Question

Position vector of a particle moving in $xy$ plane at time $t$ is $\overrightarrow{r}=a(1-\cos \omega t)\hat{i}+a\sin \omega t\hat{j}$.
The path of the particle is
­

• A. neither a circle nor an ellipse
  
• B. an ellipse
  
• C. a source of a radius $a$ and a center at $(a,0)$
• D. a source of a radius $a$ and a center at $(0,0)$

Answer 1

The correct option is C a source of a radius $a$ and a center at $(a,0)$

Write $x$ and $y$ co-ordinate separately
$x=a(1-\cos \omega t)$
$y=a\sin \omega t$

Compare with standard equation of circle

Circle having center as $(0,0)$ have $x$ co-ordinate as $a\cos \omega t$ and $y$ co-ordinate as $a\sin \omega t$ and here we have $x$ as $a(1-\cos \omega t)$ which suggest that center is shifted to some other $x$ co-ordinate.

If particle is rotating with angular speed $\omega$ then
$\theta =\omega t$
$x=a(1-\cos \omega t)$ $y=a\sin \omega t$
$\Rightarrow \cos \omega t=\frac{x-a}{a}$ $\sin \omega t=\frac{y}{a}$
${\cos }^2\omega t={\sin }^2\omega t=1$
${\left(\frac{x-a}{a}\right)}^2+{\left(\frac{y}{a}\right)}^2=1$
${(x-a)}^2+y^2=a^2$
This is an equation of circle of radius $a$ with centre at $(a,0)$.

Final Answer: (a)