A particle is moving with a uniform speed v in a circular path of radius r with the centre at 0. When the particle moves from a point P to Q on the circle such that `anglePOQ = theta` , then the magnitude of the change in velocity is

Options

A. `2v sin (2 theta)`
B. zero
C. `2v sin (theta)/2`
D. `2v cos theta (theta/2)`

Explanation

To find the magnitude of the change in velocity of a particle moving along a circular path when it moves from point P to point Q, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
The particle is moving with a uniform speed \( v \) in a circular path of radius \( r \). The angle \( \angle POQ = \theta \) is given, where O is the center of the circle, P is the initial position, and Q is the final position.

2. **Identify Initial and Final Velocities**:
Since the particle is moving in a circular path, at point P, the velocity vector \( \vec{v_1} \) is tangent to the circle at that point. Similarly, at point Q, the velocity vector \( \vec{v_2} \) is also tangent to the circle at that point. Both velocities have the same magnitude \( v \).

3. **Determine the Angle Between the Velocity Vectors**:
The angle between the two velocity vectors \( \vec{v_1} \) and \( \vec{v_2} \) is equal to \( \theta \) because both vectors are tangent to the circle and the angle \( \angle POQ \) subtended at the center is \( \theta \).

4. **Use the Formula for Change in Velocity**:
The change in velocity \( \Delta \vec{v} \) can be calculated using the formula:
\[ |\Delta \vec{v}| = |\vec{v_2} - \vec{v_1}| \]
Using the law of cosines:
\[ |\Delta \vec{v}| = \sqrt{|\vec{v_1}|^2 + |\vec{v_2}|^2 - 2|\vec{v_1}||\vec{v_2}|\cos(\theta)} \]
Since \( |\vec{v_1}| = |\vec{v_2}| = v \):
\[ |\Delta \vec{v}| = \sqrt{v^2 + v^2 - 2v^2\cos(\theta)} \]
\[ |\Delta \vec{v}| = \sqrt{2v^2(1 - \cos(\theta))} \]

5. **Simplify the Expression**:
Factor out \( v^2 \):
\[ |\Delta \vec{v}| = v\sqrt{2(1 - \cos(\theta))} \]

6. **Use the Trigonometric Identity**:
We can use the identity \( 1 - \cos(\theta) = 2\sin^2(\frac{\theta}{2}) \):
\[ |\Delta \vec{v}| = v\sqrt{2 \cdot 2\sin^2\left(\frac{\theta}{2}\right)} \]
\[ |\Delta \vec{v}| = v\cdot 2\sin\left(\frac{\theta}{2}\right) \]

7. **Final Result**:
Therefore, the magnitude of the change in velocity is:
\[ |\Delta \vec{v}| = 2v\sin\left(\frac{\theta}{2}\right) \]