Two coaxial discs, having moments of inertia `l_(1) and (I_(1))/(2)` are rotating with respective angular velocities `omega and (omega_(1))/(2)` about their common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If `E_(f) and E_(i)` are the final and initial total energies, then `(E_(f)-E_(i))` is:

Options

A. `(-I_(1)omega_(1)^(2))/(12)`
B. `(I_(1)omega_(1)^(2))/(6)`
C. `(3)/(8)I_(1) omega_(1)^(2)`
D. `(-I_(1)omega_(1)^(2))/(24)`

Answer

To solve the problem, we need to analyze the system of two coaxial discs and apply the principles of conservation of angular momentum and energy.

### Step 1: Identify the initial conditions
We have two discs with the following properties:
- Disc 1: Moment of inertia \( I_1 \) and angular velocity \( \omega \)
- Disc 2: Moment of inertia \( I_2 = \frac{I_1}{2} \) and angular velocity \( \frac{\omega}{2} \)

### Step 2: Calculate the initial angular momentum
The initial angular momentum \( L_i \) of the system can be calculated as:
\[ L_i = I_1 \cdot \omega + I_2 \cdot \left(\frac{\omega}{2}\right) \]
Substituting \( I_2 \):
\[ L_i = I_1 \cdot \omega + \frac{I_1}{2} \cdot \left(\frac{\omega}{2}\right) = I_1 \cdot \omega + \frac{I_1 \cdot \omega}{4} = I_1 \cdot \omega \left(1 + \frac{1}{4}\right) = I_1 \cdot \omega \cdot \frac{5}{4} \]

### Step 3: Calculate the final moment of inertia
When the discs come into contact and rotate together, the total moment of inertia \( I_f \) is:
\[ I_f = I_1 + I_2 = I_1 + \frac{I_1}{2} = \frac{3I_1}{2} \]

### Step 4: Apply conservation of angular momentum
According to the conservation of angular momentum:
\[ L_i = L_f \]
Thus,
\[ I_1 \cdot \omega \cdot \frac{5}{4} = I_f \cdot \omega_f \]
Substituting \( I_f \):
\[ I_1 \cdot \omega \cdot \frac{5}{4} = \frac{3I_1}{2} \cdot \omega_f \]
Dividing both sides by \( I_1 \):
\[ \frac{5\omega}{4} = \frac{3}{2} \cdot \omega_f \]
Solving for \( \omega_f \):
\[ \omega_f = \frac{5\omega}{4} \cdot \frac{2}{3} = \frac{5\omega}{6} \]

### Step 5: Calculate the final energy \( E_f \)
The final kinetic energy \( E_f \) is given by:
\[ E_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} \cdot \frac{3I_1}{2} \cdot \left(\frac{5\omega}{6}\right)^2 \]
Calculating \( E_f \):
\[ E_f = \frac{1}{2} \cdot \frac{3I_1}{2} \cdot \frac{25\omega^2}{36} = \frac{75I_1\omega^2}{144} = \frac{25I_1\omega^2}{48} \]

### Step 6: Calculate the initial energy \( E_i \)
The initial kinetic energy \( E_i \) is:
\[ E_i = \frac{1}{2} I_1 \omega^2 + \frac{1}{2} I_2 \left(\frac{\omega}{2}\right)^2 \]
Substituting \( I_2 \):
\[ E_i = \frac{1}{2} I_1 \omega^2 + \frac{1}{2} \cdot \frac{I_1}{2} \cdot \frac{\omega^2}{4} = \frac{1}{2} I_1 \omega^2 + \frac{I_1 \omega^2}{16} \]
Combining terms:
\[ E_i = \frac{8I_1 \omega^2}{16} + \frac{I_1 \omega^2}{16} = \frac{9I_1 \omega^2}{16} \]

### Step 7: Calculate the difference in energy \( E_f - E_i \)
Now, we find the difference:
\[ E_f - E_i = \frac{25I_1\omega^2}{48} - \frac{9I_1\omega^2}{16} \]
Finding a common denominator (48):
\[ E_i = \frac{9I_1\omega^2}{16} = \frac{27I_1\omega^2}{48} \]
Thus,
\[ E_f - E_i = \frac{25I_1\omega^2}{48} - \frac{27I_1\omega^2}{48} = -\frac{2I_1\omega^2}{48} = -\frac{I_1\omega^2}{24} \]

### Final Answer
\[ E_f - E_i = -\frac{I_1\omega^2}{24} \]