JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A wheel is rotaing 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 momet 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 \(0\)
- B \(\frac{1}{4}\)
- C \(\frac{3}{4}\)
- D \(\frac{5}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{4}\)
Step-by-step Solution
Detailed explanation
By anglar momentum conservation \(\omega I+3 I \times 0=4 I \omega^{\prime} \Rightarrow \omega^{\prime}=\frac{\omega}{4}\) \(( KE )_{ i }=\frac{1}{2} I \omega^{2}\) \(( KE )_{ f }=\frac{1}{2} \times(4 I ) \times\left(\frac{\omega}{4}\right)^{2}=\frac{ I \omega^{2}}{8}\)…
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