A particle moves so that its position vector is given by $\vec{r}=\cos \omega t \hat{x}+\sin \omega

Question

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

Professor PC Thomas Classes · Answer

(a) : Given, $\vec{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y}$\ $	herefore \quad \vec{v}=\frac{d \vec{r}}{d t}=-\omega \sin \omega t \hat{x}+\omega \cos \omega t \hat{y}$\ $\vec{a}=\frac{d \vec{v}}{d t}=-\omega^{2} \cos \omega t \hat{x}-\omega^{2} \sin \omega t \hat{y}=-\omega^{2} \vec{r}$\ Since position vector ( $\vec{r}$ ) is directed away from the origin, so, acceleration $\left(-\omega^{2} \vec{r}ight)$ is directed towards the origin.\ Also,\ $\vec{r} \cdot \vec{v}=(\cos \omega t \hat{x}+\sin \omega t \hat{y}) \cdot(-\omega \sin \omega t \hat{x}+\omega \cos \omega t \hat{y})$\ $=-\omega \sin \omega t \cos \omega t+\omega \sin \omega t \cos \omega t=0$\ $\Rightarrow \quad \vec{r} \perp \vec{v}$