Answer to Question #219606 in Mechanics | Relativity for Anand

Question

For a particle undergoing uniform circular motion, show that
(i) the velocity of the particle is perpendicular to the position vector, and
(ii) the acceleration of the particle is perpendicular to its velocity.

Answer 1

Let the position vector of the particle at any instant be $(\overrightarrow{r})=r_o \cos \theta \hat{i}+r_o \sin\theta\hat{j}$

i)

We know that, $\overrightarrow{v}=\frac{\overrightarrow{dr}}{dt}$

So, we can write it as $\overrightarrow{v}=r_o\frac{\overrightarrow{d\theta}}{dt} (-\sin\theta \hat{i}+\cos\theta\hat{j})$

Now, we know that if two vectors are perpendicular to each other then their dot product will be zero.

$\overrightarrow{r}.\overrightarrow{v}=0$

$\Rightarrow \frac{\overrightarrow{d\theta}}{dt}(r_o \cos \theta \hat{i}+r_o \sin\theta\hat{j}).r_o (-\sin\theta \hat{i})+r_o(\cos\theta\hat{j})=0$

$\Rightarrow \frac{\overrightarrow{d\theta}}{dt}(r_o)^2(-\cos\theta . \sin \theta+\sin \theta. \cos\theta )=0$

$\Rightarrow \frac{\overrightarrow{d\theta}}{dt}(r_o)^2.0=0$

$\Rightarrow 0=0$

Hence, from the above, we can conclude that the position vector of the particle and the velocity vector is perpendicular to each other.

ii)

We know that, the relation between the velocity of the particle and the acceleration is

$\overrightarrow{a}=\frac{\overrightarrow{dv}}{dt}$

so, we can write it as $(\overrightarrow{a})=r_o(\frac{\overrightarrow{d\theta}}{dt})^2(-\cos\theta\hat{i}-\sin\theta\hat{j})$

$\Rightarrow (\overrightarrow{a})=-r_o(\frac{\overrightarrow{d\theta}}{dt})^2(\cos\theta\hat{i}+\sin\theta\hat{j})$

Now, if the scalar product of the acceleration and velocity is zero, it means acceleration and velocity is perpendicular to each other.

$\overrightarrow{a}.\overrightarrow{v}=0$

Now,

$=-r_o(\frac{\overrightarrow{d\theta}}{dt})^2(\cos\theta\hat{i}+\sin\theta\hat{j})r_o\frac{\overrightarrow{d\theta}}{dt} (-\sin\theta \hat{i}+\cos\theta\hat{j})$

$=(r_o)^2(\frac{\overrightarrow{d\theta}}{dt})^3(-\cos\theta . \sin \theta+\sin \theta. \cos\theta )$

Hence, we can conclude that acceleration and velocity is perpendicular to each other.