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A uniform rod AB of length l and mass...

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

A

`(2g)/(3l)`

B

`mg(l)/(2)`

C

`(3)/(2)gl`

D

`(3g)/(2l)`

Text Solution

Verified by Experts

The correct Answer is:
D
`tau=mgxx(l)/(2)`
`tau=Ialpha=(ml^(2))/(3)alpha`
`"mg"(l)/(2)=(ml^(2))/(3)alphaimpliesalpha=(3g)/(2l)`
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