Looking for classes? Ksquare Career Institute, Bengaluru →
Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio -
A
1 : 2
B
$\sqrt{2} : 1$
C
$1 : \sqrt{2}$
D
2 : 1
Detailed Solution
Rotational kinetic energy: $K = \dfrac{1}{2}I\omega^2 = \dfrac{L^2}{2I}$, since $L = I\omega$
Given $K_1 = K_2$: $\dfrac{L_1^2}{2I_1} = \dfrac{L_2^2}{2I_2}$
$\dfrac{L_1^2}{2(I)} = \dfrac{L_2^2}{2(2I)}$
$\dfrac{L_1^2}{L_2^2} = \dfrac{1}{2}$
$\dfrac{L_1}{L_2} = \dfrac{1}{\sqrt{2}}$
So the angular momenta are in the ratio $1 : \sqrt{2}$.
Given $K_1 = K_2$: $\dfrac{L_1^2}{2I_1} = \dfrac{L_2^2}{2I_2}$
$\dfrac{L_1^2}{2(I)} = \dfrac{L_2^2}{2(2I)}$
$\dfrac{L_1^2}{L_2^2} = \dfrac{1}{2}$
$\dfrac{L_1}{L_2} = \dfrac{1}{\sqrt{2}}$
So the angular momenta are in the ratio $1 : \sqrt{2}$.