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 $\frac{29}{6 K} π R^{5} ρ$ . Find K.

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TestHub · Accepted Answer

∵ M c y l i n d e r = π R 2 × 3 R ρ . l c y l i n d e r = 1 2 3 π R 3 ρ R 2 M s p h e r e = 2 3 π R 3 ρ , l s p h e r e = 2 5 2 3 π R 3 ρ R 2 ∴ l r e m a i n i n g = l c y l i n d e r - l s p h e r e = 29 30 π R 5 ρ .

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