A circular disc of mass M and radius R rotating with speed of `omega` and another disc of same mass M and radius R/2 placed co axially on first disc, after sometime disc-2 also attain constant speed `w_2` . Find loss of percentage in energy

Answer

To solve the problem, we need to find the percentage loss of energy when a circular disc of mass \( M \) and radius \( R \) is rotating with an angular speed \( \omega \), and another disc of the same mass \( M \) and radius \( \frac{R}{2} \) is placed coaxially on the first disc. After some time, the second disc attains a constant angular speed \( \omega_2 \).

### Step-by-Step Solution:

1. **Calculate the Moment of Inertia of Each Disc:**
- The moment of inertia \( I_1 \) of the first disc (radius \( R \)):
\[ I_1 = \frac{1}{2} M R^2 \]
- The moment of inertia \( I_2 \) of the second disc (radius \( \frac{R}{2} \)):
\[ I_2 = \frac{1}{2} M \left(\frac{R}{2}\right)^2 = \frac{1}{2} M \cdot \frac{R^2}{4} = \frac{1}{8} M R^2 \]

2. **Initial Angular Momentum:**
- The initial angular momentum \( L_i \) of the system (only the first disc is rotating initially):
\[ L_i = I_1 \omega = \left(\frac{1}{2} M R^2\right) \omega \]

3. **Final Moment of Inertia of the Combined System:**
- The total moment of inertia \( I_f \) when both discs are rotating together:
\[ I_f = I_1 + I_2 = \frac{1}{2} M R^2 + \frac{1}{8} M R^2 = \frac{4}{8} M R^2 + \frac{1}{8} M R^2 = \frac{5}{8} M R^2 \]

4. **Final Angular Momentum:**
- The final angular momentum \( L_f \) when both discs are rotating with the final angular speed \( \omega_2 \):
\[ L_f = I_f \omega_2 = \left(\frac{5}{8} M R^2\right) \omega_2 \]

5. **Conservation of Angular Momentum:**
- Since angular momentum is conserved:
\[ L_i = L_f \]
\[ \left(\frac{1}{2} M R^2\right) \omega = \left(\frac{5}{8} M R^2\right) \omega_2 \]
- Cancel \( M R^2 \) from both sides:
\[ \frac{1}{2} \omega = \frac{5}{8} \omega_2 \]
- Solving for \( \omega_2 \):
\[ \omega_2 = \frac{4}{5} \omega \]

6. **Initial and Final Kinetic Energy:**
- Initial kinetic energy \( E_i \):
\[ E_i = \frac{1}{2} I_1 \omega^2 = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \omega^2 = \frac{1}{4} M R^2 \omega^2 \]
- Final kinetic energy \( E_f \):
\[ E_f = \frac{1}{2} I_f \omega_2^2 = \frac{1}{2} \left(\frac{5}{8} M R^2\right) \left(\frac{4}{5} \omega\right)^2 \]
\[ E_f = \frac{1}{2} \left(\frac{5}{8} M R^2\right) \left(\frac{16}{25} \omega^2\right) = \frac{5}{16} M R^2 \omega^2 \]

7. **Calculate the Loss of Energy:**
- Loss of energy \( \Delta E \):
\[ \Delta E = E_i - E_f = \frac{1}{4} M R^2 \omega^2 - \frac{5}{16} M R^2 \omega^2 \]
\[ \Delta E = \left(\frac{4}{16} - \frac{5}{16}\right) M R^2 \omega^2 = -\frac{1}{16} M R^2 \omega^2 \]

8. **Percentage Loss of Energy:**
- Percentage loss:
\[ \text{Percentage Loss} = \frac{\Delta E}{E_i} \times 100 = \frac{-\frac{1}{16} M R^2 \omega^2}{\frac{1}{4} M R^2 \omega^2} \times 100 = \frac{-\frac{1}{16}}{\frac{1}{4}} \times 100 = -\frac{1}{16} \times 4 \times 100 = -25\% \]
- Since we are interested in the loss, we take the absolute value:
\[ \text{Percentage Loss} = 25\% \]

### Final Answer:
The percentage loss of energy is **25%**.