# 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. 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 $\mathbf{v}_p - \mathbf{v}_c$? 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. May 10, 2021 at 20:30