Let the moment of inertia of a hollow cylinde | vedclass

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(B) The moment of inertia of a hollow cylinder about its central axis is given by the formula:$I = \frac{1}{2} M (R^2 + r^2)$where $M$ is the mass,$R$

Vedclass · Accepted Answer

$16$

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(B) The moment of inertia of a hollow cylinder about its central axis is given by the formula:$I = \frac{1}{2} M (R^2 + r^2)$where $M$ is the mass,$R$ is the outer radius,and $r$ is the inner radius.Given $r = 10\, cm$ and $R = 20\, cm$,we have:$I = \frac{1}{2} M (20^2 + 10^2) = \frac{1}{2} M (400 + 100) = \frac{1}{2} M (500) = 250 M$For a thin cylinder (or a hoop) of the same mass $M$ and radius $r_0$,the moment of inertia about its axis is:$I = M r_0^2$Equating the two expressions for $I$:$M r_0^2 = 250 M$$r_0^2 = 250$$r_0 = \sqrt{250} \approx 15.81\, cm \approx 16\, cm$Thus,the radius of the thin cylinder is approximately $16\, cm$.

Let the moment of inertia of a hollow cylinder of length $30\, cm$ (inner radius $r = 10\, cm$ and outer radius $R = 20\, cm$),about its axis be $I$. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also $I$,is ......... $cm$.

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