Question

In three dimensional system, the position coordinates of a particle (in motion) are given below
$x=a\cos \omega t$
$y=a\sin \omega t$
$z=a\omega t$
The velocity of particle will be

• A. $\sqrt{2}a\omega$
• B. $2a\omega$
• C. $a\omega$
• D. $\sqrt{3}a\omega$

Answer 1

Answer: (A)

Given that the position coordinates of a particle

So, the position vector of the particle is
$\hat{r}=x\hat{i}+y\hat{j}+z\hat{k}$
$\Rightarrow \hat{r}=a\cos \omega t\hat{i}+a\sin \omega t\hat{j}+a\omega t\hat{k}$
$\hat{r}=a[\cos \omega t\hat{i}+\sin \omega t\hat{j}+\omega t\hat{k}]$
therefore, the velocity of the particle is
$\because \hat{v}=\frac{dr}{dt}=\frac{d[a][\cos \omega t\hat{i}+\sin \omega t\hat{j}+\omega t\hat{k}]}{dt}$
$\Rightarrow \hat{v}=-a\omega \sin \omega t\hat{i}+a\omega \cos \omega t\hat{j}+a\omega \hat{k})$
The magnitude of velocity is
$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$
or $|v|=\sqrt{(-a\omega \sin \omega t{)}^2+(a\omega \cos \omega t{)}^2+(a\omega {)}^2}$
$=\omega a\sqrt{(-\sin \omega t{)}^2+(\cos \omega t{)}^2+(1{)}^2}$
$=\sqrt{2}\omega a$