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Question

(B) The moment of inertia of a solid cylinder of mass $M$,length $L$,and radius $R$ about an axis passing through its centre of mass and perpendicular

Vedclass · Accepted Answer

$I_2 - I_1 = M R^2$

Vedclass · Answer

(B) The moment of inertia of a solid cylinder of mass $M$,length $L$,and radius $R$ about an axis passing through its centre of mass and perpendicular to its longitudinal axis is given by $I_{CM} = M(\frac{L^2}{12} + \frac{R^2}{4})$.Given $L = 2R$,we substitute this into the formula:$I_1 = M(\frac{(2R)^2}{12} + \frac{R^2}{4}) = M(\frac{4R^2}{12} + \frac{R^2}{4}) = M(\frac{R^2}{3} + \frac{R^2}{4}) = M(\frac{7R^2}{12})$.The moment of inertia about an axis passing through one end and perpendicular to the longitudinal axis is given by the parallel axis theorem: $I = I_{CM} + M d^2$,where $d = \frac{L}{2} = R$.$I_2 = I_1 + M R^2 = M(\frac{7R^2}{12}) + M R^2 = M(\frac{7R^2 + 12R^2}{12}) = M(\frac{19R^2}{12})$.Thus,$I_2 - I_1 = M R^2$.

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