Question:
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:-
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:-
Updated On: Jul 13, 2024
- $\frac{1}{4} I (\omega_1 - \omega_2)^2$
- $I (\omega_1 - \omega_2)^2$
- $\frac{1}{8} I (\omega_1 - \omega_2)^2$
- $\frac{1}{2} I (\omega_1+ \omega_2)^2$
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The Correct Option is A
Solution and Explanation
$\Delta KE =\frac{1}{2} \frac{I_{1} I_{2}}{I_{1}+I_{2}}\left(\omega_{1}-\omega_{2}\right)^{2}$
$=\frac{1}{2} \frac{I^{2}}{(2 I)}\left(\omega_{1}-\omega_{2}\right)^{2}$
$=\frac{1}{4} l\left(\omega_{1}-\omega_{2}\right)^{2}$
$=\frac{1}{2} \frac{I^{2}}{(2 I)}\left(\omega_{1}-\omega_{2}\right)^{2}$
$=\frac{1}{4} l\left(\omega_{1}-\omega_{2}\right)^{2}$
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Rotational Motion
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