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A uniform rod of length L and mass M is...

A uniform rod of length L and mass M is held vertical, with its bottom end pivoted to the floor. The rod falls under gravity, freely turning about the pivot. If acceleration due to gravity is g, what is the instantaneous angular speed of the rod when it makes an angle `60^(@)` with the vertical

A

`((g)/(L))^(1//2)`

B

`((3g)/(4L))^(1//2)`

C

`((3 sqrt3g)/(2l))^(1//2)`

D

`((3g)/(2l))^(1//2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`cos 60^(@)= ((L)/(2)- Delta h)/(L//2)`
`rArr (L)/(2)- Delta h= (L)/(4)`
`rArr Delta h= (L)/(4)`

Decreases is PE= `Mg Delta h= "Mg" (L)/(4)`
K.E. of rotation `=(1)/(2) I omega^(2)= (1)/(2) .(ML^(2))/(3) omega^(2)`
`rArr "Mg" (L)/(4) = (ML^(2))/(6) omega^(2)`
`rArr omega^(2) = (6g)/(4L) rArr omega= sqrt((3g)/(2L))`
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