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 an acceleration both are perpendicular to $\overrightarrow{r}$
• B. Velocity and acceleration both are parallel 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

$\text{Step 1: Velocity and acceleration calculation}$

$\overrightarrow{r}=\cos \omega t\hat{x}+\sin \omega t\hat{y}$ $....\left(1\right)$

Velocity is given by $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

$\Rightarrow$ $\overrightarrow{v}=-\omega \sin \omega t\hat{x}+\omega \cos \omega t\hat{y}m/s$ $....\left(2\right)$

Acceleration is given by $\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$

$\Rightarrow$ $\overrightarrow{a}=-{\omega }^2\cos \omega t\hat{x}-{\omega }^2\sin \omega t\hat{y}m/s^2$ $....\left(3\right)$

$\text{Step 2: Analysing}$ $\overrightarrow{r},\overrightarrow{v}$ and $\overrightarrow{a}$

If the dot product between two vectors is zero that means the vectors are perpendicular to each other.

$\overrightarrow{v}\cdot \overrightarrow{r}=-\omega \sin \omega t\cos \omega t+\omega \sin \omega t\cos \omega t$$=0$

$\therefore \overrightarrow{v}$ is perpendicular to $\overrightarrow{r}$

$\overrightarrow{a}\cdot \overrightarrow{r}=-{\omega }^2{\cos }^2\omega t-{\omega }^2{\sin }^2\omega t\ne 0$

$\therefore \overrightarrow{a}$ is not perpendicular to $\overrightarrow{r}$

Negative sign in $e{q}^n\left(3\right)$ indicates acceleration is towards the origin (Opposite to $\overrightarrow{r}$)

Hence , $\overrightarrow{v}$ is perpendicular to $\overrightarrow{r}$ and the acceleration is directed towards the origin

$\therefore$ Option C is the correct answer.