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Q.

A solid cylinder is made of radius R and height 3R having mass density $ρ$ . Now two half spheres of radius R are removed from both ends. The moment of inertia of remaining portion about axis ZZ’ can be calculated as $\frac{29}{6 K} {πR}^{5} ρ$ . Find K.

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Detailed Solution

$\begin{array}{l} ∵ M_{cylinder} = ({πR}^{2} \times 3 R) ρ \Rightarrow l_{cylynder} = \frac{1}{2} (3 {πR}^{3} ρ) R^{2} \\ M_{sphere} = \frac{2}{3} {πR}^{3} ρ \Rightarrow l_{spheres} = \frac{2}{5} (2 \times \frac{2}{3} {πR}^{3} ρ) R^{2} (Two halves) \\ ∴ l_{remaining} = l_{cylinder} - l_{sphere} = \frac{29}{30} {πR}^{5} ρ \end{array}$

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A solid cylinder is made of radius R and height 3R having mass density ρ. Now two half spheres of radius R are removed from both ends. The moment of inertia of remaining portion about axis ZZ’ can be calculated as 296KπR5ρ. Find K.