Question

What is the value of linear velocity, if angular velocity is $3 \hat{ i }-4 \hat{ j }+\hat{ k }$ and distance from the centre is $5 \hat{ i }-6 \hat{ j }+6 \hat{ k }$ ?

• A. $6\hat{i} + 6\hat{j} - 3\hat{k}$
• B. $-18\hat{i} - 13\hat{j} +2 \hat{k}$
• C. $4\hat{i} - 13\hat{j} -6 \hat{k}$
• D. $6\hat{i} - 2\hat{j} +8 \hat{k}$

Answer 1

Answer: (B)

$-18\hat{i} - 13\hat{j} +2 \hat{k}$

Answer 2

The linear velocity of a particle is given by
$v=\omega \,r$ As shown in figure,
the direction of velocity $\vec{v}$ is tangential to the circular path.
Both the magnitude and direction of $\vec{v}$ can be accounted for by using the cross product of $cd$ and $\vec{r}$ .
Hence, $\vec{v}=\vec{\omega }\times \vec{r}$
Given, $\vec{\omega }=3\hat{i}-4\hat{j}+\hat{k}$
and $\vec{r}=5\hat{i}-6\hat{j}+6\hat{k}$
$\therefore$ $\vec{v}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 3 & -4 & 1 \\\ 5 & -6 & 6 \\\ \end{matrix} \right|$
$=\hat{i}(-24+6)-\hat{j}(18-5)+\hat{k}(-18+20)$
$=-18\hat{i}-13\hat{j}+2\hat{k}$
Note: Greater the distance of the particle from the centre, greater will be its linear velocity.