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Physics Question on rotational motion

Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω1\omega_1 and ω2\omega_2 . They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is:-

A

14I(ω1−ω2)2\frac{1}{4} I (\omega_1 - \omega_2)^2

B

I(ω1−ω2)2I (\omega_1 - \omega_2)^2

C

18I(ω1−ω2)2\frac{1}{8} I (\omega_1 - \omega_2)^2

D

12I(ω1+ω2)2\frac{1}{2} I (\omega_1+ \omega_2)^2

Answer

14I(ω1−ω2)2\frac{1}{4} I (\omega_1 - \omega_2)^2

Explanation

Solution

ΔKE=12I1I2I1+I2(ω1−ω2)2\Delta KE =\frac{1}{2} \frac{I_{1} I_{2}}{I_{1}+I_{2}}\left(\omega_{1}-\omega_{2}\right)^{2}
=12I2(2I)(ω1−ω2)2=\frac{1}{2} \frac{I^{2}}{(2 I)}\left(\omega_{1}-\omega_{2}\right)^{2}
=14l(ω1−ω2)2=\frac{1}{4} l\left(\omega_{1}-\omega_{2}\right)^{2}