Question:
A wheel is rotating freely with an angular speed \(\omega\) on a shaft. The moment of inertia of the wheel is \(I\) and the moment of inertia of the shaft is negligible. Another wheel of moment of inertia \(3I\) initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is :
A wheel is rotating freely with an angular speed \(\omega\) on a shaft. The moment of inertia of the wheel is \(I\) and the moment of inertia of the shaft is negligible. Another wheel of moment of inertia \(3I\) initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is :
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Conserve angular momentum to get the final angular speed, then compare initial and final rotational kinetic energies.
Updated On: Nov 5, 2026
- 0
- \(\frac{5}{6}\)
- \(\frac{1}{4}\)
- \(\frac{3}{4}\)
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The Correct Option is D
Solution and Explanation
Angular momentum is conserved: \(I\omega=(I+3I)\omega_f\Rightarrow\omega_f=\frac{\omega}{4}\).
Initial kinetic energy: \(K_i=\frac{1}{2}I\omega^{2}\).
Final kinetic energy: \(K_f=\frac{1}{2}(4I)\left(\frac{\omega}{4}\right)^{2}=\frac{1}{8}I\omega^{2}\).
Fractional loss \(=\frac{K_i-K_f}{K_i}=1-\frac{1/8}{1/2}=1-\frac{1}{4}=\frac{3}{4}\).
Answer: (4) \(\frac{3}{4}\)
Initial kinetic energy: \(K_i=\frac{1}{2}I\omega^{2}\).
Final kinetic energy: \(K_f=\frac{1}{2}(4I)\left(\frac{\omega}{4}\right)^{2}=\frac{1}{8}I\omega^{2}\).
Fractional loss \(=\frac{K_i-K_f}{K_i}=1-\frac{1/8}{1/2}=1-\frac{1}{4}=\frac{3}{4}\).
Answer: (4) \(\frac{3}{4}\)
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