Two discs, each having moment of inertia `5 kg m^(2)`. about its central axis, rotating with speeds `10 rad s^(-1)` and `20 rad s^(-1)`, are brought in contact face to face with their axes of rotation coincided. The loss of kinetic energy in the process is

Options

A. `2J`
B. `5J`
C. `125J`
D. `0J`

Explanation

To solve the problem, we need to calculate the loss of kinetic energy when two discs are brought into contact. Here’s the step-by-step solution:

### Step 1: Calculate the Initial Angular Momentum
The angular momentum \( L \) of a rotating body is given by the formula:
\[ L = I \omega \]
Where:
- \( I \) is the moment of inertia
- \( \omega \) is the angular velocity

For the two discs:
- Moment of inertia \( I_1 = I_2 = 5 \, \text{kg m}^2 \)
- Angular velocity of disc 1 \( \omega_1 = 10 \, \text{rad/s} \)
- Angular velocity of disc 2 \( \omega_2 = 20 \, \text{rad/s} \)

Calculating the initial angular momentum:
\[ L_{\text{initial}} = I_1 \omega_1 + I_2 \omega_2 = 5 \cdot 10 + 5 \cdot 20 = 50 + 100 = 150 \, \text{kg m}^2/\text{s} \]

### Step 2: Calculate the Final Angular Speed
When the discs come into contact, they will rotate together with a common angular speed \( \omega_f \). The total moment of inertia when they are combined is:
\[ I_{\text{total}} = I_1 + I_2 = 5 + 5 = 10 \, \text{kg m}^2 \]

Using the conservation of angular momentum:
\[ L_{\text{initial}} = L_{\text{final}} \]
\[ 150 = I_{\text{total}} \cdot \omega_f \]
\[ 150 = 10 \cdot \omega_f \]
\[ \omega_f = \frac{150}{10} = 15 \, \text{rad/s} \]

### Step 3: Calculate the Initial Kinetic Energy
The initial kinetic energy \( KE_{\text{initial}} \) of the system is the sum of the kinetic energies of both discs:
\[ KE_{\text{initial}} = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 \]
Calculating:
\[ KE_{\text{initial}} = \frac{1}{2} \cdot 5 \cdot 10^2 + \frac{1}{2} \cdot 5 \cdot 20^2 \]
\[ = \frac{1}{2} \cdot 5 \cdot 100 + \frac{1}{2} \cdot 5 \cdot 400 \]
\[ = \frac{5}{2} \cdot 100 + \frac{5}{2} \cdot 400 \]
\[ = 250 + 1000 = 1250 \, \text{J} \]

### Step 4: Calculate the Final Kinetic Energy
The final kinetic energy \( KE_{\text{final}} \) when both discs are rotating together is:
\[ KE_{\text{final}} = \frac{1}{2} I_{\text{total}} \omega_f^2 \]
Calculating:
\[ KE_{\text{final}} = \frac{1}{2} \cdot 10 \cdot 15^2 \]
\[ = \frac{1}{2} \cdot 10 \cdot 225 \]
\[ = 5 \cdot 225 = 1125 \, \text{J} \]

### Step 5: Calculate the Loss of Kinetic Energy
The loss of kinetic energy \( \Delta KE \) is given by:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} \]
Calculating:
\[ \Delta KE = 1250 - 1125 = 125 \, \text{J} \]

### Final Answer
The loss of kinetic energy in the process is \( \boxed{125 \, \text{J}} \).

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