Question

Angular velocity of a body moving on a circular track is given by $\overrightarrow{\omega }=\hat{i}+\hat{j}+\hat{k}$ and the radius vector $\overrightarrow{r}=2\hat{i}-2\hat{j}-2\hat{k}$ at the instant of time $‘t^{\prime }$. Then the magnitude of linear velocity of the body at that instant is.

• A. $2.01\text{m/s}$
• B. $4.23\text{m/s}$
• C. $6.21\text{m/s}$
• D. $5.65\text{m/s}$

Answer 1

The correct option is D $5.65\text{m/s}$

Given $\overrightarrow{\omega }=\hat{i}+\hat{j}+\hat{k}$ and
$\overrightarrow{r}=2\hat{i}-2\hat{j}-2\hat{k}$
$\therefore \overrightarrow{v}=\overrightarrow{\omega }\times \overrightarrow{r}$
$=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& -2& -2\end{array}\right]$
$=4\hat{j}-4\hat{k}=\sqrt{4^2+4^2}=5.65m/s$