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From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is : |
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Answer» `(MR^(2))/(16sqrt(2pi))` `implies a=(2)/(sqrt(3))R` `(M)/(M.)=((4)/(3)piR^(3))/(((2)/(sqrt(3))R)^(3))=sqrt(3)/(2)pi` `implies M.=(2M)/(sqrt(3pi))` `I=(M.a^(2))/(6)=(2M)/(sqrt(3pi))xx(4)/(3)R^(2)xx(1)/(6)` `I=(4MR^(2))/(9sqrt(3pi))` `v=(3000)/(60)=50Hz therefore omega=2piv=100pi" rad/s"` Angular acceleration `a=(omega-omega_(0))/(t)=(100pi-60pi)/(20)=2omega" rad/s"` 
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