Rotation Ques 59
- A mass $m$ supported by a massless string wound around a uniform hollow cylinder of mass $m$ and radius $R$. If the string does not slip on the cylinder, with what acceleration will the mass fall on release? (2014 Main)
(a) $2 g / 3$
(b) $g / 2$
(c) $5 g / 6$
$g$
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Answer:
Correct Answer: 59.(b)
Solution:
- For the mass $m, m g-T=m a$
As we know, $\quad a=R \alpha$
So, $\quad m g-T=m R \alpha$
Torque about centre of pulley
$$ T \times R=m R^{2} \alpha $$
From Eqs. (i) and (ii), we get, $a=g / 2$
Hence, the acceleration of a body in free fall is $g$.
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