A light rope is wound around a hollow cylinde | vedclass

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(B) The torque $	au$ acting on the cylinder is given by $	au = F 	imes R$. For a hollow cylinder,the moment of inertia $I$ about its central axis i

Vedclass · Accepted Answer

$15$

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(B) The torque $	au$ acting on the cylinder is given by $	au = F 	imes R$. For a hollow cylinder,the moment of inertia $I$ about its central axis is $I = mR^2$. Using the rotational form of Newton's second law,$	au = I \alpha$,where $\alpha$ is the angular acceleration. Substituting the expressions,we get $F 	imes R = (mR^2) \alpha$. Simplifying for $\alpha$,we have $\alpha = \frac{F}{mR}$. Given $F = 52.5\,N$,$m = 5\,kg$,and $R = 70\,cm = 0.7\,m$. Calculating the value: $\alpha = \frac{52.5}{5 	imes 0.7} = \frac{52.5}{3.5} = 15\,rad\,s^{-2}$.

$A$ light rope is wound around a hollow cylinder of mass $5\,kg$ and radius $70\,cm$. The rope is pulled with a force of $52.5\,N$. The angular acceleration of the cylinder will be.....$rad\,s^{-2}$.

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