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You CAN Ace Calculus

17calculus > integrals > moments, center of mass

### Calculus Main Topics

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Applied Integration - Moment, Center of Mass and 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.

### Full Video Lecture

 Prof Leonard - Calculus 3 Lecture 14.4: Center of Mass (and Moments of Mass and Inertia) for Lamina in 2-D [1hr-16min-43sec]

### Guidelines For Working Word Problems

Word 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.

I will tell you up front that figuring out how to work word problems is not easy and takes some independent work on your part to master them. But once you do, you will find them enjoyable and, since so many students struggle with them, most teachers give pretty easy problems, even on exams. So you should be able to breeze your way through them.

First, what doesn't work. Most books try to lump all word problems together and give you general guidelines on how to work them. I have NEVER found that helpful. It wasn't until I was able separate out the different types of word problems, that I came to understand how to work them. Since there are different types of word problems, there are different ways to work them.

Here is what you need to do.

1. Find plenty of problems with worked out solutions. Here are some suggested resources.
- Get a good book with examples and worked out solutions of the type of word problems you are studying. We have posted several suggestions on the books page.
- Check out the solution manual for problems in your textbook.
- 17calculus practice problems

2. Once you have a good selection of worked out solutions, go through them carefully and pick up patterns on how they set up the problems, solve them and give the final answer. Pick the ones that are similar to ones in your textbook that you are working on for your class.

3. Key - - Group the problems into categories that make sense to you. Some examples might be problems with triangles, problems with right circular cylinders, problems asking you to find areas or volumes. A single problem can go into multiple categories based on configuration or type of question or any other category that makes sense to you.

4. Work the problems yourself before looking at the solutions. Then compare your solutions with the book. Determine what you did wrong and what you need to learn in order to work the problems correctly. At first, this will be slow and painful but once your brain catches on, it will start to be fun. Be patient with yourself, work hard and don't give up. [ In the case of videos, stop the video after the presenter has given the problem statement and work it yourself before watching them solve it. ]

6. Make sure you understand every single step and, when looking at the solution, figure out why they do things the way they do. If you made a mistake, try to understand what your mistake was and what you need to understand in order to not make the same mistake again. [ Also remember that no textbook or video is always 100% correct. If you can not figure out your mistake, find someone to ask and see if the solution manual is incorrect. ]

7. Pick up patterns and general ideas from each group of problems by working the same type of problems all together. Don't jump around to different types. Stay with one type for several problems. I won't tell you exactly how many. You need to determine that by how difficult the problems are, how well you think you understand the current type, how much time you have and how well you want to do on your homework and exams. Sometimes you can go on after working 5 of the same type, sometimes it takes 10 or more.

8. Find a friend to work with and go over the problems with them AFTER you have worked them on your own. Remember, at exam time you will be on your own. So don't rely on someone else too much. If you know more than the other person, explain your work to them. Communicating your work to someone else helps you understand it better. If you know less, ask lots of questions and ask them to explain their solution to you.

9. Do NOT do shortcuts. Shortcuts are good AFTER you have learned the material, not while you are learning the material. Do it the long way for a while until you are know it really well.

10. 2nd Key - - Do not just look at the solutions or watch someone else work the problems. You need to get out a pencil and paper and work them yourself. You are going to get frustrated. You are going to want to quit, but don't quit. Use that feeling to motivate yourself and show yourself that you can do it. It feels great to master something that is difficult. If you have never pushed through something difficult before, try it now. It is not easy but it is worth it. I know because I went through this same process myself.

11. Finally, do not skip ANYTHING and NEVER GIVE UP. Make sure you understand every single step in every single problem. Here's why: Chances are, if you skip something, it will show up on an exam precisely because the part you don't understand is probably the most difficult part of the problem and teachers expect you to skip it. So they put it on exams to see if you understand the difficult parts.

So far, I have found that implementing these ideas as the best way to figure out how to work word problems. There are tons of general guidelines in books (most likely in your textbook too) that never really helped me. Give this technique a try. Remember, you are now in charge of your own learning. No one is going to help you from here on out. You need to do it.

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.

 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/sec2.
'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 $$x_1+x_2=4 \to x_1=4-x_2$$. Substituting this into the equation for moment equation for equilibrium allows us to solve for $$x_2$$ as follows.
$$\begin{array}{rcl} -m_1gx_1 + m_2gx_2 & = & 0 \\ -m_1g(4-x_2) + m_2gx_2 & = & 0 \\ -20g(4-x_2) + 30gx_2 & = & 0 \\ -80g+20gx_2 + 30gx_2 & = & 0 \\ 50gx^2 & = & 80g \\ x_2 & = & 8/5 \end{array}$$
$$x_1=4-x_2= 4-8/5=12/5$$
Final Answer: $$x_1=12/5m$$ $$x_2=8/5m$$

Does this make sense? Let's calculate the moment. $$M=(8/5)30g+20g(-12/5) = 48g-48g=0$$

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
$$m_1g(x_1-\bar{x}) +$$ $$m_2g(x_2-\bar{x}) +$$ $$m_3g(x_3-\bar{x}) + \cdots +$$ $$m_{n-1}g(x_{n-1}-\bar{x}) +$$ $$gm_n(x_n-\bar{x}) = 0$$.
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 $$x_2=2-\bar{x}=2-2/5=8/5$$. 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).

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 y-axis is $$\displaystyle{M_y = \sum_{i=1}^{n}{x_im_i}}$$ where each $$x_i$$ is the directed distance to the y-axis.
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 x-direction is $$\bar{x}=M_y/m=44/20=2.2$$.

We can do the same calculations about the x-axis.
The moment about the x-axis is $$\displaystyle{M_x=\sum_{i=1}^{n}{y_im_i}}$$ where $$y_i$$ is the directed distance to the x-axis. For our example, $$M_x=(-2)6+0{3}+3(2)+2(9)=12$$.
The center of mass in the y-direction 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

$$\displaystyle{M_x=\rho\int_{a}^{b}{(1/2)[f(x)+g(x)][f(x)-g(x)]~dx}}$$

$$\displaystyle{M_y=\rho\int_{a}^{b}{x[f(x)-g(x)]~dx}}$$

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 [6min-1sec]

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.