The maximum velocity at the lowest point, so that the string just slack at the highest point in a vertical circle of radius l.

Options

A. `sqrt(gl)`
B. `sqrt(3gl)`
C. `sqrt(5gl)`
D. `sqrt(7 gl)`

Explanation

To solve the problem of finding the maximum velocity at the lowest point of a vertical circle of radius \( L \) such that the string just slackens at the highest point, we can follow these steps:

### Step 1: Understand the Energy Conservation
At the lowest point of the vertical circle, the object has maximum kinetic energy and minimum potential energy. At the highest point, the object has minimum kinetic energy and maximum potential energy. We will use the principle of conservation of mechanical energy.

### Step 2: Write the Energy Equations
Let:
- \( m \) = mass of the object
- \( v_L \) = velocity at the lowest point
- \( v_H \) = velocity at the highest point
- \( g \) = acceleration due to gravity

At the lowest point (point A):
- Kinetic Energy (KE) = \( \frac{1}{2} mv_L^2 \)
- Potential Energy (PE) = 0 (taking this point as reference)

At the highest point (point B):
- Kinetic Energy (KE) = \( \frac{1}{2} mv_H^2 \)
- Potential Energy (PE) = \( mg(2L) \) (since the height is \( 2L \))

Using conservation of energy:
\[ \frac{1}{2} mv_L^2 = \frac{1}{2} mv_H^2 + mg(2L) \]

### Step 3: Simplify the Energy Equation
Cancel \( m \) from the equation:
\[ \frac{1}{2} v_L^2 = \frac{1}{2} v_H^2 + 2gL \]

Multiply through by 2:
\[ v_L^2 = v_H^2 + 4gL \]

### Step 4: Analyze Forces at the Highest Point
At the highest point, for the string to just slack, the tension in the string must be zero. The only forces acting on the mass at the highest point are the gravitational force and the centripetal force required to keep it moving in a circle.

The equation for the forces at the highest point (where tension \( T = 0 \)) is:
\[ mg = \frac{mv_H^2}{L} \]

Cancel \( m \):
\[ g = \frac{v_H^2}{L} \]

Rearranging gives:
\[ v_H^2 = gL \]

### Step 5: Substitute \( v_H^2 \) into the Energy Equation
Now, substitute \( v_H^2 \) back into the energy equation:
\[ v_L^2 = gL + 4gL \]
\[ v_L^2 = 5gL \]

### Step 6: Solve for \( v_L \)
Taking the square root:
\[ v_L = \sqrt{5gL} \]

### Conclusion
Thus, the maximum velocity at the lowest point so that the string just slackens at the highest point is:
\[ v_L = \sqrt{5gL} \]