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My book says that "it is clear that if you replace the system of forces with a second system having the same resulting force and the same resulting moment, with the same initial conditions the rigid body moves exactly in the same way".  Unfortunately it doesn't seem so obvious to me. Can someone help me?

I am absolutely certain that the statement is completely correct because the book is excellent. I just can't understand why "it's clear". Haven't we to prove that the resulting equations of motion are the same if you take another equivalent system of forces in the sense thus expressed?

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  • $\begingroup$ Maybe my traduction was bad. I edited. $\endgroup$
    – Haumea
    May 4 at 12:35

2 Answers 2

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The equations of motion can be thought of given a specified motion, provided by the translation acceleration vector of the center of mass $\boldsymbol{a}_C$ and the rotation acceleration vector of the body $\boldsymbol{\alpha}$, to provide the required force vector $\boldsymbol{F}$ and moment vector about the center of mass $\boldsymbol{M}_C$.

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

This tells you nothing about the actual loading on a body, just that the combined loading and moments about the COM be $\boldsymbol{F}$ and $\boldsymbol{M}_C$ respectively.

The total of 6 loading components (3 force components, 3 moment components) map to the 6 motion components (3 translational, 3 rotational) in a one-to-one mapping. This is with specified initial conditions (positions and velocities). This mapping occurs through the mass $m$ scalar, and mass moment of inertia $\mathcal{I}_C$ tensor.

The above situation leads to the following conclusions

  1. So the motion of the body ($\boldsymbol{a}_C$, $\boldsymbol{\alpha}$) is completely specified by the loading components ($\boldsymbol{F}$, $\boldsymbol{M}_C$).

  2. Multiple configurations of forces and torques at different positions can yield the same loading components

    $$ \begin{aligned} \boldsymbol{F} & = \sum_i (\boldsymbol{F}_i) \\ \boldsymbol{M}_C &= \sum_i (\boldsymbol{\tau}_i + \boldsymbol{r}_i \times \boldsymbol{F}_i) \end{aligned}$$

  3. All the different loading combinations possible, the ones with the same loading components ($\boldsymbol{F}$, $\boldsymbol{M}_C$) will results in the same motion components ($\boldsymbol{a}_C$, $\boldsymbol{\alpha}$).

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  • $\begingroup$ Thank you. The tensor of inertia with respect to the center of mass can be considered constant in a rigid body, so I follow you if you says that the total force and the total torque about center of mass determine $\vec{a}_{center of mass}$ and $\dot{\vec{\omega}}$ (there is a lot of physics behind this statement and I will not dwell, but I think I have understood so far, also reflecting on your post). $\endgroup$
    – Haumea
    May 4 at 14:41
  • $\begingroup$ Only one doubt was left. Why knowing these two quantities is equivalent to claiming to know completely the motion of the rigid body? But maybe It is solved in this way: clearly the motion of c.m. is know, and $\vec{\omega}=\vec{\omega}(t)$ too, so $\vec{r}_P=\vec{r}_{c.m.}+ \vec{\omega}\times(\vec{r}_{P}-\vec{r}_{c.m.} )$ do the rest: the motion for any point is given. Correct? $\endgroup$
    – Haumea
    May 4 at 14:43
  • $\begingroup$ The inertia tensor is constant with respect to the rotating body axes; it is not necessarily constant with respect to the fixed space axes $\endgroup$
    – John Darby
    May 4 at 14:47
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    $\begingroup$ @Haumea indeed $$ \frac{\rm d}{{\rm d}t} \mathcal{I} = \vec{\omega} \times \mathcal{I}$$ which comes from the derivative on a rotating frame. Each column of $\mathcal{I}$ can be considered a rotating vector (since MMOI is fixed on the body frame). i did not prove the equations of motion in my answer, because the proof is found easily elsewhere in any intro to dynamics book and many online resources have it. $\endgroup$
    – JAlex
    May 4 at 17:25
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    $\begingroup$ It seemed difficult to keep these discussions with the comments, and it seemed interesting to me, so I asked a new question. $\endgroup$
    – Haumea
    May 4 at 18:38
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A theorem for a rigid body is: "Every system of forces is equivalent to a single force through an arbitrary point, plus a couple (either or both of which may be zero.)".

A couple is defined as a system of forces whose sum is zero. A couple exerts the same torque about every point. The couple is characterized by a single vector, the total torque, and all couples with the same torque are equivalent.

The theorem can be proven by finding the equivalent force and couple. Let $P$ be an arbitrary point, let the sum of forces be $\vec F$, and let the total torque from the forces about $P$ be $\vec N$. Let the single force $\vec F$ act at P, and add a couple whose torque is $\vec N$; this is a system equivalent to the original system. Specifically, the single force $\vec F$ produces the same translational motion as the original set of forces, and the couple produces the same rotational about the arbitrary point $P$ as the original total torque.

It is assumed that the initial conditions are the same; either at rest or with a given translational/rotational motion. The progression from the initial state is the same using the theorem. Equivalent force provides equivalent translational motion. Equivalent torque about any point provides equivalent rotational motion about the point. Equivalent force and torque give the same changes in linear and angular momentum, respectively, from the initial conditions, so the subsequent motion is the same.

Note $P$ is typically taken as the center of mass for unconstrained motion, or a point fixed in space for rotation about that point. The overall motion is the translational motion of the center of mass due to the total external force, plus the rotational motion about the point $P$ due to the total external torque.

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  • $\begingroup$ Thanks but I can't understand how these words can be seen as a proof. You have enunciated the theorem again. $\endgroup$
    – Haumea
    May 4 at 13:18
  • $\begingroup$ This is a proof because "it finds the equivalent force and couple" for any point. This is the same proof as given in Mechanics by Symon for the stated theorem. Symon also states and proves other associated theorems. $\endgroup$
    – John Darby
    May 4 at 13:24
  • $\begingroup$ But saying that there are systems of forces with the same resultant force and the same momentum is not the same as claiming that the evolution of the rigid body will be the same. "Equivalent" can have 2 meaning: "same Ftot and Ntot", and "same e evolution of rigid body with same initial conditions". Why these two meaning are two sides of the same coin? $\endgroup$
    – Haumea
    May 4 at 13:39
  • $\begingroup$ In Italy we say "momento della (della=of the) forza" to say torque, sorry I was not clear. $\endgroup$
    – Haumea
    May 4 at 14:18
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    $\begingroup$ I updated my response to address the initial conditions. Yes, I thought you meant torque. Your English is very good; much better than my Italian! Good discussion. To really understand rotational motion beyond using the equations is not easy. I still get confused sometimes. $\endgroup$
    – John Darby
    May 4 at 14:44

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