A uniform solid cylinder with radius R and le | vedclass

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(C) Let $\rho$ be the density of the material of the cylinder.The mass of the original cylinder is $m_1 = \rho \cdot \pi R^2 L$.The moment of inertia

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) Let $\rho$ be the density of the material of the cylinder.The mass of the original cylinder is $m_1 = \rho \cdot \pi R^2 L$.The moment of inertia of the original cylinder about its axis is $I_1 = \frac{1}{2} m_1 R^2 = \frac{1}{2} (\rho \pi R^2 L) R^2 = \frac{1}{2} \rho \pi R^4 L$.The mass of the carved-out cylinder is $m_2 = \rho \cdot \pi (R')^2 L' = \rho \cdot \pi (\frac{R}{2})^2 (\frac{L}{2}) = \rho \cdot \pi \frac{R^2}{4} \cdot \frac{L}{2} = \frac{1}{8} \rho \pi R^2 L$.The moment of inertia of the carved-out cylinder about its axis is $I_2 = \frac{1}{2} m_2 (R')^2 = \frac{1}{2} (\frac{1}{8} \rho \pi R^2 L) (\frac{R}{2})^2 = \frac{1}{2} \cdot \frac{1}{8} \rho \pi R^2 L \cdot \frac{R^2}{4} = \frac{1}{64} \rho \pi R^4 L$.Therefore,the ratio is $\frac{I_1}{I_2} = \frac{\frac{1}{2} \rho \pi R^4 L}{\frac{1}{64} \rho \pi R^4 L} = \frac{64}{2} = 32$.

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