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Two discs of same moment of inertia rotate about their regular axis passing through the centre and perpendicular to the plane of disc with angular velocities ${w_1}$ and ${w_2}$. They are brought into contact face to face to coincide the axis of rotation. The expression for loss coinciding of energy during this process is-
A) $\dfrac{1}{8}{\left( {{\omega_1} - {\omega_2}} \right)^2}$
B) $\dfrac{1}{2}I{\left( {{\omega_1} + {\omega_2}} \right)^2}$
C) $\dfrac{1}{4}I{\left( {{\omega_1} - {\omega_2}} \right)^2}$
D) $I{\left( {{\omega_1} - {\omega_2}} \right)^2}$

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Last updated date: 20th Jun 2024
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Answer
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Hint:As the total energy loss during this process is the change in kinetic energy so first find out the kinetic energy at initial and then at final when the two discs are brought into face to face coincide the axis of rotation.

Complete step by step solution:
As we have given the angular velocity of two discs ${\omega_1},{\omega_2}$ respectively. They rotate with the same moment of inertia about their regular axis passing through the centre and perpendicular to the plane of the disc. They are brought face to face to coincide the axis of rotation.
We know that the angular moment of any system remains constant. So as the two discs are brought face to face coincide the axis of rotation we can assume it’s a complete system.
Let us assume the angular velocity of the complete system be $\omega$ and the moment of inertia is $I$.
So, from the law of conservation of angular momentum,
$I{\omega_1} + I{\omega_2} = \omega\left( {I + I} \right)$
$ \Rightarrow w = \dfrac{{{\omega_1} + {\omega_2}}}{2}$
Now we know rotational kinetic energy can be written as,
$K = \dfrac{1}{2}I{\omega^2}$ [where $K$ is kinetic energy, $I$ is moment of inertia, $\omega$ is angular momentum].
Thus, initial kinetic energy is ${K_i} = \dfrac{1}{2}{\omega_1}^2 + \dfrac{1}{2}{\omega_2}^2$
As the two discs are brought face to face so their angular momentum will be$w$as they become a complete system.
Final kinetic energy, ${K_f} = \dfrac{1}{2}\left( {2I} \right){\omega^2}$
Now the loss of energy is the change in kinetic energy, hence $\Delta K = {K_i} - {K_f}$
Now, we substitute the values in the above equation to obtain,
$\Delta K = \dfrac{1}{2}I{\omega_1}^2 + \dfrac{1}{2}I{\omega_2}^2 - \dfrac{1}{2}\left( {2I} \right){\omega^2} = \dfrac{1}{2}I\left( {{\omega_1}^2 + {\omega_2}^2 - {\omega^2}} \right)$
Put the value of $\omega = \dfrac{{{\omega_1} + {\omega_2}}}{2}$
$ \Rightarrow \Delta K = \dfrac{I}{2}\left[ {{\omega_1}^2 + {\omega_2}^2 - {{\left( {\dfrac{{{\omega_1} + {\omega_2}}}{2}} \right)}^2}} \right]$
$\therefore \Delta K = \dfrac{I}{4}{\left( {{\omega_1} - {\omega_2}} \right)^2}$
During this process the loss of energy will be$\dfrac{I}{4}{\left( {{\omega_1} - {\omega_2}} \right)^2}$.

Hence, option (C) is correct.

Note:When the two discs are brought face to face their moment of inertia is same and they rotates at same angular velocity as they become a complete cycle so to find the final kinetic energy we use the moment of inertia as $2I$, angular velocity as $\omega$.