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

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Detailed Solution

∵Mcylinder =πR2×3Rρ.Icylinder =123πR3ρR2Msphere =23πR3ρ,Isphere =2523πR3ρR2∴Iremaining portion =Icyinder −2Isphere =2930πR5ρ.
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