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In my quest for getting comfortable with the mathematical modelling of a system of rigid bodies, I am posing the following problem:

Problem:

Suppose in outer space (no gravity) there are two rigid bodies as in the following figure: system of 2 rigid bodies; one dof joint The rigid bodies are connected with one degree of freedom rotational joint about $\vec{i}_1$ axis. There is a coordinate system attached in the center of mass of each rigid body. Suppose the following: $\begin{bmatrix} \vec{i}_1\\ \vec{j}_1\\ \vec{k}_1\end{bmatrix} = R_{i\psi} R_{j\chi} R_{k\theta} \cdot \begin{bmatrix} \vec{i}_0\\ \vec{j}_0\\ \vec{k}_0\end{bmatrix}$ and $\begin{bmatrix} \vec{i}_2\\ \vec{j}_2\\ \vec{k}_2\end{bmatrix} = R_{i\phi} \cdot \begin{bmatrix} \vec{i}_1\\ \vec{j}_1\\ \vec{k}_1\end{bmatrix}$, where $R = R_{i\psi} R_{j\chi} R_{k\theta}$ is a rotation matrix and $R_{i\phi}$ is the rotation matrix about $\vec{i}$ axis with angle $\phi$. On the first rigid body is a attached a motor in point $S$ in the figure. The motor's shaft is attached by the second rigid body, thus the motor's bearings are forming that "1 d.o.f" rotational joint. There are no other forces or torques acting on the system. I want to model the dynamics of this system.

Modelling:

I am using D'Alembert principle. The system has $3 + 3 + 1$ degrees of freedom. The variables I am choosing are $x,y,z$ the coordinates (in world frame) of the center of mass of first rigid body, $\psi, \chi, \theta$ the angles giving the orientation of the first rigid body and finally the angle $\phi$. Suppose the first rigid body has the inertia matrix $I_{01}$ in it's coordinate system and the second rigid body has the inertia matrix $I_{02}$ in it's coordinate system. D'Alembert's equations for a system of $N$ rigid bodies with $q_1, ... , q_n$ degrees of freedom are: \begin{align} \begin{cases} \sum_{i=1}^N \left( F_i^a(t) - \dot{P}_i(t)\right) \cdot \frac{\partial \bar{v}_i}{\partial \dot{q}_1} + \sum_{i=1}^m \left( T_i^a(t) - \dot{L}_i(t)\right) \cdot \frac{\partial \omega_i}{\partial \dot{q}_1} = 0 \\ \vdots \\ \sum_{i=1}^N \left( F_i^a(t) - \dot{P}_i(t)\right) \cdot \frac{\partial \bar{v}_i}{\partial \dot{q}_n} + \sum_{i=1}^m \left( T_i^a(t) - \dot{L}_i(t)\right) \cdot \frac{\partial \omega_i}{\partial \dot{q}_n} = 0 \end{cases} \end{align} where $\bar{v}_i$ and $\omega_i$ are the linear respectively angular velocity of each rigid body and $F_i^a$ is the total applied force on the $i'th$ rigid body whereas $T_i^a$ is the total applied torque (about the center of mass) on the $i'th$ rigid body, all in world frame.

Question:

Suppose the motor in point $S$ applies the torque $R^T \cdot \begin{bmatrix} T_m\\0\\0\end{bmatrix}$ on the second rigid body. Am want to know if in the above presented case:

  1. $F_1^a = F_2^a = \begin{bmatrix} 0\\0\\0 \end{bmatrix}$ ?
  2. $T_1^a = -R^T \cdot \begin{bmatrix} T_m\\0\\0\end{bmatrix} = -T_2^a$ ?

Conclusion:

Therefore the equations of the motion will result upon replacing everything in D'Alembert's equations ... am I correct? I've seen (I think) solutions using the parallel axis theorem, which I do not understand. Why is the parallel axis theorem needed? Where is it used? It seems that every thing can be modeled without it ... Am I wrong?

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  • $\begingroup$ Parallel axis theorem is used because you are modeling the motion about the joint which is not on either center of mass and you need a way to connect the moments at the joint with the motion of the bodies. $\endgroup$ – ja72 Mar 5 '17 at 15:06
  • $\begingroup$ Even though I have a Masters in Multibody Dynamics I am not familiar with D'Alembert's equations. Can you provide a link or more information, specifically on how this method is used to model the joint constraints. $\endgroup$ – ja72 Mar 7 '17 at 0:21
  • $\begingroup$ en.wikipedia.org/wiki/Rigid_body_dynamics especially at the paragraph "Virtual work of forces acting on the rigid body" $\endgroup$ – C Marius Mar 9 '17 at 12:49
  • $\begingroup$ This is what I have understood: for each rigid body in the system you write the forces acting on it, including the constraint forces (which I think model the 1 d.o.f joint). For static equilibrium the work done by these forces is zero. But the constraint forces are anyway having zero work (related to the fact that the constraint forces act orthogonal on the constraint ( ? I do not remember exactly ...1 d.o.f joint ...) ) hence it will result in a system of equations involving just the applied forces on the body. For D'Alembert principle in dynamics, the core idee is $F - m\cdot a$ does no work $\endgroup$ – C Marius Mar 9 '17 at 12:58
  • $\begingroup$ Yes I am familiar with this idea, just didn't recognize the equations. If the step in velocity across the joint is $\Delta \vec{v}$ and $\Delta \vec{\omega}$ then I would model the joint with $$\vec{F} \cdot \Delta \vec{v} + \vec{T} \cdot \Delta \vec{\omega} = 0$$ Is this what the equations do above? Is it a power balance? $\endgroup$ – ja72 Mar 9 '17 at 13:29
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Consider a joint connecting bodies [1] and [2]. I use following expressions for the joint kinematics

$$ \begin{align} \mathbf{v}_2 & = \mathbf{v}_1 + \mathbf{x} \dot{q} \\ \boldsymbol{\omega}_2 & = \boldsymbol{\omega}_1 + \mathbf{z} \dot{q} \end{align} $$

where $\mathbf{x} = \left( \frac{\partial \mathbf{v}}{\partial \dot{q}} \right) $ and $\mathbf{z} = \left( \frac{\partial \boldsymbol{\omega}}{\partial \dot{q}} \right)$, and all linear velocities are measured at the top of the joint (on body [2]).

Equal and opposite forces $\mathbf{F}$ and moments $\mathbf{T}$ act on the two bodies from the joint. If the moments are expressed on the top of the joint also (just a velocity) then the following equation is used to either find the joint acceleration or the joint force

$$ Q = \mathbf{F} \cdot \mathbf{x} + \mathbf{T} \cdot \mathbf{z} $$

For example consider a revolute joint with $\mathbf{x}=0$ and $\mathbf{z} = \hat{k}$. Then any reaction force $\mathbf{F}$ is allowed, since $\mathbf{F}\cdot 0 = 0$. The reaction moment has to have a torque along the $\hat{k}$ direction of value $Q$ since $\mathbf{T} \cdot \hat{k} = Q$.

In addition, the reaction forces are not directly specified, but a result of a free body diagram and the equations of motion

$$ \left. \begin{align} \mathbf{F}_1^a - \mathbf{F} & = \dot{\mathbf{p}}_1 & \mathbf{F}_2^a + \mathbf{F} & = \dot{\mathbf{p}}_2 \\ \mathbf{T}_1^a - \mathbf{T} - \mathbf{r}_1 \times \mathbf{F} & = \dot{\mathbf{L}}_1 & \mathbf{T}_2^a + \mathbf{T} + \mathbf{r}_2 \times \mathbf{F} & = \dot{\mathbf{L}}_2 \end{align} \right\} $$

where $\mathbf{r}_i$ is the position of the joint relative to the body center of mass, and the superscript $\square^a$ indicates applied forces/moments.

In order for all to work you have to follow these rules

  • All vectors in an expression need to be expressed on the same point and along the same coordinate system
  • Equations involving momentum (and derivatives) need to be expressed at the center of mass only and along the same coordinate system as forces (typically the global system)
  • Care must be taken to include any necessary transformations when velocities, angular momentum or moments are used at a point different from where they are defined.
  • Body MMOI matrix needs to be expressed at the center of mass, but along the global coordinates in each equation of motion.
  • Joint kinematics need to be differentiated in order to find the dependency of momentum to joint acceleration.
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  • $\begingroup$ Why would somebody, in this framework, use velocities, or angular momentum in a different point then the center of mass? I do not understand, and I am somehow afraid that I should have but I haven't cared of any velocity or angular momentum of any point except for the center of masses. $\endgroup$ – C Marius Mar 9 '17 at 19:06
  • $\begingroup$ You do need to consider the velocities on both ends of the joint. It is only natural because the kinematics are simple when done on the joint. $\endgroup$ – ja72 Mar 9 '17 at 20:16
  • $\begingroup$ Can you sir, have a look, if you please at another question of mine: physics.stackexchange.com/questions/317981/… That is about linearisation of equations of motion $\endgroup$ – C Marius Mar 11 '17 at 7:48

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