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

Answer

B

Explanation

Rotational kinetic energy is converted into potential energy and vice versa.
`(1)/(2)omega^(2)=mgh`
`(1)/(2)((ml^(2))/(3))omega^(2)=mgh`
`h=(omega^(2)l^(2))/(6g)`