Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Triangle defined by points OA, OB and OC : (-i,3 j,-4 k), (i,2 j,2 k) and (3 i,7 j,- k) where i, j, k are unit vectors along x,y,z axis. Point mass is placed at OA. Triangle rotates with angular velocity ω about BC axis. Calculate linear velocity of mass as it passes OA.

My Work: $$v=\omega \times r $$ Does r have to be the vector perpendicular to BC to the point OA or can it be any vector from the line BC to OA??? If so can you explain? I know the answer as my tutor told me I got it right but I didn't need to find the vector perpendicular to BC and I forgot to ask why...

share|improve this question
add comment

2 Answers 2

up vote 0 down vote accepted

It could be any vector from BC to OA.

Let's assume that $r$ is the vector perpendicular to BC to the point OA. Any vector from BC to OA (whether or not it is perpendicular to BC) has the form $r + s$, where $s$ is some vector parallel to BC. Of course, you've told us that $\omega$ is also parallel to BC, so we could also write that as $r+\alpha\, \omega$, for some number $\alpha$. So let's take the cross product: \begin{equation} \omega \times (r+\alpha\, \omega) = \omega \times r + \omega \times (\alpha\, \omega) = \omega \times r + \alpha\, \omega \times \omega = \omega \times r~, \end{equation} since $\omega \times \omega = 0$. So $\omega$ cross any vector from BC to OA will equal $\omega$ cross the perpendicular vector. So you don't need to specifically find the perpendicular.

share|improve this answer
add comment

$$ \vec{v}_A = \vec{v}_B + \vec{\omega} \times (\vec{r}_A -\vec{r}_B ) $$

with $\vec{v}_B =0$.

Any components of $(\vec{r}_A -\vec{r}_B )$ along $\vec{\omega}$ cancel out, so you don't have to worry about finding the perpendicular.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.