A thin horizontal uniform rod `AB` of mass `m` and length `l` can rotate freely about a vertical axis passing through its end `A`. At a certain moment, the end `B` starts experiencing a constant force `F` which is always perpendicular to the original position of the stationary rod and directed in a horizontal plane. The angular velocity of the rod as a function of its rotation angle `theta` measured relative to the initial position should be.

Options

A. `sqrt((6 F sin theta)/(m l))`
B. `sqrt((2 F sin theta)/(m l))`
C. `sqrt((3 F sin theta)/(m l))`
D. `sqrt((5 F sin theta)/(m l))`

Answer

A

Explanation

(a) work done by the torque
`Delta W = int tau d theta = int_0^(theta) F l cos theta d theta`
`Delta W = F l sin theta`
Now using work energy theorem,
`Delta W = Delta k`
:. `F l sin theta = [(1)/(2)((m l^2)/(3)) omega^2 - 0]`
Which gives, `omega = sqrt((6 F sin theta)/(m l))`.
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