# The Magic of Balance Pt. 2 (Center of Mass)

Continued from my last post, I further researched into the calculation for center of mass. Today’s post is going to be about how we derive the equations for center of mass.

Mass

First we need to define the mass of an object. It is easy if the object is of uniform density. However, if it has a density function $\rho(x,y)$, we define the mass by cutting the object into infinitely small areas where the density is approximately uniform.

Mass = Density $\times$ Area

When we add all the infinitely small areas together, we get the mass for the entire object, which is expressed by a double integral:

$\int\int_R\rho(x,y)dydx$

Moments

The moment of mass of a system of particles is equal to the sum of the product of each particle’s mass and its distance from an arbitrary reference. The moments about an axis are defined by the product of the mass times the distance from the axis.

So $M_x = (Mass)(y)$

$M_y = (Mass)(x)$

Plugging in our definition for mass, we get

$M_x = \int\int_R\rho(x,y)ydydx$

$M_y = \int\int_R\rho(x,y)xdydx$

Dividing the moment of mass by mass, we are able to find the center of mass

$(\frac{M_y}{M}, \frac{M_x}{M})$

Uniform Density

If the shape has uniform density, we can cross out the density from the denominator and the numerators. The center of mass equation simplifies into

$(\frac{\int\int_Rydydx}{Area}, \frac{\int\int_Rxdydx}{Area})$

We can further simplify the equations into single integrals:

$x=\frac{\int\int_Rxdydx}{Area}=\frac{1}{A}\int{x}\int{1dy} dx = \frac{1}{A}\int{xy}dx$

$y=\frac{\int\int_Rydydx}{Area}=\frac{1}{A}\int\frac{1}{2}y^2dx$

Example

So for a shape like above, we can just plug in the functions f(x) and g(x)

$x=\frac{1}{A}\int_a^b{x(f(x)-g(x))}dx$

$y=\frac{1}{A}\int_a^b\frac{1}{2}(f(x)^2-g(x)^2)dx$

Let’s look at an actual problem:

Determine the center of mass for the region bounded by $y=2sin(2x), y=0$ on the interval $[0,\frac{\pi}{2}]$

We first get the area of the region:

$A=\int_0^{\frac{\pi}{2}}2sin(2x)dx = -cos(2x)\vert_0^{\frac{\pi}{2}}=2$

Then we can find the center of mass equation:

$x=\frac{1}{A}\int_a^b{x(f(x)-g(x))}dx=\frac{1}{2}\int_0^{\frac{\pi}{2}}2xsin(2x)dx$

$y=\frac{1}{A}\int_a^b\frac{1}{2}(f(x)^2-g(x)^2)dx=\frac{1}{2}\int_0^{\frac{\pi}{2}}2sin^2(2x)dx$

(The calculations are left as an exercise to the reader. Hint: For the x coordinate, try integrating by parts; for the y coordinate, you may need to look up what $sin^2$ is equal to),

After calculations, we get the answers:

$x=\frac{\pi}{4}$

$y=\frac{\pi}{4}$

Checking on the graph, we have found the center of mass of this shape assuming uniform density.

References:

http://tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx

http://ltcconline.net/greenl/courses/202/multipleintegration/MassMoments.htm

http://en.wikipedia.org/wiki/Moment_(mathematics)

http://www.coastal.edu/mathcenter/HelpPages/Handouts/moments.PDF

Tagged with: , , , ,
Posted in Multivariable Calculus
###### One comment on “The Magic of Balance Pt. 2 (Center of Mass)”
1. Excellent work Zfu71! I can’t wait to see how you use this approach to construct the centroids of three dimensional objects!

Popular