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Step-by-step Solution
Step 1: Identify the Physical Principles
The problem involves two rotating disks brought into contact along their common axis. Since no external torque acts on the system, angular momentum is conserved. However, kinetic energy is not necessarily conserved due to friction between the disks, which causes some energy to be lost.
Step 2: Apply Conservation of Angular Momentum
Before the second disk is placed on the first, the total angular momentum is
$I_t \omega_i$,
where
$I_t$
is the moment of inertia of the top (initially rotating) disk, and
$\omega_i$
is its initial angular velocity.
After both disks stick together and rotate with a common angular velocity
$\omega_f$,
the combined moment of inertia is
$(I_t + I_b)$.
Hence, by conservation of angular momentum:
$I_t \,\omega_i \;=\; (I_t + I_b)\,\omega_f$
Solving for
$\omega_f$:
$$
\omega_f = \frac{I_t\,\omega_i}{I_t + I_b}
$$
Step 3: Compute the Initial Rotational Kinetic Energy
The initial kinetic energy
$E_i$
comes solely from the top disk rotating at
$\omega_i$.
Thus:
$$
E_i = \frac{1}{2}\,I_t\,\omega_i^2
$$
Step 4: Compute the Final Rotational Kinetic Energy
Once the disks rotate together, their total moment of inertia is
$(I_t + I_b)$,
and their common angular velocity is
$\omega_f$.
The final kinetic energy
$E_f$
is:
$$
E_f = \frac{1}{2}(I_t + I_b)\,\omega_f^2
$$
Substituting
$\displaystyle \omega_f = \frac{I_t\,\omega_i}{I_t + I_b}$
gives:
$$
E_f
= \frac{1}{2} \,(I_t + I_b) \left(\frac{I_t\,\omega_i}{I_t + I_b}\right)^2
= \frac{1}{2}\,\frac{I_t^2\,\omega_i^2}{I_t + I_b}
$$
Step 5: Find the Energy Lost
The energy lost due to friction,
$\Delta E$,
is the difference between the initial and final rotational kinetic energies:
$$
\Delta E
= E_i - E_f
= \frac{1}{2}\,I_t\,\omega_i^2
- \frac{1}{2} \,\frac{I_t^2\,\omega_i^2}{I_t + I_b}
= \frac{1}{2} \,\omega_i^2
\left(I_t
- \frac{I_t^2}{I_t + I_b}\right)
$$
Factor out
$I_t$:
$$
\Delta E
= \frac{1}{2} \,\frac{I_b\,I_t}{I_t + I_b}\,\omega_i^2
$$
Step 6: State the Final Answer
The energy lost due to friction is
$$
\Delta E
= \frac{1}{2}\,\frac{I_b\,I_t}{I_t + I_b}\,\omega_i^2,
$$
which matches the provided correct choice.
