System Of Particles And Rotational Motion Ques 28
28. A solid cylinder of mass $50$ $ kg$ and radius $0.5 $ $m$ is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of $2$ revolutions $s^{-2}$ is :
[2014]
(a) $25 $ $N$
(b) $50 $ $N$
(c) $78.5 $ $N$
(d) $157 $ $N$
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Answer:
Correct Answer: 28.(d)
Solution:
- (d) Here $\alpha=2$ revolutions $/ s^{2}=4 \pi$ $ rad / s^{2}$ (given)
$I _{\text{cylinder }}=\frac{1}{2} MR^{2}=\frac{1}{2}(50)(0.5)^{2}=\frac{25}{4}$ $ Kg-m^{2}$
As $\tau=I \alpha \quad$ so $TR=I \alpha$
$\Rightarrow T=\frac{I \alpha}{R}=\frac{(\frac{25}{4})(4 \pi)}{(0.5)} N=50$ $ \pi N=157 $ $N$
BETA
