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Consider the problem of a point p rotating around a given axis. The axis of rotation is ω , |ω| = 1 , and q is a point on the axis. Assuming that the point rotates around the axis with unit angular velocity, then the velocity of the point p(t) is

p'(t) = ω x ( p(t) - q )

I'm looking for a more detailed explanation for this equation, as it is not the same with the classic equation of angular velocity. I'm confused with the term p(t) - q in the right hand side.

Thanks

2 Answers 2

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Normally, $\omega$ is a vector, pointing along the axis with magnitude the rotation rate. The x you have is the cross product of two vectors, not a simple multiply. It only depends upon perpendicular components of the vectors. So the projection of $\vec{p(t)}-\vec{q}$ perpendicular to $\vec{\omega}$ is just the radius and your expression is the usual $v=\omega r$ of circular motion

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The point ${\bf q}$ on the axis is the natural origin pertaining to your problem, it does not have to be the origin of the coordinate system. Choosing another point $\bar{\bf q}$ on the axis does not change the value of the vector product since ${\bf q}$ and $\bar{\bf q}$ differ by a vector parallel to ${\vec \omega}$. The equation expresses the fact that {\bf p}'(t) is perpendicular to ${\vec \omega}$ and to ${\bf p}(t)-{\bf q}$, and has the right absolute value $r \thinspace |{\vec \omega}|$ where $r$ denotes the distance of the point ${\bf p}(t)$ from the axis.