A thin uniform rod AB of mass m and length l is hinged at one end to the level floor and stands vertically. If it s allowed to fall, with what angular velocity will it strike the floor?

Options

A. `((mg)/(L))`
B. `((mg)/(3L))^(1//2)`
C. `((g)/(L))`
D. `((3g)/(L))^(1//2)`

Answer

D

Explanation

As, the rod is higged at one end, its moment of inertia about this end is, `l=(ML^(2))/(3)`
Total energy is upright position = total energy on striking the floor
`0+(MgL)/(2)=(1)/(2)l omega^(2)+0=(1)/(2)(ML^(2))/(3)omega^(2)`
`rArr" "g=(Lomega^(2))/(3) or omega=sqrt((3g)/(L))`