Question

A child is standing with folded hands at the centre of the platform rotating about its central axis. The kinetic energy of the system is '$K$'. The child now stretches his arms so that the moment of inertia of the system becomes double. The kinetic energy of the system now is

• A. $\frac{K}{2}$
• B. $2K$
• C. $4K$
• D. $\frac{K}{4}$

Hint:

Use the direct rule for angular momentum systems: if the moment of inertia increases by a factor of $n$ due to internal mass redistribution, the rotational speed drops to $\frac{1}{n}$ and the kinetic energy drops to $\frac{1}{n}$ of its original value. Since $n = 2$, the energy drops to $\frac{K}{2}$ instantly!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A child stands at the center of a friction-free turntable system rotating at an initial velocity. The initial rotational kinetic energy is $K$. When the child extends their arms outward, the total moment of inertia of the system doubles ($I_2 = 2I_1$). We need to find the new rotational kinetic energy $K_2$.

Step 2: Key Formula or Approach:
Because there are no external forces or torques acting on the system ($\tau_{\text{ext}} = 0$), the total angular momentum $L$ is perfectly conserved: $$L = I_1 \omega_1 = I_2 \omega_2 = \text{constant}$$ The rotational kinetic energy can be expressed in terms of angular momentum and the moment of inertia using the formula: $$K = \frac{L^2}{2I}$$ This shows that kinetic energy is inversely proportional to the moment of inertia: $K \propto \frac{1}{I}$.

Step 3: Detailed Explanation:
Let's write down the ratio of the initial and final kinetic energies using our proportional relationship: $$\frac{K_2}{K_1} = \frac{I_1}{I_2}$$ We are given that the new moment of inertia is double the original value ($I_2 = 2I_1$). Substituting this into the ratio gives: $$\frac{K_2}{K} = \frac{I_1}{2I_1}$$ The individual moment of inertia terms $I_1$ cancel out perfectly: $$\frac{K_2}{K} = \frac{1}{2} \implies K_2 = \frac{K}{2}$$

Step 4: Final Answer:
The rotational kinetic energy of the system becomes $\frac{K}{2}$, which corresponds to option (A).