A uniform rod of length 2L 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)/(2L))`
B. `omega= sqrt((2L)/(3g sin alpha))`
C. `omega= sqrt((6g sin alpha)/(L))`
D. `omega= sqrt((L)/(g sin alpha))`

Answer

A

Explanation

By law of conservation of energy
`P.E.= K.E_(R)`
`rArr mgL sin alpha= (1)/(2) I omega^(2)`
`rArr omega^(2)= (2mgL sin alpha)/(I)` ...(i)
For a rod, moment of inertia `rArr I= (1)/(12) m(2L)^(2) + mL^(2) = (4mL^(2))/(3)` ...(ii)
By eq. (i) & (ii)
`rArr omega^(2)= (2mgL sin alpha)/(4mL^(2)//3)`
`rArr omega^(2)= (3g sin alpha)/(2L) rArr omega= sqrt((3g sin alpha)/(2L))`