Question

A particle moves in $xy$-plane. The position vector of particle at any time $t$ is $\overrightarrow{r}=a(1-\cos \omega t)\hat{i}+a\sin \omega t\hat{j}m$. The rate of change of $\theta$ at time $t=2$ second. ( where $\theta$ is the angle which its velocity vector makes with positive x-axis) is:

• A. $-\omega$
• B. $-\omega /2$
  
• C. $-2\omega$
  
• D. $1-\omega$
  

Answer 1

Answer: (A)

$-\omega$

we need to find rate of change of $\theta$ i.e $\frac{d\theta }{dt}$

and $\theta$ is the angle which its velocity vector makes with positive x-axis

given $\overrightarrow{r}=a(1-\cos (\omega t\left)\right)\hat{i}+a\sin \left(\omega t\right)\hat{j}$

so $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

$\overrightarrow{v}=a(0+\omega \sin (\omega t\left)\right)\hat{i}+a\omega \cos \left(\omega t\right)\hat{j}$

$\tan \theta =\frac{a\omega \cos \left(\omega t\right)}{a\omega \sin \left(\omega t\right)}$

$\tan \theta =\cot \omega t$

$\tan \theta =\tan (\frac{\pi }{2}-\omega t)$

$\theta =\frac{\pi }{2}-\omega t$

$\frac{d\theta }{dt}=-\omega$