Rotation Ques 103
- A hoop of radius $r$ and mass $m$ rotating with an angular velocity $\omega _0$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?
(2013 Main)
(a) $r \omega _0 / 4$
(b) $r \omega _0 / 3$
(c) $r \omega _0 / 2$
(d) $r \omega _0$
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Answer:
Correct Answer: 103.(c)
Solution:
Formula:
$\omega=v / r$
From conservation of angular momentum about bottom most point
$$ \begin{aligned} m r^{2} \omega _0 & =m v r+m r^{2} \times v / r \\ \Rightarrow \quad v & =\frac{\omega _0 r}{2} \end{aligned} $$
BETA
