Rigorous proof that a net force of zero guarantees zero linear acceleration in rigid bodies

Question

I've never found a rigorous proof of this fact.

The center of mass' acceleration is not necessarily the linear acceleration, specially if the body is attached to a pin or another geometric constrain, then the center of mass spins like the rest of the body. So how can we find the linear acceleration of a body?

EDIT: ok, the pin or constrain seems to add an external force and thus is a bad example to ilustrate the zero transltational acceleration derived from zero net force.

Yet, the result is still seemly true.

Answer 1

Suppose we have a system of point masses $m_i$ at positions $\vec{r}_i$. The center of mass position is defined such that $$ \vec{R} = \frac{1}{M} \sum_i m_i \vec{r}_i, $$ where $M$ is the total mass of the system. Suppose further that each mass $m_i$ experiences a net force $$ \vec{F}_i = \vec{F}_{i\text{ext}} + \sum_{j \neq i} \vec{F}_{ij} $$ where $\vec{F}_{i\text{ext}}$ is the external force on $m_i$ (i.e., exerted by objects outside of the system) and $\vec{F}_{ij}$ is the force on $m_i$ due to mass $m_j$. Then we have \begin{align*} M \ddot{\vec{R}} &= \sum_i m_i \ddot{\vec{r}}_i && \text{(by definition)} \\ &= \sum_i \left( \vec{F}_{i\text{ext}} + \sum_{j \neq i} \vec{F}_{ij} \right) && \text{(Newton 2 on each point mass)} \\ &= \underbrace{\sum_i \vec{F}_{i\text{ext}}}_{\displaystyle {} \equiv \vec{F}_\text{ext}} + \sum_i \sum_{j \neq i} \vec{F}_{ij} \end{align*} The double summation will vanish by Newton's Third Law (for any $i,j$, the sum will contain $\vec{F}_{ij}$ and $\vec{F}_{ji}$, which cancel); and the first summation is the net external force on the system. Thus, we have $$ M \ddot{\vec{R}} = \vec{F}_\text{ext}, $$ and if the system experiences no net force ($\vec{F}_\text{ext} = 0$), then $\ddot{\vec{R}} = 0$, i.e., the center of mass does not accelerate. None of the above requires that the system of point masses be a rigid body, but it is true for rigid bodies.

It is also important to note that a net force of zero does not guarantee that all points within the body will have zero acceleration, even for a rigid body. It is perfectly possible for a rigid body to rotate while no net force is exerted on it, if the net force is zero but the net torque is non-zero. In such a situation, the linear velocities of various points of the body will change while the center of mass remains at constant velocity. However, in a frame in which the center of mass is at rest, the motion of the body will be a pure rotation; in this sense there is "no translational motion" for the body, only rotational.

Answer 2

The center of mass' acceleration is not necessarily the linear acceleration

This is your problem. An object that is not point-like does not have a single linear acceleration, rather each point on the object has its own linear acceleration given by $$\vec a_P = \vec a_0 + \dot{\vec\omega} \times (\vec r_P - \vec r_0) + \vec\omega \times (\vec v_P - \vec v_0),$$ where $\vec a_P$ is the linear acceleration of the point $P$, $\vec a_0$ is the linear acceleration of some arbitrary reference point (often the centre of mass for convenience), $\dot{\vec\omega}$ is the angular acceleration, $\vec r_P$ and $\vec r_0$ are the positions of $P$ and the reference point, respectively, and $\vec v_P$ and $\vec v_0$ are their velocities. For convenience, it is common to refer to the linear acceleration of the centre of mass as that of the object. Given this, it is trivial that the centre of mass' acceleration is equal to the linear acceleration of the centre of mass.

However, one must note that zero linear acceleration for the centre of mass does not mean that each point on the object has zero linear acceleration.

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Also, you mentioned that

the center of mass spins like the rest of the body

The problem here is that the centre of mass is a point, so it doesn't make sense to say it spins.

Answer 3

I may be misinterpreting this question. But if not, then the statement is true by definition. If there is some other constraint which redirects the center-of-mass motion, such as a pin at the corner or whatever, then that constraint has by definition imparted a force. For any acceleration of the center of mass, the corresponding force may be found with Newton’s second law. To some extent this is a matter of semantics, but I think we all agree on what we’re talking about.

Answer 4

The statement in question is a direct result of the following two statements

1. Linear momentum of a rigid body is defined as the total mass times the velocity of the center of mass.

$$ \boldsymbol{p} = m \, \boldsymbol{v} \tag{1}$$

2. The net force acting on a body equals the rate of change of linear momentum (Newton's 2nd Law).

   $$ \boldsymbol{F} = \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} \tag{2} $$

As far as a rigorous proof of the above,

1. Can be obtained by summing up the individual linear momentum of each particle in a body $\boldsymbol{p} = \sum_i m_i \boldsymbol{v}_i$. But this depends on a) the definition of momentum which is taken at face value, and b) the kinematics of a rigid body and how any affect due to rotation cancels themselves out and only the motion of the center of mass matters.

2. While I have not read Principia myself I am hoping there is sufficient evidence and reasoning behind this law. Maybe someone else can chime in here and point me and the op in the correct direction in proving (2).

Note that the corollary to your statement is that

A rigid body under the influence of a pure torque will rotate about its center of mass.

Answer 5

real answer: it uses three facts:

• the arbirtrary movement of a rigid can be seen as a translation through any of it's points $P$, whose image is $P'$, and a rotation by some axis passing through $P'$. Valid to any point $P$ on the body. (Chasles theorem)

• the center of mass of a rigid body can be seen as a point of the body (with mass zero, only kinematically), meaning it's distance from the rest of points remain unchanged (easily checkable with previous fact and definition of COM).

• If a rigid body movement (continuous in time) is such that it keeps one of its points in the same position, then it must be equivalent to a rotation around some axis that passes through the fixed point.

So here it is: an infinitesimal movement of a rigid body can be understood, in particular, as a translation and a rotation through (an instantaneous axis that passes by) the center of mass. If the net external force is zero, then the total acceleration of the COM is zero (famous result), but, as the center of mass lies on the instanteneous rotation axis, it only has translational acceleration. Thus, the translation acceleration of the system is zero with respect to the COM. If the net external force is not zero, this argument guarantees that the COM will transladate.

If the net external force is zero, we know that the center of mass is not moving (accelerating). From facts 2 and 3, the system will be then rotating about some axis that passes through the COM, and thus the system as a whole must have ZERO translational acceleration.

Thus it is necessary and sufficient that the net external force is null in order for a rigid body to not transladate.

EDIT: this argument only works for static equilibrium, not dynamic, but maybe an analysis over a inertial frame with same speed as the center of mass in the moment of equilibrium may do the trick.