Question

Two rigid bodies A and B rotate with rotational kinetic energies $E_A$ and $E_B$ respectively. The moments of inertia of A and B about the axis of rotation are $I_A$ and $I_B$ respectively. If $I_A=\frac{I_B}{4}$ and $E_A=100E_B$ the ratio of angular momentum $(L_A)$ of A to the angular momentum $(L_B)$ of B is:

• A. 25
• B. 5/4
• C. 5
• D. 1/4

Answer 1

Answer: (C)

From the formula, rotational K.E. $E_{rotational}=\frac{1}{2}I{\omega }^2=\frac{L^2}{2I}$ $(\because L=I\omega )$ Therefore, $L^2=2EI$ Hence $L_A^2=2E_A{I}_A$ ?(i) and $L_B^2=2E_B{I}_B$ ?(ii) From equation (i) and (ii) $\frac{L_A}{L_B}=\sqrt{\frac{2E_A{I}_B}{2E_B{I}_A}}=\sqrt{\frac{100}{4}}=5$ $\left(\begin{array}{rl}& \because \frac{E_B}{E_A}=4\\ & \text{and}\frac{I_A}{I_B}=100\end{array}\right)$