1.

A uniform rod AB of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is ml^(2)//3, the initial angular acceleration of the rod will be

Answer»

`(MGL)/2`
`3/2gl`
`(3g)/(2L)`
`(2g)/(3l)`

Solution :Torque about A,
`tau=mgxxl/2=(mgl)/2`
Also `tau=Ialpha`
`therefore` Angular ACCELERATION, `alpha=(tau)/I=(mgl//2)/(ml^(2)//3)=3/2g/l`


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