A stone is tied to one end of a light inexensible string of length l and made to roate on a vertical circle keeping other end of the spring at the centre. If speed of stone at the highest point is `v(v gt sqrt(gl))` then its speed at the lowest point is

Options

A. `sqrt(4gl)`
B. `sqrt(v^(2)+2gl)`
C. `sqrt(v^(2)+gl)`
D. `sqrt(v^(2)+4gl)`

Answer

To find the speed of the stone at the lowest point of its vertical circular motion, we can use the principle of conservation of energy or the work-energy theorem. Here’s the step-by-step solution:

### Step 1: Understand the Problem
- A stone is tied to a string of length \( l \) and rotates in a vertical circle.
- The speed of the stone at the highest point is given as \( v \), where \( v > \sqrt{g l} \).

### Step 2: Identify the Forces at the Highest Point
At the highest point of the circle:
- The forces acting on the stone are the tension in the string \( T \) and the weight of the stone \( mg \).
- The net centripetal force required to keep the stone in circular motion is provided by the sum of these forces:
\[ T + mg = \frac{mv^2}{l} \]

### Step 3: Rearranging the Equation
From the equation above, we can express the tension \( T \):
\[ T = \frac{mv^2}{l} - mg \]

### Step 4: Determine the Conditions for Tension
For the stone to complete the circular motion, the tension must be greater than or equal to zero:
\[ \frac{mv^2}{l} - mg \geq 0 \implies v^2 \geq g l \implies v \geq \sqrt{g l} \]
This condition is satisfied as given in the problem.

### Step 5: Apply the Work-Energy Theorem
To find the speed at the lowest point, we will apply the work-energy theorem:
- The work done by gravity when the stone moves from the highest point to the lowest point is equal to the change in kinetic energy.

### Step 6: Calculate the Change in Potential Energy
The change in height from the highest point to the lowest point is \( 2l \):
- The change in potential energy \( \Delta U \) is:
\[ \Delta U = -mg(2l) = -2mgl \]

### Step 7: Calculate the Change in Kinetic Energy
Let \( u \) be the speed of the stone at the lowest point. The change in kinetic energy \( \Delta K \) is:
\[ \Delta K = K_f - K_i = \frac{1}{2} mu^2 - \frac{1}{2} mv^2 \]

### Step 8: Set Up the Work-Energy Equation
According to the work-energy theorem:
\[ \Delta U + \Delta K = 0 \]
Substituting the expressions for \( \Delta U \) and \( \Delta K \):
\[ -2mgl + \left(\frac{1}{2} mu^2 - \frac{1}{2} mv^2\right) = 0 \]

### Step 9: Simplify the Equation
Rearranging gives:
\[ \frac{1}{2} mu^2 = 2mgl + \frac{1}{2} mv^2 \]
Dividing through by \( m \) and multiplying by 2:
\[ u^2 = 4gl + v^2 \]

### Step 10: Solve for \( u \)
Taking the square root:
\[ u = \sqrt{v^2 + 4gl} \]

### Conclusion
Thus, the speed of the stone at the lowest point is:
\[ u = \sqrt{v^2 + 4gl} \]