A uniform solid cylinder of length L and radi | vedclass

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(A) Let $\rho$ be the density of the material of the cylinder.The mass of the original cylinder is $M = \rho \cdot \pi R^2 L$.The moment of inertia of

Vedclass · Accepted Answer

$162$

Vedclass · Answer

(A) Let $ho$ be the density of the material of the cylinder.The mass of the original cylinder is $M = ho \cdot \pi R^2 L$.The moment of inertia of the original cylinder about its axis is $I_1 = \frac{1}{2} M R^2 = \frac{1}{2} (ho \pi R^2 L) R^2 = \frac{1}{2} ho \pi R^4 L$.The mass of the carved cylinder is $m = ho \cdot \pi (R/3)^2 \cdot (L/2) = ho \cdot \pi (R^2/9) \cdot (L/2) = \frac{ho \pi R^2 L}{18} = \frac{M}{18}$.The moment of inertia of the carved cylinder about its axis is $I_2 = \frac{1}{2} m (R/3)^2 = \frac{1}{2} (\frac{M}{18}) (\frac{R^2}{9}) = \frac{1}{324} M R^2$.Now,the ratio $I_1/I_2$ is:$\frac{I_1}{I_2} = \frac{\frac{1}{2} M R^2}{\frac{1}{324} M R^2} = \frac{1}{2} 	imes 324 = 162$.

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