A bar of length l canying 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 same `omega`. The mass moves ve1tically up, comes back and reaches the bar at the same point. At that place, the acceleration due to gravity is g.

Options

A. this is possible if the quantity `(omega^(2)l)/(2pig)` is an integer
B. the total time of flight of the mass is proportional to `omega^(2)`
C. the total distance travelled by the mass in air is proportional to `omega^(2)`
D. the total distance travelled by the mass in air and its total time of flight are both independent of its mass.

Answer

A, C, D

Explanation

We have, `v=1/2omegal`
So, time period, T`=(2v)/g=(omegal)/g`
or `nxx(2pi)/omega=(omegal)/g`
(as the bar completes .n. rotations within time period T.)
`thereforen=(lomega^(2))/(2pig)=(omega^(2)l)/(2pig)` will be an integer.
So, distance travelled = `2h=(2v^(2))/(2g)=(l^(2)omega^(2))/(4g)`
`therefore` Distance travelled `prop omega^(2)`
So, option (a), (c) and (d) are correct.