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Step-by-Step Solution
Step 1: Identify the given physical quantities
• Disc A: Moment of inertia = $I_{1}$, Angular velocity = $\omega_{1}$
• Disc B: Moment of inertia = $\dfrac{I_{1}}{2}$, Angular velocity = $\dfrac{\omega_{1}}{2}$
These two discs share the same axis (coaxial). They are brought into contact and, due to friction, finally rotate together with a common angular velocity $\omega$.
Step 2: Write down the initial rotational kinetic energy
The total initial energy $E_{i}$ is the sum of the individual rotational kinetic energies.
• Kinetic energy of Disc A:
\[
K_{1} \;=\; \tfrac{1}{2}\;I_{1}\;\omega_{1}^{2}
\]
• Kinetic energy of Disc B:
\[
K_{2} \;=\; \tfrac{1}{2}\;\left(\dfrac{I_{1}}{2}\right)\;\left(\dfrac{\omega_{1}}{2}\right)^{2}
\;=\;\tfrac{1}{2}\;\times\dfrac{I_{1}}{2}\;\times\dfrac{\omega_{1}^{2}}{4}
\;=\;\dfrac{I_{1}\,\omega_{1}^{2}}{16}
\]
Hence, the initial total energy:
\[
E_{i} \;=\; K_{1} + K_{2}
\;=\;\tfrac{1}{2}\,I_{1}\,\omega_{1}^{2} \;+\;\dfrac{I_{1}\,\omega_{1}^{2}}{16}
\;=\;\dfrac{8}{16}I_{1}\,\omega_{1}^{2} \;+\;\dfrac{1}{16}I_{1}\,\omega_{1}^{2}
\;=\;\dfrac{9}{16}I_{1}\,\omega_{1}^{2}.
\]
Step 3: Conserve angular momentum to find the final angular velocity
Before contact, the total angular momentum $L_{i}$ is:
\[
L_{i}
\;=\; I_{1}\,\omega_{1}
\;+\;\left(\dfrac{I_{1}}{2}\right)\left(\dfrac{\omega_{1}}{2}\right)
\;=\; I_{1}\,\omega_{1} + \dfrac{I_{1}\,\omega_{1}}{4}
\;=\;\dfrac{5}{4} I_{1}\,\omega_{1}.
\]
After contact, both discs rotate together with a combined moment of inertia:
\[
I_{\text{total}} \;=\; I_{1} + \dfrac{I_{1}}{2}
\;=\;\dfrac{3}{2}\,I_{1}.
\]
Let the final common angular velocity be $\omega$. By conservation of angular momentum:
\[
I_{\text{total}}\;\omega
\;=\; \dfrac{5}{4}\;I_{1}\,\omega_{1}.
\]
So,
\[
\left(\dfrac{3}{2}\,I_{1}\right)\omega
\;=\;\dfrac{5}{4}I_{1}\,\omega_{1}
\;\;\Longrightarrow\;\;
\omega
\;=\;\dfrac{\dfrac{5}{4}I_{1}\,\omega_{1}}{\dfrac{3}{2}I_{1}}
\;=\;\dfrac{5}{4}\;\times\dfrac{2}{3}\;\omega_{1}
\;=\;\dfrac{5}{6}\;\omega_{1}.
\]
Step 4: Find the final rotational kinetic energy
The final kinetic energy $E_{f}$ of the two-discs system (rotating together) is:
\[
E_{f}
\;=\;\tfrac{1}{2}\;\Bigl(I_{1} \;+\;\dfrac{I_{1}}{2}\Bigr)\,\omega^{2}
\;=\;\tfrac{1}{2}\;\Bigl(\dfrac{3}{2}I_{1}\Bigr)\,\bigl(\dfrac{5}{6}\omega_{1}\bigr)^{2}.
\]
First simplify the factor in parentheses:
\[
\dfrac{3}{2} \;=\; 1.5,
\quad
\Bigl(\dfrac{5}{6}\Bigr)^{2} \;=\;\dfrac{25}{36}.
\]
So,
\[
E_{f}
\;=\;\tfrac{1}{2}\;\times\dfrac{3}{2}I_{1}\;\times\dfrac{25}{36}\,\omega_{1}^{2}
\;=\;\dfrac{3}{2}\,\times\dfrac{25}{36}\,\times\dfrac{1}{2}\,I_{1}\,\omega_{1}^{2}
\;=\;\dfrac{3}{4}\;\times\dfrac{25}{36}\;I_{1}\,\omega_{1}^{2}
\;=\;\dfrac{75}{144}\;I_{1}\,\omega_{1}^{2}
\;=\;\dfrac{25}{48}\;I_{1}\,\omega_{1}^{2}.
\]
Step 5: Calculate the change in energy $(E_{f} - E_{i})$
\[
E_{f} - E_{i}
\;=\;\dfrac{25}{48}I_{1}\,\omega_{1}^{2} \;-\;\dfrac{9}{16}I_{1}\,\omega_{1}^{2}.
\]
Convert $\tfrac{9}{16}$ to a denominator of 48:
\[
\dfrac{9}{16}
\;=\;\dfrac{9 \times 3}{16 \times 3}
\;=\;\dfrac{27}{48}.
\]
Thus,
\[
E_{f} - E_{i}
\;=\;\Bigl(\dfrac{25}{48} - \dfrac{27}{48}\Bigr)I_{1}\,\omega_{1}^{2}
\;=\;-\dfrac{2}{48}\,I_{1}\,\omega_{1}^{2}
\;=\;-\dfrac{I_{1}\,\omega_{1}^{2}}{24}.
\]
Because the final energy is less than the initial energy (some energy is lost to friction), the numerical value of the change is $\dfrac{I_{1}\,\omega_{1}^{2}}{24}$ but it is negative.
In many problems, the question’s options give the magnitude of energy change (or energy lost). If the question states the answer as $+\dfrac{I_{1}\,\omega_{1}^{2}}{24}$, it is typically referring to the magnitude of the energy difference (i.e., the energy lost). Strictly, from conservation of energy with friction involved, the system’s rotational kinetic energy decreases by $\dfrac{I_{1}\,\omega_{1}^{2}}{24}.$
Final Answer
• The magnitude of $(E_{f} - E_{i})$ is $\boxed{\dfrac{I_{1}\,\omega_{1}^{2}}{24}}$, and the sign is negative because energy is lost.
