## 17Calculus Integrals - Moments, Center of Mass and Centroids

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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 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.

If you want a full lecture on moments and center of mass, we recommend this video from one of our favorite lecturers.

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

video by Prof Leonard

Here is a playlist containing a full discussion of centroids and center of gravity.

### Michel vanBiezen - Mechanical Engineering: Centroids & Center of Gravity Playlist

video by Michel vanBiezen

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 the picture in the example above 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 check by calculating the moment.
$$M=(8/5)30g+20g(-12/5) = 48g-48g=0$$

Center of Mass and Center of Gravity

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 $$(\bar{x},\bar{y})=(2.2,0.6)$$.

Okay, let's work some practice problems before we go on.

Practice - Discrete Objects

A 5kg mass is placed at the origin and a 9kg mass is placed at $$x=2$$m. What is the center to mass?

Problem Statement

A 5kg mass is placed at the origin and a 9kg mass is placed at $$x=2$$m. What is the center to mass?

Solution

### 3599 video

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An 8kg mass is placed at y=3m. Where should a 10kg mass be placed along the y-axis so that the center of mass will be located at y=4.5m?

Problem Statement

An 8kg mass is placed at y=3m. Where should a 10kg mass be placed along the y-axis so that the center of mass will be located at y=4.5m?

Solution

### 3600 video

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We have three masses at various points in the plane as follows.
2lbs at $$(3,2)$$, 6lbs at $$(-1,4)$$ and 1lb at $$(0,6)$$.
(a) What is the total moment with respect to the x-axis?
(b) What is the total moment with respect to the y-axis?
(c) What is the center of mass of the system?

Problem Statement

We have three masses at various points in the plane as follows.
2lbs at $$(3,2)$$, 6lbs at $$(-1,4)$$ and 1lb at $$(0,6)$$.
(a) What is the total moment with respect to the x-axis?
(b) What is the total moment with respect to the y-axis?
(c) What is the center of mass of the system?

Solution

### 3598 video

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We have four masses at various points in the plane as follows.
4kg at $$(0,0)$$, 6kg at $$(4,0)$$, 7kg at $$(0.5)$$ and 10kg at $$(5,6)$$. Determine the center of mass of this system.

Problem Statement

We have four masses at various points in the plane as follows.
4kg at $$(0,0)$$, 6kg at $$(4,0)$$, 7kg at $$(0.5)$$ and 10kg at $$(5,6)$$. Determine the center of mass of this system.

Solution

### 3601 video

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A 10m long 4000kg truck has its center of mass 4m behind the front of the truck. Where should a 1400kg load be placed so that the center of mass will be right in the middle of the truck?

Problem Statement

A 10m long 4000kg truck has its center of mass 4m behind the front of the truck. Where should a 1400kg load be placed so that the center of mass will be right in the middle of the truck?

Solution

### 3602 video

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Centroids and Center of Mass of a Wire

Interestingly enough, we can calculate the center of mass/gravity and the centroid of a wire. Here is a video explaining this in detail.

### Michel vanBiezen - Mechanical Engineering: Centroids & Center of Gravity - Center of Gravity of a Wire

video by Michel vanBiezen

Before we go on, work these practice problems involving the centroids and center of mass of a wire.

Practice - Center of Mass of a Wire

Unless otherwise instructed, solve these problems giving your answers in exact form.

Find the center of gravity of a semi-circle of radius $$R$$ of very thin wire.

Problem Statement

Find the center of gravity of a semi-circle of radius $$R$$ of very thin wire.

Hint

Center the circle at the origin, which, because of symmetry, causes $$\bar{x} = 0$$

Problem Statement

Find the center of gravity of a semi-circle of radius $$R$$ of very thin wire.

Hint

Center the circle at the origin, which, because of symmetry, causes $$\bar{x} = 0$$

Solution

### 3576 video

video by Michel vanBiezen

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Find the center of gravity of a thin wire shaped in the arc of circle with radius $$R$$ and angle $$2\alpha$$. Position it like problem 3575.

Problem Statement

Find the center of gravity of a thin wire shaped in the arc of circle with radius $$R$$ and angle $$2\alpha$$. Position it like problem 3575.

Solution

### 3577 video

video by Michel vanBiezen

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Before discussing the moment of a wire, let's look at the center of mass of an area, usually called a planar lamina.

Moment, Center of Mass, Center of Gravity and Centroid of a Planar Lamina

As mentioned above, the moment is the tendency an object to rotate. The other three terms are closely related.

For a Region (Thin Plate, Planar Lamina)

 The center of mass a region is the point in which the region will be perfectly balanced horizontally if suspended from that point. [src-Paul's Online Notes] The center of mass is affected by it's shape, it's density and gravity. The center of gravity is the same as the center of mass if gravity is the same everywhere in the region. This is so common that many people will use the terms center of mass and center of gravity interchangably. The centroid is the same as center of mass if the density is the same everywhere in the region. Again, this is extremely common, so some people just talk about the centroid.

Before we go on, let's watch a video discussing the center of gravity for a plate-like object. We usually call this type of object a planar lamina.

### Michel vanBiezen - Mechanical Engineering: Centroids & Center of Gravity - What is Center of Gravity?

video by Michel vanBiezen

Here is a related video discussing centroid and center of gravity. His figure actually has a thickness but since it is the same thickness everywhere, it doesn't come into play.

### Michel vanBiezen - Mechanical Engineering: Centroids & Center of Gravity - Centroids

video by Michel vanBiezen

Okay, now we are ready to discuss the moments of a plane area and a wire or line. Here is a great video.

### Michel vanBiezen - Mechanical Engineering: Centroids & Center of Gravity - 1st Moments of Areas and Lines

video by Michel vanBiezen

Table of Equations

The moment and center of mass equations for a planar lamina are given in Table 1. In this table, ρ is the density.

Table 1 - Planar Lamina Moments and Center of Mass Equations

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

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

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]

video by David Lippman

Okay, with all that information you should be able to succesfully work these practice problems.

Practice - Planar Lamina

Unless otherwise instructed, solve these problems giving your answers in exact form.

Find the centroid of the region bounded by $$y = \sqrt{x}$$, $$y = 0$$ and $$x = 4$$. Also, find the moments $$M_x$$ and $$M_y$$ for the planar laminar of uniform density ρ.

Problem Statement

Find the centroid of the region bounded by $$y = \sqrt{x}$$, $$y = 0$$ and $$x = 4$$. Also, find the moments $$M_x$$ and $$M_y$$ for the planar laminar of uniform density ρ.

$$M_x = 4\rho$$, $$M_y = 64\rho/5$$, centroid = $$(12/5, 3/4)$$

Problem Statement

Find the centroid of the region bounded by $$y = \sqrt{x}$$, $$y = 0$$ and $$x = 4$$. Also, find the moments $$M_x$$ and $$M_y$$ for the planar laminar of uniform density ρ.

Solution

### 2445 video

$$M_x = 4\rho$$, $$M_y = 64\rho/5$$, centroid = $$(12/5, 3/4)$$

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Find the centroid of the region bounded by $$y = \sqrt{x}$$ and $$y = x$$.

Problem Statement

Find the centroid of the region bounded by $$y = \sqrt{x}$$ and $$y = x$$.

Solution

### 3595 video

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Find the centroid of the region bounded by $$y = 4-x^2$$ and $$y = x+2$$.

Problem Statement

Find the centroid of the region bounded by $$y = 4-x^2$$ and $$y = x+2$$.

Solution

### 3596 video

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Find the centroid of the region bounded by $$y = x^2$$ and $$y = 8-x^2$$.

Problem Statement

Find the centroid of the region bounded by $$y = x^2$$ and $$y = 8-x^2$$.

Solution

This problem is solved in two consecutive videos.

### 3566 video

video by PatrickJMT

### 3566 video

video by PatrickJMT

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Find the centroid of the region bounded by $$y = 1-x^2$$ in the first quadrant.

Problem Statement

Find the centroid of the region bounded by $$y = 1-x^2$$ in the first quadrant.

Solution

### 3567 video

video by blackpenredpen

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Find the center of gravity of part of a circle that lies in the first quadrant with radius $$R$$.

Problem Statement

Find the center of gravity of part of a circle that lies in the first quadrant with radius $$R$$.

Hint

Because of the symmetry of the area, $$\bar{x} = \bar{y}$$.

Problem Statement

Find the center of gravity of part of a circle that lies in the first quadrant with radius $$R$$.

Hint

Because of the symmetry of the area, $$\bar{x} = \bar{y}$$.

Solution

### 3568 video

video by Michel vanBiezen

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Find the center of gravity of a semi-circle that lies above the x-axis with radius $$R$$.

Problem Statement

Find the center of gravity of a semi-circle that lies above the x-axis with radius $$R$$.

Hint

Because of the symmetry of the area, $$\bar{x} = 0$$.

Problem Statement

Find the center of gravity of a semi-circle that lies above the x-axis with radius $$R$$.

Hint

Because of the symmetry of the area, $$\bar{x} = 0$$.

Solution

### 3569 video

video by Michel vanBiezen

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Find the center of gravity of a triangle with height $$h$$ and width $$b$$. Draw the triangle so that the base is on the x-axis and y-axis cuts the triangle in half, going through the third point, which means that the other two sides are equal.

Problem Statement

Find the center of gravity of a triangle with height $$h$$ and width $$b$$. Draw the triangle so that the base is on the x-axis and y-axis cuts the triangle in half, going through the third point, which means that the other two sides are equal.

Hint

Because of the symmetry of the area, $$\bar{x} = 0$$.

Problem Statement

Find the center of gravity of a triangle with height $$h$$ and width $$b$$. Draw the triangle so that the base is on the x-axis and y-axis cuts the triangle in half, going through the third point, which means that the other two sides are equal.

Hint

Because of the symmetry of the area, $$\bar{x} = 0$$.

Solution

### 3570 video

video by Michel vanBiezen

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Find the center of gravity of the top half of an ellipse, centered at the origin. The equation of the ellipse is $$\displaystyle{ \frac{x^2}{a} + \frac{x^2}{b} = 1 }$$ with $$a$$ along the x-axis.

Problem Statement

Find the center of gravity of the top half of an ellipse, centered at the origin. The equation of the ellipse is $$\displaystyle{ \frac{x^2}{a} + \frac{x^2}{b} = 1 }$$ with $$a$$ along the x-axis.

Solution

### 3571 video

video by Michel vanBiezen

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Find the center of gravity of the parabolic area $$y = kx^2$$ with width $$2a$$ and height $$h$$.

Problem Statement

Find the center of gravity of the parabolic area $$y = kx^2$$ with width $$2a$$ and height $$h$$.

Hint

$$k = h/a^2$$

Problem Statement

Find the center of gravity of the parabolic area $$y = kx^2$$ with width $$2a$$ and height $$h$$.

Hint

$$k = h/a^2$$

Solution

### 3572 video

video by Michel vanBiezen

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Find the center of gravity of the area bounded by $$y = kx^2$$, $$y = 0$$ and $$x = a$$.

Problem Statement

Find the center of gravity of the area bounded by $$y = kx^2$$, $$y = 0$$ and $$x = a$$.

Hint

In this video solution, he labels the height of the area $$h$$ and gives his answer in terms of $$h$$. However, $$h=ka^2$$.

Problem Statement

Find the center of gravity of the area bounded by $$y = kx^2$$, $$y = 0$$ and $$x = a$$.

Hint

In this video solution, he labels the height of the area $$h$$ and gives his answer in terms of $$h$$. However, $$h=ka^2$$.

Solution

### 3573 video

video by Michel vanBiezen

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Find the center of gravity of the area bounded by $$y = kx^n$$, $$y = 0$$ and $$x = a$$.

Problem Statement

Find the center of gravity of the area bounded by $$y = kx^n$$, $$y = 0$$ and $$x = a$$.

Hint

In this video solution, he labels the height of the area $$h$$ and gives his answer in terms of $$h$$. However, $$h=ka^n$$.

Problem Statement

Find the center of gravity of the area bounded by $$y = kx^n$$, $$y = 0$$ and $$x = a$$.

Hint

In this video solution, he labels the height of the area $$h$$ and gives his answer in terms of $$h$$. However, $$h=ka^n$$.

Solution

This problem is solved in two consecutive videos.

### 3574 video

video by Michel vanBiezen

### 3574 video

video by Michel vanBiezen

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Find the center of gravity of a section of a circle with radius $$R$$ and angle $$2\alpha$$. Center the circle so that the origin is at the center and the positive x-axis cuts the sector into two equal sections, $$\alpha$$ above the x-axis and $$\alpha$$ below the x-axis.

Problem Statement

Find the center of gravity of a section of a circle with radius $$R$$ and angle $$2\alpha$$. Center the circle so that the origin is at the center and the positive x-axis cuts the sector into two equal sections, $$\alpha$$ above the x-axis and $$\alpha$$ below the x-axis.

Hint

Because of symmetry, $$\bar{y} = 0$$

Problem Statement

Find the center of gravity of a section of a circle with radius $$R$$ and angle $$2\alpha$$. Center the circle so that the origin is at the center and the positive x-axis cuts the sector into two equal sections, $$\alpha$$ above the x-axis and $$\alpha$$ below the x-axis.

Hint

Because of symmetry, $$\bar{y} = 0$$

Solution

### 3575 video

video by Michel vanBiezen

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Find the center of mass of the region bounded by the triangle in the figure.

Problem Statement

Find the center of mass of the region bounded by the triangle in the figure.

Solution

This problem is solved in two consecutive videos.

### 3597 video

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Centroid of Composite Shapes

Similar to how you found the centroid of individual points, we can find the centroid of composite shapes. The equations are

$$\displaystyle{ \bar{x} = \frac{\sum{ x_iA_i }}{\sum{A_i}} }$$

$$\displaystyle{ \bar{y} = \frac{\sum{ y_iA_i }}{\sum{A_i}} }$$

These equations may look complicated but they are quite understandable once you know what all the symbols mean. Here is a video that goes through an example and explains how this works in detail.

### Engineer4Free - How to find the centroid of simple composite shapes [8min-54sec]

video by Engineer4Free

Okay, that ought to be enough to get you started on these practice problems.

Practice - Composite Shapes

Unless otherwise instructed, solve these problems giving your answers in exact form.

Find the center of gravity of a composite plate as shown in this figure with units in meters.

Problem Statement

Find the center of gravity of a composite plate as shown in this figure with units in meters.

Solution

### 3578 video

video by Michel vanBiezen

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Find the center of gravity of a composite plate as shown in this figure with units in meters.

Problem Statement

Find the center of gravity of a composite plate as shown in this figure with units in meters.

Solution

### 3579 video

video by Michel vanBiezen

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Find the center of gravity of a composite plate as shown in this figure with units in centimeters. The full circle on the right is a hole with center at $$(22,0)$$

Problem Statement

Find the center of gravity of a composite plate as shown in this figure with units in centimeters. The full circle on the right is a hole with center at $$(22,0)$$

Hint

Of course, because of symmetry, $$\bar{y} = 0$$

Problem Statement

Find the center of gravity of a composite plate as shown in this figure with units in centimeters. The full circle on the right is a hole with center at $$(22,0)$$

Hint

Of course, because of symmetry, $$\bar{y} = 0$$

Solution

### 3580 video

video by Michel vanBiezen

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Find the center of gravity of a composite plate as shown in this figure with units in centimeters. The full circle is a hole.

Problem Statement

Find the center of gravity of a composite plate as shown in this figure with units in centimeters. The full circle is a hole.

Solution

### 3581 video

video by Michel vanBiezen

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Calculate the first moment about the y-axis of the composite plate shown in the figure.

Problem Statement

Calculate the first moment about the y-axis of the composite plate shown in the figure.

Solution

### 3584 video

video by Michel vanBiezen

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Calculate the first moment about the y-axis of the composite plate shown in the figure. The circle is a hole.

Problem Statement

Calculate the first moment about the y-axis of the composite plate shown in the figure. The circle is a hole.

Solution

### 3585 video

video by Michel vanBiezen

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Centroid of Composite Wires

For composite wires, the calculations are very similar but instead of areas like we did in the previous section, we use lengths. If you understood those equations, these should be fairly intuitive.

$$\displaystyle{ \bar{x} = \frac{\sum{ x_iL_i }}{\sum{L_i}} }$$

$$\displaystyle{ \bar{y} = \frac{\sum{ y_iL_i }}{\sum{L_i}} }$$

Here are some practice problems to test your understanding.

Practice - Composite Wires

Unless otherwise instructed, solve these problems giving your answers in exact form.

Find the center of gravity of a composite wire in the shape of a right triangle where the two legs of the right angle are 10cm and 30cm.

Problem Statement

Find the center of gravity of a composite wire in the shape of a right triangle where the two legs of the right angle are 10cm and 30cm.

Solution

### 3582 video

video by Michel vanBiezen

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Considering the wire as shown in this figure, calculate the force labeled in red in the figure. We assume that the wire is strong enough to hold the semi-circular shape.

Problem Statement

Considering the wire as shown in this figure, calculate the force labeled in red in the figure. We assume that the wire is strong enough to hold the semi-circular shape.

Hint

For a semi-circular wire, $$\bar{x} = 2R/\pi$$

Problem Statement

Considering the wire as shown in this figure, calculate the force labeled in red in the figure. We assume that the wire is strong enough to hold the semi-circular shape.

Hint

For a semi-circular wire, $$\bar{x} = 2R/\pi$$

Solution

### 3583 video

video by Michel vanBiezen

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Do you want more on moments and center of mass? Check out the Mechanical Engineering - Moments, Center of Mass page where we discuss the two theorems of Pappus-Guldinus.

You CAN Ace Calculus

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

### Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

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 Definition of Moment Moment About A Point Center of Mass and Center of Gravity Moment About A Line Practice - Discrete Objects Centroids and Center of Mass of a Wire Practice - Center of Mass of a Wire Moment, Center of Mass of a Planar Lamina Practice - Planar Lamina Table of Equations Centroid of Composite Shapes Practice - Composite Shapes Centroid of Composite Wires Practice - Composite Wires

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