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Applied Integration  Moment, Center of Mass and Centroid  

on this page: ► moment about a point ► center of mass ► moment about a line ► moment of a planar lamina ► centroid  
A moment measures the tendency of a region or point to rotate. On this page, we discuss two types of moments, moment about a point and moment about a line. The term moment is shortened from moment of force and is also called torque.  
Guidelines For Working Word ProblemsWord problems are what math students dread the most. I completely understand. I had a lot of difficulty too. However, I found a technique so that I was not only able to successfully work word problems but, eventually, I came to like them and am now able to teach them.

Definition of Moment 

A moment measures the tendency of a region or point to rotate. 
Equation: \(M=F\cdot l\) 
The length l is a directed distance from some fixed reference point or axis.
Moment About A Point 

First, let's talk about moment about a point.
Example 1 
Consider the configuration in Figure 1. Calculate the moment each mass produces and the total moment about the fulcrum.
Figure 1  Example 1 

We have a two masses, one at each end of 4 meter plank with a fulcrum in the middle. For now, we will ignore the weight of the plank and consider only the two masses. Mass 1 is 20kg and mass 2 is 30kg. Assume that gravity is 9.81m/sec^{2}.
'Left' moment: \(M_l = m_1g(2m) =\) \(20kg(2m)g =\) \((40kg\text{}m)(9.81m/sec^2) = 392.4~N\text{}m\)
'Right' moment: \(M_r = m_2g(2m) =\) \(30kg(2m)g =\) \((60kg\text{}m)(9.81m/sec^2) = 588.6~N\text{}m\)
[Left and right are designated with respect to the fulcrum.]
We don't usually talk about left and right moments. Instead we assign positive and negative directions. If we say that → is the positive side of the fulcrum (most common case), then
 the moment due to mass 1 is \(M_1=20kg(2m)9.81m/sec^2 = \) \(392.4~N\text{}m\)
 the moment due to mass 2 is \(M_2=30kg(2m)9.81m/sec^2 = \) \(588.6~N\text{}m\)
The moment due to mass 1 is negative because it is on the negative (left) side of the fulcrum.
Now we can calculate the total moment due to both mass 1 and mass 2 as \(M=M_1+M_2=588.6~N\text{}m  392.4~N\text{}m =\) \( 196.2~N\text{}m\)
Equilibrium 
You know from experience that the plank in this picture will have a tendency to rotate. This is due \(\abs{M_2} > \abs{M_1}\).
To keep the system in equilibrium, we need \(M_2 = M_1\), or equivalently the total moment \(M=M_1+M_2=0\).
Now we are going to work with the system above to get the system into equilibrium.
Example 2 
Leaving mass 1 in the same place in Example 1, where would we need to locate mass 2 so that the resulting system is in equilibrium?
We want \(M=M_1+M_2=0\).
\(M_1=m_1gx_1=(20kg)g(2m) =\) \(40g~N\text{}m\)
\(M_2=m_2gx_2=(30kg)gx_2 = \) \(30gx_2~N\text{}m\)
\(
\begin{array}{rcl}
M=M_1+M_2 & = & 0 \\
40g+30gx_2 & = & 0 \\
30gx_2 & = & 40g \\
x_2 & = & 4/3
\end{array}
\)
So if we move \(m_2\) closer to the fulcrum so that \(x_2=4/3\), then the system is in equilibrium.
Okay, so that is one way to get the system in equilibrium. However, usually, we won't move either of the masses. Instead, we calculate where we need to move the fulcrum to achieve stability.
Example 3 
In this example, we leave the masses where they were originally in Example 1 and move the fulcrum so that the system is in equilibrium. Our system looks like this. In this figure, we assumed that the fulcrum will need to be moved to the left. This makes sense, since \(m_2 > m_1\).
Figure 2  Example 3 

The moment due to mass 1 is \(M_1=m_1gx_1\). The negative sign indicates that the mass is on the negative (left) side of the fulcrum.
Similarly, the moment due to mass 2 is \(M_2=m_2gx_2\).
The total moment is \(M=M_1+M_2=m_1gx_1+m_2gx_2\). This must be equal to zero for equilibrium, i.e. \(m_1gx_1+m_2gx_2=0\).
Notice in this equation we have 2 unknowns. So we need to remove one. Remember from Example 1, that the total distance between the masses was 4. So \(x_1+x_2=4 \to x_1=4x_2\). Substituting this into the equation for moment equation for equilibrium allows us to solve for \(x_2\) as follows.
\(
\begin{array}{rcl}
m_1gx_1 + m_2gx_2 & = & 0 \\
m_1g(4x_2) + m_2gx_2 & = & 0 \\
20g(4x_2) + 30gx_2 & = & 0 \\
80g+20gx_2 + 30gx_2 & = & 0 \\
50gx^2 & = & 80g \\
x_2 & = & 8/5
\end{array}
\)
\(x_1=4x_2= 48/5=12/5\)
Final Answer: \(x_1=12/5m\) \(x_2=8/5m\)
Does this make sense? Let's calculate the moment. \( M=(8/5)30g+20g(12/5) = 48g48g=0\) ✓
Center of Mass 

The fulcrum point that we calculated in Example 3 where the total moment is zero is called the center of mass. When gravity is the same everywhere in the problem, the center of mass is equal to the center of gravity.
When there are more than 2 masses, we just add up all the moments due to each mass. In general we have a configuration represented in Figure 3, where \(\bar{x}\) is the center of mass and each \(x_i\) is the distance from the origin (which we specify in each problem) to \(m_i\).
Figure 3 

Since \(\bar{x}\) is the center of mass then the sum of the all the moments about \(\bar{x}\) is equal to zero. In equation form
\(m_1g(x_1\bar{x}) + \) \( m_2g(x_2\bar{x}) + \) \( m_3g(x_3\bar{x}) + \cdots +\) \( m_{n1}g(x_{n1}\bar{x}) + \) \(gm_n(x_n\bar{x}) = 0\).
We can write this in summation form and simplify as follows.
\(
\begin{array}{rcl}
\displaystyle{\sum_{i=1}^{n}{m_ig(x_i\bar{x})}} & = & 0 \\
\displaystyle{\sum_{i=1}^{n}{m_igx_i}  \sum_{i=1}^{n}{m_ig\bar{x}}} & = & 0 \\
\displaystyle{g\sum_{i=1}^{n}{m_ix_i}} & = & \displaystyle{g\bar{x}\sum_{i=1}^{n}{m_i}} \\
\bar{x} & = & \displaystyle{\frac{\sum_{i=1}^{n}{m_ix_i}}{\sum_{i=1}^{n}{m_i}}}
\end{array}
\)
We can use this last equation to calculate the center of mass \(\bar{x}\) of a system of masses. Notice that the numerator is the moment of the system about the origin and the denominator is the total mass of the system.
Example 4 
Let's use this last equation to calculate the center of mass in Example 3. We will set the origin at the center of the plank (where the fulcrum is).
\(\displaystyle{ \bar{x} = \frac{m_1x_1+m_2x_2}{m_1+m_2} = }\) \(\displaystyle{ \frac{20(2)+30(2)}{20+30} = }\) \(\displaystyle{\frac{40+60}{50}=\frac{2}{5}}\)
So \(\bar{x}=2/5m\) means the center of mass is \(2/5m\) to the right (since it is positive) of the origin. In this problem, the origin is the center of the plank, so that means \(x_2=2\bar{x}=22/5=8/5\). This is the same value we got in Example 3.
Keys 
1. We set up a zero point or origin and it can be anywhere along the plank.
2. We define up a positive direction, we used positive to the right.
3. Our answer is also with respect to the origin, positive to the right, negative to the left (based on our choice in key 2).
Moment About A Line 

The moment about a line is very similar to a moment about a point. The \(x_i\)'s are just the perpendicular distance from the mass to the line. Here is an example.
Example 5 
\(m_1=6, m_2=3, m_3=2, m_4=9\) 

Figure 4  Example 5 
The moment about the yaxis is \(\displaystyle{M_y = \sum_{i=1}^{n}{x_im_i}}\) where each \(x_i\) is the directed distance to the yaxis.
In this example, \(M_y = 3(6)+0(3)+(5)(2)+4(9) = 44\).
Just like we did for center of mass about a point, the center of mass about a line is the moment divided by the total mass. In this example, total mass is \(m=6+3+2+9=20\) and the center of mass in the xdirection is \(\bar{x}=M_y/m=44/20=2.2\).
We can do the same calculations about the xaxis.
The moment about the xaxis is \(\displaystyle{M_x=\sum_{i=1}^{n}{y_im_i}}\) where \(y_i\) is the directed distance to the xaxis. For our example, \(M_x=(2)6+0{3}+3(2)+2(9)=12\).
The center of mass in the ydirection is \(\bar{y}=M_x/m=12/20=0.6\)
So the center of mass in the plane is \((\bar{x},\bar{y})=(2.2,0.6)\).
Moment of a Planar Lamina 

The moment and center of mass for a planar lamina is given by the equations in Table 1.
Table 1  Moments and Center of Mass Equations for Planar Lamina  

moment about the xaxis 
\(\displaystyle{M_x=\rho\int_{a}^{b}{(1/2)[f(x)+g(x)][f(x)g(x)]~dx}}\) 
moment about the yaxis 
\(\displaystyle{M_y=\rho\int_{a}^{b}{x[f(x)g(x)]~dx}}\) 
total mass 
\(\displaystyle{m=\rho\int_{a}^{b}{f(x)g(x)~dx}}\) 
center of mass \(\bar{x},\bar{y}\) 
\(\displaystyle{\bar{x}=\frac{M_y}{m}, \bar{y}=\frac{M_x}{m}}\) 
\(f(x) \geq g(x), [a,b]\); \(\rho\) is the planar mass density and is a constant in these equations 
Here is a short video explaining how to derive these equations.
David Lippman  Deriving center of mass equations for a lamina [6min1sec]  
Centroid 

When ρ is constant, it will cancel in the equations for the center of mass in Table 1. When this happens, we call the center of mass, the centroid.