Two disc having moment of inertias `I1` & `I2` and angle velocities `omega_1` & `omega_2` are placed coaxially find total kinetic energy when they rotates with same angular velocity `I1 = 0.10 Kgm^(2)` `I2 = 0.20 Kgm^(2)` `omega_1 = 10 (rad)/(sec)` `omega_2 = 5 (rad)/(sec)`

Options

A. 0 J
B. 5 J
C. 10 J
D. 20/3 J

Answer

To find the total kinetic energy of the two discs when they rotate with the same angular velocity, we can follow these steps:

### Step 1: Calculate the Initial Angular Momentum
The initial angular momentum \( L_{\text{initial}} \) of the system is the sum of the angular momenta of both discs. The formula for angular momentum is given by:

\[ L = I \cdot \omega \]

Thus,

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

Substituting the given values:

\[ I_1 = 0.10 \, \text{kgm}^2, \quad \omega_1 = 10 \, \text{rad/s} \]
\[ I_2 = 0.20 \, \text{kgm}^2, \quad \omega_2 = 5 \, \text{rad/s} \]

Calculating:

\[ L_{\text{initial}} = (0.10 \times 10) + (0.20 \times 5) = 1 + 1 = 2 \, \text{kgm}^2/\text{s} \]

### Step 2: Determine the Final Angular Velocity
When the discs are brought together, they will rotate with a common angular velocity \( \omega \). The final angular momentum \( L_{\text{final}} \) can be expressed as:

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

Setting the initial and final angular momentum equal gives:

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

Thus,

\[ 2 = (0.10 + 0.20) \cdot \omega \]

Calculating the total moment of inertia:

\[ 0.10 + 0.20 = 0.30 \, \text{kgm}^2 \]

Now, substituting this into the equation:

\[ 2 = 0.30 \cdot \omega \]

Solving for \( \omega \):

\[ \omega = \frac{2}{0.30} = \frac{20}{3} \, \text{rad/s} \]

### Step 3: Calculate the Total Kinetic Energy
The total kinetic energy \( KE_f \) when both discs are rotating with the same angular velocity is given by:

\[ KE_f = \frac{1}{2} I_1 \omega^2 + \frac{1}{2} I_2 \omega^2 \]

This can be simplified to:

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

Substituting the values:

\[ KE_f = \frac{1}{2} (0.10 + 0.20) \left(\frac{20}{3}\right)^2 \]

Calculating:

\[ KE_f = \frac{1}{2} (0.30) \left(\frac{400}{9}\right) = \frac{0.30 \times 400}{18} = \frac{120}{18} = \frac{20}{3} \, \text{Joules} \]

### Final Answer
The total kinetic energy when both discs rotate with the same angular velocity is:

\[ KE_f = \frac{20}{3} \, \text{Joules} \]