A rope of negligible mass is wound round a ho | vedclass

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(C) Given: Mass of the hollow cylinder $M = 3 \, kg$. Radius of the hollow cylinder $R = 40 \, cm = 0.4 \, m$. Force applied $F = 30 \, N$. The torque

Vedclass · Accepted Answer

$25$

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(C) Given: Mass of the hollow cylinder $M = 3 \, kg$. Radius of the hollow cylinder $R = 40 \, cm = 0.4 \, m$. Force applied $F = 30 \, N$. The torque $	au$ produced by the force is given by $	au = F 	imes R$. $	au = 30 \, N 	imes 0.4 \, m = 12 \, Nm$. The moment of inertia $I$ of a hollow cylinder about its central axis is $I = MR^2$. $I = 3 \, kg 	imes (0.4 \, m)^2 = 3 	imes 0.16 = 0.48 \, kg \, m^2$. Using the relation $	au = I \alpha$,where $\alpha$ is the angular acceleration: $\alpha = \frac{	au}{I} = \frac{12 \, Nm}{0.48 \, kg \, m^2} = 25 \, rad \, s^{-2}$.

$A$ rope of negligible mass is wound round a hollow cylinder of mass $3 \, kg$ and radius $40 \, cm$. If the rope is pulled with a force of $30 \, N$,then the angular acceleration produced in the cylinder is ........ $rad \, s^{-2}$.

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