Two coaxial discs, having moments of inertie `I_1` and `I_1/2` are rotating with respective angular volecities `omega_1` and `(omega1)/2`, about their common axis.They are brought in contact with each other and thereafter they rotate with a common angular volecity.If `E_f` and `E_i` are the final and initial total energies, then `(E_f-E_1)` is:

Options

A. `-(I_1omega_1^2)/12`
B. `-(I_1omega_1^2)/6`
C. `3/8I_1omega_1^2`
D. `-(I_1omega_1^2)/24`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine Initial Energies
We have two discs with moments of inertia \( I_1 \) and \( I_2 = \frac{I_1}{2} \), rotating with angular velocities \( \omega_1 \) and \( \omega_2 = \frac{\omega_1}{2} \) respectively.

The initial kinetic energy \( E_i \) of the system is the sum of the kinetic energies of both discs:
\[ E_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 \]
Substituting \( I_2 \) and \( \omega_2 \):
\[ E_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} \left(\frac{I_1}{2}\right) \left(\frac{\omega_1}{2}\right)^2 \]
Calculating the second term:
\[ E_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} \cdot \frac{I_1}{2} \cdot \frac{\omega_1^2}{4} = \frac{1}{2} I_1 \omega_1^2 + \frac{I_1 \omega_1^2}{16} \]
Combining the terms:
\[ E_i = \frac{8}{16} I_1 \omega_1^2 + \frac{1}{16} I_1 \omega_1^2 = \frac{9}{16} I_1 \omega_1^2 \]

### Step 2: Apply Conservation of Angular Momentum
The total angular momentum before contact must equal the total angular momentum after contact:
\[ L_i = L_f \]
Calculating initial angular momentum:
\[ L_i = I_1 \omega_1 + I_2 \omega_2 = I_1 \omega_1 + \frac{I_1}{2} \cdot \frac{\omega_1}{2} = I_1 \omega_1 + \frac{I_1 \omega_1}{4} = \frac{5}{4} I_1 \omega_1 \]
After contact, the combined moment of inertia is:
\[ I_f = I_1 + I_2 = I_1 + \frac{I_1}{2} = \frac{3}{2} I_1 \]
Let \( \omega \) be the common angular velocity after contact:
\[ L_f = I_f \omega = \frac{3}{2} I_1 \omega \]
Setting \( L_i = L_f \):
\[ \frac{5}{4} I_1 \omega_1 = \frac{3}{2} I_1 \omega \]
Solving for \( \omega \):
\[ \omega = \frac{5}{4} \cdot \frac{2}{3} \omega_1 = \frac{5}{6} \omega_1 \]

### Step 3: Calculate Final Energy
Now we can find the final energy \( E_f \):
\[ E_f = \frac{1}{2} I_f \omega^2 = \frac{1}{2} \cdot \frac{3}{2} I_1 \left(\frac{5}{6} \omega_1\right)^2 \]
Calculating:
\[ E_f = \frac{1}{2} \cdot \frac{3}{2} I_1 \cdot \frac{25}{36} \omega_1^2 = \frac{3}{4} I_1 \cdot \frac{25}{36} \omega_1^2 = \frac{75}{144} I_1 \omega_1^2 = \frac{25}{48} I_1 \omega_1^2 \]

### Step 4: Find the Difference in Energies
Now, we find \( E_f - E_i \):
\[ E_f - E_i = \frac{25}{48} I_1 \omega_1^2 - \frac{9}{16} I_1 \omega_1^2 \]
Finding a common denominator (48):
\[ E_f - E_i = \frac{25}{48} I_1 \omega_1^2 - \frac{27}{48} I_1 \omega_1^2 = -\frac{2}{48} I_1 \omega_1^2 = -\frac{1}{24} I_1 \omega_1^2 \]

### Final Answer
Thus, the difference in energies is:
\[ E_f - E_i = -\frac{1}{24} I_1 \omega_1^2 \]