System Of Particles And Rotational Motion Ques 55
55. 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: 55.(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. $=(K . E) _{f}-(K . E) _{i}=\frac{1}{4} I(\omega_1-\omega_2)^{2}$
BETA
