A thin uniform rod of length `l` and mass `m` is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is `omega`. Its centre of mass rises to a maximum height of -

Options

A. `(1)/(6)(lomega)/(g)`
B. `(1)/(2)(l^(2)omega^(2))/(g)`
C. `(1)/(6)(l^(2)omega^(2))/(g)`
D. `(1)/(3)(l^(2)omega^(2))/(g)`

Answer

C

Explanation

The moment of inertia of the rod about O is `(1)/(3)ml^(2)`. The maximum angular speed of the rod is when the rod is instantaneously vertical. The energy of the rod in this condition is `(1)/(2)Iomega^(2)` where I is the moment of inertia of the rod about O. When the rod is in its extreme portion, its angular velocity is zero momentarily. In this case, the energy of the rod is mgh where h is the maximum height to which the centre of mass (C.M) rises
`therefore mgh=(1)/(2)Iomega^(2)=(1)/(2)((1)/(3)ml^(2))omega^(2)impliesh=(l^(2)omega^(2))/(6g)`