Center of mass

The center of mass (also focus or sometimes to distinguish it from the geometric center of gravity and center of gravity referred to) of a body is the mass weighted average of the positions of its mass points . For continuous mass distributions , the local mean of the density is defined as the center of mass. In the case of a homogeneous body (i.e. with the same density everywhere) the center of mass coincides with the geometric center of gravity . The stand-up man is an example of an inhomogeneous body.

The concept of the center of mass is used in physics to reduce a complex, extended rigid body to a single mass point for easier calculation of its trajectory when a force is applied . Many calculations are also simplified in the center of gravity system , in which the center of mass is used as the origin of the coordinates (see also multi-body system ). External forces acting in the center of mass can not change the rotational state of the object because they do not exert any torque due to the lack of a lever arm in the center of gravity . Axes through the center of gravity are also referred to as axes of gravity .

In celestial mechanics , the center of mass of a system of several celestial bodies is called the barycenter .

The center of mass of a body does not have to be inside the body. Examples are the torus , a boomerang , a cup or the center of gravity of a high jumper . But if the body is convex , the center of gravity is never outside.

Center of mass of two point masses on a rod

Given a rod of length . The two point masses and are located on this rod at the locations and . ${\displaystyle a}$ ${\displaystyle m_{1}}$ ${\displaystyle m_{2}}$ ${\displaystyle x_{1}}$ ${\displaystyle x_{2}}$

Figure 1: Rod with two point masses and a center of mass (here marked with ) ${\displaystyle x_{s}}$ ${\displaystyle x_{m}}$

The center of mass (center of mass) can then be calculated as follows: ${\displaystyle x_{s}}$

${\displaystyle x_{s}={\frac {m_{2}}{m_{1}+m_{2}}}\cdot a}$

The mass ratio is, so to speak, a percentage factor too . If the mass becomes infinitely large, the center of gravity shifts to the place . However, if the mass becomes infinitely large in relation to it, the center of mass shifts to the location . ${\displaystyle a}$ ${\displaystyle m_{2}}$ ${\displaystyle x_{2}}$ ${\displaystyle m_{1}}$ ${\displaystyle m_{2}}$ ${\displaystyle x_{1}}$

Something more general:

Image 2: Center of mass a little more general

From Figure 1 it can be seen that applies. In Figure 2 the point masses are no longer at the start and end point of the member. Since the scale runs from left to right in the pictures, you have to add the distance between the starting point of the rod and the mass point . This leads to the following formula: ${\displaystyle a=x_{2}-x_{1}}$ ${\displaystyle x_{1}}$

${\displaystyle x_{s}={\frac {m_{2}}{m_{1}+m_{2}}}\cdot (x_{2}-x_{1})+x_{1}={\frac {x_{1}\cdot m_{1}+x_{2}\cdot m_{2}}{m_{1}+m_{2}}}}$

Center of mass of several point masses on a rod

To continue from the previous section, we will now place 3 point masses on a rod.

Figure 3: Rod with three point masses

In order to determine the center of mass, we split this construct into 2 partial rods. To do this, we cut the rod on site and divide the mass halfway on one part of the rod and the other half on the other part of the rod. First we calculate the centers of gravity of the partial rods as follows, as known from the previous section: ${\displaystyle x_{2}}$ ${\displaystyle m_{2}}$

${\displaystyle x_{s1}={\frac {0{,}5\cdot m_{2}}{m_{1}+0{,}5\cdot m_{2}}}\cdot (x_{2}-x_{1})+x_{1}}$

${\displaystyle x_{s2}={\frac {m_{3}}{0{,}5\cdot m_{2}+m_{3}}}\cdot (x_{3}-x_{2})+x_{2}}$

With the total mass of the partial rods and the center of gravity, the partial rods can now be combined as a new point mass:

${\displaystyle m_{xs1}=m_{1}+0{,}5\cdot m_{2}}$

${\displaystyle m_{xs2}=0{,}5\cdot m_{2}+m_{3}}$

With these new values, one now calculates another center of mass, which is ultimately the center of mass of the three point masses:

${\displaystyle x_{s}={\frac {m_{xs2}}{m_{xs1}+m_{xs2}}}\cdot (x_{s2}-x_{s1})+x_{s1}}$

When used, it looks like this:

${\displaystyle {x_{s}={\frac {0{,}5\cdot m_{2}+m_{3}}{m_{1}+m_{2}+m_{3}}}\cdot \left({\frac {m_{3}\cdot (x_{3}-x_{2})}{0{,}5\cdot m_{2}+m_{3}}}+x_{2}-{\frac {0{,}5\cdot m_{2}\cdot (x_{2}-x_{1})}{m_{1}+0{,}5\cdot m_{2}}}-x_{1}\right)+{\frac {0{,}5\cdot m_{2}\cdot (x_{2}-x_{1})}{m_{1}+0{,}5\cdot m_{2}}}+x_{1}}}$

If you reformulate this equation a bit, you get the following result:

${\displaystyle x_{s}={\frac {x_{1}\cdot m_{1}+x_{2}\cdot m_{2}+x_{3}\cdot m_{3}}{m_{1}+m_{2}+m_{3}}}}$

If you compare this result with that from the previous section, a regularity can be seen. If you now distribute n many point masses on a rod, the center of mass can be determined as follows:

${\displaystyle x_{s}={\frac {1}{M}}\cdot \sum _{i=1}^{n}{x_{i}\cdot m_{i}}}$

It is the total mass, ie the sum of all point masses: ${\displaystyle M}$

${\displaystyle M=\sum _{i=1}^{n}{m_{i}}}$

Center of mass with continuous mass distribution along a rod

Here we use the formula from the previous section and form the limit value. This gives an integral representation.

Center of mass:

${\displaystyle x_{s}={\frac {1}{M}}\cdot \int _{x_{1}}^{x_{2}}{x\cdot \mathrm {d} m}={\frac {1}{M}}\cdot \int _{x_{1}}^{x_{2}}{x\cdot \lambda (x)\mathrm {d} x}}$

Density function:

${\displaystyle {\frac {\mathrm {d} m}{\mathrm {d} x}}=\lambda (x)}$

Total mass:

${\displaystyle M=\int _{x_{1}}^{x_{2}}{\lambda (x)\mathrm {d} x}}$

Sample calculation

Given a rod of length . The density increases proportionally with the length of the rod. Now calculate the center of mass of the rod! ${\displaystyle l=1\;\mathrm {m} }$

Density function:

${\displaystyle {\frac {\mathrm {d} m}{\mathrm {d} x}}=\lambda (x)=c\;x}$

The proportionality factor here is chosen arbitrarily as . ${\displaystyle c=1{\frac {\mathrm {kg} }{\mathrm {m^{2}} }}}$

Total mass:

${\displaystyle M=\int _{0}^{l}cx\;\mathrm {d} x={\frac {c}{2}}\cdot \left[x^{2}\right]_{0}^{l}=0{,}5\;\mathrm {kg} }$

Center of mass:

${\displaystyle x_{s}={\frac {1}{M}}\cdot \int _{0}^{l}{x\cdot cx\;\mathrm {d} x}={\frac {1}{M}}\cdot \left[{\frac {1}{3}}x^{3}\right]_{0}^{l}\approx 0{,}667\;\mathrm {m} }$

Mathematical definition

The center of mass is the weighted mean of the position vectors of all mass points of a body: ${\displaystyle {\vec {r}}_{s}}$ ${\displaystyle {\vec {r}}}$ ${\displaystyle \mathrm {d} m}$

${\displaystyle {\vec {r}}_{s}={\frac {1}{M}}\int _{K}{{\vec {r}}\,\mathrm {d} m}={\frac {1}{M}}\int _{K}{{\vec {r}}\,\rho ({\vec {r}})\,\mathrm {d} V}}$

It is the density at the site and a volume element . The denominator of these terms is the total mass. ${\displaystyle \rho ({\vec {r}})}$ ${\displaystyle {\vec {r}}}$ ${\displaystyle dV}$ ${\displaystyle M}$

In the case of a homogeneous body, the density can be taken as a factor in front of the integral; the center of mass then coincides with the volume center (the geometric center of gravity). In many cases the calculation can then be simplified; for example, if the volume center lies on an axis of symmetry of the body, for example in the case of a sphere in the center. ${\displaystyle \rho }$

In discrete systems , the volume integral can be replaced by a sum of the position vectors of all mass points: ${\displaystyle {\vec {r}}_{i}}$

${\displaystyle {\vec {r}}_{s}={\frac {1}{M}}\sum _{i}m_{i}\,{\vec {r}}_{i}}$

where the sum of all individual masses is: ${\displaystyle M}$ ${\displaystyle m_{i}}$

${\displaystyle M=\sum _{i}m_{i}}$

The term center of mass compared to the center of gravity

→ Main article : Gravizentrum

The gravitation acts on all mass points of a body. Only in a homogeneous gravitational field is the overall effect as if the gravitational force were acting in the center of mass. Since the gravitational field can often be assumed to be homogeneous, e.g. B. in the vicinity of the earth's surface, the terms center of gravity and center of mass are often both undifferentiated referred to as the center of gravity . In an inhomogeneous field, this effective point is different from the center of mass and centers of gravity in non-uniform fields mentioned. In such a case, tidal forces occur.

The term center of mass compared to the center of mass

→ Main article : Geometric focus

If a body is homogeneous (i.e. if it consists of a material that has the same density everywhere), its center of mass coincides with its geometric center of gravity. If the body consists of parts of different densities, the center of mass can deviate from the center of gravity of the volume. If the distribution of mass within the body is known, the center of mass can be calculated by integration . This was the occasion that led Isaac Newton to develop calculus (at the same time as Leibniz ).

Determination of the center of mass

The focus is below the suspension point on the "center of gravity".

The center of gravity is also under a different suspension point. The position of the center of gravity can thus be determined from the intersection of the two lines.

The above explanations lead to a simple method for the approximate determination of the center of mass of any rigid body. The approximation consists in disregarding the deviations from the center of gravity and the center of mass and thus also the changes in the position of the center of gravity when the body rotates: If you hang the body at any point, the (approximate) center of mass lies (at rest) on the vertical line (= "median line") through the suspension point (blue line in the picture on the right).

If you repeat this with a different suspension point, you will find (approximately) the center of mass as the intersection of two such straight lines ("centroid"). The fact that such an intersection actually exists and is independent of the choice of suspension points is, however, less trivial than the first impression suggests.

The following method for determining the center of mass of a narrow and elongated object (e.g. ruler or broom) is astonishing: Place the object across the two forefingers stretched forward at the same height, which is easily possible as long as the fingers are still far apart are. Now slowly bring your index fingers closer together until they touch, always keeping them at the same height as possible. If you do this slowly enough, the object glides slowly over your fingers without tilting to one side. The finger, which is closer to the center of mass, is subject to greater pressure, which leads to greater friction. This means that the object primarily slides over the other finger. This regulates the system in such a way that both fingers have roughly the same friction and the center of mass is in their center. Finally, the index fingers touch, the object is still horizontal and the center of gravity is above the two fingers. However, if the object is bent too much, the above-mentioned effect occurs and the center of gravity is below the support point.

See also

• Mass distribution
• Mass line (method)

literature

• Physics: a lexicon of all school physics . Schülerduden, Bibliographisches Institut, Mannheim 1974, ISBN 3-411-01122-X , pp. 367–368.

Individual evidence

1. ↑ D. Gross, W. Hauger, J. Schröder, and WA Wall: Technische Mechanik 1: Statik. Springer textbook 2011, ISBN 9783642138058 , p. 114.
2. ↑ John McLester, Peter St. Pierre: Applied biomechanics: concepts and connections . Cengage Learning, 2008, ISBN 978-0-495-10586-2 , p. 28.
3. ^ John Harris, Walter Benenson, Horst Stöcker: Handbook of physics . Springer, 2002, ISBN 978-0-387-95269-7 , p. 94.
4. ↑ Theo Koupelis, Karl F. Kuhn: In quest of the universe . Jones & Bartlett Learning, April 13, 2007, ISBN 978-0-7637-4387-1 , p. 86.
5. ↑ Philip Ball: Life's matrix: a biography of water . University of California Press, 2001, ISBN 978-0-520-23008-8 , p. 37.