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Hello I am working on a project in which I need to predict the movement of an object given the torques and forces applied on it. My problem is that these torques and forces are not necessarily applied at the bodies center of mass.

So the question is: Assuming I know the torques and forces applied on a rigid body, not necessarily on its center of mass, and know it's mass and moment of inertia is there a way to describe it's dynamics?

I have tried a couple of things to solve this problem.

First I came to the conclusion that this motion must be equivalent to a single unique force and torque being applied to the center of mass. This is because the position of the center of mass and rotation around it can be measured and by taking the second derivative of both of them we can find the translational and angular acceleration. Then using the mass and moment of inertia we can convert these into the force and torque that were applied. This means that there is only one possible solution and not many options for the bodies movement.

The next thing I tried was finding a way to "move" forces to the center of mass. I found a way to do this in which you can create helper forces of equal size and opposite direction. One of these will create a torque which is not around the center of mass and the other will just be a force around the center of mass. By doing this we have changed the problem into forces on the center of mass, and torques not on the center of mass.

At this point I have gotten stuck. I have not been able to find a way to change these torques so that they will be around the center of mass.

If anyone has any ideas about how to solve this problem, either by continuing my method or in a completely different way I would greatly appreciate it.

Amit
  • 6,024
Kfir Neuman
  • 11

2 Answers2

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$\def \c {\boldsymbol} \def \p {\dot} \def \pp {\ddot} \def \b {\mathbf}$

the equation of motion at the center of mass in body frame

6 DoF rigid body equations and tensor of inertia

Translation \begin{align*} &m\,\b{\dot{v}}_c=\b F_c-m\,\left(\c\omega\times \b v_c\right)\tag 1 \end{align*}

Rotation \begin{align*} &\b I_c\,\c{\dot{\omega}}+\c\omega\times\left(\b I_c\,\c\omega\right)=\c\tau_c\tag 2 \end{align*}

at arbitrary point $~p~$ \begin{align*} &\b v_p=\b v_c+\c\omega\times\b r\quad\Rightarrow\\ &\b v_c=\b v_p-\c\omega\times\b r\quad, \b{\dot{v}}_c=\b{\dot{v}}_p-\c{\dot{\omega}}\times\b r \end{align*}

Translation \begin{align*} &m\,\left(\b{\dot{v}}_p-\c{\dot{\omega}}\times\b r\right)=\b F_c-m\,\left(\c\omega\times \left(\b v_p-\c\omega\times\b r\right)\right) \end{align*}

Rotation \begin{align*} &\b I_p\,\c{\dot{\omega}}+\c\omega\,\times\left(\b I_p\,\c\omega\right)=\c\tau_c+\b r\times\,\b F_c\\ &\b I_p=\b I_c-m\,\b r^\times\,\b r^\times \end{align*}

Translation and Rotation \begin{align*} &\underbrace{\begin{bmatrix} \b m& \b r^\times \\ \c 0 & \b I_p \\ \end{bmatrix}}_{\b A}\, \underbrace{ \begin{bmatrix} \b{\dot{v}}_p \\ \c{\dot{\omega}} \\ \end{bmatrix}}_{\b x}= \underbrace{\begin{bmatrix} \b F_c-m\,\left(\c\omega\times \left(\b v_p-\c\omega\times\b r\right)\right) \\ \c\tau_c+\b r\times\,\b F_c-\left(\c\omega\,\times\left(\b I_p\,\c\omega\right)\right)\\ \end{bmatrix}}_{\b b} \end{align*}

solution \begin{align*} &\b x=\b A^{-1}\,\b b\quad, \b A^{-1}=\underbrace{\begin{bmatrix} \frac{1}{m}\,\b I_3 & - \frac{1}{m}\,\left[\b r^\times\,\right]\,\b I_p^{-1} \\ \c 0 & \b I_p^{-1} \\ \end{bmatrix}}_{6\times 6} \end{align*}

  • $~\b v~$ Velocity
  • $~\c \omega~$ Angular velocity
  • $~\b r~$ Vector from the center of mass to point $~p~$
  • $$~\b r^\times=\left[ \begin {array}{ccc} 0&-r_{{z}}&r_{{y}}\\ r_{ {z}}&0&-r_{{x}}\\ -r_{{y}}&r_{{x}}&0\end {array} \right] $$
  • $~c~$ Center of mass
  • $~p~$ Rigid body arbitrary point
Eli
  • 13,829
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Start from the vector equations summed at the center of mass

$$ \begin{aligned} \boldsymbol{F} & = m \boldsymbol{a}_C \\ \boldsymbol{\tau}_C &= \mathcal{I}_C \boldsymbol{\alpha} + \boldsymbol{\omega} \times \mathcal{I}_C \boldsymbol{\omega} \\ \end{aligned} \tag{1}$$

where $\boldsymbol{a}_C$ is the acceleration of the center of mass, $\mathcal{I}_C$ is the mass moment of inertia tensor aligned to the interial coordinate directions, $\boldsymbol{F}$ is the net force applied on the body, and $\boldsymbol{\tau}_C$ the net torque applied on the body summed about the center of mass (which includes all effects from applied forces offset from the center of mass)

Now take an arbitrary point A, not at the center of mass, and apply the following transformations

$$ \begin{aligned} \boldsymbol{a}_C &= \boldsymbol{a}_A + \boldsymbol{\alpha}\times \boldsymbol{c} + \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \boldsymbol{c}) \\ \boldsymbol{\tau}_A &= \boldsymbol{\tau}_C + \boldsymbol{c}\times \boldsymbol{F} \end{aligned} \tag{2}$$

where $\boldsymbol{c}$ is the instantaneous vector position of the center of mass relative to the arbitrary point A, $\boldsymbol{a}_A$ the acceleration of the arbitrary point, and $\boldsymbol{\tau}_A$ is the net torque also summed about the arbitrary point.

$$\begin{aligned} \boldsymbol{F} & = m \left( \boldsymbol{a}_A + \boldsymbol{\alpha}\times \boldsymbol{c} + \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \boldsymbol{c}) \right) \\ \boldsymbol{\tau}_A & = \mathcal{I}_C \boldsymbol{\alpha} + \boldsymbol{\omega} \times \mathcal{I}_C \boldsymbol{\omega} + \boldsymbol{c} \times \boldsymbol{F} \end{aligned} \tag{3}$$

And now we can solve for the translational and rotational acceleration vectors, given the force vector $\boldsymbol{F}$ and torque vector $\boldsymbol{\tau}_A$

$$ \begin{aligned} \boldsymbol{\alpha} & =\mathcal{I}_{C}^{-1}\left(\boldsymbol{\tau}_{A}-\boldsymbol{\omega}\times\mathcal{I}_{C}\boldsymbol{\omega}-\boldsymbol{c}\times\boldsymbol{F}\right)\\ \boldsymbol{a}_{A} & =\tfrac{1}{m}\boldsymbol{F}-\boldsymbol{c}\times\mathcal{I}_{C}^{-1}\left(\boldsymbol{c}\times\boldsymbol{F}\right)+\boldsymbol{c}\times\boldsymbol{\alpha}-\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\boldsymbol{c}) \end{aligned} \tag{4} $$

Please note that (3) can be expanded out as $$\begin{aligned}\boldsymbol{\tau}_{A} & =\mathcal{I}_{C}\boldsymbol{\alpha}+\boldsymbol{c}\times m\left(\boldsymbol{a}_{A}+\boldsymbol{\alpha}\times\boldsymbol{c}+\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\boldsymbol{c})\right)+\boldsymbol{\omega}\times\mathcal{I}_{C}\boldsymbol{\omega}\\ & =\boldsymbol{c}\times m\boldsymbol{a}_{A}+\mathcal{I}_{C}\boldsymbol{\alpha}+\boldsymbol{c}\times m\left(\boldsymbol{\alpha}\times\boldsymbol{c}\right)+\boldsymbol{c}\times m\left(\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\boldsymbol{c})\right)+\boldsymbol{\omega}\times\mathcal{I}_{C}\boldsymbol{\omega}\\ & =\boldsymbol{c}\times m\boldsymbol{a}_{A}+\underbrace{\mathcal{I}_{C}\boldsymbol{\alpha}-\boldsymbol{c}\times m\left(\boldsymbol{c}\times\boldsymbol{\alpha}\right)}_{\mathcal{I}_{A}\boldsymbol{\alpha}}+\boldsymbol{\omega}\times m\left(\boldsymbol{c}\times(\boldsymbol{\omega}\times\boldsymbol{c})\right)+\boldsymbol{\omega}\times\mathcal{I}_{C}\boldsymbol{\omega}\\ \boldsymbol{\tau}_{A} & =\boldsymbol{c}\times m\boldsymbol{a}_{A}+\mathcal{I}_{A}\boldsymbol{\alpha}+\boldsymbol{\omega}\times\mathcal{I}_{A}\boldsymbol{\omega} \end{aligned} \tag{5}$$

which is used to produce the 6-vector form of the Newton-Euler equations of motion

$$\begin{bmatrix}\boldsymbol{F}\\ \boldsymbol{\tau}_{A} \end{bmatrix}=\begin{bmatrix}m & \text{-}m[\boldsymbol{c}\times]\\ m[\boldsymbol{c}\times] & \mathcal{I}_{A} \end{bmatrix}\begin{bmatrix}\boldsymbol{a}_{A}\\ \boldsymbol{\alpha} \end{bmatrix}+\begin{bmatrix}\boldsymbol{\omega}\times m(\boldsymbol{\omega}\times\boldsymbol{c})\\ \boldsymbol{\omega}\times\mathcal{I}_{A}\boldsymbol{\omega} \end{bmatrix} \tag{6} $$ where $$ [\boldsymbol{c}\times] = \pmatrix{0 & -c_z & c_y \\ c_z & 0 & -c_x \\ -c_y & c_x & 0}$$ is the 3×3 skew-symmetric cross-product matrix operator, and $\mathcal{I}_{A}=\mathcal{I}_{C}-m[\boldsymbol{c}\times][\boldsymbol{c}\times]$ is the mass moment of inertia tensor summed about point A produced by the parallel axis theorem in vector form.

The inverse of (6) is a bit awkward as it involves both $\mathcal{I}_C$ and $\mathcal{I}_A$ at the same time

$$\begin{bmatrix}\boldsymbol{a}_{A}\\ \boldsymbol{\alpha} \end{bmatrix}=\begin{bmatrix}\tfrac{1}{m}-[\boldsymbol{c}\times]\mathcal{I}_{C}^{-1}[\boldsymbol{c}\times] & \mathcal{I}_{C}^{-1}[\boldsymbol{c}\times]\\ -[\boldsymbol{c}\times]\mathcal{I}_{C}^{-1} & \mathcal{I}_{C}^{-1} \end{bmatrix}\left(\begin{bmatrix}\boldsymbol{F}\\ \boldsymbol{\tau}_{A} \end{bmatrix}-\begin{bmatrix}\boldsymbol{\omega}\times m(\boldsymbol{\omega}\times\boldsymbol{c})\\ \boldsymbol{\omega}\times\mathcal{I}_{A}\boldsymbol{\omega} \end{bmatrix}\right) \tag{7}$$

John Alexiou
  • 40,139