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From figure above, suppose a three-dimensional vector $p(0)$ is rotated by an angle $\theta$ about the fixed rotation axis $\omega$ to $p(\theta)$. Here we assume all quantities are expressed fixed frame coordinates and $\Vert \omega \Vert = 1$. This rotation can be achieved by imagining that $p(0)$ is subject to a rotation about $\omega$ at a constant rate of $1$ rad/sec, from time $t = 0$ to $t = \theta$. Let $p(t)$ denote this path. The velocity of $p(t)$, denoted $\dot{p}$, is then given by

$$ \begin{equation} \dot{p} = \omega \times p \end{equation} $$

Please explain in detail as to how to derive the above equation. Thank you.

  • $\begingroup$ is this a school assignment? ... also, using capitals equates to YELLING $\endgroup$
    – jsotola
    Commented Feb 3, 2018 at 21:16
  • $\begingroup$ @jsotola It is not a school assignment but I learnt it in my robotics class and I am finding it difficult to derive (1) . $\endgroup$
    – npkp
    Commented Feb 3, 2018 at 21:37

2 Answers 2


First note that

  1. $p(0)$ travels along an arc of the circle of radius $r = \Vert p \Vert \sin(\phi)$ centered at a point on the axis of $\omega$; and

  2. the velocity $\dot{p}$ is perpendicular to the arc of the circle; and

  3. (from the definition) $\omega = \dot{\theta} u$, where $u$ is a unit vector perpendicular to the plane of rotation.

Now we try to relate $\dot{p}$ to the known quantities.

We have $$ \begin{align} \Vert \dot{p} \Vert &= r\dot{\theta} & \text{(from 2.)}\\ &= (\Vert p \Vert \sin(\phi))\dot{\theta} & \text{(from 1.)}\\ &= \Vert p \Vert \Vert \omega \Vert \sin(\phi) & \text{(from 3.)} \end{align} $$ From the above, we can therefore conclude that $\dot{p} = \omega \times p$.


The author expects a background that includes a course in physics or mechanics where this equation is taught. When that is the case, this equation gives you instantaneous velocity of a particle (point) moving on a circular path.

The $\times$ in $\dot{p} = \omega \times p$ is the cross product. (This may already be obvious to you, it's hard to tell from the question).

The vector result of the cross product will always be perpendicular to the plane that the operand vectors lie in. The magnitude of the vector result will be $\Vert \omega \Vert\Vert p \Vert \sin(\phi)$. How to do the cross product operation is in the wikipedia article.

A break down of the equation that may answer your question is on wikipedia.

For a book that does a good job teaching working in exponential coordinates I highly recommend A mathematical introduction to robotic manipulation by Murray, Li, and Sastry.


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