Hello I want to ask you a question about rotational movement. Say I would consider a system and the center of mass is moving with a velocity $\dot{\mathbf{R}}$ and the system is rotating with an angular velocity $\boldsymbol{\Omega}$. How does the movement look if $\boldsymbol{\Omega} \vert\vert \dot{\mathbf{R}}$? Is this a screw motion?


2 Answers 2


Yes this is by definition screw motion with pitch value $ h = \frac{ \| \boldsymbol{\dot{R}} \|}{\| \boldsymbol{\Omega} \|}$. Pitch is the scalar ratio of translational velocity to rotational velocity in units of length.


Helical motion figure from wikipedia

Now the theory (Chasle's Theorem, wikipedia) goes that any motion of a 3D solid body can be decomposed as screw motion along an arbitrary axis of rotation. The direction of the axis of rotation is parallel to $\boldsymbol{\Omega}$.

This means that if the center of mass translates with $\boldsymbol{\dot{R}}$, and rotates by $\boldsymbol{\Omega}$ not necessarily parallel to each other, then the rotation axis position $\boldsymbol{R}_{\rm axis}$ and pitch $h$ are found by

$$ \begin{aligned} \boldsymbol{R}_{\rm axis} & = \boldsymbol{R} + \frac{ \boldsymbol{\Omega} \times \boldsymbol{\dot{R}}}{ \| \boldsymbol{\Omega} \|^2} & h & = \frac{ \boldsymbol{\Omega} \cdot \boldsymbol{\dot{R}} }{ \| \boldsymbol{\Omega} \|^2} \end{aligned} $$

Here $\cdot$ is the dot product and $\times$ is the cross product

In reverse, given the rotation $\boldsymbol{\Omega}$, the rotation axis position $\boldsymbol{R}_{\rm axis}$ and the pitch $h$ then the translational velocity is

$$ \boldsymbol{\dot{R}} = h\,\boldsymbol{\Omega} + ( \boldsymbol{R} - \boldsymbol{R}_{\rm axis} ) \times \boldsymbol{\Omega} $$

Note that $\boldsymbol{R}_{\rm axis}$ is the point on the rotation axis closest to $\boldsymbol{R}$.


  • $\begingroup$ Added a link to Chasle's theorem. $\endgroup$
    – JAlex
    Commented Mar 11, 2021 at 20:17

Yes, the trajectory will look screw-like. You are right about that. The direction of the angular velocity vector is the axis of the rotation. And if this axis is aligned with the COM motion superposing both movements yield a screw like motion.

  • $\begingroup$ Do you have a picture of this? $\endgroup$
    – Babu
    Commented Mar 2, 2021 at 15:30
  • $\begingroup$ You have to be careful that the angular velocity vector points in the direction of the AXIS of the rotation. So when the axis of the rotation is aligned with the linear motion this yields a spiral (at least for points not on the axis itself). If the vector is perpendicular this would not yield a screw like trajectory $\endgroup$
    – bluesky
    Commented Mar 2, 2021 at 15:53
  • $\begingroup$ @ Buraian No, sorry, I am searching for the picture of it :) $\endgroup$
    – Q.stion
    Commented Mar 2, 2021 at 19:41
  • $\begingroup$ @bluesky Thanks for your comment! $\endgroup$
    – Q.stion
    Commented Mar 2, 2021 at 19:41

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