System of Particles and Rotational Motion - Result Question 52
56. 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 $\omega_1$ and $\omega_2$. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is:-
[2017]
(a) $\frac{1}{4} I(\omega_1-\omega_2)^{2}$
(b) $I(\omega_1-\omega_2)^{2}$
(c) $\frac{1}{8}(\omega_1-\omega_2)^{2}$
(d) $\frac{1}{2} I(\omega_1+\omega_2)^{2}$
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Answer:
Correct Answer: 56. (a)
Solution:
- (a) Here, $I \omega_1+I \omega_2=2 I \omega$
$\Rightarrow \omega=\frac{\omega_1+\omega_2}{2}$
$(K . E) _{i}=\frac{1}{2} I \omega_1^{2}+\frac{1}{2} I \omega_2^{2}$
(K.E. $) _{f}=\frac{1}{2} \times 2 I \omega^{2}=I(\frac{\omega_1+\omega_2}{2})^{2}$
Loss in K.E. $=(K . E) _{f}-(K . E) _{i}=\frac{1}{4} I(\omega_1-\omega_2)^{2}$
