Why can we model a body to have a center of mass?

Question

I accept that bodies have a center of mass. I also accept that if I were to suspend a body by only supporting it at its center of mass, then the body wouldn't rotate. In the scenario that we suspend a body (and only in this scenario), it is as if we were supporting just a point a point particle. Here, we can model mass to be concentrated at one point.

I cannot convince myself, however, that we can model mass to be concentrated at a point in other scenarios. Consequently, I cannot accept that a body has a center of mass whose acceleration is given by $\frac{F}{m}$, where $F$ is a force applied at any point on the body.

On a similar vein, I can accept that, were a body capable of being modeled to have all its mass concentrated at one point, then its momentum would be $\frac {\Sigma m_{i}v_{i}}{\Sigma m_{i}}$. I cannot see, however, why such a possibility exists -- that a body can be modeled to have its mass concentrated at one point, the suspension scenario aside?

Answer 1

A rigid body can not in general be modelled as a mass point. This is possible in celestial mechanics, as the forces encountered there act uniformly and therefore can be effectively described as forces acting on the centre of mass.

In general, you have to consider the orientation of the body as well, then one gets the equations of motion for the centre of mass $\vec R$ and for the angular momentum $\vec L$: $$M \ddot{\vec R} = \sum_i \vec F_i,$$ $$ \dot{\vec L} = \sum_i \vec \tau_i,$$ where $F_i$ are the applied forces and $\vec \tau_i = \vec r_i \times \vec F_i$ are the applied torques. It requires some work to work out a useful version of the second equation (one common form is called Euler's equation).

The solution of the second equation can be quite intricate, as the complex motion of the rigid body has to be handled (to calculate the torques, especially if the forces depend on the position).

Addendum

Proof of the centre-of-mass-equation. Assume a set of $N$ point particles with masses $m_i$ at the coordinates $\vec r_i$. The equations of motion for this system are: $$m_i \ddot{\vec r}_i = \vec F_i(\vec r_1, \ldots, \vec r_N, \dot{\vec r}_1, \ldots, \dot{\vec r}_N, t)$$ Where $\vec F_i$ is the force applied to particle $i$ at time $t$ (which is allowed to depend on the other particle's positions and velocities).

Now we just use the definition of the centre of mass $\vec R = \frac{\sum_i m_i\vec r_i}{M}$ with total mass $M = \sum_i m_i$ to derive its equation of motion (this is an ansatz at the moment, we do not know yet this point is special/follows a simple equation, but we can just try): $$ M\ddot{\vec R} = \frac M M \sum_i m_i\ddot{\vec r}_i = \sum_i \vec F_i. $$ What happened here is that we first used the linearity of the derivative (i.e. $(A+B)' = A' + B'$) and then the equation of motion for the single particles.

So it follows that for an arbitrary collection of point masses the centre of mass moves as if it were a point mass of mass $M$ on which the sum of all forces act.

But this does not give the result for the rigid body directly. However, a rigid body can be approximated arbitrarily well by a collection of point particles plus constraint forces between them that keep them in place relative to each other. So the forces are written as two parts, external forces $\vec F_{i,\text{ext}}$ and constraint forces $\vec F_{i,\text{cons}}$ with $\vec F_i = \vec F_{i,\text{ext}} + \vec F_{i,\text{cons}}$. Restated in that language your question is why the equation $M\ddot{\vec R} = \sum_i F_{i,\text{ext}}$ holds.

To obtain that result it only remains to be shown, that the constraint forces sum to zero. This follows directly from a simple assumption for the constraint forces: The requirement they obey Newton's third law.