A circular disc of moment of inertia `I_(t)` is rotating in a horizontal plane about its symmetry axis with a constant angular velocity `omega_(i)`. Another disc of moment of inertia `I_(b)` is dropped co-axially onto the rotating disc. Initially, the second disc has zero angular speed. Eventually, both the discs rotate with a constant angular speed `omega_(f)`. Calculate the energy lost by the initially rotating disc due to friction.

Options

A. `1/2 I_(b)^(2)/((I_(t)+I_(b))) omega_(i)^(2)`
B. `1/2 I_(t)^(2)/((I_(t)+I_(b))) omega_(i)^(2)`
C. `1/2 (I_(b)-I_(t))/((I_(t)+I_(b))) omega_(i)^(2)`
D. `1/2 (I_(b) I_(t))/((I_(t)+I_(b))) omega_(i)^(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the System
We have two discs:
- Disc A (the initially rotating disc) with moment of inertia \( I_t \) and initial angular velocity \( \omega_i \).
- Disc B (the disc being dropped) with moment of inertia \( I_b \) and initial angular velocity \( 0 \).

### Step 2: Apply Conservation of Angular Momentum
Since there are no external torques acting on the system, angular momentum is conserved. The initial angular momentum \( L_i \) of the system is given by:
\[ L_i = I_t \omega_i \]
After disc B is dropped, both discs rotate together with a final angular velocity \( \omega_f \). The final angular momentum \( L_f \) is:
\[ L_f = (I_t + I_b) \omega_f \]
Setting the initial and final angular momentum equal gives:
\[ I_t \omega_i = (I_t + I_b) \omega_f \]

### Step 3: Solve for Final Angular Velocity
Rearranging the equation for \( \omega_f \):
\[ \omega_f = \frac{I_t \omega_i}{I_t + I_b} \]

### Step 4: Calculate Initial and Final Energy
The initial kinetic energy \( E_i \) of the system is:
\[ E_i = \frac{1}{2} I_t \omega_i^2 \]
The final kinetic energy \( E_f \) of the system is:
\[ E_f = \frac{1}{2} (I_t + I_b) \omega_f^2 \]
Substituting \( \omega_f \) from Step 3 into the equation for \( E_f \):
\[ E_f = \frac{1}{2} (I_t + I_b) \left( \frac{I_t \omega_i}{I_t + I_b} \right)^2 \]
Simplifying this:
\[ E_f = \frac{1}{2} (I_t + I_b) \cdot \frac{I_t^2 \omega_i^2}{(I_t + I_b)^2} = \frac{I_t^2 \omega_i^2}{2(I_t + I_b)} \]

### Step 5: Calculate Energy Lost
The energy lost \( \Delta E \) due to friction is the difference between the initial and final energies:
\[ \Delta E = E_i - E_f \]
Substituting the expressions for \( E_i \) and \( E_f \):
\[ \Delta E = \frac{1}{2} I_t \omega_i^2 - \frac{I_t^2 \omega_i^2}{2(I_t + I_b)} \]
Factoring out \( \frac{1}{2} \omega_i^2 \):
\[ \Delta E = \frac{1}{2} \omega_i^2 \left( I_t - \frac{I_t^2}{I_t + I_b} \right) \]
Finding a common denominator:
\[ \Delta E = \frac{1}{2} \omega_i^2 \left( \frac{I_t(I_t + I_b) - I_t^2}{I_t + I_b} \right) = \frac{1}{2} \omega_i^2 \left( \frac{I_t I_b}{I_t + I_b} \right) \]

### Final Result
Thus, the energy lost by the initially rotating disc due to friction is:
\[ \Delta E = \frac{1}{2} \frac{I_t I_b}{I_t + I_b} \omega_i^2 \]

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