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. `1/3.(I^(2)omega^(2))/g`
B. `1/6.(I omega)/g`
C. `1/2.(I^(2)omega^(2))/g`
D. `1/6.(I^(2)omega^(2))/g`

Answer

D

Explanation

The uniform rod of length l and mass m can swing about an axis passing through the end. When it is given angular velocity it starts rotating in vertical plane. When the centre of mass is raised through h, the increase in potential energy is mgh. This is equal to the decrease in kinetic energy.
Using conservation of mechanical energy `DeltaK + DeltaU=0`
`(0-1/2Iomega^(2)) + (mgh) =0 rArr mgh =1/2(m I^(2)/3).omega^(2)`