Question

Two rotating bodies $P$ and $Q$ of masses $m$ and $2m$ with moment of inertia $I_P$ and $I_Q$ ($I_Q > I_P$) have equal Kinetic energy of rotation. If $L_P$ and $L_Q$ be their angular momenta respectively, then

• A. $L_Q = 0$
• B. $L_Q = L_P$
• C. $L_Q < L_P$
• D. $L_Q > L_P$

Hint:

Think of this using a simple direct proportionality concept: since $E_k = \frac{L^2}{2I}$ is constant, it means $L^2 \propto I$, or simply $L \propto \sqrt{I}$. Therefore, the body with the larger moment of inertia must automatically carry the larger angular momentum to keep the kinetic energies balanced. Since $I_Q > I_P$, it immediately follows that $L_Q > L_P$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem compares two rotating objects, $P$ and $Q$, with different masses and moments of inertia. We are told that their rotational kinetic energies are completely equal, and we need to determine the correct relationship between their respective angular momenta, $L_P$ and $L_Q$.

Step 2: Key Formula or Approach:
The rotational kinetic energy ($E_k$) of a rotating body can be related directly to its angular momentum ($L$) and its moment of inertia ($I$) using the formula: $$E_k = \frac{L^2}{2I}$$ Rearranging this equation allows us to express angular momentum explicitly as: $$L = \sqrt{2I \cdot E_k}$$

Step 3: Detailed Explanation:
We are given that the rotational kinetic energies of both bodies are equal: $$E_{k, P} = E_{k, Q} = E_k$$ Using our key relation, we write the angular momenta for both bodies: $$L_P = \sqrt{2I_P \cdot E_k}$$ $$L_Q = \sqrt{2I_Q \cdot E_k}$$ Let's take the ratio of the two angular momenta to eliminate common factors: $$\frac{L_Q}{L_P} = \frac{\sqrt{2I_Q \cdot E_k}}{\sqrt{2I_P \cdot E_k}} = \sqrt{\frac{I_Q}{I_P}}$$ The problem explicitly states that the moment of inertia of body $Q$ is strictly greater than that of body $P$: $$I_Q > I_P \implies \frac{I_Q}{I_P} > 1$$ Taking the square root on both sides maintains the inequality direction: $$\sqrt{\frac{I_Q}{I_P}} > 1 \implies \frac{L_Q}{L_P} > 1 \implies L_Q > L_P$$ Note that the masses of the bodies ($m$ and $2m$) do not alter this relationship, as the moment of inertia already accounts for the mass distribution during rotation.

Step 4: Final Answer:
The angular momentum of $Q$ is greater than that of $P$ ($L_Q > L_P$), which corresponds to option (D).