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Lesson Video: Center of Gravity of Particles Mathematics

In this video, we will learn how to find the position of the center of gravity of a set of particles arranged in a two-dimensional plane.

18:13

Video Transcript

Center of Gravity of Particles

In this video, we will learn how to find the position of the center of gravity of a set of particles arranged in a two-dimensional plane.

We know that the weight of a particle on Earth acts from its center directly downwards towards the center of the Earth. This makes calculations with particles or individual point masses very easy to work with. But what happens when we have a collection or system of particles and we want to consider all of these particles in one go? This is where the center of gravity comes in handy. The center of gravity is essentially the point in space, which is the weighted average of all the particles in the system.

Let’s now consider two particles in the π‘₯𝑦-plane.

Here, we have our two particles. Let’s suppose that the weight of the particle on the left is 𝑀 one and that the weight of the particle on the right is 𝑀 two. We can also add in the position vectors of 𝑀 one and 𝑀 two. We can label the position vector of 𝑀 one as 𝐫 one, and the position vector of 𝑀 two is 𝐫 two. Now, the position of the center of gravity of the system containing these two weights will lie somewhere along the line joining center of the two particles. Now, if the weight of the two particles are equal β€” that is, that 𝑀 one is equal to 𝑀 two β€” then the center of gravity will be directly in the middle of the line segment joining the center of the two particles. However, in the cases where the two weights are not equal, the center of gravity will shift along this line towards the larger weight.

If 𝑀 one is larger than 𝑀 two, then the center of gravity will shift along this line towards 𝑀 one. Similarly, if 𝑀 two is larger than 𝑀 one, then the center of gravity will shift along this line towards 𝑀 two. In order to actually find the position of this center of gravity, weβ€²ll need to use a formula. This formula tells us that the position of the center of gravity 𝐑 is equal to 𝑀 one 𝐫 one plus 𝑀 two 𝐫 two over 𝑀 one plus 𝑀 two. Now, in most systems where we’re trying to find the position of the center of gravity, weβ€²ll be considering more than two particles. Hence, we will need to expand this formula in order to cater for any number of particles.

Here we have the definition for the center of gravity of a system of particles in two dimensions. Suppose we have a system of 𝑛 particles with weights of 𝑀 one, 𝑀 two, 𝑀 three, and so on up to 𝑀 𝑛, which are located at the position vectors 𝐫 one, 𝐫 two, 𝐫 three, and so on up to 𝐫 𝑛, respectively. Then, the center of gravity of the system can be found using the formula 𝐫 is equal to 𝑀 one 𝐫 one plus 𝑀 two 𝐫 two plus 𝑀 three 𝐫 three and so on all the way up to 𝑀 𝑛 𝐫 𝑛 over 𝑀 one plus 𝑀 two plus 𝑀 three plus all the way up to 𝑀 𝑛.

Using this formula, weβ€²re able to find the center of gravity of a system of particles in two dimensions when we’re given the weights and respective position vectors of each of the particles. We can also rewrite this formula using summation notation. We can see that these two formulas are identical. However, the one using summation notation is a lot more compact. We have that 𝐑 is equal to the sum from 𝑖 equals one to 𝑛 of 𝑀 𝑖 𝐫 𝑖 over the sum from 𝑖 equals one to 𝑛 of 𝑀 𝑖.

Next, we will be rearranging this formula slightly so we can see how we can find the center of gravity of a system when we’re given the masses of the particles instead of the weight.

We know that the weight of any particle is equal to its mass π‘š multiplied by the acceleration due to gravity 𝑔. Now, in a uniform gravitational field, 𝑔 is a constant. On Earth, 𝑔 is roughly equal to 9.8 meters per second squared.

Using this, we can rewrite 𝑀 one as π‘š one 𝑔, 𝑀 two as π‘š two 𝑔, all the way up to 𝑀 𝑛 as π‘š 𝑛 𝑔. Now, if we’re in a uniform gravitational field, the value of 𝑔 in each of these weights will be the same. Therefore, we can say that for any of our weights, 𝑀 𝑖 is equal to π‘š 𝑖 𝑔. We can substitute this equation for 𝑀 𝑖 into our formula for finding the center of gravity. We have that 𝐑 is equal to the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖 𝑔𝐫 𝑖 over the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖 𝑔. Since 𝑔 is just a constant in both the numerator and denominator, we can take it out to the front in both summations. Next, we can cancel this factor of 𝑔. Here, we have found another formula for finding the center of gravity of a system of particles. It tells us that 𝐑 is equal to the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖 𝐫 𝑖 over the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖.

This formula will be useful for when we’re given the masses of the particles rather than their weight. It is worth noting that this formula for the center of gravity, which uses masses instead of weight, is sometimes called the center of mass of a system. However, as we can see from our calculations, it is equivalent to the center of gravity of a system of particles when the particles are in a uniform gravitational field.

In this video, we’ll only be considering systems which meet this condition of a uniform gravitational field. Therefore, weβ€²ll be able to use these formulas interchangeably for finding the center of gravity. Now, when we’re actually finding the center of gravity of a system, we may find it easier to break down this formula even further. We will now consider the π‘₯- and 𝑦-coordinates of 𝐑.

Since 𝐑 is a vector in two dimensions, we can say that 𝐑 is equal to π‘₯𝐒 plus 𝑦𝐣, where 𝐒 and 𝐣 are the fundamental unit vectors. If we were to draw the vector 𝐑 on a grid, then the π‘₯- and 𝑦-coordinates of the endpoint of 𝐑 can be found by looking at the coefficients of 𝐒 and 𝐣. So that is what we have labeled π‘₯ and 𝑦 here. We can also rewrite the position vectors for each of the particles in our system. We’ll have that the vector 𝐫 𝑖 is equal to π‘₯ 𝑖 𝐒 plus 𝑦 𝑖 𝐣.

Now, when we look back at the formula for the center of gravity of a system, we can consider just the horizontal or just the vertical part. Considering just the horizontal parts of these vectors, we can say that the π‘₯-coordinate of the position of the center of gravity is equal to the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖 π‘₯ 𝑖 over the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖. Similarly, the 𝑦-coordinate is equal to the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖 𝑦 𝑖 over the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖.

It’s important to note that these formulas also work when we’re given the weights of the particles instead of their masses. We can obtain these formulas by multiplying the numerator and denominator of the original formulas by 𝑔 and then substituting 𝑀 𝑖 for π‘š 𝑖 𝑔. And this will give us the second line of formulas for finding the center of gravity when weβ€²re given the weights of the particles instead of their masses.

Let’s now move on to look at an example.

In the given figure, three weights of magnitudes two newtons, five newtons, and three newtons are placed on the vertices of an equilateral triangle of side length eight centimeters. Find the center of gravity of the system.

In order to solve this problem, we’re first going to need to find the π‘₯- and the 𝑦-coordinates of each of the weights. We have been given that the weights lie on the vertices of an equilateral triangle. Hence, each of these side lengths will also be eight centimeters. We can already see that the weight of two newtons lies at the origin. So its coordinates will be zero, zero. The weight of five newtons lies on the 𝑦-axis and is eight centimeters from the weight of two newtons. So, its coordinates will be zero, eight.

In order to find the position of the weight of three newtons, we can add in this horizontal line to help us. This line will make a right angle with the vertical axis. Also, since this is an equilateral triangle, this line will bisect the vertical side. So we can see that this length will be four centimeters. So this is the 𝑦-coordinate of the weight of three newtons.

In order to find the π‘₯-coordinate, we’re going to use the Pythagorean theorem. The Pythagorean theorem tells us that this π‘₯-coordinate is equal to the square root of eight squared minus four squared, which is equal to the square root of 64 minus 16, which is also equal to the square root of 48. And this simplifies to give four root three. So the coordinate of our weight of three newtons is four root three four.

We can now write a table which will show us the positions of each of the weights. We’ll also need the formulas for finding the π‘₯- and 𝑦-coordinates of the center of gravity. Let’s start by finding the π‘₯-coordinate. In the numerator, we need to sum the products of the weight and their π‘₯-coordinates. So that’s two multiplied by zero plus five multiplied by zero plus three multiplied by four root three. And then in the denominator, we simply need to sum the weights. This gives us two multiplied by zero plus five multiplied by zero plus three multiplied by four root three over two plus five plus three. The first two terms in the numerator are multiplied by zero. So we can ignore these. Simplifying the rest of the fraction, we find that π‘₯ is equal to 12 root three over 10 centimeters.

Let’s now move on to find the 𝑦-coordinate. We follow a similar process to finding the π‘₯-coordinate, except this time in the numerator, we need to sum the products of the weights with their 𝑦-coordinates. So thatβ€²s two multiplied by zero plus five multiplied by eight plus three multiplied by four. Then, we need to divide this by the sum of the weights, giving us that 𝑦 is equal to two multiplied by zero plus five multiplied by eight plus three multiplied by four over two plus five plus three. Since the first term in the numerator is multiplied by zero, we can again ignore this term. Simplifying the rest of the fraction, and weβ€²re left with 𝑦 is equal to 52 over 10 centimeters. Now that we have both the π‘₯- and 𝑦-coordinates, we found the center of gravity of the system. And it’s located at the coordinates 12 root three over 10, 52 over 10.

We’ll now move on to our second example, where we’ll see how we can find the center of gravity of a system when we’re given the masses of the particles instead of their weights.

Suppose three masses of one kilogram, four kilograms, and six kilograms are located at points whose positions are negative six 𝐒 minus 𝐣, two 𝐒 minus nine 𝐣, and seven 𝐒 plus eight 𝐣. Determine the position vector of the center of gravity of the system of masses.

We can start by forming a table of the three masses and each of their π‘₯- and 𝑦-coordinates of their positions. Next, we need to recall the formulas for finding the π‘₯- and 𝑦-coordinates of the center of gravity of a system in terms of their masses. We have that π‘₯ is equal to the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖 π‘₯ 𝑖 over the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖. And 𝑦 is equal to the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖 𝑦 𝑖 over the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖.

We will start by finding the π‘₯-coordinate of the center of gravity. For the numerator, we need to sum the products of the masses with their π‘₯-coordinates. And in the denominator, we simply have the sum of the masses. This gives us one multiplied by negative six plus four multiplied by two plus six multiplied by seven over one plus four plus six. Simplifying both the numerator and denominator, we have that π‘₯ is equal to 44 over 11. Since 44 is a multiple of 11, we can cancel through by this factor. Hence, we have that π‘₯ is equal to four.

Next, we’ll find the 𝑦-coordinate of the center of gravity. Using the formula, we have that 𝑦 is equal to one multiplied by negative one plus four multiplied by negative nine plus six multiplied by eight over one plus four plus six. Simplifying this fraction, we have that 𝑦 is equal to 11 over 11. And of course, this simplifies to one. So we found that the 𝑦-coordinate of the center of gravity is equal to one.

So we’ve now found the coordinates of the center of gravity. However, the question has asked us to find the position vector of the center of gravity. The π‘₯-coordinate of the center of gravity will be the coefficient of the 𝐒 in the position vector. Similarly, the 𝑦-coordinate will be the coefficient of 𝐣. Here, we reach our solution, which is that the position vector of the center of gravity of the system of masses is four 𝐒 plus 𝐣.

In the next example, we’ll see how the center of gravity can be used to find a missing mass in a system of masses.

The points zero, six; zero, nine; and zero, four on the 𝑦-axis are occupied by three solid masses of nine kilograms, six kilograms, and 𝑀 kilograms, respectively. Determine the value of 𝑀 given the center of gravity of the system is at the point zero, seven.

The first thing that we notice about this question is that all of the masses along with the center of mass are located on the 𝑦-axis. Hence, each of the π‘₯-coordinates are equal to zero. If we were to try and perform any calculations with the π‘₯-coordinate, we would end up with every term being multiplied by zero. And so, this would not tell us anything useful. Hence, we should focus on the 𝑦-coordinates here. Let’s recall the formula for finding the 𝑦-coordinate of the center of gravity of a system of masses. We have that 𝑦 is equal to the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖 𝑦 𝑖 over the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖. In order to make our lives a little easier, let’s form a table consisting of the masses and their relative 𝑦-coordinates.

Now, we are ready to use the formula. We know that the 𝑦 in our formula is the 𝑦-coordinate of the center of gravity, which we’ve been given as seven. Using our formula, we have that seven is equal to nine multiplied by six plus six multiplied by nine plus four multiplied by 𝑀 over nine plus six plus 𝑀. Simplifying both the numerator and denominator, we have that seven is equal to 108 plus four 𝑀 over 15 plus 𝑀. Next, we can multiply through on the left and right by 15 plus 𝑀. We have that seven multiplied by 15 plus 𝑀 is equal to 108 plus four 𝑀.

Next, we can expand the parentheses on the left-hand side, giving us 105 plus seven 𝑀 is equal to 108 plus four 𝑀. Next, we can subtract 105 from both sides and four 𝑀 from both sides. We are left with three 𝑀 is equal to three. Finally, we divide both sides by three to find our solution, which is that 𝑀 is equal to one. Therefore, the mass of our final solid is one kilogram.

For the final example of this video, we’ll see how we can find the center of gravity of a system where the masses are placed on the vertices of a square.

A square 𝐴𝐡𝐢𝐷 has side length 𝐿. Three masses of 610 grams are placed at 𝐴, 𝐡, and 𝐷. Find the coordinates of the center of gravity of the system.

In order to solve this problem, weβ€²ll need to use the formulas for finding the π‘₯- and 𝑦-coordinates of the center of gravity of a system of masses. We have that the π‘₯-coordinate of the center of gravity is equal to the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖 π‘₯ 𝑖 over the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖. And the 𝑦-coordinate is equal to the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖 𝑦 𝑖 over the sum from 𝑖 equals one to 𝑛 of π‘š 𝑖. We can now form a table to show the masses and their corresponding coordinates. Noting that the side length of the square is 𝐿, we have that the coordinates of the masses are zero, zero at 𝐴; 𝐿, zero at 𝐡; and zero, 𝐿 at 𝐷.

We are now ready to substitute these values into our formulas to find the coordinates of the center of gravity. For the π‘₯-coordinate, we have that π‘₯ is equal to 610 multiplied by zero plus 610 multiplied by 𝐿 plus 610 multiplied by zero over 610 plus 610 plus 610. Now, the first and last term in the numerator are multiplied by zero. So we can ignore these. Simplifying what we have left, we have 610𝐿 over three multiplied by 610. So we can cancel this factor of 610. Here, we have found the π‘₯-coordinate, which is 𝐿 over three.

Next, we can find the 𝑦-coordinate of the center of gravity in a similar fashion. We have that the 𝑦-coordinate is equal to the sum of the masses multiplied by their respective 𝑦-coordinates divided by the sum of the masses. We have that 𝑦 is equal to 610 multiplied by zero plus 610 multiplied by zero plus 610 multiplied by 𝐿 divided by 610 plus 610 plus 610. Since the first two terms in the numerator are multiplied by zero, we can ignore them. We can then simplify this to 610𝐿 over three multiplied by 610. And we can simplify it further by canceling the factor of 610. So now, we have found our 𝑦-coordinate of the center of gravity, which is 𝐿 over three. Combining this with our π‘₯-coordinate, and we have found the coordinates of the center of gravity of the system. And this is at 𝐿 over three, 𝐿 over three.

We have now covered a variety of examples. Let’s recap some key points of the video.

Key Points

The center of gravity of a system of particles is the average position of the particles weighted according to their weights. We can find the center of gravity of a system using 𝐑 is equal to the sum from 𝑖 equals one to 𝑛 of 𝑀 𝑖 𝐫 𝑖 over the sum from 𝑖 equals one to 𝑛 of 𝑀 𝑖. To find the π‘₯- and 𝑦-coordinates of the center of gravity, we can use π‘₯ is equal to the sum from 𝑖 equals one to 𝑛 of 𝑀 𝑖 π‘₯ 𝑖 over the sum from 𝑖 equals one to 𝑛 of 𝑀 𝑖. And 𝑦 is equal to the sum from 𝑖 equals one to 𝑛 of 𝑀 𝑖 𝑦 𝑖 over the sum from 𝑖 equals one to 𝑛 of 𝑀 𝑖.

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