Question

If a particle is moving along a circle of radius \( r \) with constant speed \( v \), then the magnitude of the change in velocity when the particle moves from \( P \) to \( Q \), where \( PQ \) makes an angle \( \theta \) at the center of the circle is:

• A. \( 2v\sin\frac{\theta}{2} \)
• B. \( 2v\cos\frac{\theta}{2} \)
• C. \( 2v\sin\theta \)
• D. \( 2v\cos\theta \)

Hint:

For problems involving velocity change in circular motion, use the formula: \[ \Delta v = 2v \sin \frac{\theta}{2} \] where \( \theta \) is the angle subtended at the center by the path segment \( PQ \).

Answer 1

The Correct Option is A

In uniform circular motion, the velocity of the particle is always tangential to the path. The change in velocity is given by the vector difference between the velocities at points \( P \) and \( Q \). The velocity vectors at these points subtend an angle \( \theta \) at the center. The magnitude of the change in velocity is given by: \[ \Delta v = 2v \sin \frac{\theta}{2} \] Thus, the correct answer is: \[ 2v \sin \frac{\theta}{2} \]