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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 :
$0$
$\frac{1}{4}$
$\frac{3}{4}$
$\frac{5}{6}$
Solution
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 $=\frac{\Delta KE }{ KE _{1}}=\frac{\frac{3}{8} I \omega^{2}}{\frac{1}{2} I \omega^{2}}=\frac{3}{4}$
