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A bar of length L carrying a small mass $m$ at one of its ends rotates with a uniform angular speed $\omega$ in a vertical plane about the mid point of the bar. During the rotation, at some instant of time when the bar is horizontal, the mass is detached from the bar but the bar continues to rotate with some $\omega .$ The mass moves vertically up, comes back and reaches the bar at the same point. At that place, the acceleration due to gravity is $g$
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This possible if the quantity $\frac{\omega^{2} I}{2 \pi g}$ is an integer, The total distance travelled by the mass n in air is proportional to $\omega^{2}$, The total distance travelled by the mass in air and its total time of flight are both independent on its mass.
We have, $v=\frac{1}{2} \omega l$ 
$$
T=\frac{2 v}{g}=\frac{\omega l}{g} \text { or } n \frac{2 \pi}{\omega }=\frac{\omega l}{g}
$$
(as completes $n$ rotations with in $I$ ) $\therefore \quad n=\frac{/ \omega^{2}}{2 \pi g}$
Distance travelled $=2 h=\frac{2 v^{2}}{2 g}=\frac{l^{2} \omega^{2}}{4 g}$
$$
T=\frac{2 v}{g}=\frac{\omega l}{g} \text { or } n \frac{2 \pi}{\omega }=\frac{\omega l}{g}
$$
(as completes $n$ rotations with in $I$ ) $\therefore \quad n=\frac{/ \omega^{2}}{2 \pi g}$
Distance travelled $=2 h=\frac{2 v^{2}}{2 g}=\frac{l^{2} \omega^{2}}{4 g}$
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