Two uniform circular rough disc of moment of inertia `I_(1)` and `(I_(1))/(2)` are rotating with angular velocity `omega_(1)` and `(omega_(1))/(2)` respectively in same direction. Now one disc is placed the other disc co-axially. The change in kinetic energy of the system is :

Options

A. `-(1)/(24)I_(1)omega_(1^(2))`
B. `(1)/(24)I_(1)omega_(1^(2))`
C. `(1)/(12)I_(1)omega_(1^(2))`
D. `-(1)/(12)I_(1)omega_(1^(2))`

Answer

To solve the problem step by step, we will analyze the situation involving two discs and calculate the change in kinetic energy when they are placed coaxially.

### Step 1: Identify the initial kinetic energies of the discs
The initial kinetic energy (KE) of a rotating disc is given by the formula:
\[ KE = \frac{1}{2} I \omega^2 \]
For the first disc with moment of inertia \(I_1\) and angular velocity \(\omega_1\):
\[ KE_1 = \frac{1}{2} I_1 \omega_1^2 \]
For the second disc with moment of inertia \(\frac{I_1}{2}\) and angular velocity \(\frac{\omega_1}{2}\):
\[ KE_2 = \frac{1}{2} \left(\frac{I_1}{2}\right) \left(\frac{\omega_1}{2}\right)^2 = \frac{1}{2} \cdot \frac{I_1}{2} \cdot \frac{\omega_1^2}{4} = \frac{I_1 \omega_1^2}{16} \]

### Step 2: Calculate the total initial kinetic energy
The total initial kinetic energy of the system is:
\[ KE_{\text{initial}} = KE_1 + KE_2 = \frac{1}{2} I_1 \omega_1^2 + \frac{I_1 \omega_1^2}{16} \]
To combine these, we need a common denominator:
\[ KE_{\text{initial}} = \frac{8I_1 \omega_1^2}{16} + \frac{I_1 \omega_1^2}{16} = \frac{9I_1 \omega_1^2}{16} \]

### Step 3: Find the final angular velocity after they are placed coaxially
The total angular momentum before they are combined must equal the total angular momentum after they are combined (since no external torques are acting):
\[ L_{\text{initial}} = I_1 \omega_1 + \frac{I_1}{2} \cdot \frac{\omega_1}{2} = I_1 \omega_1 + \frac{I_1 \omega_1}{4} = \frac{5I_1 \omega_1}{4} \]
Let \(\omega_f\) be the final angular velocity of the combined system. The total moment of inertia after combining is:
\[ I_{\text{total}} = I_1 + \frac{I_1}{2} = \frac{3I_1}{2} \]
Setting the initial angular momentum equal to the final angular momentum:
\[ \frac{5I_1 \omega_1}{4} = \frac{3I_1}{2} \omega_f \]
Solving for \(\omega_f\):
\[ \omega_f = \frac{5 \omega_1}{6} \]

### Step 4: Calculate the final kinetic energy
Now we calculate the final kinetic energy of the combined system:
\[ KE_{\text{final}} = \frac{1}{2} I_{\text{total}} \omega_f^2 = \frac{1}{2} \cdot \frac{3I_1}{2} \left(\frac{5 \omega_1}{6}\right)^2 \]
Calculating this:
\[ KE_{\text{final}} = \frac{3I_1}{4} \cdot \frac{25 \omega_1^2}{36} = \frac{75 I_1 \omega_1^2}{144} = \frac{25 I_1 \omega_1^2}{48} \]

### Step 5: Calculate the change in kinetic energy
The change in kinetic energy is given by:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = \frac{25 I_1 \omega_1^2}{48} - \frac{9 I_1 \omega_1^2}{16} \]
Finding a common denominator (48):
\[ \Delta KE = \frac{25 I_1 \omega_1^2}{48} - \frac{27 I_1 \omega_1^2}{48} = -\frac{2 I_1 \omega_1^2}{48} = -\frac{I_1 \omega_1^2}{24} \]

### Final Answer:
The change in kinetic energy of the system is:
\[ \Delta KE = -\frac{I_1 \omega_1^2}{24} \]

---