VC1.11

Given, $r=\cos \omega t i+\sin \omega t j\,$

i).Velocity=v= $\frac{dr}{dt}=-\omega \sin \omega t i+\omega \cos \omega t j\,$

Therefore, $r\cdot\frac{dr}{dt}=(\cos \omega t i+\sin \omega t j)\cdot (-\omega \sin \omega t i+\omega \cos \omega t j)\,$

or $r\cdot\frac{dr}{dt}=-\omega \cos \omega t +\omega \sin \omega t \cos \omega t=0\,$

Showing that the velocity of the particle v is perpendicular to r.

ii).Acceleration a= $\frac{d^2 r}{dt^2}=-\omega^{2} \cos \omega t i-\omega^{2} \sin \omega t j=-\omega^{2} r\,$

Showing that the acceleration of the particle a is opposite to that of r and so it is directed towards the origin.

Again,the magnitude of acceleration =modulus of $-\omega^{2} r\,$ is

$\omega^{2} r \,$ which is proportional to r. i.e the distance of the particle from the origin.