Question

In the figure shown below, $\overrightarrow{\omega }$ is angular velocity and $\overrightarrow{r}$ is position vector. What is the direction and magnitude of linear velocity.

• A. $\hat{k},20\text{m/s}$
• B. $-\hat{k},20\text{m/s}$
• C. $\hat{k},\frac{20}{\sqrt{2}}\text{m/s}$
• D. $-\hat{k},\frac{20}{\sqrt{2}}\text{m/s}$

Answer 1

The correct option is A $\hat{k},20\text{m/s}$

It is given that
$\overrightarrow{\omega }=(2\hat{i}+\hat{j})$ and $\overrightarrow{r}=(-4\hat{i}+8\hat{j})$
So, $\overrightarrow{v}=\overrightarrow{\omega }\times \overrightarrow{r}$
i.e $\overrightarrow{v}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 0\\ -4& 8& 0\end{array}\right|$
$=(0-0)\hat{i}-(0-0)\hat{j}+(16+4)\hat{k}=20\hat{k}$
Hence, linear velocity $\overrightarrow{v}=20\hat{k}m/s$

Alternate solution:
We know that $\overrightarrow{v}=\overrightarrow{\omega }\times \overrightarrow{r}$
Applying Right hand rule, the direction of $\overrightarrow{v}$ is along $\hat{k}$ direction.
Checking the angle between the vectors,
$\overrightarrow{\omega }.\overrightarrow{r}=(2\hat{i}+\hat{j}).(-4\hat{i}+8\hat{j})=8-8=0$
$\therefore \overrightarrow{\omega }\perp \overrightarrow{r}$

Hence, $\overrightarrow{v}=\overrightarrow{\omega }\times \overrightarrow{r}=r\omega \sin {90}^{\circ }$
i.e $v=\left(\sqrt{(-4{)}^2+(8{)}^2}\right)\left(\sqrt{(2{)}^2+(1{)}^2}\right)\sin {90}^{\circ }$
$=\sqrt{400}=20\text{m/s}$