Are internal forces necessary for particles to form a system?

[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.

 

[mfb]

You can always call it "system". The system is just boring if there is no interaction between its components.

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.

Sometimes you can split the analysis up like that, sometimes (e. g. if the external force depends on the position) you cannot.

 

[Chandra Prayaga]

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.

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

 

[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 $F_1 + F_2 = (m_1 + m_2) \frac{d^2r}{dt^2}$ ?

 

[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) = F_{1 on 2}

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

 

[ognik]

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

in the absence of external forces

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?

 

[mfb]

Reduced mass (at centre of mass?)

The reduced mass doesn't have a position.

And don't the internal forces cancel?

Not for the relative position.

 

[Chandra Prayaga]

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?

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

 

[Chandra Prayaga]

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

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

 

[ognik]

Thanks - good summary doc.