Let $\mathbf{p}$ denote a point on a three-dimensional rotating rigid body, and consider a body-fixed frame of coordinates centered on $\mathbf{p}$ rotating with the body. Assume that the body is solely subject to gravity, so that the acceleration of the center of mass is given by the gravitational acceleration vector $\mathbf{g}$. I have seen in many places online that we can write, with $\mathbf{v}$ denoting the velocity of $\mathbf{p}$ expressed in a space-fixed inertial frame,

$$\dot{\mathbf{v}} = \mathbf{g} + \dot{\boldsymbol{\omega}} \times \mathbf{r} + \boldsymbol{\omega} \times \boldsymbol{\omega} \times \mathbf{r}$$

where $\mathbf{r}$ is a vector expressed in the body frame from the point $\mathbf{p}$ to the center of mass. My question is: in what frame is $\boldsymbol\omega$ expressed? Am I to interpret $\boldsymbol\omega$ as the angular velocity in the body frame or in the space-fixed frame?

  • $\begingroup$ For future reference add subscripts for the location of where vectors are summed at (when it makes sense). This can help clarify what is what. $\endgroup$ May 10, 2021 at 20:09

1 Answer 1


You have to be consistent with the choice of basis vectors.


All of the quantities should have components oriented with the inertial coordinate frame. This includes the relative position vector $\boldsymbol{r}$ of point P to the center of mass C.

$$ \boldsymbol{r}= \boldsymbol{r}_P - \boldsymbol{r}_C \tag{1}$$

$$ \boldsymbol{v}_P = \boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{r} \tag{2}$$

$$ \boldsymbol{\dot{v}}_P = \boldsymbol{\dot{v}}_C + \boldsymbol{\alpha} \times \boldsymbol{r} + \boldsymbol{\omega} \times ( \boldsymbol{v}_P - \boldsymbol{v}_C ) \tag{3}$$

where $\boldsymbol{\dot{v}}_C = \boldsymbol{g}$ and $\boldsymbol{\alpha} = \boldsymbol{\dot{\omega}}$.

If you know the location of P in body riding coordinate system $\boldsymbol{p}$ and you know the 3×3 rotation matrix of the body $\mathbf{R}$ then the location of point P in the inertial frame is

$$ \boldsymbol{r}_P = \boldsymbol{r}_C + \mathbf{R}\,\boldsymbol{p} \tag{4}$$

  • $\begingroup$ Thanks. Is there an expression for $\mathbf{v}_p - \mathbf{v}_c$? $\endgroup$ May 10, 2021 at 20:24
  • $\begingroup$ @TheWind-UpBird yes. see (2) above and move $\boldsymbol{v}_C$ to the other side of the =. $$\boldsymbol{v}_P - \boldsymbol{v}_C = \boldsymbol{\omega} \times \boldsymbol{r}$$ I just to not like to nest a lot of cross products together. $\endgroup$ May 10, 2021 at 20:30

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