A stone is tied to a string of length l and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is

Options

A. `sqrt(2(u^(2)-gl)`
B. `sqrt(u^(2)=gl)`
C. `u-sqrt(u^(2)=2gl)`
D. `sqrt(2gl)`

Explanation

To solve the problem, we need to determine the change in velocity of the stone as it moves from the lowest position of the vertical circle to the horizontal position. Let's break this down step by step.

### Step 1: Understand the Initial Conditions
At the lowest point of the vertical circle, the stone has a speed \( u \). The forces acting on the stone at this point are the tension in the string and the weight of the stone.

### Step 2: Identify the Final Position
When the stone reaches the horizontal position, it will have a different speed, which we will denote as \( v \).

### Step 3: Apply Conservation of Mechanical Energy
The total mechanical energy at the lowest point will be equal to the total mechanical energy at the horizontal position.

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

- At the horizontal position:
- Kinetic Energy (KE) = \( \frac{1}{2} m v^2 \)
- Potential Energy (PE) = \( mgL \) (where \( L \) is the length of the string)

Setting the initial mechanical energy equal to the final mechanical energy gives us:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mgL \]

### Step 4: Simplify the Equation
We can cancel \( m \) from the equation (assuming \( m \neq 0 \)):
\[ \frac{1}{2} u^2 = \frac{1}{2} v^2 + gL \]

Multiplying through by 2:
\[ u^2 = v^2 + 2gL \]

### Step 5: Solve for \( v^2 \)
Rearranging the equation gives:
\[ v^2 = u^2 - 2gL \]

### Step 6: Find the Change in Velocity
The change in velocity \( \Delta v \) can be found using vector subtraction. The initial velocity \( \vec{u} \) is directed upwards at the lowest point, while the final velocity \( \vec{v} \) is directed horizontally. The angle between these two velocities is \( 90^\circ \).

Using the Pythagorean theorem:
\[ |\Delta \vec{v}| = \sqrt{|\vec{u}|^2 + |\vec{v}|^2} \]
Substituting for \( v \):
\[ |\Delta \vec{v}| = \sqrt{u^2 + (u^2 - 2gL)} \]
\[ |\Delta \vec{v}| = \sqrt{u^2 + u^2 - 2gL} = \sqrt{2u^2 - 2gL} \]

### Final Answer
Thus, the magnitude of the change in velocity as the stone reaches the horizontal position is:
\[ |\Delta \vec{v}| = \sqrt{2(u^2 - gL)} \]