A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is

Options

A. `(2g)/(3L)`
B. `(3g)/(2L)`
C. `(g)/(L)`
D. `(2g)/(L)`

Answer

B

Explanation

Moment of inertia of rod at point O `l = (mL^(2))/( 3 )`
Now `tau= Ialpha`
`(L )/(2 ) F = (mL^(2))/( 3 ) alpha`

`:. , (mg L)/( 2 ) = (mL^(2))/( 3) alpha`
`alpha= (3)/(2) (g )/(L)`