I know that in fluid dynamics, we use Lagrangian description of acceleration. That is, a material derivative $$\frac{dv}{dt}=\frac{\partial v}{\partial t}+(v\cdot\nabla )v .$$ My question is can we use the same formulation for rigid body kinematics because it seems quite general and if we can, why don't we see it being used anywhere in classical mechanics?


In most classical mechanics problems, the velocity $v$ of a particle is usually a function of $t$ exclusively. If it were to be a function of both $t$ and its position vector $\vec{x}$, then $v=v(t,x,y)$. (In the 2D case, of course)

To find $\frac{dv}{dt}$, we can apply the chain rule, which results in:

$$\frac{dv}{dt}=\frac{\partial v}{\partial t}+\frac{dx}{dt}\frac{\partial v}{\partial x}+\frac{dy}{dt}\frac{\partial v}{\partial y}$$ This is the same as the expression you provide. The second and third terms are usually zero in problems that are commonly studied.


In rigid body dynamics there is the concept of spatial acceleration $ \vec{\psi} = \dot{\vec{v}}$ as well as material acceleration $\vec{a}$. This is the Lagrangian description. The two are related as $$ \vec{a} = \vec{\psi} + \vec{\omega} \times \vec{v} $$

Consider a moving rigid body, and at some instant a point A riding on the body has position $\vec{r}_A$, as well as the center of mass C with position $\vec{r}_C$.

  • Material acceleration of $\vec{a}_A$ is the acceleration of the particle of mass A as it moves about. Its transformation law to the center of mass is $$\vec{a}_C = \vec{a}_A + \vec{\alpha} \times (\vec{r}_C-\vec{r}_A)+\vec{\omega}\times (\vec{v}_C-\vec{v}_A) $$

  • Spatial acceleration of $\vec{\psi}_A$ is the acceleration of whatever mass particle happens to pass under the fixed location defined by $\vec{r}_A$ (where A is now). Its transformation law to the center of mass is $$\vec{\psi}_C = \vec{\psi}_A + \vec{\alpha} \times (\vec{r}_C-\vec{r}_A)$$

  • You can prove the above once you define the spatial accelerations from the material accelerations.

    $$\matrix{\vec{\psi}_A = \vec{a}_A - \vec{\omega} \times \vec{v}_A \\ \vec{\psi}_C = \vec{a}_C - \vec{\omega} \times \vec{v}_C}$$

  • Spatial acceleration transforms similarly with velocities and torques. It is thus used in the equations of motion (in spatial form) to simplify the transformations from location to location. From now on don't think of particles of mass moving, but of fixed locations in space.

I am skipping the vector decorations from now on


Consider the equations of motion at the center of mass $$ \begin{bmatrix}F\\ M_{C} \end{bmatrix}=\begin{bmatrix}m\\ & I_{C} \end{bmatrix}\begin{bmatrix}a_{C}\\ \alpha \end{bmatrix}+\begin{bmatrix}0\\ \omega\times I_{C}\omega \end{bmatrix} $$

We formulate the spatial equation of motion by using $$\begin{bmatrix}a_{C}\\ \alpha \end{bmatrix}=\begin{bmatrix}\psi_{C}\\ \alpha \end{bmatrix}+\begin{bmatrix}\omega\times v_{C}\\ 0 \end{bmatrix}$$ to arrive at

The equation of motion of rigid body with spatial notation

$$\begin{bmatrix}F\\ M_{C} \end{bmatrix}=\begin{bmatrix}m\\ & I_{C} \end{bmatrix}\begin{bmatrix}\psi_{C}\\ \alpha \end{bmatrix}+\begin{bmatrix}\omega\times & \\ v_{C}\times & \omega\times \end{bmatrix}\begin{bmatrix}m\\ & I_{C} \end{bmatrix}\begin{bmatrix}v_{C}\\ \omega \end{bmatrix} $$

Note that $v_C \times m v_C =0$ and thus $F = m a_C + v_C \times (m v_C)$. Also $\omega\times$ and $v_C \times$ are the 3×3 skew symmetric matrix cross product operator (see note below also).

The above is easier to deal with because all terms transform from location to location using standard linear transformation rules. For example formulating the equations of motion at a point A away from the center of mass is complex in vector form. But in spatial form it is easier because we follow the simple rules below:

$$ \begin{bmatrix}v_{C}\\ \omega \end{bmatrix}=\begin{bmatrix}1 & -[c\times]\\ & 1 \end{bmatrix}\begin{bmatrix}v_{A}\\ \omega \end{bmatrix}$$

$$\begin{bmatrix}\psi_{C}\\ \alpha \end{bmatrix}=\begin{bmatrix}1 & -[c\times]\\ & 1 \end{bmatrix}\begin{bmatrix}\psi_{A}\\ \alpha \end{bmatrix}$$

$$\begin{bmatrix}F\\ M_{C} \end{bmatrix}=\begin{bmatrix}1\\ -[c\times] & 1 \end{bmatrix}\begin{bmatrix}F\\ M_{A} \end{bmatrix}$$

Note: $[c\times]$ denotes the 3×3 skew symmetric cross product operator for the vector $c = r_C-r_A$.

The equations of motion at a point not at the center of mass are:

$$\begin{align} \begin{bmatrix}F\\ M_{A} \end{bmatrix} & =\begin{bmatrix}m & -m[c\times]\\ m[c\times] & I_{C}-m[c\times][c\times] \end{bmatrix}\begin{bmatrix}\psi_{A}\\ \alpha \end{bmatrix} \\ & +\begin{bmatrix}\omega\times\\ v_{A}\times & \omega\times \end{bmatrix}\begin{bmatrix}m & -m[c\times]\\ m[c\times] & I_{C}-m[c\times][c\times] \end{bmatrix}\begin{bmatrix}v_{A}\\ \omega \end{bmatrix} \end{align}$$

The above is often condensed into $$ \hat{f}_A = {\rm I}_A \dot{\hat{v}}_A + \hat{v}_A \times {\rm I}_A \hat{v}_A $$

  • $\hat{f}_A$ is the 6×1 spatial loading (called a force wrench).
  • $\hat{v}_A$ is the 6×1 spatial velocity (called a velocity twist).
  • $\dot{\hat{v}}_A$ is the 6×1 spatial acceleration (called an acceleration twist).
  • ${\rm I}_A$ is the 6×6 spatial inertia matrix (with includes the parallel axis theorem in the lower right 3×3 block).
  • $\hat{v}_A\times$ is the spatial derivative operator (similar to the vector $\dot{\vec{A}} = \vec{\omega} \times \vec{A}$)

Spatial Form of Newton-Euler Laws of Motion

The equations of motion can be explained from the spatial momentum $\hat{\ell}_A = {\rm I}_A \hat{v}_A$ and the extension to Newton's law:

$$\hat{f}_A = \frac{{\rm d}}{{\rm d}t} \hat{\ell}_A = \frac{{\rm d}}{{\rm d}t}( {\rm I}_A \hat{v}_A )$$

The spatial derivative of momentum is

$$ \frac{{\rm d}}{{\rm d}t} \hat{\ell}_A = \frac{\partial}{\partial t} \hat{\ell}_A + \hat{v}_A\times \hat{\ell}_A$$

$$\hat{f}_A = {\rm I}_A \dot{\hat{v}}_A + \hat{v}_A \times {\rm I}_A \hat{v}_A$$

References - Newton-Euler combined equations of motion - Spatial Acceleration

  • $\begingroup$ thanks for your reply.I sought of understand your reply. Basically,material acceleration is the total derivative whereas spatial derivative is partial derivative with respect to space.In a normal circulation motion.you will get the spatial acceleration by removing the centripetal force. Am i coreect? Also,can you suggest some link,video or books for additional in depth knowledge?@ja72 $\endgroup$ Jun 14 '17 at 14:45
  • $\begingroup$ Spatial acceleration is the partial with respect to time. If you held a magnifying glass over a fixed point in space, the spatial acceleration is the acceleration of whatever material happens to be visible in the lens. In fluids steady flow has zero spatial acceleration because at any point the velocity is fixed. $\endgroup$ Jun 14 '17 at 14:48
  • 1
    $\begingroup$ Look at a point on a rotating disk (with constant rotational speed). The velocity under the fixed location remains constant, even though each mass particle is under centripetal acceleration. The spatial acceleration is simply $\psi = \dot{\omega} \times r = 0$ whereas the material acceleration is $a = \dot{\omega} \times r + \omega \times \omega \times r \neq 0 $ $\endgroup$ Jun 14 '17 at 14:50

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