Two discs of moments of inertia `I_1` and `I_2` about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed `omega_1` and `omega_2` are brought into contact face to face with their axes of rotation coincident. What is the loss in kinetic energy of the system in the process?

Options

A. `I_(1)I_(2)(omega_1 - omega_2)^(2)/2(I_1 + I_2)`
B. `I_(1)I_(2)(omega_1 - omega_2)^(2)/(2(I_1 + I_2))`
C. `I_(1)I_(2)(omega_1 + omega_2)^(2)/(I_1 - I_2)`
D. `I_(1)I_(2)(omega_1 + omega_2)^(2)/2(I_1 - I_2)`

Answer

To find the loss in kinetic energy when two discs are brought into contact, we can follow these steps:

### Step 1: Calculate Initial Kinetic Energy
The initial kinetic energy of the system is the sum of the kinetic energies of both discs. The formula for rotational kinetic energy is given by:

\[ KE = \frac{1}{2} I \omega^2 \]

Thus, the initial kinetic energy \( KE_{initial} \) can be expressed as:

\[ KE_{initial} = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 \]

### Step 2: Determine Final Angular Velocity
When the discs are brought into contact, they will rotate together with a common angular velocity \( \omega \). To find this angular velocity, we use the conservation of angular momentum:

\[ L_{initial} = L_{final} \]

The initial angular momentum is:

\[ L_{initial} = I_1 \omega_1 + I_2 \omega_2 \]

The final angular momentum when both discs are rotating together is:

\[ L_{final} = (I_1 + I_2) \omega \]

Setting these equal gives:

\[ I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega \]

From this, we can solve for \( \omega \):

\[ \omega = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} \]

### Step 3: Calculate Final Kinetic Energy
The final kinetic energy \( KE_{final} \) of the system when both discs are rotating together is:

\[ KE_{final} = \frac{1}{2} (I_1 + I_2) \omega^2 \]

Substituting the expression for \( \omega \):

\[ KE_{final} = \frac{1}{2} (I_1 + I_2) \left( \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} \right)^2 \]

This simplifies to:

\[ KE_{final} = \frac{1}{2} \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{I_1 + I_2} \]

### Step 4: Calculate Loss in Kinetic Energy
The loss in kinetic energy \( \Delta KE \) is given by:

\[ \Delta KE = KE_{initial} - KE_{final} \]

Substituting the expressions we derived:

\[ \Delta KE = \left( \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 \right) - \frac{1}{2} \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{I_1 + I_2} \]

### Step 5: Simplify the Expression
To simplify, we can factor out \( \frac{1}{2} \):

\[ \Delta KE = \frac{1}{2} \left( I_1 \omega_1^2 + I_2 \omega_2^2 - \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{I_1 + I_2} \right) \]

This can be further simplified to:

\[ \Delta KE = \frac{1}{2} \frac{I_1 I_2 (\omega_1 - \omega_2)^2}{I_1 + I_2} \]

### Final Result
Thus, the loss in kinetic energy when the two discs come into contact is:

\[ \Delta KE = \frac{I_1 I_2 (\omega_1 - \omega_2)^2}{2(I_1 + I_2)} \]