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 same `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

Options

A. This is possible if the quantity `(omega^(2) l)/(2pi g)` 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 on its mass

Answer

A, C, D

Explanation

Velocity
`v= (l)/(2) omega`
`T= (2v)/(g)= (l omega)/(g)`

Let no. of rotation = n (within T times)
`nT= (I omega)/(g)`
`rArr n.(2pi)/(omega)= (l omega)/(g) rArr n= (l omega^(2))/(2pi g)`