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

Options

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

Explanation

To solve the problem of energy lost by the initially rotating disk when a second disk is dropped onto it, we will follow these steps:

### Step 1: Understand the Initial Conditions
- We have two disks:
- Disk 1 (rotating) with moment of inertia \( I_t \) and initial angular speed \( \omega_i \).
- Disk 2 (dropped) with moment of inertia \( I_b \) and initial angular speed \( 0 \).

### Step 2: Apply Conservation of Angular Momentum
- Since there are no external torques acting on the system, the angular momentum before and after the second disk is dropped must be conserved.
- The initial angular momentum \( L_i \) is given by:
\[ L_i = I_t \cdot \omega_i \]
- The final angular momentum \( L_f \) when both disks are rotating together at angular speed \( \omega_f \) is:
\[ L_f = (I_t + I_b) \cdot \omega_f \]
- Setting the initial and final angular momentum equal gives us:
\[ I_t \cdot \omega_i = (I_t + I_b) \cdot \omega_f \]

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

### Step 4: Calculate Initial Kinetic Energy
- The initial kinetic energy \( KE_i \) of the rotating disk is:
\[ KE_i = \frac{1}{2} I_t \cdot \omega_i^2 \]

### Step 5: Calculate Final Kinetic Energy
- The final kinetic energy \( KE_f \) of both disks rotating together is:
\[ KE_f = \frac{1}{2} (I_t + I_b) \cdot \omega_f^2 \]
- Substituting \( \omega_f \) from Step 3:
\[ KE_f = \frac{1}{2} (I_t + I_b) \cdot \left( \frac{I_t \cdot \omega_i}{I_t + I_b} \right)^2 \]
- Simplifying this:
\[ KE_f = \frac{1}{2} \cdot \frac{(I_t^2 \cdot \omega_i^2)}{I_t + I_b} \]

### Step 6: Calculate Energy Lost
- The energy lost due to friction \( E_{lost} \) is the difference between the initial and final kinetic energies:
\[ E_{lost} = KE_i - KE_f \]
- Substituting the expressions for \( KE_i \) and \( KE_f \):
\[ E_{lost} = \frac{1}{2} I_t \cdot \omega_i^2 - \frac{1}{2} \cdot \frac{(I_t^2 \cdot \omega_i^2)}{I_t + I_b} \]
- Factoring out \( \frac{1}{2} \omega_i^2 \):
\[ E_{lost} = \frac{1}{2} \omega_i^2 \left( I_t - \frac{I_t^2}{I_t + I_b} \right) \]
- Simplifying further gives:
\[ E_{lost} = \frac{1}{2} \cdot \frac{I_t \cdot I_b \cdot \omega_i^2}{I_t + I_b} \]

### Final Answer
The energy lost by the initially rotating disk to friction is:
\[ E_{lost} = \frac{1}{2} \cdot \frac{I_t \cdot I_b \cdot \omega_i^2}{I_t + I_b} \]