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

Two rotating bodies A and B of masses m and 2 m with moments of inerti

1.	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
Answer» `L_(A) = (L_(B))/(2)` `L_(A) = 2 L_(B)` `L_(B) gt L_(A)` `L_(A) gt L_(B)` Solution :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)`