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. $2v\sin \left(\theta /2\right)$
• D. $2v\cos \left(\theta /2\right)$

Answer 1

The correct option is C $2v\sin \left(\theta /2\right)$


Let the $x$ axis passes through the point $P$. The velocity of the particle at the point $P$, the velocity at the point $Q$ and the change in velocity as particle moves from $P$ to $Q$ are given by
${\overrightarrow{v}}_P=v\hat{j}$,
${\overrightarrow{v}}_Q=-v\sin \theta \hat{i}+v\cos \theta \hat{j}$,
$\Delta \overrightarrow{v}={\overrightarrow{v}}_Q-{\overrightarrow{v}}_P=-v\sin \theta \hat{i}+v(\cos \theta -1)\hat{j}$
The magnitude of the change in velocity is given by
$\left|\Delta \overrightarrow{v}\right|=\sqrt{(-v\sin \theta {)}^2+(v(\cos \theta -1){)}^2}$
$=v\sqrt{2(1-\cos \theta )}=v\sqrt{2\times 2{\sin }^2\left(\theta /2\right)}$
$=2v\sin \left(\theta /2\right)$

Alternative:
Using law of vector subtraction,
$\left|\Delta \overrightarrow{v}\right|=\sqrt{v_P^2+v_Q^2-2v_P{v}_Q\cos \theta }$
$=\sqrt{v^2+v^2-2v^2\cos \theta }$
$=\sqrt{2v^2(1-\cos \theta }$
$=2v\sin \left(\theta /2\right)$