A solid sphere of mass M and a disc of mass M | vedclass

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(B) $1$. Moment of inertia of a disc of mass $M_d = M/2$ and radius $R$ about a tangent in its plane:Using the parallel axis theorem, $I_{tangent} = I

Vedclass · Accepted Answer

$25 : 56$

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(B) $1$. Moment of inertia of a disc of mass $M_d = M/2$ and radius $R$ about a tangent in its plane:Using the parallel axis theorem, $I_{tangent} = I_{cm} + M_d R^2$.For a disc, $I_{cm} = (1/4) M_d R^2$.So, $I_{disc} = (1/4) (M/2) R^2 + (M/2) R^2 = (1/8) M R^2 + (4/8) M R^2 = (5/8) M R^2$.$2$. Moment of inertia of a solid sphere of mass $M$ and radius $R$ about its tangent:Using the parallel axis theorem, $I_{tangent} = I_{cm} + M R^2$.For a solid sphere, $I_{cm} = (2/5) M R^2$.So, $I_{sphere} = (2/5) M R^2 + M R^2 = (7/5) M R^2$.$3$. Ratio of the moment of inertia of the disc to the sphere:Ratio $= I_{disc} / I_{sphere} = [(5/8) M R^2] / [(7/5) M R^2] = (5/8) * (5/7) = 25/56$.Thus, the ratio is $25 : 56$.

$A$ solid sphere of mass $M$ and a disc of mass $M/2$ have the same radius $R$. The ratio of the moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent is:

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