Question

A particle moves so that its position vector varies with time as $\overrightarrow{r}=Acos\omega t\hat{i}+Asin\omega t\hat{j}$. Find the
a. initial velocity of the particle,
b. angle between the positon vector and velocity of the particle at any time, and
c. speed at any instant

Answer 1

a. Position at time t

$\overrightarrow{r}=Acos\omega t\hat{i}+Acos\omega t\hat{j}$

Instantaneous velocity, $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

We have $\vec{v} = A\dfrac{d}{dt}(cos \omega t)\hat{i} + A\dfrac{d}{dt}(sin \omega t)\hat{j}

= - A\omega sin\omega t\hat{i} + A\omega sin\omega t\hat{j}$

At t=0,v = $-A\omega sin0\hat{i}+A\omega sin0\hat{j}$

b.For calculating the angle between two vectors, we use the concept of dot product of the vectors. The angle $\Theta$ between $\overrightarrow{r}$ and $\overrightarrow{v}$ can be given as

$\Theta =co{s}^{-1}\frac{\overrightarrow{r}.\overrightarrow{v}}{|\overrightarrow{r}||\overrightarrow{v}|}$

where

$\overrightarrow{r}.\overrightarrow{v}=(Acos\omega t\hat{i}+Asin\omega t\hat{j}).(Asin\omega t\hat{i}+Acos\omega t\hat{j})$

Hence, $\Theta =cos-10=\pi /2$

That means $\overrightarrow{v}\perp \overrightarrow{r}$.

c. Speed at any time is the magnitude of instantaneous velocity, i.e.,

$v=|\overrightarrow{v}|=\sqrt{(-A\omega sin\omega t{)}^2+A\omega cos\omega t{)}^2}=A\omega$