center of mass examples | Physics Library

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 x₁,x₂,x₃,...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

x1-+-x2-+-x3 +-...+-xn- 1-∑ xcm = n = n 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:

(m1 + m2 + ...+ mn )xcm = m1x1 + m2x2 + ...+ mnxn

∑ ---mx-- xcm = ∑ m

∑ ---my- ycm = ∑ m

(1)

∑ mz zcm = ∑----- m

Evidently x_cm,y_cm,z_cm are the coordinates of the center of mass.

0.1 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

xcm = 0 +-nma--= --n--a m + nm n + 1

ycm = zcm = 0

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

0 + 2ma + 3ma cos60o 7 xcm = -----------------------= --a m + 2m + 3m 12

0 + 0 + 3ma sin 60o 1 √-- ycm = ------------------- = -- 3a 6m 4

zcm = 0

0.2 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:

∫m xcm = -0∫mxdm-- 0 dm

∫m y = ∫0-ydm-- cm m dm 0

(2)

∫ m0 zdm zcm = ∫-m----- 0 dm

where m is the mass of the body.

0.3 Illustrative Examples

1. Find the center of mass of the parabolic lamina bounded by the curves y² = 2px and x = a, Fig. 74.

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

∘ ---- dm = σ2ydx = 2σ 2pxdx

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

∫ a √ ---- 2-σ-0-x--2pxdx- xcm = 2σ ∫a √2pxdx 0

∫ ax3∕2dx xcm = ∫0a------- 0 x1∕2dx

xcm = 3a- 5

2. Find the center of mass of the lamina bounded by the curves y² = 4ax and y = bx, Fig. 75. Let dxdy be the area of the element of mass, then

dm = σdxdy

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

∫ 4a2-∫2√ax- -0b--bx---xdydx- xcm = ∫ 4ba2∫ 2√ax 0 bx dydx

∫ 4a √ --- 0b2 (2 ax − bx)xdx xcm = ∫-4a--√-------------- 0b2(2 ax − bx )dx

8a xcm = --2 5b

∫ 4ab2-∫2√ax ycm = -0---bx√--ydydx- ∫ 4ba2∫ 2 axdydx 0 bx

∫ 4a2-( b2 2) 0b 2ax − 2 x dx ycm = ∫-4a--√------------- 0b2(2 ax − bx )dx

2a ycm = --- 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

dm = σ ⋅ ρd 𝜃 ⋅ dρ

∫π ∫a ycm = -0∫-π0∫-ya ⋅-σρdρd-𝜃 0 0 σ ρdρd𝜃

∫ π∫ aρ2 sin 𝜃dρd𝜃 ycm = -0-∫0π∫a---------- 0 0 ρdρd𝜃

ycm = 4a- 3π

xcm = 0

0.4 References

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

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