A ring of mass M and radius R is at rest at the top of an incline as shown. The ring rolls down the plane without slipping. When the ring reaches bottom, its angular momentum about its centre of mass is:

Options

A. `MR sqrt(gh)`
B. `MR sqrt((gh)/(2))`
C. `MR sqrt(2gh)`
D. none

Answer

A

Explanation

By work - energy theorem,
`Mgh = (1)/(2) Mv^(2) + (1)/(2) I omega^(2) = (1)/(2) M (R omega)^(2) + (1)/(2) (MR^(2)) omega^(2)`
`:. omega = sqrt((gh)/(R))`
Angular momentum about C.M.
`L = (MR^(2)) omega = MR sqrt(gh)`