A rod of length L is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod, when it is in vertical position, is

Options

A. `sqrt(g/l)`
B. `sqrt((2g)/l)`
C. `sqrt((3g)/l)`
D. `sqrt(g/(2l))`

Answer

C

Explanation

Mass of the rod is concentrated at its centre of mass. In the vertical position of the rod, the centre of mass is at C.. Now loss in PE = Gain in Rotational KE

`mg l/2 = 1/2 Iomega^(2)`
where `I = (ml^(2))/3`
(The moment of inertia of the rod about axis passing through O and `bot` to the plane of paper)
`therefore (mgl)/2 = 1/2(ml^(2))/3 omega^(2)`
`omega = sqrt((3g)/l)`