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Step-by-Step Solution
Step 1: Identify the Physical Principle
When two rotating discs are brought into contact along their axes, no external torque acts on the system. Hence, the total angular momentum of the system is conserved.
Step 2: Write the Conservation of Angular Momentum
Let each disc have moment of inertia $I$ and initial angular velocities $\omega_1$ and $\omega_2$. After contact, they rotate together with a common angular velocity $\omega_f$. By conservation of angular momentum:
$ I \omega_1 + I \omega_2 = (I + I)\,\omega_f \quad \Longrightarrow \quad I(\omega_1 + \omega_2) = 2I\,\omega_f. $
Solving for the common angular velocity $\omega_f$:
$ \omega_f = \frac{\omega_1 + \omega_2}{2}. $
Step 3: Calculate the Initial Kinetic Energy
The initial kinetic energy $K_i$ is the sum of the rotational kinetic energies of the two discs:
$ K_i = \tfrac{1}{2} I \omega_1^2 + \tfrac{1}{2} I \omega_2^2 \;=\; \tfrac{I}{2}(\omega_1^2 + \omega_2^2). $
Step 4: Calculate the Final Kinetic Energy
After the discs come into contact, they rotate as one combined object with moment of inertia $2I$ (because the two discs have the same $I$ and now rotate together) and common angular velocity $\omega_f$. Thus,
$ K_f = \tfrac{1}{2} \,(2I)\, \omega_f^2 \;=\; I \,\left(\frac{\omega_1 + \omega_2}{2}\right)^2 \;=\; \frac{I}{4} \, (\omega_1 + \omega_2)^2. $
Step 5: Compute the Loss in Kinetic Energy
The loss in energy $\Delta K$ is the difference between the initial and final kinetic energies:
$ \Delta K = K_i - K_f
= \frac{I}{2}(\omega_1^2 + \omega_2^2)
\;-\;
\frac{I}{4}(\omega_1 + \omega_2)^2. $
Expand and simplify:
$
(\omega_1 + \omega_2)^2 = \omega_1^2 + 2 \omega_1 \omega_2 + \omega_2^2.
$
Thus,
$
K_f = \frac{I}{4} \bigl(\omega_1^2 + 2\omega_1 \omega_2 + \omega_2^2\bigr).
$
So,
$
\Delta K
= \frac{I}{2}(\omega_1^2 + \omega_2^2)
\;-\;
\frac{I}{4}(\omega_1^2 + 2 \omega_1 \omega_2 + \omega_2^2)
= \frac{I}{4}\bigl(2(\omega_1^2 + \omega_2^2) - (\omega_1^2 + 2\omega_1 \omega_2 + \omega_2^2)\bigr).
$
$
\Delta K
= \frac{I}{4}(\omega_1^2 - 2\omega_1 \omega_2 + \omega_2^2)
= \frac{I}{4}(\omega_1 - \omega_2)^2.
$
Step 6: State the Final Answer
Hence, the loss of kinetic energy when the two discs come into contact is
$ \displaystyle \Delta K = \frac{1}{4}\,I\,(\omega_1 - \omega_2)^2. $
