Question

A man on a turn-table folds his arms. If his moment of inertia with folded arms is 75% of original, his rotational kinetic energy will

• A. increase by 33.3%
• B. decrease by 33.3%
• C. increase by 25.0%
• D. decrease by 25.0%

Hint:

Since $K = L^2/2I$, kinetic energy is inversely proportional to moment of inertia when $L$ is constant.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Since no external torque acts on the system, angular momentum is conserved. We need to find the change in kinetic energy when the moment of inertia changes.

Step 2: Key Formula or Approach:
1. Conservation of Angular Momentum: \( L = I \omega = \text{constant} \).
2. Rotational Kinetic Energy: \( K = \frac{L^2}{2I} \).

Step 3: Detailed Explanation:
Initial moment of inertia = \( I \).
Final moment of inertia \( I' = 75% \text{ of } I = 0.75 I = \frac{3}{4} I \).
From conservation of angular momentum, \( L \) is constant.
Ratio of final to initial kinetic energy:
\[ \frac{K'}{K} = \frac{L^2 / 2I'}{L^2 / 2I} = \frac{I}{I'} \] \[ \frac{K'}{K} = \frac{I}{0.75 I} = \frac{1}{3/4} = \frac{4}{3} \] Percentage change in K.E.:
\[ % \Delta K = \left( \frac{K' - K}{K} \right) \times 100 = \left( \frac{K'}{K} - 1 \right) \times 100 \] \[ % \Delta K = \left( \frac{4}{3} - 1 \right) \times 100 = \frac{1}{3} \times 100 = 33.3% \] Since the final energy is greater than initial, it increases.

Step 4: Final Answer:
The rotational kinetic energy will increase by 33.3%.