Answer to Question #203435 in Geometry for Nestor Freeman

Question

A particle moves so that its position vector is given by r =cosωti+sinωtj. where ω is a constant.

a)Show that the velocity v of the particle is perpendicular to r, and the velocity acceleration a is directed towards the origin and has magnitude proportional to the distance from the origin.

b) Show also that r × r is a constant vector.

Answer 1

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

Velocity, $\overrightarrow{v} = \dfrac{d\overrightarrow{r}}{dt}= - \omega sin\omega t\hat{x} + \omega cos\omega\hat{y}.........(2)$

Acceleration, $\overrightarrow{a} = \dfrac{d^2\overrightarrow{r}}{dt^2} = -\omega^2 cos\omega t \hat{x} - \omega^2sin\omega t \hat{y} = -\omega^2\overrightarrow{r} ............(3)$

Now using (1),(2) and (3) we get, $\overrightarrow{v}.\overrightarrow{r} =0$ Its indicate that velocity is perpendicular to $\overrightarrow{r}$ and $\overrightarrow{a}.\overrightarrow{r}\neq 0$ So acceleration is neither parallel nor perpendicular to $\overrightarrow{r}$

The equation (3) indicates the acceleration is directed toward the origin.

(b) Ans:-

$\overrightarrow{r}= cos\omega \hat{x} + sin\omega\hat{y}$

$|\overrightarrow{r}|=\sqrt{(Cos\omega)^2+(Sin\omega)^2}=1$

As we know that the angle between itself will be $0\degree$

Hence,

$\overrightarrow{r}\times \overrightarrow{r}= |\overrightarrow{r}| .|\overrightarrow{r}|.Sin0\degree=0$

Hence, $\overrightarrow{r}\times \overrightarrow{r}$ is a constant vector.