A uniform rod of length `L` is placed with one end in contact with the horizontal and is then inclined at an angle `alpha` to the horizontal and allowed to fall without slipping at contact point. When it becomes horizontal, its angular velocity will be

Options

A. `omega=sqrt((3g sin alpha)/(L))`
B. `omega=sqrt((2L)/(3gsin alpha))`
C. `omega=sqrt((6g sin alpha)/(L))`
D. `omega=sqrt((L)/(gsin alpha))`

Explanation

Energy consevation between (`1`) and (`2`)
`mg((L)/(2))sinalpha=(1)/(2)xx(mL^(2))/(3)omega`
`omega=sqrt((3gsinalpha)/(L))`