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 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
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Since $K = L^2/2I$, kinetic energy is inversely proportional to moment of inertia when $L$ is constant.
Updated On: Jun 19, 2026
- increase by 33.3%
- decrease by 33.3%
- increase by 25.0%
- decrease by 25.0%
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Angular momentum $L = I \omega$ is conserved. $L_1 = L_2$.
Step 2: Analysis
Rotational K.E. $K = \frac{L^2}{2I}$.
$K_2/K_1 = I_1/I_2$.
Given $I_2 = 0.75 I_1$, so $K_2/K_1 = 1/0.75 = 4/3 \approx 1.333$.
Step 3: Calculation
% change $= (1.333 - 1) \times 100 = 33.3%$.
Step 4: Conclusion
Hence, kinetic energy increases by 33.3%. Final Answer: (A)
Angular momentum $L = I \omega$ is conserved. $L_1 = L_2$.
Step 2: Analysis
Rotational K.E. $K = \frac{L^2}{2I}$.
$K_2/K_1 = I_1/I_2$.
Given $I_2 = 0.75 I_1$, so $K_2/K_1 = 1/0.75 = 4/3 \approx 1.333$.
Step 3: Calculation
% change $= (1.333 - 1) \times 100 = 33.3%$.
Step 4: Conclusion
Hence, kinetic energy increases by 33.3%. Final Answer: (A)
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