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

Answer

D

Explanation

Loss in KE = Gain in PE
`(1)/(2)l omega^(2)="mgh and the Ml of the rod l "=(ml^(2))/(3)`
`(1)/(2)((ml^(2))/(3))omega^(2)=mgh rArr h=(1)/(6)(l^(2)omega^(2))/(g)`