question_answer

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

A) \[6\mathbf{\hat{i}}+6\mathbf{\hat{j}}-3\mathbf{\hat{k}}\] 

B)  \[-18\mathbf{\hat{i}}-13\mathbf{\hat{j}}+2\mathbf{\hat{k}}\]

C)  \[4\mathbf{\hat{i}}-13\mathbf{\hat{j}}-6\mathbf{\hat{k}}\]

D)  \[6\mathbf{\hat{i}}-2\mathbf{\hat{j}}+8\mathbf{\hat{k}}\]

Correct Answer: B

Solution :

 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.