_1}{32}$. Therefore, $I_1/I_2 = 32$ or $32 : 1$.

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