Two discs of moment of inertia `9kgm^2` and `3kgm^2`were rotating with angular velocity 6 rad/sec and 10 rad/sec respectively in same direction. They are brought together gently to move with same angular velocity. The loss of kinetic energy in Joules is______.

Explanation

To solve the problem, we need to calculate the loss of kinetic energy when two discs with different moments of inertia and angular velocities are brought together to rotate with a common angular velocity. Here's the step-by-step solution:

### Step 1: Calculate the Initial Kinetic Energy of Each Disc

The formula for the kinetic energy (KE) of a rotating body is given by:

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

Where:
- \( I \) is the moment of inertia
- \( \omega \) is the angular velocity

For the first disc:
- Moment of inertia \( I_1 = 9 \, \text{kg m}^2 \)
- Angular velocity \( \omega_1 = 6 \, \text{rad/s} \)

Calculating the kinetic energy of the first disc:

\[ KE_1 = \frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} \times 9 \times (6^2) = \frac{1}{2} \times 9 \times 36 = 162 \, \text{J} \]

For the second disc:
- Moment of inertia \( I_2 = 3 \, \text{kg m}^2 \)
- Angular velocity \( \omega_2 = 10 \, \text{rad/s} \)

Calculating the kinetic energy of the second disc:

\[ KE_2 = \frac{1}{2} I_2 \omega_2^2 = \frac{1}{2} \times 3 \times (10^2) = \frac{1}{2} \times 3 \times 100 = 150 \, \text{J} \]

### Step 2: Calculate the Total Initial Kinetic Energy

Now, we can find the total initial kinetic energy:

\[ KE_{\text{initial}} = KE_1 + KE_2 = 162 + 150 = 312 \, \text{J} \]

### Step 3: Apply Conservation of Angular Momentum

When the two discs are brought together, angular momentum is conserved. The initial angular momentum \( L_{\text{initial}} \) is given by:

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

Calculating the initial angular momentum:

\[ L_{\text{initial}} = (9 \times 6) + (3 \times 10) = 54 + 30 = 84 \, \text{kg m}^2/\text{s} \]

The total moment of inertia when both discs are combined is:

\[ I_{\text{total}} = I_1 + I_2 = 9 + 3 = 12 \, \text{kg m}^2 \]

Let \( \omega_f \) be the final angular velocity when both discs are rotating together. By conservation of angular momentum:

\[ L_{\text{initial}} = L_{\text{final}} \implies 84 = I_{\text{total}} \omega_f \]

Solving for \( \omega_f \):

\[ \omega_f = \frac{84}{12} = 7 \, \text{rad/s} \]

### Step 4: Calculate the Final Kinetic Energy

Now we can calculate the final kinetic energy when both discs rotate together:

\[ KE_{\text{final}} = \frac{1}{2} I_{\text{total}} \omega_f^2 = \frac{1}{2} \times 12 \times (7^2) = \frac{1}{2} \times 12 \times 49 = 294 \, \text{J} \]

### Step 5: Calculate the Loss of Kinetic Energy

The loss of kinetic energy is given by:

\[ \text{Loss in KE} = KE_{\text{initial}} - KE_{\text{final}} = 312 - 294 = 18 \, \text{J} \]

### Final Answer

The loss of kinetic energy is **18 Joules**.

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