Question

A particle starts from rest, travelling on a circle with constant tangential acceleration. The angle between velocity vector and acceleration vector, at the moment when the particle completes half the circular track is

• A. ${\tan }^{-1}\left(2\pi \right)$
• B. ${\tan }^{-1}\left(\pi \right)$
• C. ${\tan }^{-1}\left(3\pi \right)$
• D. ${\tan }^{-1}\left(2\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(2\pi \right)$

Let tangential acceleration $a_t=am/s^2$
Using $v^2=u^2+2a_{t}s$ and $u=0$,
we get, $v=\sqrt{2as}$ where $s$ is distance covered.
For half circular path, $s=\pi R$
i.e $v=\sqrt{\left(2a\right)\left(\pi R\right)}(\because a$ is constant)

Radial (or normal) acceleration,
$a_n=\frac{v^2}{R}=\frac{2\pi aR}{R}=2\pi am/s^2$
$\theta ={\tan }^{-1}\left(\frac{a_n}{a_t}\right)={\tan }^{-1}\left(\frac{2\pi a}{a}\right)={\tan }^{-1}\left(2\pi \right)$