Lesson Explainer: Center of Mass of Particles

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

Often in mechanics, we consider the motion of an object as though that object were a point massโ€”an object that has mass, but no length, width, or height. Even when we solve problems involving real-world objects such as tennis balls or boxes, we often model them as point masses. Doing this makes solving a lot of problems simpler.

However, real-world objects do have length, width, and height, and the shape of an object can affect its motion. But, depending on the scenario, we can often still apply many of the formulae and methods we use for point masses to more complex objects. We can do this because, for many problems, we can model a complex object as though it were a point mass at what is called its center of mass, which is also sometimes called its center of gravity.

The center of mass of a set of particles can be thought of as the average of the positions of all of the particles weighted by their masses. So, for example, imagine two particles with position vectors โƒ‘๐‘Ÿ๏Šง and โƒ‘๐‘Ÿ๏Šจ and masses ๐‘š๏Šง and ๐‘š๏Šจ, as shown in the diagram of a two-dimensional system.

The position vector of the center of mass of the two particles, โƒ‘๐‘…, is given by โƒ‘๐‘…=๐‘šโƒ‘๐‘Ÿ+๐‘šโƒ‘๐‘Ÿ๐‘š+๐‘š.๏Šง๏Šง๏Šจ๏Šจ๏Šง๏Šจ

Notice that this is different from the geometric center of the particles. The geometric center of the particles would just be equal to โƒ‘๐‘Ÿ+โƒ‘๐‘Ÿ2๏Šง๏Šจ. The center of mass would only be in the same position as the geometric center in the special case where all of the particles have the same mass.

The center of mass is weighted by the masses of the particles. This means that if the mass of the first particle, ๐‘š๏Šง, were less than that of the second particle, the center of mass would be closer to the second particle than the first, which is the case in the diagram.

If a third particle were added to the system, with a position vector โƒ‘๐‘Ÿ๏Šฉ and a mass ๐‘š๏Šฉ, we would have to add a third term to the numerator of the fraction that is the product of the mass and the position vector. The denominator of the fraction is just the total mass of all of the particles, so this formula would become โƒ‘๐‘…=๐‘šโƒ‘๐‘Ÿ+๐‘šโƒ‘๐‘Ÿ+๐‘šโƒ‘๐‘Ÿ๐‘š+๐‘š+๐‘š.๏Šง๏Šง๏Šจ๏Šจ๏Šฉ๏Šฉ๏Šง๏Šจ๏Šฉ

If we call the total mass ๐‘€, we can write this as โƒ‘๐‘…=1๐‘€๏€น๐‘šโƒ‘๐‘Ÿ+๐‘šโƒ‘๐‘Ÿ+๐‘šโƒ‘๐‘Ÿ๏….๏Šง๏Šง๏Šจ๏Šจ๏Šฉ๏Šฉ

We can see a pattern forming here. As we add more particles to the system, we keep adding terms of the mass of each particle times its position vector to the bracketed expression.

Definition: Center of Mass of a System of Particles

For a system of ๐‘› particles, where the ๐‘–th particle in the system has position vector โƒ‘๐‘Ÿ๏ƒ and mass ๐‘š๏ƒ, the position vector of the center of mass of the system, โƒ‘๐‘…, is given by โƒ‘๐‘…=1๐‘€๏„š๐‘šโƒ‘๐‘Ÿ,๏Š๏ƒ๏Šฒ๏Šง๏ƒ๏ƒ where ๐‘€ is the total mass of all of the particles.

Let us now consider the center of mass of a rigid body about the origin point of a coordinate system. Suppose we are given the weights of the particles forming the rigid body, which are ๐‘Š,๐‘Š,๐‘Š,โ‹ฏ,๐‘Š๏Šง๏Šจ๏Šฉ๏Š, situated at the position vectors โƒ‘๐‘Ÿ,โƒ‘๐‘Ÿ,โƒ‘๐‘Ÿ,โ‹ฏ,๏ƒŸ๐‘Ÿ๏Šง๏Šจ๏Šฉ๏Š, respectively, relative to the origin. We can find the position of the center of mass, โƒ‘๐‘…, relative to the origin using the formula โƒ‘๐‘…=๐‘Šโƒ‘๐‘Ÿ+๐‘Šโƒ‘๐‘Ÿ+๐‘Šโƒ‘๐‘Ÿ+โ‹ฏ+๐‘Š๏ƒŸ๐‘Ÿ๐‘Š+๐‘Š+๐‘Š+โ‹ฏ+๐‘Š,๏Šง๏Šง๏Šจ๏Šจ๏Šฉ๏Šฉ๏Š๏Š๏Šง๏Šจ๏Šฉ๏Š which can also be rewritten using summation notation as โƒ‘๐‘…=โˆ‘๐‘Šโƒ‘๐‘Ÿโˆ‘๐‘Š.๏Š๏ƒ๏Šฒ๏Šง๏ƒ๏ƒ๏Š๏ƒ๏Šฒ๏Šง๏ƒ

We can see how this formula links back to the original formula involving just the masses. We know that we can write each of the weights as the product of their mass and the gravitational constant, ๐‘”: ๐‘Š=๐‘š๐‘”,๐‘Š=๐‘š๐‘”,๐‘Š=๐‘š๐‘”,โ‹ฎ๐‘Š=๐‘š๐‘”.๏Šง๏Šง๏Šจ๏Šจ๏Šฉ๏Šฉ๏Š๏Š

If we were to substitute each of these into the formula we have just found for finding the position of the center of mass, then each of the terms would be a multiple of ๐‘”. If we were to cancel this factor, then we would end up with โƒ‘๐‘…=๐‘šโƒ‘๐‘Ÿ+๐‘šโƒ‘๐‘Ÿ+๐‘šโƒ‘๐‘Ÿ+โ‹ฏ+๐‘š๏ƒŸ๐‘Ÿ๐‘š+๐‘š+๐‘š+โ‹ฏ+๐‘š,๏Šง๏Šง๏Šจ๏Šจ๏Šฉ๏Šฉ๏Š๏Š๏Šง๏Šจ๏Šฉ๏Š which is the same as the original formula we found. We can now consider ๐‘ฅ- and ๐‘ฆ-components of the position of the center of mass. We can let ๐‘ฅ be the horizontal component of โƒ‘๐‘… and ๐‘ฆ be the vertical component of โƒ‘๐‘…. Then, if ๐‘ฅ,๐‘ฅ,๐‘ฅ,โ‹ฏ,๐‘ฅ๏Šง๏Šจ๏Šฉ๏Š are the horizontal components and ๐‘ฆ,๐‘ฆ,๐‘ฆ,โ‹ฏ,๐‘ฆ๏Šง๏Šจ๏Šฉ๏Š are the vertical components of โƒ‘๐‘Ÿ,โƒ‘๐‘Ÿ,โƒ‘๐‘Ÿ,โ‹ฏ,๏ƒŸ๐‘Ÿ๏Šง๏Šจ๏Šฉ๏Š, respectively, we can form the following two formulas by equating the horizontal and vertical components of our vector for the center of mass: ๐‘ฅ=๐‘š๐‘ฅ+๐‘š๐‘ฅ+๐‘š๐‘ฅ+โ‹ฏ+๐‘š๐‘ฅ๐‘š+๐‘š+๐‘š+โ‹ฏ+๐‘š,๐‘ฆ=๐‘š๐‘ฆ+๐‘š๐‘ฆ+๐‘š๐‘ฆ+โ‹ฏ+๐‘š๐‘ฆ๐‘š+๐‘š+๐‘š+โ‹ฏ+๐‘š.๏Šง๏Šง๏Šจ๏Šจ๏Šฉ๏Šฉ๏Š๏Š๏Šง๏Šจ๏Šฉ๏Š๏Šง๏Šง๏Šจ๏Šจ๏Šฉ๏Šฉ๏Š๏Š๏Šง๏Šจ๏Šฉ๏Š

This formula would also work if we were given the weights of the particles instead of the mass, since if we multiply each term in both of the fractions by ๐‘”, then can rewrite each ๐‘š๐‘”๏ƒ as ๐‘Š๏ƒ.

If you are presented with a problem where you are given either the coordinates, or the position vectors for a system of particles, and you are asked to find the center of mass of the system, you can solve this directly using the formula, or it can be helpful to present the information in a table to help you form the equations that you need to solve.

For example, consider particles ๐ด and ๐ต whose masses are 3 kg and 4 kg, respectively, and whose position vectors are (3,4) and (2,5) respectively. We can present this information in a table as follows.

Mass347
๐‘ฅ-coordinates32๐‘ฅ
๐‘ฆ-coordinates45๐‘ฆ

Here, 7 is the total mass of the system and (๐‘ฅ,๐‘ฆ) represents the position vector of the center of mass of the system. From the table, we can then form two equations allowing us to calculate the coordinates for the center of mass: 3ร—3+4ร—2=7๐‘ฅ,3ร—4+4ร—5=7๐‘ฆ.

Solving these equations gets us to a position vector of ๏€ผ177,327๏ˆ for the center of mass of the system. In this explainer, we will predominantly solve the examples directly using the formula.

We will now look at an example of how we can use these formulas to find the position of the center of mass of a system in two dimensions.

Example 1: Finding the Center of Mass of a Two-Dimensional System

In the given figure, three weights of magnitudes 2 N, 5 N, and 3 N are placed on the vertices of an equilateral triangle of side length 8 cm.

Find the center of mass of the system.

Answer

We can start by putting the data for the coordinates of each of the weights in a table.

Weight2 N5 N3 N
๐‘ฅ-Coordinate004โˆš3
๐‘ฆ-Coordinate084

In order to find the ๐‘ฅ-coordinate of the center of mass, we can use the formula ๐‘ฅ=๐‘Š๐‘ฅ+๐‘Š๐‘ฅ+๐‘Š๐‘ฅ+โ‹ฏ+๐‘Š๐‘ฅ๐‘Š+๐‘Š+๐‘Š+โ‹ฏ+๐‘Š.๏Šง๏Šง๏Šจ๏Šจ๏Šฉ๏Šฉ๏Š๏Š๏Šง๏Šจ๏Šฉ๏Š

Substituting our values into this formula, we have ๐‘ฅ=2ร—0+5ร—0+3ร—4โˆš32+5+3=6โˆš35.cm

We can now consider the ๐‘ฆ-coordinate, using a similar formula: ๐‘ฆ=๐‘Š๐‘ฆ+๐‘Š๐‘ฆ+๐‘Š๐‘ฆ+โ‹ฏ+๐‘Š๐‘ฆ๐‘Š+๐‘Š+๐‘Š+โ‹ฏ+๐‘Š.๏Šง๏Šง๏Šจ๏Šจ๏Šฉ๏Šฉ๏Š๏Š๏Šง๏Šจ๏Šฉ๏Š

Substituting from the table, we have ๐‘ฆ=2ร—0+5ร—8+3ร—42+5+3=265.cm

Hence the position of the center of mass of the system is ๏€ฟ6โˆš35,265๏‹.

Now, let us look at another example.

Example 2: Determining the Position Vector of the Center of Mass of Three Discrete Masses

Suppose three masses of 1 kg, 4 kg, and 6 kg are located at points whose position vectors are ๏€บโˆ’6โƒ‘๐‘–โˆ’โƒ‘๐‘—๏†, ๏€บ2โƒ‘๐‘–โˆ’9โƒ‘๐‘—๏†, and ๏€บ7โƒ‘๐‘–+8โƒ‘๐‘—๏†. Determine the position vector of the center of mass for this system of masses.

Answer

We can use the formula โƒ‘๐‘…=1๐‘€๏„š๐‘šโƒ‘๐‘Ÿ๏Š๏ƒ๏Šฒ๏Šง๏ƒ๏ƒ to find the position vector for the center of mass of the system, โƒ‘๐‘…, where ๐‘€ is the total mass of the system, ๐‘š๏ƒ is the mass of object ๐‘–, and โƒ‘๐‘Ÿ๏ƒ is the position vector of object ๐‘–. ๐‘€ is the total mass, so ๐‘€=1+4+6๐‘€=11.kgkgkgkg

Now, let us substitute the values for the masses and position vectors of the objects into the above formula: โƒ‘๐‘…=111๏€บ1ร—๏€บโˆ’6โƒ‘๐‘–โˆ’โƒ‘๐‘—๏†+4ร—๏€บ2โƒ‘๐‘–โˆ’9โƒ‘๐‘—๏†+6ร—๏€บ7โƒ‘๐‘–+8โƒ‘๐‘—๏†๏†.

Now, we just need to simplify the expression: โƒ‘๐‘…=111๏€บโˆ’6โƒ‘๐‘–โˆ’โƒ‘๐‘—+8โƒ‘๐‘–โˆ’36โƒ‘๐‘—+42โƒ‘๐‘–+48โƒ‘๐‘—๏†โƒ‘๐‘…=111๏€บ44โƒ‘๐‘–+11โƒ‘๐‘—๏†โƒ‘๐‘…=4โƒ‘๐‘–+โƒ‘๐‘—.

The position vector of the center of mass is 4โƒ‘๐‘–+โƒ‘๐‘—.

With questions involving finding the position vector, or coordinates of the the center of mass of a system, it is often worth considering the geometry of the system prior to finding the solution. For example, in a two-dimensional system, if the system of particles all lie on a horizontal or vertical line, then we know that the center of mass must also lie on this line, and, therefore, we only have one unknown coordinate.

This is demonstrated in our next example.

Example 3: Finding the Position of the Center of Mass for a System of Particles

Four particles of masses 9 kg, 10 kg, 4 kg, and 7 kg are placed on the ๐‘ฅ-axis at the points (4,0), (3,0), (8,0), and (1,0) respectively. What is the position of the center of mass of the four particles?

Answer

Here, we can start by representing the coordinates of each of the particles as vectors and then solving the problem directly using the formula, or we can simplify the problem by first considering its geometry. For completeness, let us demonstrate both methods here.

Method 1

First, we convert the coordinates to vectors: ๏€บ4โƒ‘๐‘–+0โƒ‘๐‘—๏†, ๏€บ3โƒ‘๐‘–+0โƒ‘๐‘—๏†, ๏€บ8โƒ‘๐‘–+0โƒ‘๐‘—๏†, and ๏€บ1โƒ‘๐‘–+0โƒ‘๐‘—๏†.

We can use the formula โƒ‘๐‘…=1๐‘€๏„š๐‘šโƒ‘๐‘Ÿ๏Š๏ƒ๏Šฒ๏Šง๏ƒ๏ƒ to find the center of mass of the system, โƒ‘๐‘…, where ๐‘€ is the total mass of the system, ๐‘š๏ƒ is the mass of object ๐‘–, and โƒ‘๐‘Ÿ๏ƒ is the position vector of object ๐‘–. ๐‘€ is the total mass, so ๐‘€=9+10+4+7๐‘€=30.kgkgkgkgkg

Now, let us substitute the values for the masses and position vectors of the objects into the above formula: โƒ‘๐‘…=130๏€บ9ร—๏€บ4โƒ‘๐‘–+0โƒ‘๐‘—๏†+10ร—๏€บ3โƒ‘๐‘–+0โƒ‘๐‘—๏†+4ร—๏€บ8โƒ‘๐‘–+0โƒ‘๐‘—๏†+7ร—๏€บ1โƒ‘๐‘–+0โƒ‘๐‘—๏†๏†.

Now, we just need to simplify the expression: โƒ‘๐‘…=130๏€บ36โƒ‘๐‘–+30โƒ‘๐‘–+32โƒ‘๐‘–+7โƒ‘๐‘–๏†โƒ‘๐‘…=130๏€บ105โƒ‘๐‘–๏†โƒ‘๐‘…=72โƒ‘๐‘–.

The position vector of the center of mass is 72โƒ‘๐‘–, which is (3.5,0) when written as a set of coordinates.

Method 2

Notice first that all of the points lie on the ๐‘ฅ-axis, and, therefore, the center of mass of the system must also lie on the ๐‘ฅ-axis and must have coordinates (๐‘ฅ,0).

Second, we need to calculate the total mass of the system: ๐‘€=9+10+4+7๐‘€=30.kgkgkgkgkg

Now, we can present the information in a table, as follows.

Mass9104730
๐‘ฅ-coordinates4381๐‘ฅ

From the table, we can form the following equation: 9ร—4+10ร—3+4ร—8+7ร—1=30๐‘ฅ.

Simplifying gives us 36+30+32+7=30๐‘ฅ, which we can solve as follows: 30๐‘ฅ=105๐‘ฅ=10530=3.5.

Therefore, the coordinates of the center of mass are (3.5,0).

Sometimes, we will be asked to calculate an unknown mass in a system given that we know its center of mass. We will demonstrate this in our next example.

Example 4: Finding Unknown Discrete Masses given the Coordinates of Their Center of Mass

The points (0,6), (0,9), and (0,4) on the ๐‘ฆ-axis are occupied by three solids of masses 9 kg, 6 kg, and ๐‘š kg respectively. Determine the value of ๐‘š given the center of mass of the system is at the point (0,7).

Answer

There are two methods we could use to solve this problem. The first is to form and complete a table of the masses in the system and the second is to convert the positions of the masses to vector form and solve using the formula. We will be showing both methods here.

Method 1

We can start by forming a table containing the mass, ๐‘ฅ-coordinate, and ๐‘ฆ-coordinate of each of the solids in the system. This is fairly straightforward since the positions of the masses have been given as coordinates. We will call the solid at (0,6) solid A, the solid at (0,9) solid B, and the solid at (0,4) solid C. The final column of the table contains the information for the system as a whole.

SolidABCSystem
Mass (kg)96๐‘š15+๐‘š
๐‘ฅ-Coordinate0000
๐‘ฆ-Coordinate6947

We know that the sum of the products of the masses and their respective ๐‘ฅ-coordinate is equal to the product of the total mass and the ๐‘ฅ-coordinate of the center of mass. Similarly, the sum of the products of the masses and their respective ๐‘ฆ-coordinate is equal to the product of the total mass and the ๐‘ฆ-coordinate of the center of mass. Therefore, we can now form two equations using the table: 9ร—0+6ร—0+๐‘šร—0=(15+๐‘š)ร—0,9ร—6+6ร—9+๐‘šร—4=(15+๐‘š)ร—7.

In each element of the first equation, we are multiplying by zero, so it will simplify to 0=0. Therefore, only the second of these equations will be useful to us. Let us simplify this equation: 54+54+4๐‘š=105+7๐‘š.

Now, we can move all the multiples of ๐‘š to one side and everything else to the other side as follows: 3=3๐‘š.

Finally, we divide both sides by 3 to reach our solution of ๐‘š=1.kg

Method 2

In this question, we have been given the coordinates of three masses, as well as the coordinates of their center of mass. We can use the formula for the center of mass to relate these quantities, and then rearrange it to make the unknown mass, ๐‘š, the subject.

Let us first write the positions of the three masses as vectors: ๏€บ0โƒ‘๐‘–+6โƒ‘๐‘—๏†, ๏€บ0โƒ‘๐‘–+9โƒ‘๐‘—๏†, and ๏€บ0โƒ‘๐‘–+4โƒ‘๐‘—๏†. The position vector of the center of mass is ๏€บ0โƒ‘๐‘–+7โƒ‘๐‘—๏†.

The formula for the center of mass of a set of objects is โƒ‘๐‘…=1๐‘€๏„š๐‘šโƒ‘๐‘Ÿ.๏Š๏ƒ๏Šฒ๏Šง๏ƒ๏ƒ

Let us substitute the values that we have been given: ๏€บ0โƒ‘๐‘–+7โƒ‘๐‘—๏†=19+6+๐‘š๏€บ9ร—๏€บ0โƒ‘๐‘–+6โƒ‘๐‘—๏†+6ร—๏€บ0โƒ‘๐‘–+9โƒ‘๐‘—๏†+๐‘šร—๏€บ0โƒ‘๐‘–+4โƒ‘๐‘—๏†๏†7โƒ‘๐‘—=115+๐‘š๏€บ54โƒ‘๐‘—+54โƒ‘๐‘—+4๐‘šโƒ‘๐‘—๏†7โƒ‘๐‘—=108โƒ‘๐‘—+4๐‘šโƒ‘๐‘—15+๐‘š.

Now, let us rearrange the equation to make ๐‘š the subject: 105โƒ‘๐‘—+7๐‘šโƒ‘๐‘—=108โƒ‘๐‘—+4๐‘šโƒ‘๐‘—7๐‘šโƒ‘๐‘—=3โƒ‘๐‘—+4๐‘šโƒ‘๐‘—3๐‘šโƒ‘๐‘—=3โƒ‘๐‘—๐‘š=1.

So, the mass of the third object is 1 kg.

In our final examples, let us look at problems where systems are presented geometrically.

Example 5: Finding the Center of Mass of a System of Three Masses Placed on the Sides of a Square

A square ๐ด๐ต๐ถ๐ท has side length ๐ฟ. Three masses of 610 g are placed at ๐ด, ๐ต, and ๐ท. Find the coordinates of the center of mass of the system.

Answer

There are two methods we could use to solve this problem. The first is to form and complete a table of the masses in the system and the second is to convert the positions of the masses to vector form and solve using the formula. We will be showing both methods here.

Method 1

For the first method, we need to find the coordinates of the three masses. We know that they lie on the corners of a square with side length ๐ฟ. From the diagram, we can see that the mass ๐ด is on the origin, the mass ๐ต is on the ๐‘ฅ-axis, and the mass ๐ท is on the ๐‘ฆ-axis. Therefore, we can say that the coordinates of mass ๐ด are (0,0), the coordinates of mass ๐ต are (๐ฟ,0), and the coordinates of mass ๐ท are (0,๐ฟ).

We can now form a table containing the mass, ๐‘ฅ-coordinate, and ๐‘ฆ-coordinate of each of the masses in the system. The final column of the table contains the information for the system as a whole.

Mass๐ด๐ต๐ทSystem
Mass (g)6106106101โ€Žโ€‰โ€Ž830
๐‘ฅ-Coordinate0๐ฟ0๐‘ฅ
๐‘ฆ-Coordinate00๐ฟ๐‘ฆ

Now, we can form two equations, one relating the masses to the ๐‘ฅ-coordinate and the other relating the masses to the ๐‘ฆ-coordinates, as follows: 610ร—0+610ร—๐ฟ+610ร—0=1830ร—๐‘ฅ,610ร—0+610ร—0+610ร—๐ฟ=1830ร—๐‘ฆ.

Simplifying these equations, we obtain 610๐ฟ=1830๐‘ฅ,610๐ฟ=1830๐‘ฆ.

Finally, if we divide by 610, we will see that ๐‘ฅ=๐ฟ3,๐‘ฆ=๐ฟ3.

Hence the coordinates of the center of mass are ๏€ผ๐ฟ3,๐ฟ3๏ˆ.

Method 2

In this question, we have been given the masses of three objects and their positions in terms of a constant, ๐ฟ. This means that when we calculate the center of mass of the system, it is also going to be in terms of ๐ฟ.

We can represent the positions of the objects at ๐ด, ๐ต, and ๐ท as vectors. The object at ๐ด has a position vector of ๏€บ0โƒ‘๐‘–+0โƒ‘๐‘—๏†; the object at ๐ต has a position vector of ๏€บ๐ฟโƒ‘๐‘–+0โƒ‘๐‘—๏†; the object at ๐ท has a position vector of ๏€บ0โƒ‘๐‘–+๐ฟโƒ‘๐‘—๏†.

We can use the formula โƒ‘๐‘…=1๐‘€๏„š๐‘šโƒ‘๐‘Ÿ๏Š๏ƒ๏Šฒ๏Šง๏ƒ๏ƒ to find the center of mass of the system. Let us substitute in the values: โƒ‘๐‘…=1610+610+610๏€บ610ร—๏€บ0โƒ‘๐‘–+0โƒ‘๐‘—๏†+610ร—๏€บ๐ฟโƒ‘๐‘–+0โƒ‘๐‘—๏†+610ร—๏€บ0โƒ‘๐‘–+๐ฟโƒ‘๐‘—๏†๏†.

A factor of 610 can be taken out of the numerator and the denominator on the right-hand side: โƒ‘๐‘…=11+1+1๏€บ๏€บ0โƒ‘๐‘–+0โƒ‘๐‘—๏†+๏€บ๐ฟโƒ‘๐‘–+0โƒ‘๐‘—๏†+๏€บ0โƒ‘๐‘–+๐ฟโƒ‘๐‘—๏†๏†โƒ‘๐‘…=13๏€บ๐ฟโƒ‘๐‘–+๐ฟโƒ‘๐‘—๏†โƒ‘๐‘…=๐ฟ3๏€บ1โƒ‘๐‘–+1โƒ‘๐‘—๏†.

This is the position vector of the center of mass. We can also write it as a set of coordinates: ๏€ผ๐ฟ3,๐ฟ3๏ˆ.

Example 6: Finding the Center of Gravity of Three Equal Discrete Masses Placed on the Sides of a Triangle

A triangle ๐ด๐ต๐ถ, where ๐ด๐ต=33cm, ๐ต๐ถ=44cm, ๐ถ๐ด=55cm, and ๐ท and ๐ธ are the midpoints of ๐ด๐ต and ๐ด๐ถ, respectively, is located in the first quadrant of a Cartesian plane such that ๐ต is at the origin and the point ๐ถ is on the ๐‘ฅ-axis. Three equal masses are placed at points ๐ต, ๐ท, and ๐ธ. Determine the coordinates of the center of gravity of the system.

Answer

There are two methods we could use to solve this problem. The first is to form and complete a table of the masses in the system and the second is to convert the positions of the masses to vector form and solve using the formula. We will be showing both methods here.

Method 1

Firstly, we need to work out the coordinates of the three masses. Mass ๐ต is on the origin, so its coordinates are (0,0). Mass ๐ต is on the midpoint of ๐ด๐ต, and we know ๐ด๐ต=33cm, so its coordinates are ๏€ผ0,332๏ˆ. It is a little more difficult to find the coordinates of mass ๐ธ. However, since ๐ด๐ต๐ถ is a right triangle and ๐ธ is at the midpoint of ๐ด๐ถ, the hypotenuse, that means its ๐‘ฅ-coordinate will be half the width of the triangle and its ๐‘ฆ-coordinate will be half of the height the triangle. The width of the triangle is 44 cm and the height is 33 cm, so its coordinates are ๏€ผ22,332๏ˆ.

We can now form a table containing the mass, ๐‘ฅ-coordinate, and ๐‘ฆ-coordinate of each of the masses in the system. The final column of the table contains the information for the system as a whole.

Mass๐ต๐ท๐ธSystem
Mass (g)๐‘š๐‘š๐‘š3๐‘š
๐‘ฅ-Coordinate0022๐‘ฅ
๐‘ฆ-Coordinate0332332๐‘ฆ

Now, we can form two equations, one relating the masses to the ๐‘ฅ-coordinate and another relating the masses to the ๐‘ฆ-coordinate, as follows: ๐‘šร—0+๐‘šร—0+๐‘šร—22=3๐‘šร—๐‘ฅ,๐‘šร—0+๐‘šร—332+๐‘šร—332=3๐‘šร—๐‘ฆ.

Since each term is a multiple of ๐‘š, we can cancel out this factor and simplify the terms in the two equations to obtain 22=3๐‘ฅ,33=3๐‘ฆ.

Next, we divide both equations by 3: ๐‘ฅ=223,๐‘ฆ=11.

Hence, we have found the coordinate of the center of mass. Our solution is ๏€ผ223,11๏ˆ.

Method 2

Let us start by finding position vectors for the points ๐ต, ๐ท, and ๐ธ, where the masses are.

๐ต is at the origin, so its position vector is ๏€บ0โƒ‘๐‘–+0โƒ‘๐‘—๏†. ๐ด is 33 cm away from ๐ต along the ๐‘ฆ-axis, so its position vector is ๏€บ0โƒ‘๐‘–+33โƒ‘๐‘—๏†. ๐ท is at the midpoint between ๐ด and ๐ต, so its position vector is ๏€ผ0โƒ‘๐‘–+332โƒ‘๐‘—๏ˆ.

We can find the position vector of ๐ธ by adding the vectors ๏ƒŸ๐ต๐ถ and 12๏ƒ ๐ถ๐ด. ๏ƒ ๐ถ๐ด is equal to ๏ƒŸ๐ถ๐ต+๏ƒ ๐ต๐ด, so the position vector of ๐ธ, equal to ๏ƒŸ๐ต๐ธ, is given by ๏ƒŸ๐ต๐ถ+12๏€บ๏ƒŸ๐ถ๐ต+๏ƒ ๐ต๐ด๏†. Since ๏ƒŸ๐ถ๐ต=โˆ’๏ƒŸ๐ต๐ถ, the position vector of ๐ธ is given by 12๏€บ๏ƒŸ๐ต๐ถ+๏ƒ ๐ต๐ด๏†. ๏ƒŸ๐ต๐ถ is equal to ๏€บ44โƒ‘๐‘–+0โƒ‘๐‘—๏† and ๏ƒ ๐ต๐ด is equal to ๏€บ0โƒ‘๐‘–+33โƒ‘๐‘—๏†, so the position vector of ๐ธ is ๏€ผ442โƒ‘๐‘–+332โƒ‘๐‘—๏ˆ.

We can now use the formula โƒ‘๐‘…=1๐‘€๏„š๐‘šโƒ‘๐‘Ÿ๏Š๏ƒ๏Šฒ๏Šง๏ƒ๏ƒ to find the center of mass of the system in terms of ๐‘š. Let us substitute the values: โƒ‘๐‘…=13๐‘š๏€ผ๐‘š๏€บ0โƒ‘๐‘–+0โƒ‘๐‘—๏†+๐‘š๏€ผ0โƒ‘๐‘–+332โƒ‘๐‘—๏ˆ+๐‘š๏€ผ442โƒ‘๐‘–+332โƒ‘๐‘—๏ˆ๏ˆ=13๏€ผ๏€บ0โƒ‘๐‘–+0โƒ‘๐‘—๏†+๏€ผ0โƒ‘๐‘–+332โƒ‘๐‘—๏ˆ+๏€ผ442โƒ‘๐‘–+332โƒ‘๐‘—๏ˆ๏ˆ=13๏€ผ332โƒ‘๐‘—+442โƒ‘๐‘–+332โƒ‘๐‘—๏ˆ=112โƒ‘๐‘—+223โƒ‘๐‘–+112โƒ‘๐‘—=223โƒ‘๐‘–+11โƒ‘๐‘—.

So, the center of mass of the system is at ๏€ผ223โƒ‘๐‘–+11โƒ‘๐‘—๏ˆ, or ๏€ผ223,11๏ˆ in coordinate form.

Let us finish by recapping some key points.

Key Points

  • For a system of ๐‘› particles, where the ๐‘–th particle in the system has position vector โƒ‘๐‘Ÿ๏ƒ and mass ๐‘š๏ƒ, the position vector of the center of mass of the system, โƒ‘๐‘…, is given by โƒ‘๐‘…=1๐‘€๏„š๐‘šโƒ‘๐‘Ÿ,๏Š๏ƒ๏Šฒ๏Šง๏ƒ๏ƒ where ๐‘€ is the total mass of all of the particles.
  • Always consider the geometry of the system of particles as this can sometimes allow you to simplify the problem. If, for example, all of the particles lie on a horizontal, or vertical line then the center of mass of the system must also lie on the same line.
  • It can be helpful to convert all points and coordinates into position vectors before attempting to find the center of mass.

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