1.

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

Answer»

`(1)/(2)(l^(2)omega^(2))/(g)`
`(1)/(6)(l^(2)omega^(2))/(g)`
`(1)/(3)(l^(2)omega^(2))/(g)`
`(1)/(6)(lomega)/(g)`

Solution :Here K.E. of rotation = P.E.
`(1)/(2)IOMEGA^(2)=mgh`
`(1)/(2)XX(1)/(3)ML^(2)omega^(2)=mgh`
implies `h=(l^(2)omega^(2))/(6g)`


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