The angular momentum of a particle describing uniform circular motion is L. If its kinetic energy is halved and angular velocity doubled, its new angular momentum is

Options

A. 4L
B. `(L)/(4)`
C. `(L)/(2)`
D. 2L

Answer

B

Explanation

We knows `L = I omega" "...(i)`
`L^(2) = 2KI`
From Eq. (i)
`L^(2) = 2K (L)/(omega)`
`L = (2K)/(omega)`
`L. = (2 ((K)/(2)))/(2 omega) = (L)/(4)`