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 ${\omega }^2={\omega }_0^2+2\alpha \theta$ and ${\omega }_0=0$,
we get, ${\omega }^2=2\times \frac{a}{r}\times \pi$, where $r$ is the radius of circular path and for half circular path, $\theta =\pi$
$\Rightarrow \omega =\sqrt{\frac{2a\pi }{r}}(\because a$ is constant)
Also, velocity $v=r\omega \Rightarrow v=\sqrt{2ar\pi }$

Radial (or normal) acceleration,
$a_n=\frac{v^2}{R}=\frac{2ar\pi }{r}=2a\pi m/s^2$
Therefore, angle between velocity vector and net acceleration vector is given by
$\theta ={\tan }^{-1}\left(\frac{a_n}{a_t}\right)={\tan }^{-1}\left(\frac{2a\pi }{a}\right)={\tan }^{-1}\left(2\pi \right)$