A stone tied to a string of length `L` is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at lowest position and has a speed `u` . Find the magnitude of the change in its velocity as it reaches a position, where the string is horizontal.

Explanation

To find the magnitude of the change in velocity of a stone tied to a string as it moves from the lowest position to the horizontal position in a vertical circle, we can follow these steps:

### Step 1: Understand the initial and final positions
- The stone is initially at the lowest position of the vertical circle, where it has a speed `u`.
- The final position is when the stone reaches the horizontal position, where it will have a different speed `V`.

### Step 2: Identify the initial and final velocity vectors
- The initial velocity vector \( \vec{u} \) is directed horizontally to the right at the lowest position.
- The final velocity vector \( \vec{V} \) is directed horizontally to the left at the horizontal position.

### Step 3: Calculate the change in velocity
- The change in velocity \( \Delta \vec{V} \) is given by:
\[ \Delta \vec{V} = \vec{V} - \vec{u} \]
- The magnitude of the change in velocity can be found using the Pythagorean theorem since the initial and final velocities are perpendicular to each other:
\[ |\Delta \vec{V}| = \sqrt{V^2 + u^2} \]

### Step 4: Determine the final speed \( V \)
- To find \( V \), we can use the principle of conservation of energy. The work done by gravity as the stone moves from the lowest position to the horizontal position is equal to the change in kinetic energy.
- The work done by gravity \( W \) when the stone moves a height \( L \) is:
\[ W = -mgL \]
- The change in kinetic energy is:
\[ \Delta KE = \frac{1}{2}mV^2 - \frac{1}{2}mu^2 \]
- Setting the work done equal to the change in kinetic energy:
\[ -mgL = \frac{1}{2}mV^2 - \frac{1}{2}mu^2 \]
- Simplifying this equation (the mass \( m \) cancels out):
\[ -gL = \frac{1}{2}V^2 - \frac{1}{2}u^2 \]
\[ V^2 = u^2 - 2gL \]

### Step 5: Substitute \( V^2 \) back into the change in velocity equation
- Now substitute \( V^2 \) into the change in velocity magnitude equation:
\[ |\Delta \vec{V}| = \sqrt{(u^2 - 2gL) + u^2} \]
\[ |\Delta \vec{V}| = \sqrt{2u^2 - 2gL} \]
\[ |\Delta \vec{V}| = \sqrt{2(u^2 - gL)} \]

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