Question

A particle moves so that its position vector is given by $\overrightarrow{r}=cos\omega t\hat{x}+sin\omega t\hat{y}$. Where $\omega$ is a constant. Which of the following is true?

• A. Velocity and acceleration both are perpendicular to $\overrightarrow{r}$
• B. Velocity and acceleration both are paralles to $\overrightarrow{r}$
• C. Velocity is perpendicular to $\overrightarrow{r}$ and acceleration is directed towards the origin
• D. Velocity is perpendicular to $\overrightarrow{r}$ and acceleration is directed away from the origin.

Answer 1

The correct option is C Velocity is perpendicular to $\overrightarrow{r}$ and acceleration is directed towards the origin

Position vector$⟹r=(\cos \omega t\hat{x}+\sin \omega t\hat{y})$

Now, velocity of particle$⟹v=\frac{dr}{dt}=\frac{d}{dt}(\cos \omega t\hat{x}+\sin \omega t\hat{y})$

$⟹v=(-\sin \omega t)\omega \hat{x}+\left(\cos \omega t\right)\omega \hat{y}⟹v=-\omega (\sin \omega t\hat{x}-\cos \omega t\hat{y})$

Acceleration of the particle$=a=\frac{dv}{dt}=\frac{d}{dt}[-\omega (\sin \omega t\hat{x}-\cos \omega t\hat{y}\left)\right]$

$⟹a=-{\omega }^2\cos \omega t\hat{x}-{\omega }^2\sin \omega t\hat{y}⟹a=-{\omega }^2(\cos \omega t\hat{x}+\sin \omega t\hat{y})⟹a=-{\omega }^{2}r={\omega }^2(-r)$

The particle is at point $P.$

We have,

$\overrightarrow{v}\cdot \overrightarrow{r}=-\omega (\sin \omega t\hat{x}-\cos \omega t\hat{y})\cdot (\cos \omega t\hat{x}+\sin \omega t\hat{y})=-\omega [\sin \omega t\cos \omega t+0+0-\sin \omega t\cos \omega t]=-\omega \left(0\right)=0⟹v\perp r$