The moment of inertia of a solid cylinder abo | vedclass

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(B) The moment of inertia of a solid cylinder of mass $M$,radius $R$,and length $L$ about its own axis is given by $I_1 = \frac{1}{2}MR^2$.The moment

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$L=\sqrt{3}R$

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(B) The moment of inertia of a solid cylinder of mass $M$,radius $R$,and length $L$ about its own axis is given by $I_1 = \frac{1}{2}MR^2$.The moment of inertia of the same cylinder about an axis passing through its centre of gravity and perpendicular to its length is given by $I_2 = \frac{ML^2}{12} + \frac{MR^2}{4}$.According to the problem,$I_1 = I_2$,so:$\frac{1}{2}MR^2 = \frac{ML^2}{12} + \frac{MR^2}{4}$.Subtracting $\frac{1}{4}MR^2$ from both sides:$\frac{1}{4}MR^2 = \frac{ML^2}{12}$.Multiplying both sides by $12/M$:$3R^2 = L^2$.Taking the square root of both sides:$L = \sqrt{3}R$.

The moment of inertia of a solid cylinder about its own axis is the same as its moment of inertia about an axis passing through its centre of gravity and perpendicular to its length. The relation between its length $L$ and radius $R$ is

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