Question

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

• A. $2v\sin \left(2\theta \right)$
• B. Zero
• C. $2v\sin \left(\begin{array}{c}\frac{\theta }{2}\end{array}\right)$
• D. $2v\cos \left(\begin{array}{c}\frac{\theta }{2}\end{array}\right)$

Answer 1

The correct option is C $2v\sin \left(\begin{array}{c}\frac{\theta }{2}\end{array}\right)$

Velocity of particle at point P $v_p=-v\sin (90-\frac{\theta }{2})\hat{x}+v\cos (90-\frac{\theta }{2})\hat{y}$

$\therefore$ $v_p=-vcos\left(\frac{\theta }{2}\right)\hat{x}+v\sin \left(\frac{\theta }{2}\right)\hat{y}$

Velocity of particle at point Q $v_Q=-v\sin (90-\frac{\theta }{2})\hat{x}-v\cos (90-\frac{\theta }{2})\hat{y}$

$\therefore$ $v_Q=-vcos\left(\frac{\theta }{2}\right)\hat{x}-v\sin \left(\frac{\theta }{2}\right)\hat{y}$

Thus change in velocity $\Delta v=v_Q-v_p$

$\therefore$ We get $\Delta v=-2v\sin \left(\frac{\theta }{2}\right)$

$⟹$ $|\Delta v|=2v\sin \left(\frac{\theta }{2}\right)$