A particle is moving in a circular orbit of radius `r_(1)` with an angular velocity `omega_(1)`. It jumps to another circular orbit of radius `r_(2)` and attains an angular velocity `omega_(2)`. IF `r_(2) = 0.5 r_(1)` and assuming that no external torque is applied to that system, then, the angular velocity `omega_(2)` is given by

Options

A. `omega_(2) = 4omega_(1)`
B. `omega_(2) = 3omega_(1)`
C. `omega_(2) = 2omega_(1)`
D. `omega_(2) = omega_(1)`

Answer

To solve the problem, we need to use the principle of conservation of angular momentum. The angular momentum of a particle moving in a circular orbit is given by the formula:

\[ L = m \cdot r^2 \cdot \omega \]

where:
- \( L \) is the angular momentum,
- \( m \) is the mass of the particle,
- \( r \) is the radius of the circular orbit,
- \( \omega \) is the angular velocity.

### Step-by-Step Solution:

1. **Identify Initial and Final Conditions**:
- Initial radius: \( r_1 \)
- Initial angular velocity: \( \omega_1 \)
- Final radius: \( r_2 = 0.5 r_1 \)
- Final angular velocity: \( \omega_2 \)

2. **Write the Expression for Initial Angular Momentum**:
The initial angular momentum \( L_1 \) when the particle is at radius \( r_1 \):
\[ L_1 = m \cdot r_1^2 \cdot \omega_1 \]

3. **Write the Expression for Final Angular Momentum**:
The final angular momentum \( L_2 \) when the particle is at radius \( r_2 \):
\[ L_2 = m \cdot r_2^2 \cdot \omega_2 \]

4. **Set Initial and Final Angular Momentum Equal**:
Since no external torque is applied, we can set the initial and final angular momentum equal:
\[ L_1 = L_2 \]
Therefore,
\[ m \cdot r_1^2 \cdot \omega_1 = m \cdot r_2^2 \cdot \omega_2 \]

5. **Cancel the Mass**:
The mass \( m \) appears on both sides of the equation, so we can cancel it out:
\[ r_1^2 \cdot \omega_1 = r_2^2 \cdot \omega_2 \]

6. **Substitute \( r_2 \)**:
Substitute \( r_2 = 0.5 r_1 \) into the equation:
\[ r_1^2 \cdot \omega_1 = (0.5 r_1)^2 \cdot \omega_2 \]
Simplifying the right side:
\[ r_1^2 \cdot \omega_1 = 0.25 r_1^2 \cdot \omega_2 \]

7. **Cancel \( r_1^2 \)**:
Since \( r_1^2 \) is common on both sides, we can cancel it:
\[ \omega_1 = 0.25 \cdot \omega_2 \]

8. **Solve for \( \omega_2 \)**:
Rearranging gives:
\[ \omega_2 = \frac{\omega_1}{0.25} = 4 \cdot \omega_1 \]

### Final Result:
Thus, the angular velocity \( \omega_2 \) is given by:
\[ \omega_2 = 4 \cdot \omega_1 \]