Question

Two rotating bodies $A$ and $B$ of masses $m$ and $2m$ with moments 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. $L_A=\frac{L_B}{2}$
• B. $L_A=2L_B$
• C. $L_B>L_A$
• D. $L_A>L_B$

Answer 1

The correct option is C $L_B>L_A$

Rotational kinetic energy, $\begin{array}{cc}{K}_{rot}=\frac{1}{2}I{w}^2& ;L=Iw\\ K_{rot}=\frac{1}{2}I\times \frac{L^2}{I^2}& (\because w^2=\frac{L^2}{I^2})\end{array}$
Given ${}_{K_A=K_B}$
$\begin{array}{cc}{K}_{rot}& =\frac{1}{2}\frac{L^2}{I}\\ & \\ K_A& =\frac{1}{2}\frac{L_A^2}{I_A}=K_B=\frac{1}{2}\frac{L_B^2}{I_B}\\ & \\ \frac{L_A^2}{L_B^2}& =\frac{I_A}{I_B}\end{array}$

Hence, $L^2\propto I$

Given $I_B>I_A$

So, $L_B>L_A$