A solid cylinder of mass 50, kg and radius 0. | vedclass

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(D) Given: Mass of the cylinder, $M = 50\, kg$. Radius of the cylinder, $R = 0.5\, m$. Angular acceleration, $\alpha = 2\, 	ext{rev } s^{-2}$.Convert

Vedclass · Accepted Answer

$157$

Vedclass · Answer

(D) Given: Mass of the cylinder, $M = 50\, kg$. Radius of the cylinder, $R = 0.5\, m$. Angular acceleration, $\alpha = 2\, 	ext{rev } s^{-2}$.Converting angular acceleration to $SI$ units: $\alpha = 2 	imes 2\pi\, 	ext{rad } s^{-2} = 4\pi\, 	ext{rad } s^{-2}$.The torque $	au$ produced by the tension $T$ in the string is $	au = T 	imes R$.The moment of inertia $I$ of a solid cylinder about its central axis is $I = \frac{1}{2}MR^2$.Using the relation $	au = I\alpha$, we have $TR = (\frac{1}{2}MR^2)\alpha$.Solving for tension $T$: $T = \frac{1}{2}MR\alpha$.Substituting the values: $T = \frac{1}{2} 	imes 50 	imes 0.5 	imes 4\pi = 50\pi$.Using $\pi \approx 3.14$, $T = 50 	imes 3.14 = 157\, N$.

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