Derivation of the center of mass

The center of mass of an object is a point at which, for the purposes of total force and total linear momentum, the mass of the object can be taken to be concentrated, for a particular configuration of the system at an instant of time. This fact can be derived in a few ways.

Derivation of center of mass

The following equations of motion assume that there is a system of particles governed by internal and external forces. An internal force is a force caused by the interaction of the particles within the system. An external force is a force that originates from outside the system, and acts on one or more particles within the system. The external force need not be due to a uniform field.

For any system with no external forces, the center of mass moves with constant velocity. This applies for all systems with classical internal forces, including magnetic fields, electric fields, chemical reactions, and so on. More formally, this is true for any internal forces that satisfy the weak form of Newton's Third Law.

The total momentum for any system of particles is given by

$\mathbf{p}=M\mathbf{v}_\mathrm{cm}$

Where M indicates the total mass, and v_cm is the velocity of the center of mass. This velocity can be computed by taking the time derivative of the position of the center of mass.

An analogue to Newton's Second Law is

$\mathbf{F} = M\mathbf{a}_\mathrm{cm}$

Where F indicates the sum of all external forces on the system, and a_cm indicates the acceleration of the center of mass.

Letting the total internal force of the system.

$\mathbf{F}=M\ddot{\mathbf{R}}$

where M is the total mass of the system and $\mathbf{R}$ is a vector yet to be defined, since:

$\mathbf{p} = \sum{m_j\dot{r}_j}$

and

$\mathbf{F}=\dot{\mathbf{p}}$

then

$\mathbf{R}=\frac{1}{M}\sum{m_jr_j}$

We therefore have a vectorial definition for center of mass in terms of the total forces in the system. This is particularly useful for two-body systems.

Alternative derivation

Consider first two bodies, with masses m₁ and m₂, and position vectors r₁ and r₂. Write M = m₁ + m₂ for the total mass of the 2-body system, and R for the position vector of the center of mass.

It is reasonable to require, for any system of masses, that the center of mass lie within the convex hull of the system. In particular, for a pair of mass points, this means that the tip of R must lie on the line segment joining the tips of r₁ and r₂. By geometry, R - r₁ = k(r₂ - R) for some positive constant k. Taking magnitudes on both sides of this equation, we get d₁ = kd₂, where d₁ is the distance from the center of mass to body 1, and d₂ is the distance from the center of mass to body 2. The constant k should obviously depend only on the masses m₁ and m₂, and we will examine the nature of this dependence.

Assuming the total mass M is nonzero, it is clear that if m₂ = 0, the center of mass should coincide with body 1, and d₁ = 0. This means d₂ = D, the total distance between the two bodies, and m₁ = M. Symmetry demands that these relations remain true when the subscripts 1 and 2 are interchanged everywhere.

The simplest model satisfying these requirements is the linear one, d₁ = (D/M)m₂ and d₂ = (D/M)m₁.

Under this model, we have k = d₁/d₂ = m₂/m₁. Therefore, after multiplying our vector equation by m₁, we find that m₁(R - r₁) = m₂(r₂ − R), or (m₁ + m₂)R = m₁r₁ + m₂r₂. Thus,

$\mathbf{R} = \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2}{m_1 + m_2}.$

Now suppose there is a third body, of mass m₃ and position r₃. Temporarily break the symmetry between the three bodies, and define the 3-body center of mass as the 2-body center of mass determined by body 3 together with a single body of mass M₀ = m₁ + m₂ placed at the center of mass of bodies 1 and 2, whose position vector we now denote by R₀. The formula derived above gives

$\mathbf{R} = \frac{M_0 \mathbf{R}_0 + m_3 \mathbf{r}_3}{M_0 + m_3} = \frac{ (m_1 + m_2) \left( \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2}{m_1 + m_2} \right) + m_3 \mathbf{r}_3}{M_0 + m_3} = \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2 + m_3 \mathbf{r}_3}{m_1 + m_2 + m_3}.$

Since R turns out to be symmetric in the m_i and r_i, it would not have mattered had we started by combining bodies 2 and 3, or bodies 1 and 3, instead of bodies 1 and 2. This kind of reasoning clearly extends to any number of masses, and yields the formula

$\mathbf{R} = \frac{\sum m_i \mathbf{r}_i}{\sum m_i}.$

So our simple model of the 2-body center of mass uniquely and consistently determines the corresponding formula in any number of mass points. Writing M = m₁ + m₂ + ... + m_n, the above formula for the center of mass may be expressed in the form

$M \mathbf{R} = \sum m_i \mathbf{r}_i.$

Differentiating both sides with respect to time yields the principle that

$M \mathbf{V} = \sum m_i \mathbf{v}_i,$

i.e., the sum of the momentum of a number of bodies is the momentum of their center of mass. It is this principle that gives precise expression to the intuitive notion that the system as a whole behaves like a mass of M placed at R, and justifies our simple linear model of the one-dimensional center of mass.

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