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

The Correct Option is B

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.