What's the difference between Spatial angular velocity and Body angular velocity?

1.8k Views Asked by At

In rigid body, Spatial angular velocity is defined by $$\dot{R}(t)R^{-1}(t)$$

where $R$ is rotation matrix, and Body angular velocity $$R^{-1}(t)\dot{R}(t)$$

Wow, it's some fuzzy.

1

There are 1 best solutions below

5
On BEST ANSWER

Let us suppose a rigid body $\mathcal{B}$ is undergoing a purely rotational motion. In this case, we say that its configuration space is the space of rotations in Euclidean three space, i.e., $SO(3)$.

Now, assume that $X \in \mathcal{B}$. An element $R \in SO(3)$ acts on $\mathbb{R}^3$ by rotations. As a result, $x = RX \in R(\mathcal{B})$.

When the rigid body is in motion, the matrix $R$ is time-dependent and the velocity of a point on the body is $\dot{x} = \dot{R}X = \dot{R}R^{-1}x$. Since $R$ is an orthogonal matrix $R^{-1}\dot{R}$ and $\dot{R}R^{-1}$ are skew-symmetric matrices, and so we can write

\begin{equation} \dot{x} = \dot{R}R^{-1}x = \omega \times x, \end{equation}

which defines the spatial angular velocity vector $\omega$. Thus, $\omega$ is given by right translation of $\dot{R}$ to the identity.

The corresponding body angular velocity is defined by $\Omega = R^{-1}\omega$, so that $\Omega$ is the angular velocity relative to a body fixed frame. Notice that

\begin{equation} R^{-1}\dot{R}X = R^{-1}\dot{R}R^{-1}x = R^{-1}(\omega \times x) = R^{-1}\omega \times R^{-1}x = \Omega \times X, \end{equation}

so that $\Omega$ is given by left translations of $\dot{R}$ to the identity (to be precise to the tangent space of the identity element of the Lie group $SO(3)$).

The physical interpretation of this mathematical jargon is the following. We fix two coordinate frames, one to the earth, $\Sigma_E$, one to the body, $\Sigma_B$. This body is purely rotating with respect to the earth. While the spatial angular velocity of the rigid body is the expression of the angular velocity of the rigid body in the inertial frame $\Sigma_E$, the body angular velocity of the rigid body is its expression in the body-fixed frame $\Sigma_B$.

Reference: Marsden and Ratiu, Introduction to Mechanics and Symmetry, Springer, 2013.