A uniform cylinder has a radius R and length | vedclass

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(B) The moment of inertia $(I_1)$ of a uniform cylinder of mass $M$,radius $R$,and length $L$ about its central longitudinal axis (passing through the

Vedclass · Accepted Answer

$L = \sqrt{3} R$

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(B) The moment of inertia $(I_1)$ of a uniform cylinder of mass $M$,radius $R$,and length $L$ about its central longitudinal axis (passing through the centre and normal to the circular face) is given by $I_1 = \frac{1}{2} M R^2$.The moment of inertia $(I_2)$ of the same cylinder about an axis passing through its centre and perpendicular to its length (transverse axis) is given by $I_2 = M \left( \frac{L^2}{12} + \frac{R^2}{4} \right)$.According to the problem,$I_1 = I_2$,so:$\frac{1}{2} M R^2 = M \left( \frac{L^2}{12} + \frac{R^2}{4} \right)$Dividing both sides by $M$:$\frac{R^2}{2} = \frac{L^2}{12} + \frac{R^2}{4}$Subtracting $\frac{R^2}{4}$ from both sides:$\frac{R^2}{4} = \frac{L^2}{12}$Multiplying by $12$:$3 R^2 = L^2$Taking the square root of both sides:$L = \sqrt{3} R$.

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