The position vector of a particle is $\vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}$.

Question

The position vector of a particle is $\vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}$. The velocity of the particle is\

Professor PC Thomas Classes · Answer

(d) : Position vector of the particle $$ \vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j} $$ velocity vector\ $\vec{v}=\frac{d \vec{r}}{d t}=(-a \omega \sin \omega t) \hat{i}+(a \omega \cos \omega t) \hat{j}$ $$ \begin{aligned} = & \omega[(-a \sin \omega t) \hat{i}+(a \cos \omega t) \hat{j}] \ & \vec{v} \cdot \vec{r}=\omega a[-\sin \omega t \hat{i}+\cos \omega t \hat{j})] \cdot[a \cos \omega t \hat{i}+a \sin \omega t \hat{j}] \ = & \omega\left[-a^{2} \sin \omega t \cos \omega t+a^{2} \cos \omega t \sin \omega tight]=0 \end{aligned} $$ Therefore velocity vector is perpendicular to the position vector.\