Two discs have moments of inertia I1 and I2 about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, ω1 and ω2 respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by
(1) \(\frac{I_1I_2}{(I_1+I_2)}\)(ω1 - ω2)2
(2) \(\frac{(I_1 - I_2)^2 \omega_1 \omega_2}{2(I_1+I_2)}\)
(3) \(\frac{I_1I_2}{2(I_1+I_2)}\)(ω1 - ω2)2
(4) \(\frac{(\omega_1 - \omega_2)^2}{2(I_1+I_2)}\)
Correct answer: (3) \(\frac{I_1I_2}{2(I_1+I_2)}\)(ω1 - ω2)2
Explanation:
From conservation of angular momentum we get
I1ω1 + I2ω2 = (I1 + I2)ω
ω = \(\frac{I_1 \omega_1 + I_2 \omega_2}{I_1+I_2}\)
ki = \(\frac{1}{2}\)I1ω12+ \(\frac{1}{2}\)I2ω22
kf = \(\frac{1}{2}\)(I1 + I2)ω2
ki - kf = \(\frac{1}{2}\)\(\Big[I_1 \omega_1^2 + I_2 \omega_2^2 - \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{I_1 + I_2}\Big]\)
Solving above we get
ki - kf = \(\frac{1}{2}\)\(\Big(\frac{I_1I_2}{I_1+I_2}\Big)\)(ω1 - ω2)2