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Calculating Center of Mass

Introduction

If I were to hand you a dinner plate and ask you to balance it on a pole, how would you go about doing it? You would probably be careful to make sure you had centered the plate perfectly on the pole before letting go. After all, most people intuitively understand that the plate will fall immediately if the pole is off-center. Why is this? Why can’t the plate be easily balanced near the rim? The answer is that the middle of the plate is the location of its center of mass. This is the point on the plate around which all of its mass is equally distributed, which, in turn, makes it the point that is the best for balancing an object. In our example, the center of mass is at the middle of the plate because our circular plate is perfectly symmetrical. However, not every shape has the same symmetry as a circle, which begs the question: how exactly is center of mass calculated? Let’s look at an example.

The Problem

Pretend you are a carpenter who is passionate about revolutionizing the table industry. The shape below is your innovative new table design:

You get to work and finish creating the tabletop, but encounter a problem: you have no idea where to place the table leg so that the table is as stable as possible. But being the brilliant mathematician that you are, you realize you need to calculate its center of mass. Let’s take a look at how this value is calculated, and then come back and calculate it for our table.

The Method

To begin, since we want to find where on a 2-dimensional plane our center of mass is, it is helpful to imagine ourselves in a coordinate plane. This way, we are calculating an \( x \)-coordinate and \( y \)-coordinate to denote our center of mass. Now, let’s consider finding the \( x \)-coordinate and \( y \)-coordinate two separate problems. The equation to find our \( x \)-coordinate is: \[ \bar{x}\ = \frac{ M_{y} }{ \textrm{Mass} }\ \] As you might expect, "Mass" is just the total mass of our object. The quantity \( M_{y} \) (or the moment of \( y \)) is a bit more confusing. If we have a set of points labeled 1 to \( n \), it can be calculated as: \[ M_{y} = x_{1}m_{1} + x_{2}m_{2} +...+x_{n}m_{n} \] Where \( m_{i} \) is the mass of point \( i \) and \( x_{1} \)is the \( x \)-coordinate of point \( i \). The goal of this is to get a measurement of where on the \( x \)-axis our mass is located. Notably, our object doesn’t consist of just points- it is a solid. So, what we can do is divide our solid into infinitely many points along the x-axis, allowing us to use calculus to find the moment of \( y \). We can use the following equation, \[ M_{y} = \rho \int_{a}^{b} x*f(x) \,dx\ \] where \( f(x) \) is height of our point (perhaps more appropriately called a wedge or slice), \( \rho \) is the density of our solid (since mass is equivalent to volume times density), and \( a \) and \( b \) denote the interval on which our solid is defined.

The calculation for our \( y \) -coordinate will work the same as the \( x \) -coordinate. Thus, the resulting equation for our center of mass will be: \[ ( \bar{x} , \bar{y} ) = \bigg( \frac{ M_{y} }{ \textrm{Mass}\ }, \frac{ M_{x} }{ \textrm{Mass} }\ \bigg) = \bigg( \frac{ \rho \int_{a}^{b} x*f(x) \,dx\ }{ \textrm{Mass} }\ , \frac{ \rho \int_{a}^{b} y*f() \,dy\ }{ \textrm{Mass} }\ \bigg) \] Great, let’s solve our table problem!

The Solution

Before we begin our calculations, we need to determine the equations that define our shape:

To calculate \( M_{y} \), we can break our shape up into four regions. The first is bounded by \( f(x) \) and \( g(x) \) on the interval \( 0 \leq x \leq 6 \), the second is bounded by \( h(x) \) and \( i(x) \) on the interval \( 4 \leq x \leq 6 \), the third is bounded by \( y=8 \) and \( i(x) \) on the interval \( 6 \leq x \leq 8 \), and the fourth is bounded by \( j(x) \) and \( k(x) \) on the interval \( 8 \leq x \leq 10 \). This gives us \[ M_{y} = \rho \int_{0}^{6} x \Big(- \frac{1}{3}\ x+10-(- \frac{2}{3}\ x+10) \Big) \,dx\ + \rho \int_{4}^{6} x \Big( 2x-6-(x-2) \Big) \,dx\ + \rho \int_{6}^{8} x \Big( 8-(x-2) \Big) \,dx\ + \rho \int_{8}^{10} x \Big( x-(2x-10) \Big) \,dx\ \] \[ = \rho \int_{0}^{6} \frac{1}{3}\ x^2 \,dx\ + \rho \int_{4}^{6} x^2-4x \,dx\ + \rho \int_{6}^{8} -x^2+10x \,dx\ + \rho \int_{8}^{10} -x^2+10x \,dx\ \] \[ = \rho \bigg( \frac{1}{9}\ x^3 \bigg\vert_{0}^{6} + \frac{1}{3}\ x^3-2x^2 \bigg\vert_{4}^{6} + - \frac{1}{3}\ x^3+5x^2 \bigg\vert_{6}^{8} + - \frac{1}{3}\ x^3+5x^2 \bigg\vert_{8}^{10} \bigg) \] \[ =\rho \bigg( \Big[ 24 - 0 \Big] + \Big[ 72-72-21.333+32 \Big] + \Big[ -170.667+320+72-180 \Big] + \Big[ -333.333+500+170.667-320 \Big] \bigg) \] \[ = \rho \Big( 24+10.667+41.333+17.333 \Big) = 93.333 \rho \] Therefore, the moment of \( y \) is 88.333 \( \rho \). Let’s do the same with \( M_{y} \). This is a little trickier, because now we need to put our shape in terms of \( y \). To do that, we can simply find the inverse of each of our equations: \[ f(y)=-3y+30 \] \[ g(y)= - \frac{3}{2}\ y+15 \] \[ h(y)= \frac{1}{2}\ y+3 \] \[ i(y)=y+2 \] \[ j(y)=y \] \[ k(y)= \frac{1}{2}\ y+5 \] Similiar to \( M_{y} \), the \( M_{x} \) can be found as the sum of integrals over 4 regions. The first is bounded by \( f(y) \) and \( g(y) \) on the interval \( 8 \leq y \leq 10 \), the second is bounded by \( k(y) \) and \( j(y) \) on the interval \( 8 \leq y \leq 10 \), the third is bounded by \( k(y) \) and \( g(y) \) on the interval \( 6 \leq y \leq 8 \), and the fourth is bounded by \( i(y) \) and \( h(y) \) on the interval \( 2 \leq y \leq 6 \). This gives us \[ M_x = \rho \int_{8}^{10} y \Big( -3y+30-(- \frac{3}{2}\ +15) \Big) \,dy\ + \rho \int_{8}^{10} y \Big( \frac{1}{2}\ y+5-(y) \Big) \,dy\ + \rho \int_{6}^{8} y \Big( \frac{1}{2}\ y+5-(- \frac{3}{2}\ y+15) \Big) \,dy\ + \rho \int_{2}^{6} y \Big( y+2-( \frac{1}{2}\ y+3) \Big) \,dy\ \] \[ = \rho \int_{8}^{10} - \frac{3}{2}\ y^2+15y \,dy\ + \rho \int_{8}^{10} - \frac{1}{2}\ y^2+5y \,dy\ + \rho \int_{6}^{8} 2y^2-10y \,dy\ + \rho \int_{2}^{6} \frac{1}{2}\ y^2-y \,dy\ \] \[ = \rho \bigg( - \frac{1}{2}\ y^3+ \frac{15}{2}\ y^2 \bigg\vert_{8}^{10} + - \frac{1}{6}\ y^3+ \frac{5}{2} y^2 \bigg\vert_{8}^{10} + \frac{2}{3}\ y^3-5y^2 \bigg\vert_{6}^{8} + \frac{1}{6}y^3- \frac{1}{2}\ y^2 \bigg\vert_{2}^{6} \bigg) \] \[ = \rho \bigg( \Big[ -500+750+256-480 \Big] + \Big[ -166.667+250+85.333-160 \Big] + \Big[ 341.333-320-144+180 \Big] + \Big[ 36-18-1.333+2 \Big] \bigg) \] \[ = \rho \Big( 26+8.667+57.333+18.667 \Big) = 110.667 \rho \] Thus, the moment of \( x \) is 110.333 \( \rho \). The last thing we need to calculate is the mass. We will split it into the same regions as when we calculated \( M_y \): \[ \rho \int_{0}^{6} - \frac{1}{3}\ x+10-(- \frac{2}{3}\ x+10) \,dx\ + \rho \int_{4}^{6} 2x-6-(x-2) \,dx\ + \rho \int_{6}^{8} 8-(x-2) \,dx\ + \rho \int_{8}^{10} x-(2x-10) \,dx\ \] \[ = \rho \int_{0}^{6} \frac{1}{3}\ x \,dx\ + \rho \int_{4}^{6} x-4 \,dx\ + \rho \int_{6}^{8} -x+10 \,dx\ + \rho \int_{8}^{10} -x+10 \,dx\ \] \[ = \rho \bigg( \frac{1}{6}\ x^2 \bigg\vert_{0}^{6} + \frac{1}{2}\ x^2-4x \bigg\vert_{4}^{6} + - \frac{1}{2}\ x^2+10x \bigg\vert_{6}^{8} + - \frac{1}{2}\ x^2+10x \bigg\vert_{8}^{10} \bigg) \] \[ = \rho \Big( 6+(18-24-8+16)+(-32+80+18-60)+(-50+100+32-80) \Big) = 16 \rho \] Using these values and the fact that the \( \rho \) s cancel, we are able to calculate the center of mass: \[ ( \bar{x} , \bar{y} )= ( \frac{93.333 \rho }{16 \rho }\ , \frac{110.667 \rho }{ 16 \rho }\ = (5.83, 6.92) \] This means that our center of mass will be found at the following point, which is where we should attach the table leg:

Why this example?

This example was chosen for the following reasons so that it would best illustrate the concepts of center of mass:

  • The equations are easy to integrate, which allows the reader to focus their attention on the center of mass concepts rather than difficult integration. Many functions (including most polynomials with more than one term) have unpleasant inverses that would’ve made calculating the moment of x an involved process. By making these functions linear, this problem was avoided.
  • The equations combined made an intriguing shape where the center of mass is somewhat non-intuitive. Even though the shape protudes in 3 directions, the leftmost protrusion has the most mass and therefore pulls the center of to the left and slighty upwards. This helps the student to understand why the center of mass is not always at the center of an object like it would be with a circular plate and reinforces the need for calculus to find the center of mass of some shapes.

Author: Sarah Bombrys
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