Question

Position vector of a particle moving in $x-y$ plane at time $t$ is $\overrightarrow{r}=a(1-\cos \omega t)\hat{i}+a\sin \omega t\hat{j}$ and the angular velocity of the particle is ${\omega }_o\hat{k}$. The magnitude of the linear velocity of particle at $t=\frac{\pi }{2\omega }$ is

• A. ${\omega }_o{a}^2$
• B. ${\omega }_{o}a$
• C. $\sqrt{{\omega }_o}a$
• D. $\sqrt{2}{\omega }_{o}a$

Answer 1

The correct option is D $\sqrt{2}{\omega }_{o}a$

We know that, $\overrightarrow{v}=\overrightarrow{w}\times \overrightarrow{r}$
Given, $\overrightarrow{\omega }={\omega }_o\hat{k}$ and $\overrightarrow{r}=a(1-\cos \omega t)\hat{i}+a\sin \omega t\hat{j}$
$\therefore \overrightarrow{v}=\overrightarrow{\omega }\times \overrightarrow{r}$
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 0& {\omega }_o\\ a(1-\cos \omega t)& a\sin \omega t& 0\end{array}\right|$
$=\hat{i}(-{\omega }_{o}a\sin \omega t)-\hat{j}(-{\omega }_{o}a(1-\cos \omega t\left)\right)+0\hat{k}$
$=-{\omega }_{o}a\sin \omega t\hat{i}+{\omega }_{o}a(1-\cos \omega t)\hat{j}$

Hence, the magnitude is given by
$|\overrightarrow{v}|=\sqrt{(-{\omega }_{o}a\sin \omega t{)}^2+({\omega }_{o}a(1-\cos \omega t){)}^2}$
$=\sqrt{\left({\omega }_{o}a{)}^2\right({\sin }^2\omega t+1+{\cos }^2\omega t-2\cos \omega t)}$
$=\sqrt{({\omega }_{o}a{)}^2\times 2(1-\cos \omega t)}$
$=2{\omega }_{o}a\sin \left(\frac{\omega t}{2}\right)$
At $t=\frac{\pi }{2\omega }$; $|\overrightarrow{v}|=2{\omega }_{o}a\sin \left(\frac{\pi }{4}\right)=\sqrt{2}{\omega }_{o}a$