Question

A child is standing with folded hands at the centre of the platform rotating about its central axis. The rotational kinetic energy of the system is '\(K\)'. The child now stretches his arms so that the moment of inertia of the system doubles. The rotational kinetic energy of the system now is '\(K_{1}\)'. The ratio \(K_{1}/K\) is

• A. \(4\)
• B. \(\frac{1}{4}\)
• C. \(\frac{1}{2}\)
• D. \(2\)

Hint:

Angular momentum is conserved when no external torque acts, so kinetic energy changes inversely with moment of inertia.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
There is no external torque, so angular momentum \(L=I\omega\) stays constant.

Step 2: Kinetic energy in terms of L
Rotational kinetic energy \(K=\dfrac{1}{2}I\omega^{2}=\dfrac{L^{2}}{2I}\). With \(L\) fixed, \(K\propto\dfrac{1}{I}\).

Step 3: Calculation
The moment of inertia becomes \(2I\), so \(K_{1}=\dfrac{K}{2}\) and \(\dfrac{K_{1}}{K}=\dfrac{1}{2}\).

Final Answer:
The ratio of the kinetic energies is one half. \[ \boxed{\text{(C) }\dfrac{1}{2}} \]