Question

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

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

Hint:

Angular momentum stays constant, so kinetic energy changes inversely with the moment of inertia.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
No external torque acts on the child and platform, so angular momentum \(L=I\omega\) is conserved.

Step 2: Kinetic energy in terms of L
\(K=\dfrac{1}{2}I\omega^{2}=\dfrac{L^{2}}{2I}\)

Step 3: Calculation
At constant \(L\), \(K\propto\dfrac{1}{I}\). Doubling \(I\) halves \(K\).

Step 4: Result
New kinetic energy \(=\dfrac{K}{2}\), option (B).

Final Answer:
The kinetic energy becomes half of its initial value. \[ \boxed{\text{(B) }\dfrac{K}{2}} \]