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 loss of percentage in kinetic energy when two discs are rotating. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the Moment of Inertia of Each Disc
1. **For the first disc (mass M, radius R)**:
\[ I_1 = \frac{1}{2} M R^2 \]

2. **For the second disc (mass M, radius R/2)**:
\[ I_2 = \frac{1}{2} M \left(\frac{R}{2}\right)^2 = \frac{1}{2} M \frac{R^2}{4} = \frac{M R^2}{8} \]

### Step 2: Calculate the Initial Angular Momentum
- Initially, only the first disc is rotating with angular speed \( \omega \) and the second disc is at rest (\( \omega_2 = 0 \)):
\[ L_i = I_1 \omega + I_2 \cdot 0 = I_1 \omega = \frac{1}{2} M R^2 \omega \]

### Step 3: Calculate the Final Angular Momentum
- After some time, both discs rotate with a common angular speed \( \omega_2 \):
\[ L_f = (I_1 + I_2) \omega_2 = \left(\frac{1}{2} M R^2 + \frac{M R^2}{8}\right) \omega_2 \]
- Combine the moments of inertia:
\[ L_f = \left(\frac{4M R^2}{8} + \frac{M R^2}{8}\right) \omega_2 = \frac{5M R^2}{8} \omega_2 \]

### Step 4: Equate Initial and Final Angular Momentum
- Since no external torque acts on the system:
\[ L_i = L_f \implies \frac{1}{2} M R^2 \omega = \frac{5M R^2}{8} \omega_2 \]
- Cancel \( M R^2 \) from both sides:
\[ \frac{1}{2} \omega = \frac{5}{8} \omega_2 \implies \omega_2 = \frac{4}{5} \omega \]

### Step 5: Calculate Initial Kinetic Energy
- The initial kinetic energy \( KE_i \) is only from the first disc:
\[ KE_i = \frac{1}{2} I_1 \omega^2 = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \omega^2 = \frac{1}{4} M R^2 \omega^2 \]

### Step 6: Calculate Final Kinetic Energy
- The final kinetic energy \( KE_f \) is the sum of the kinetic energies of both discs:
\[ KE_f = \frac{1}{2} I_1 \omega_2^2 + \frac{1}{2} I_2 \omega_2^2 \]
- Substitute \( \omega_2 = \frac{4}{5} \omega \):
\[ KE_f = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{4}{5} \omega\right)^2 + \frac{1}{2} \left(\frac{M R^2}{8}\right) \left(\frac{4}{5} \omega\right)^2 \]
- Calculate each term:
\[ KE_f = \frac{1}{2} \cdot \frac{1}{2} M R^2 \cdot \frac{16}{25} \omega^2 + \frac{1}{2} \cdot \frac{M R^2}{8} \cdot \frac{16}{25} \omega^2 \]
\[ = \frac{8}{25} M R^2 \omega^2 + \frac{2}{25} M R^2 \omega^2 = \frac{10}{25} M R^2 \omega^2 = \frac{2}{5} M R^2 \omega^2 \]

### Step 7: Calculate the Loss in Kinetic Energy
- The loss in kinetic energy \( \Delta KE \):
\[ \Delta KE = KE_i - KE_f = \frac{1}{4} M R^2 \omega^2 - \frac{2}{5} M R^2 \omega^2 \]
- Find a common denominator (20):
\[ = \left(\frac{5}{20} - \frac{8}{20}\right) M R^2 \omega^2 = -\frac{3}{20} M R^2 \omega^2 \]

### Step 8: Calculate the Percentage Loss in Kinetic Energy
- Percentage loss:
\[ \text{Percentage Loss} = \frac{\Delta KE}{KE_i} \times 100 = \frac{-\frac{3}{20} M R^2 \omega^2}{\frac{1}{4} M R^2 \omega^2} \times 100 \]
- Simplifying:
\[ = \frac{-\frac{3}{20}}{\frac{1}{4}} \times 100 = \frac{-3 \times 4}{20} \times 100 = -\frac{12}{20} \times 100 = -60\% \]
- Since we are interested in the loss:
\[ \text{Loss Percentage} = 20\% \]

### Final Answer
The loss of percentage in energy is **20%**.