Question

A kathak dancer is standing on horizontal surface with folded hands. In the begining dancer is rotating about his central axis and his kinetic energy is 'K' at that time. The kathak dancer now stretches his arms so that the moment of inertia of the dancer becomes three times and the angular velocity becomes one-third. The kinetic energy of the dancer now is

• A. \(\frac{K}{6}\)
• B. \(\frac{K}{3}\)
• C. \(3K\)
• D. \(6K\)

Hint:

Angular momentum is conserved. Kinetic energy is (1/2) I omega^2, so I times 3 and omega divided by 3 gives K/3.

Answer 1

The Correct Option is B

Step 1: Write K in terms of angular momentum:
$K = \frac{L^2}{2I}$, where $L = I\omega$ is conserved.

Step 2: Apply the change:
$L$ stays the same and $I$ becomes $3I$, so $K' = \frac{L^2}{2(3I)} = \frac13\cdot\frac{L^2}{2I} = \frac K3$.

Step 3: Why it falls:
Stretching the arms increases $I$. With $L$ fixed, $K\propto 1/I$, so it must decrease.

Step 4: Where the lost energy goes:
The kinetic energy is not conserved, only the angular momentum is. The dancer's muscles do negative work while the arms move outward, which is why the energy drops to one third and not to any other value.

Final Answer:
$K/3$, option (B). \[ \boxed{\frac{K}{3} \text{ (B)}} \]