A uniform rod of length `l` is released from the position shown in the figure. The acceleration due to gravity is `g`. There is no friction at any surface. Find the initial angular acceleration of the rod.

Options

A. `(3sqrt(3)g)/(10l)`
B. `(5sqrt(3)g)/(7l)`
C. `(3sqrt(3)g)/(11l)`
D. `(5sqrt(3)g)/(19l)`

Answer

A

Explanation

Suppose `C` is the point through which theinstantaneous axis of rotation passes and `G` is the centre of mass of the rod. From the geometry of the figure.

`CG=lcos30^(@)`
The moment of inertia about `C`
`I=(ml^(2))/12+m(l cos 30^(@))^(2)=5/6 ml^(2)`
if `alpha` is the angular acceleration, then
`(5/6ml^(2))alpha=mg(l/2 cos 30^(@))rArralpha=(3sqrt(3)g)/(10l)`