A particle will leave a vertical circle of radius `r` when its velocity at the lowest point of the circle (upsilon L) is .

Options

A. `sqrt2 gr`
B. `sqrt 5gr`
C. `sqrt`3 g r`
D. `sqrt 6 g r`

Answer

To determine the velocity at the lowest point of a vertical circle that will cause a particle to leave the circle, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have a particle moving in a vertical circle of radius \( r \).
- We need to find the minimum velocity at the lowest point (point A) such that the particle will leave the circular path.

2. **Identify Key Points**:
- The lowest point of the circle is point A.
- The highest point of the circle is point B.
- The particle will leave the circle if it does not have enough velocity to reach the highest point.

3. **Velocity at the Highest Point**:
- For a particle to just reach the highest point (point B) of the vertical circle, it must have a minimum velocity \( v_B \) at that point.
- At the highest point, the centripetal force required to keep the particle in circular motion is provided by the weight of the particle. Therefore:
\[ \frac{mv_B^2}{r} = mg \]
- Simplifying this gives:
\[ v_B^2 = gr \quad \Rightarrow \quad v_B = \sqrt{gr} \]

4. **Energy Conservation**:
- We can use the conservation of mechanical energy to find the velocity at the lowest point (point A).
- The total mechanical energy at the lowest point (potential energy + kinetic energy) must equal the total mechanical energy at the highest point.
- At the lowest point (point A):
- Potential energy \( U_A = 0 \) (taking this as the reference point)
- Kinetic energy \( K_A = \frac{1}{2} mv_A^2 \)
- At the highest point (point B):
- Potential energy \( U_B = mg(2r) \) (height is \( 2r \))
- Kinetic energy \( K_B = \frac{1}{2} mv_B^2 \)

5. **Setting Up the Energy Equation**:
- Using conservation of energy:
\[ K_A + U_A = K_B + U_B \]
\[ \frac{1}{2} mv_A^2 + 0 = \frac{1}{2} mv_B^2 + mg(2r) \]

6. **Substituting for \( v_B \)**:
- Substitute \( v_B = \sqrt{gr} \) into the energy equation:
\[ \frac{1}{2} mv_A^2 = \frac{1}{2} m(gr) + mg(2r) \]
\[ \frac{1}{2} mv_A^2 = \frac{1}{2} mgr + 2mgr \]
\[ \frac{1}{2} mv_A^2 = \frac{1}{2} mgr + \frac{4}{2} mgr = \frac{5}{2} mgr \]

7. **Solving for \( v_A \)**:
- Multiply both sides by 2:
\[ mv_A^2 = 5mgr \]
- Divide by \( m \):
\[ v_A^2 = 5gr \]
- Taking the square root:
\[ v_A = \sqrt{5gr} \]

8. **Conclusion**:
- The minimum velocity at the lowest point for the particle to leave the vertical circle is:
\[ v_A = \sqrt{5gr} \]

### Final Answer:
The particle will leave the vertical circle when its velocity at the lowest point is \( \sqrt{5gr} \).