# Forces on particle system

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1. Oct 31, 2015

### ognik

Just something, probably obvious, that I want to be sure about, please confirm or correct the following:

If we have 2 (or n) particles, they are a system only if there are internal forces between them? So, no internal forces implies forces on one won't affect the other...

If particles are a system, then the internal forces cancel and we can consider all external forces to be acting through the center of mass, and nett force = total of external forces. The eqtn of motion would be something like $M \frac{d^2 r}{dt^2}$?

The 'relative coordinate vector' would be the position vector of the center of mass? What would be the difference between the eqtn of motion in this sense, as opposed to the one above? Enlightenment much appreciated.

2. Oct 31, 2015

### Staff: Mentor

You can always call it "system". The system is just boring if there is no interaction between its components.
Sometimes you can split the analysis up like that, sometimes (e. g. if the external force depends on the position) you cannot.

3. Oct 31, 2015

### Chandra Prayaga

In a 2-particle system, the relative coordinate vector is the position vector of one particle relative to the other.

4. Nov 2, 2015

### ognik

In a 2-particle system, the relative coordinate vector is the position vector of one particle relative to the other.

Does that mean making one particle at the origin, then the relative position vector if the position vector of the other? Which would be the same as one position vector minus the other position vector?

What would be the eqtn of motion, I think $F_1 + F_2 = (m_1 + m_2) \frac{d^2r}{dt^2}$ ?

5. Nov 3, 2015

### Chandra Prayaga

The relative coordinate is indeed the position vector of one particle relative to the other. Your equation of motion for the relative coordinate is not correct. The correct equation, in the absence of external forces is,

μ(2nd derivative of r) = F1 on 2

where r = r2 - r1, and F1 on 2 is the force exerted by 1 on 2. μ is the reduced mass m1 m2 / (m1+m2)

6. Nov 3, 2015

### ognik

Reduced mass (at centre of mass?) now understood, thanks

But in this case there are external forces, $F_1$ on $m_1$, and $F_2$ on $M_2$?
And don't the internal forces cancel?

7. Nov 3, 2015

### Staff: Mentor

The reduced mass doesn't have a position.
Not for the relative position.

8. Nov 3, 2015

### Chandra Prayaga

Here is the complete derivation. Check the uploaded Word document:

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9. Nov 3, 2015

### Chandra Prayaga

Incidentally, following your last question, the internal forces cancel only in the CM equation of motion, not for the relative coordinate.

10. Nov 7, 2015

### ognik

Thanks - good summary doc.