Online Encyclopedia

Center of mass

The center of mass or center of inertia of an object is a point at which the object's mass can be assumed, for many purposes, to be concentrated. For example, an object can balance on a point only if its center of mass is directly above the point. Alternatively, if you hang an object from a string, the object's center of mass will be directly below the string. Also, the path of an object in orbit depends (to a very close approximation) only on its center of mass.

Precisely, the center of mass of a group of points is defined as the weighted mean of the points' positions, where the weight applied to each point is the point's mass.

For a body $V$ whose mass is distributed according to a density ρ(x), the center of mass is

$\bar{\mathbf{x}} = \frac{1}{M} \int_{V} \rho(\mathbf{x})\mathbf{x}\,dV$

Here $M$ is the total mass, given by

$M = \int_{V} \rho(\mathbf{x})\,dV$

In $\mathcal{R}^3$ , the components of the center of mass are given by:

$\bar{x} = \frac{1}{M} \int_{V} x\rho(x, y, z)\,dx\, dy\, dz$

$\bar{y} = \frac{1}{M} \int_{V} y\rho(x, y, z)\,dx\, dy\, dz$

$\bar{z} = \frac{1}{M} \int_{V} z\rho(x, y, z)\,dx\, dy\, dz$

For a system of point masses m₁, m₂, ..., the integrals are replaced by sums:

$\bar{\mathbf{x}} = \frac{1}{M} \sum m_i\mathbf{x}_i$

Where the total mass M is the sum of the constituent masses.

The origin from which positions are calculated has no effect on the physical position of the result. As long as the same unit is used for all the points, any length and mass unit can be used.

Motion of the 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 the famous 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.

The angular momentum vector for a system is equal to the angular momentum of all the particles around the center of mass, plus the angular momentum of the center of mass, as if it were a single particle of mass $M$ :

$\mathbf{L}_\mathrm{sys} = \mathbf{L}_\mathrm{cm} + \mathbf{L}_\mathrm{around\mbox{ }cm}$

Examples

Point A: position 2 m, mass 1 kg. Point B: position 4 m, mass 2 kg (assume positions are distances along a straight line from some origin). Center of mass:

$\frac{2\mbox{ m} \times 1\mbox{ kg} + 4\mbox{ m} \times 2\mbox{ kg}}{1\mbox{ kg}+2\mbox{ kg}} = 3.33\mbox{ m}$

Solid homogenous sphere (ideally divided in a high number of points of equal mass): each point averages with its opposite. Center of mass is at the center.

Sphere with spherically symmetric density: center of mass is at the center. This approximately applies to the Earth: the density varies considerably, but it mainly depends on depth and less on the other two coordinates.

Human being: It varies according to the body's position, but often it's somewhere inside the abdomen

A sports car: engineers try hard to make the car as light as possible, and then add weight on the bottom. This way, the center of mass is nearer to the street, and the car handles better.

When talking about celestial bodies, the center of mass has a special relevance: when a moon orbits around planet, or a planet orbits around a star, both of them are actually orbiting around their center of mass, called the barycenter, see two-body problem. Some examples:

Earth-Moon system: the Moon's mass is 0.0123 that of Earth. Put Earth in position 0, mass 1 (here we use an arbitrary mass unit. It does not matter, provided that we use the same unit for the Moon). The Moon is at an average distance of 384400 km from the Earth. Then the center of mass is at:

$\frac{0 \times 1 + 384400\mbox{ km} \times 0.0123}{1 + 0.0123} = 4671\mbox{ km}$

from the Earth's center. Thus, as opposed to the Earth standing "still" and the Moon moving, both of them move around a point about 1700 km below the Earth's surface.

Sun-Earth system: put Sun in position 0, mass=333,000 times the Earth. Earth in position 150,000,000 km, mass=1. Center of mass is 450 km from the Sun center. Here, the large mass difference between the two bodies makes the center of mass lie almost at the center of the Sun.

Sun-Jupiter system: put Sun in position 0, mass = 333,000 Earths. Jupiter in position 778,000,000 km, mass=318 Earths. Center of mass is 742,000 km from the Sun center, 96,000 km outside its surface. As Jupiter does its 11 year orbit, the Sun does a 1.5 million km orbit around the center of mass.

To calculate the actual motion of the Sun, you would need to sum all the influences from all the planets, comets, asteroids, etc. of the solar system. The influence of each

Note that the distance from the Sun's center to the center of mass of a two-body system consisting of the Sun and another celestial body, hence the size of the Sun's orbit around this center of mass, is approximately proportional to the product of the mass of that other body, and the distance between the two, even though gravity decreases with distance. That orbit is largest with Jupiter, its large mass more than compensates its smaller distance to the Sun than several other planets. If all the planets would align on the same side of the Sun, the combined center of mass would lie about 500,000 km outside the Sun surface.

Your Online Encyclopedia

Online Encylopedia and Dictionary Research Site

Online Encyclopedia

Center of mass

Motion of the center of mass

Examples

See also

	Your Online Encyclopedia
	Online Encylopedia and Dictionary Research Site
		Online Encyclopedia Search Online Encyclopedia Browse