Physics Library
 An open source physics library
Encyclopedia | Forums | Docs | Random | Template Test |  
Login
create new user
Username:
Password:
forget your password?
Main Menu
Sections

Talkback

Downloads

Information
[parent] center of mass examples (Example)

The center of mass of a system of equal particles is their average position; in other words, it is that point whose distance from any fixed plane is the average of the distances of all the particles of the system.

Let $x_1, x_2, x_3, . . . x_n$ denote the distances of the particles of a system from the yz-plane; then, by the above definition, the distance of the center of mass from the same plane is

$\displaystyle x_{cm} = \frac{ x_1 + x_2 + x_3 + . . . + x_n}{n} = \frac{1}{n} \sum x$

When the particles have different masses their distances must be weighted, that is, the distance of each particle must by multiplied by the masss of the particle before taking the average. In this case the distance of the center of mass from the yz-plane is defined by the following equation:

$\displaystyle (m_1 + m_2 + ... + m_n)x_{cm} = m_1 x_1 + m_2 x_2 + ... + m_n x_n $

or

$\displaystyle x_{cm} = \frac{ \sum mx}{\sum m}$
$\displaystyle y_{cm} = \frac{ \sum my}{\sum m}$ (1)

$\displaystyle z_{cm} = \frac{ \sum mz}{\sum m}$

Evidently $x_{cm}, y_{cm}, z_{cm}$ are the coordinates of the center of mass.

Illustrative Examples

1. Find the center of mass of two particles of masses $m$ and $nm$, which are separated by a distance $a$. Taking the origin of the axes at the particle which has the mass $m$, figure 72, and taking as the z-axis the line which joins the two particles we get

$\displaystyle x_{cm} = \frac{0 + nma}{m + nm} = \frac{n}{n+1}a$

$\displaystyle y_{cm} = z_{cm} = 0 $
\includegraphics[scale=.85]{Figure72.eps}

2. Find the center of mass of three particles of masses $m, 2m, 3m$, which are at the vertices of an equilateral triangle of sides $a$. Choosing the axes as shown if Fig. 73 we have

$\displaystyle x_{cm} = \frac{0 + 2ma + 3ma \cos 60^o}{m + 2m + 3m} = \frac{7}{12}a$

$\displaystyle y_{cm} = \frac{0 + 0 + 3ma \sin 60^o}{6m} = \frac{1}{4}\sqrt{3}a$

$\displaystyle z_{cm} = 0 $
\includegraphics[scale=.85]{figure73.eps}

Center of Mass of Continuous Bodies

When the particles form a continuous body we can replace the summation signs of equation (1) by integration signs and obtain the following expressions for the coordinates of the center of mass:

$\displaystyle x_{cm} = \frac{ \int_0^m x dm}{ \int_0^m dm}$
$\displaystyle y_{cm} = \frac{ \int_0^m y dm}{ \int_0^m dm}$ (2)

$\displaystyle z_{cm} = \frac{ \int_0^m z dm}{ \int_0^m dm}$

where $m$ is the mass of the body.

Illustrative Examples

1. Find the center of mass of the parabolic lamina bounded by the curves $y^2 = 2px$ and $x = a$, Fig. 74.

\includegraphics[scale=.85]{figure74.eps}

Obviously the center of mass lies on the x-axis. Therefore we need to determine $x_{cm}$ only. Taking a strip of width $dx$ for the element of mass we have

$\displaystyle dm = \sigma 2 y dx = 2 \sigma \sqrt{2 px} dx $

where $\sigma$ is the mass per unit area. Therefore substituting this expression of $dm$ in equation (2) nd changing the limits of integration we obtain

$\displaystyle x_{cm} = \frac{ 2 \sigma \int_0^a x\sqrt{2px} dx}{2 \sigma \int_0^a \sqrt{2px} dx} $

$\displaystyle x_{cm} = \frac{ \int_0^a x^{3/2} dx}{\int_0^a x^{1/2} dx} $

$\displaystyle x_{cm} = \frac{3a}{5} $

2. Find the center of mass of the lamina bounded by the curves $y^2 = 4ax$ and $y = bx$, Fig. 75. Let $dx \, dy$ be the area of the element of mass, then

$\displaystyle dm = \sigma dx dy $
\includegraphics[scale=.85]{figure75.eps}

Therefore substituting in equation (2) and introducing the proper limits of integration we obtain

$\displaystyle x_{cm} = \frac{\int_0^{\frac{4a}{b^2}} \int_{bx}^{2\sqrt{ax}} x dy dx}{\int_0^{\frac{4 a}{b^2}} \int_{bx}^{2\sqrt{ax}} dy dx} $

$\displaystyle x_{cm} = \frac{\int_0^{\frac{4a}{b^2}} \left ( 2\sqrt{ax} - bx \right )x dx}{\int_0^{\frac{4 a}{b^2}} \left ( 2\sqrt{ax} - bx \right ) dx} $

$\displaystyle x_{cm} = \frac{8 a}{5 b^2}$

$\displaystyle y_{cm} = \frac{\int_0^{\frac{4a}{b^2}} \int_{bx}^{2\sqrt{ax}} y dy dx}{\int_0^{\frac{4 a}{b^2}} \int_{bx}^{2\sqrt{ax}} dy dx} $

$\displaystyle y_{cm} = \frac{\int_0^{\frac{4a}{b^2}} \left ( 2ax - \frac{b^2}{2}x^2 \right ) dx}{\int_0^{\frac{4 a}{b^2}} \left ( 2\sqrt{ax} - bx \right ) dx} $

$\displaystyle y_{cm} = \frac{2 a}{b}$

3. Find the center of mass of a semicircular lamina. Selecting the coordinates and the element of mass as shown in Fig. 76 we have

$\displaystyle dm = \sigma \cdot \rho d\theta \cdot d\rho $

$\displaystyle y_{cm} = \frac{\int_0^{\pi} \int_0^a y \cdot \sigma \rho d\rho d\theta}{\int_0^{\pi} \int_0^a \sigma \rho d\rho d\theta} $

$\displaystyle y_{cm} = \frac{\int_0^{\pi} \int_0^a \rho^2 \sin \theta d\rho d\theta}{\int_0^{\pi} \int_0^a \rho d\rho d\theta} $

$\displaystyle y_{cm} = \frac{4a}{3\pi} $

$\displaystyle x_{cm} = 0 $
\includegraphics[scale=.85]{figure76.eps}

References

This article is a derivative of the public domain work, "Analytical mechanics" by Haroutune M. Dadourian, 1913. Made available by the internet archive



"center of mass examples" is owned by bloftin.

View style:

See Also: motion of the center of mass


This object's parent.

Cross-references: mechanics, work, domain, masses, position, system, center of mass

This is version 9 of center of mass examples, born on 2006-07-03, modified 2006-07-03.
Object id is 192, canonical name is CenterOfMassExamples.
Accessed 5472 times total.

Classification:
Physics Classification: 45.40.-f (Dynamics and kinematics of rigid bodies)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:

No messages.

Testing some escape charachters for html category with a generator has an injective cogenerator" now escape ” with "