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 :

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Answer & Solution

Correct Answer

(C) \(\frac{3}{4}\)

Detailed explanation

By anglar momentum conservation

\(\omega I+3 I \times 0=4 I \omega^{\prime} \Rightarrow \omega^{\prime}=\frac{\omega}{4}\)

\(( KE )_{ i }=\frac{1}{2} I \omega^{2}\)

\(( KE )_{ f }=\frac{1}{2} \times(4 I ) \times\left(\frac{\omega}{4}\right)^{2}=\frac{ I \omega^{2}}{8}\)

\(\Delta KE =\frac{3}{8} I \omega^{2}\)

fractional loss \(=\frac{\Delta KE }{ KE _{1}}=\frac{\frac{3}{8} I \omega^{2}}{\frac{1}{2} I \omega^{2}}=\frac{3}{4}\)

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