Two rotating bodies A and B of masses m and 2 m with moments of inertia `I_(A)` and `I_(B) (I_(B) gt I_(A))` have equal kinetic energy of rotation . If `L_(A)` and `L_(B)` be their angular momenta respectively , then

Options

A. `L_(A) = (L_(B))/(2)`
B. `L_(A) = 2 L_(B)`
C. `L_(B) gt L_(A)`
D. `L_(A) gt L_(B)`

Answer

C

Explanation

Here `m_(A) = m , m_(B) = 2m`
Both bodies A and B have equal kinetic energy of rotation
`k_(A) = k_(B) implies (1)/(2) I_(A) omega_(A)^(2) = (1)/(2) I_(B) omega_(B)^(2) implies (omega_(A)^(2))/(omega_(B)^(2)) = (I_(B))/(I_(A)) " " …. (i)`
Ratio of angular momenta ,
`(L_(A))/(L_(B)) = (I_(A) omega_(A))/(I_(B) omega_(B)) = (I_(A))/(I_(B)) xx sqrt((I_(B))/(I_(A))) " "` [Using eqn. (i)]
`= sqrt((I_(A))/(I_(B)) lt 1 (because I_(B) gt I_(A)) therefore L_(B) gt L_(A)`