# Motion of a point on a rotating rigid body in three dimensions

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?

• 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. Commented May 10, 2021 at 20:09

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

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

• Thanks. Is there an expression for $\mathbf{v}_p - \mathbf{v}_c$? Commented May 10, 2021 at 20:24
• @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. Commented May 10, 2021 at 20:30