Two discs of same moment of inertia rotating their regular axis passing through centre and perpendicular to the plane of disc with angular velocities `omega_(1)` and `omega_(2)`. They are brought into contact face to the face coinciding the axis of rotation. The expression for loss of enregy during this process is :

Options

A. `1/2 l (omega_(1)+omega_(2))^(2)`
B. `1/4 l(omega_(1)-omega_(2))^(2)`
C. `l (omega_(1)-omega_(2))^(2)`
D. `l/8 (omega_(1)-omega_(2))^(2)`

Answer

To solve the problem of finding the expression for the loss of energy when two discs of the same moment of inertia are brought into contact, we can follow these steps:

### Step 1: Define the Moment of Inertia
Let the moment of inertia of each disc be denoted as \( I \). Since both discs have the same moment of inertia, we will use \( I \) for both.

### Step 2: Calculate Initial Angular Momentum
The initial angular momentum of the system can be calculated by summing the angular momenta of both discs:
\[ L_{\text{initial}} = I \omega_1 + I \omega_2 = I (\omega_1 + \omega_2) \]

### Step 3: Apply Conservation of Angular Momentum
When the discs come into contact, they will rotate together with a common angular velocity \( \omega \). By the conservation of angular momentum, we have:
\[ L_{\text{initial}} = L_{\text{final}} \]
Thus,
\[ I (\omega_1 + \omega_2) = 2I \omega \]
From this, we can solve for \( \omega \):
\[ \omega = \frac{\omega_1 + \omega_2}{2} \]

### Step 4: Calculate Initial Kinetic Energy
The initial kinetic energy of the system is the sum of the kinetic energies of both discs:
\[ KE_{\text{initial}} = \frac{1}{2} I \omega_1^2 + \frac{1}{2} I \omega_2^2 = \frac{I}{2} (\omega_1^2 + \omega_2^2) \]

### Step 5: Calculate Final Kinetic Energy
The final kinetic energy, when both discs are rotating together at angular velocity \( \omega \), is given by:
\[ KE_{\text{final}} = \frac{1}{2} (2I) \omega^2 = I \omega^2 \]
Substituting \( \omega = \frac{\omega_1 + \omega_2}{2} \):
\[ KE_{\text{final}} = I \left(\frac{\omega_1 + \omega_2}{2}\right)^2 = I \cdot \frac{(\omega_1 + \omega_2)^2}{4} \]

### Step 6: Expand Final Kinetic Energy
Expanding the expression for \( KE_{\text{final}} \):
\[ KE_{\text{final}} = \frac{I}{4} (\omega_1^2 + 2\omega_1 \omega_2 + \omega_2^2) \]

### Step 7: Calculate Loss of Energy
The loss of energy during the process can be calculated as:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} \]
Substituting the expressions we derived:
\[ \Delta KE = \frac{I}{2} (\omega_1^2 + \omega_2^2) - \frac{I}{4} (\omega_1^2 + 2\omega_1 \omega_2 + \omega_2^2) \]
Simplifying this expression:
\[ \Delta KE = \frac{I}{4} (\omega_1^2 + \omega_2^2 - 2\omega_1 \omega_2) \]
This can be rewritten as:
\[ \Delta KE = \frac{I}{4} (\omega_1 - \omega_2)^2 \]

### Final Result
Thus, the expression for the loss of energy during this process is:
\[ \Delta KE = \frac{I}{4} (\omega_1 - \omega_2)^2 \]