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: Understanding the Concept:
The dancer's rotational kinetic energy depends on both the moment of inertia \(I\) and the angular velocity \(\omega\). No external torque acts, so angular momentum \(I\omega\) stays constant.

Step 2: Key Formula or Approach:
\[ K = \frac12I\omega^2 \]

Step 3: Detailed Explanation:
Initially \(K = \dfrac12I\omega^2\).
After stretching the arms, \(I' = 3I\) and \(\omega' = \dfrac\omega3\). Check that the angular momentum is conserved: \(I'\omega' = 3I\cdot\dfrac\omega3 = I\omega\). It is.
New kinetic energy:
\[ K' = \frac12(3I)\left(\frac\omega3\right)^2 = \frac12\cdot3I\cdot\frac{\omega^2}{9} = \frac13\left(\frac12I\omega^2\right) = \frac K3 \]
The kinetic energy decreases, because the dancer's arm muscles do negative work when stretching out against the spin. Option (A) K/6, (C) 3K and (D) 6K do not follow from this calculation.

Final Answer:
The new kinetic energy is \(\dfrac K3\), option (B). \[ \boxed{\frac{K}{3} \text{ (B)}} \]