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 such that the string just goes slack at the highest point, we can use the principles of conservation of energy and the dynamics of circular motion. Here’s a step-by-step solution:

### Step 1: Understand the System
We have a mass moving in a vertical circle of radius \( L \). At the lowest point, we denote the velocity as \( V_L \) and at the highest point as \( V_H \). The goal is to find \( V_L \) such that the tension in the string is zero at the highest point.

### Step 2: Apply Conservation of Energy
Using the conservation of mechanical energy, we can equate the total mechanical energy at the lowest point and the highest point.

- At the lowest point (height = 0):
- Potential Energy (PE) = 0
- Kinetic Energy (KE) = \( \frac{1}{2} m V_L^2 \)

- At the highest point (height = 2L):
- Potential Energy (PE) = \( mg(2L) = 2mgL \)
- Kinetic Energy (KE) = \( \frac{1}{2} m V_H^2 \)

Setting the total energy at the lowest point equal to the total energy at the highest point gives us:
\[ \frac{1}{2} m V_L^2 = 2mgL + \frac{1}{2} m V_H^2 \]

### Step 3: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} V_L^2 = 2gL + \frac{1}{2} V_H^2 \]

Multiplying through by 2 to eliminate the fraction:
\[ V_L^2 = 4gL + V_H^2 \tag{1} \]

### Step 4: Analyze Forces at the Highest Point
At the highest point, for the string to just go slack, the tension \( T \) must be zero. The only forces acting on the mass at the highest point are its weight \( mg \) and the centripetal force requirement.

Using Newton's second law, we have:
\[ mg = \frac{m V_H^2}{L} \]

Cancelling \( m \) (again assuming \( m \neq 0 \)):
\[ g = \frac{V_H^2}{L} \]

Rearranging gives us:
\[ V_H^2 = gL \tag{2} \]

### Step 5: Substitute \( V_H^2 \) into the Energy Equation
Now we substitute equation (2) into equation (1):
\[ V_L^2 = 4gL + gL \]
\[ V_L^2 = 5gL \]

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

### Final Answer
The maximum velocity at the lowest point such that the string just goes slack at the highest point is:
\[ V_L = \sqrt{5gL} \]

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