The center of mass of a rigid body is given by: $$ \vec{r}_c = \frac{1}{M} \sum_i m_i \cdot \vec{r}_i $$ with $M = \sum_i m_i$ the total mass or $$ \vec{r}_c = \frac{1}{M} \int \vec{r}\ ' \cdot \varrho(\vec{r}\ ')~d^3r' $$ with $M = \int \varrho(\vec{r}\ ')~d^3r'$ for a continious mass distribution. I'll stick to the discrete case for brevity.

Therefore: \begin{align} M~\frac{d^2 \vec{r}_c}{dt^2} = M~\vec{a}_c & = \sum_i m_i~\frac{d^2 \vec{r}_i}{dt^2} \\ & = \sum_i m_i~\vec{a}_i = \sum_i \vec{F}_i \tag1 \end{align} So the center of mass moves due to every force acting on the rigid body's particles. Now, what are these forces? If we set $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} $$ with $\vec{F}_{\text{ext}}$ the external forces acting on a particle, this would not account for the fact, that the body is rigid. There are indeed internal forces that guarantee the body not to deform upon the action of external forces. For exmaple, if an external force is only applied to one particle and there were no internal forces, this would allow the particle to "leave" the body. This is not what one usually means if talking about a rigid body. So one rather has $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} + \sum \vec{F}_{\text{int}} $$, this depends now largely on the specific model of internal forces one chooses (how one "point-force" is propagated through the volume).

Is there a common idealisation for rigid bodies, that allows to calculate or even ignore those internal forces and therefore solve the equation of motion (1)?

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    $\begingroup$ There is a law of motion that comes directly after the 2nd law. It is called the 3rd law. This will help you calculate the sum of internal forces (HINT: IT IS ZERO). $\endgroup$
    – hft
    Commented Apr 4, 2015 at 22:13
  • $\begingroup$ um, though out of topic, but newton's 3rd law does not come from 2nd law. It is a law of physical content, rather than definition. $\endgroup$
    – Shing
    Commented Apr 5, 2015 at 5:21
  • $\begingroup$ @Shing I think hft's "comes directly after" was meant in a listing sense only (otherwise hft would have said "comes directly from"). $\endgroup$
    – pglpm
    Commented Apr 22, 2018 at 23:45

1 Answer 1


The internal forces come in equal-and-opposite pairs (Newton's 3rd), and therefore result in no net force on the object. If you did take the vector sum of all of them, they would just cancel out. Including them therefore doesn't change the final expression.

There is no need for idealization - just the acceptance of classical mechanics.

Note that this is even true for non-rigid bodies: while the position of their center of mass will change, that center of mass will still move in accordance with the vector sum of the external forces.

  • $\begingroup$ If this were true, a rod touched on the outside would behave the same way as touched near the center of mass (in sense of translation of the center of mass). My pencil's movement of the center of mass does however depend on the position on which I apply the force. $\endgroup$
    – image357
    Commented Apr 4, 2015 at 22:11
  • $\begingroup$ I think the internal forces must not add up to zero, the only condition is, that they do no net work: $$ \sum \vec{F}_{i,\text{int}} \cdot \vec{v_i} = 0 $$ $\endgroup$
    – image357
    Commented Apr 4, 2015 at 22:19
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    $\begingroup$ And indeed, the center of mass of your pencil does behave the same way if you apply the same force for the same time. If it appears not to, it is because there are forces of friction acting. Note also that if you bounce an object off a rod, the response will be different depending on where you hit the rod - and the object will bounce back with different velocity. This was discussed at length in an earlier question $\endgroup$
    – Floris
    Commented Apr 4, 2015 at 22:20
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    $\begingroup$ @MarcelKöpke: Your seem to assume that if the vector sum of the internal forces is zero , then they have no effect. That is not true: they alter the relative motion of parts of the rigid body, but it is true that they have no effect on the movement of the centre of mass. In fact applying the same force at the centre of a pencil or off-centre (which is not really easy to achieve) does have the same effect for the motion of the centre of mass, even though the latter also imparts angular momentum leading to rotation about the centre of mass. $\endgroup$ Commented Apr 5, 2015 at 1:35
  • $\begingroup$ @MarcvanLeeuwen well said $\endgroup$
    – Floris
    Commented Apr 5, 2015 at 1:56

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