A uniform rod AB of length L and and weight Mg is hinged at one end A. The rod is kept in the horizontal position by a massless string. If the string is cut, then the angular acceleration of the rod will be
`["M.I. of the rod about A is "(ML^(2))/(3)]`

Options

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

Answer

A

Explanation

When the string is cut, the weight Mg of the rod acting through its centre, produces the torque `(tau)` about A.
`tau=Mg((L)/(2))" …(1)"`
But `tau=Ialpha` where I is the M.I. of the rod about the axis passing through A and it is given by `I=(ML^(2))/(3)`
`therefore" "tau=(ML^(2))/(3).alpha" ...(2)"`
`therefore" From (1) and (2)"`
`(ML^(2)alpha)/(3)=Mg((L)/(2))" "therefore alpha=(3g)/(2L)`