Question

Two rotating bodies A and B of masses $m$ and $2m$ with moment of inertia \[{I_a}\] and \[{I_b}\] (${I_b} $ > ${I_a}$ ) have equal kinetic energy of rotation. If $L_a$ and $L_b$ be their angular momenta respectively, then
A. \[{L_a} > {L_b}\]
B. \[{L_a} = \dfrac{{{L_b}}}{2}\]
C. \[{L_a} = 2{L_b}\]
D. \[{L_b} > {L_a}\]

Answer 1

Hint:
Moment of inertia of a rotating body is given by:
\[I = M{R^2}\]
The relationship between moment of inertia and kinetic energy is given by:
\[K.E. = \dfrac{1}{2}I{\omega ^2}\]
Angular momentum and moment of inertia are related as in below:
\[L = I\omega \]


Complete step by step solution:
We are given that the masses of the two bodies are m and 2m respectively, with moment of inertia \[{I_a}\] and \[{I_b}\] \[\;\] where (\[{I_b}\]> \[{I_a}\] ).
Given that both the body have same Kinetic energy, If is the angular velocity then we can write:
\[\dfrac{1}{2}{I_a}{\omega _a}^2 = \dfrac{1}{2}{I_b}{\omega _b}^2\]
\[{I_b} = \dfrac{{{I_a}{\omega _a}^2}}{{{\omega _b}^2}}\]
Now, we will write the angular momentum for body respectively,
\[{L_a} = {I_a}{\omega _a}\]
\[{L_b} = {I_b}{\omega _b}\]
\[K{E_a} = \dfrac{{{L_a}^2}}{{2{I_a}}},and.......K{E_b} = \dfrac{{{L_b}^2}}{{2{I_b}}}\]
Equating the above equations,
\[\dfrac{{{L_b}^2}}{{2{I_b}}} = \dfrac{{{L_a}^2}}{{2{I_a}}}\]
\[{L_b} = {\left( {\dfrac{{{I_b}}}{{{I_a}}}} \right)^{1/2}}{L_a}\]
Since ${I_b} $ is greater than $ {I_a}$, $ {\dfrac{{{I_b}}}{{{I_a}}}}$ will be greater than 1. which implies that \[{L_b} > {L_a}\]

Hence, option (D) is correct.

Note.
1. Note that, even though mass is given it wasn’t used, as we are also given the resulting moment of inertia.
2. If we want we can also approach the question by first getting the relationship between both bodies angular momentum, and then go to inertia of the bodies, if the question is posed like that.