Lesson Explainer: The Negative Mass Method | Nagwa Lesson Explainer: The Negative Mass Method | Nagwa

Lesson Explainer: The Negative Mass Method

In this explainer, we will learn how to find the center of mass (sometimes called center of gravity) of a lamina that contains holes using the negative mass method.

The center of mass of a system of bodies relative to the origin of a coordinate system is determined by the position of the center of mass of each body relative to the origin of the system.

In the case of a simple one-dimensional two-body system, we can define this as follows.

Definition: Position of the Center of Mass of a One-Dimensional Two-Body System

The position of the center of mass of a system of two bodies in one dimension is given by 𝑥=𝑚𝑥+𝑚𝑥𝑚+𝑚, where 𝑚 and 𝑚 are the masses of the bodies and 𝑥 and 𝑥 are the positions of these bodies relative to the origin of the system.

Note that this method can be generalized to consider a system of 𝑛 bodies by adding subsequent terms, as required: 𝑥=𝑚𝑥+𝑚𝑥+𝑚𝑥++𝑚𝑥𝑚+𝑚+𝑚++𝑚.

We can further generalize our method to work in 2 or more dimensions by specifying the position of each individual body using a position vector.

If we define the position vector for the center of mass of our system as 𝑟, we reach the following definition.

Definition: Position Vector of the Center of Mass of a System

The position vector of the center of mass of a system of 𝑛 bodies in space is given by 𝑟=𝑚𝑟+𝑚𝑟+𝑚𝑟++𝑚𝑟𝑚+𝑚+𝑚++𝑚, where 𝑚,,𝑚 are the masses of the bodies and 𝑟,,𝑟 are the position vectors of these bodies relative to the origin of the system.

Note that the equation stated above can be expressed in a tidier format using sigma notation: 𝑟=𝑚𝑟𝑚.

In practice, it is common to calculate the center of mass of a system where the positions of objects are defined in an orthogonal coordinate system. This means the vector relationship stated above can be expressed in terms of the base coordinates of the system in order to simplify our calculations.

For this explainer, we will be working in a 2-dimensional Cartesian coordinate system, and therefore we will be using the 𝑥- and 𝑦-directions. The individual 𝑥- and 𝑦-coordinates for the center of mass of our system can hence be found using the following equations: 𝑥=𝑚𝑥𝑚=𝑚𝑥+𝑚𝑥+𝑚𝑥++𝑚𝑥𝑚+𝑚+𝑚++𝑚,𝑦=𝑚𝑦𝑚=𝑚𝑦+𝑚𝑦+𝑚𝑦++𝑚𝑦𝑚+𝑚+𝑚++𝑚.

Note that solving these equations in a 2-dimensional system is equivalent to finding the position vector of the center of mass since 𝑟=𝑥,𝑦.

A typical method you may be familiar with to find the center of mass of a system of bodies is to create a table denoting the masses of the bodies and their positions in both the 𝑥- and 𝑦-directions relative to the origin. This method allows us to display information about our system in an easily viewable format before we delve into solving the equations.

The two-dimensional systems that we will focus on in this explainer will consist entirely of laminas and masses. Let us explore how the methods we have discussed can be applied to such systems.

We begin by considering a square uniform lamina.

In order to find the center of mass of this lamina, we simply need to find the center of the square (since it is a uniform lamina), which we can do by finding where the diagonals of the square intersect.

Now, let us add a mass to one of the corners of the lamina.

This will move the center of mass toward the added mass, along the line joining the center of mass of the lamina and the center of mass of the mass, 𝑀.

Next, let us consider what would happen if we cut a triangle-shaped hole in our original lamina, as shown in the following figure.

One way in which we could find the center of mass of this new lamina is to cut it up into rectangles and triangles, find the center of mass of each of of these shapes, and combine them together to find the center of mass of the lamina.

However, this method could take quite a long time since we need to calculate the centers of mass of many laminas. This is where the negative mass method is useful. We can consider this problem in terms of two laminas: the original square lamina and the triangle lamina that has been removed. Since we have removed the mass from the system by removing the triangle, we can consider this triangular lamina to have “negative mass.”

Now, we know that the center of mass of the square lamina is at a point where the diagonals of the square intersect. Similarly, the center of mass of the triangle is also at its geometric center, which is the point where the lines joining each corner to the midpoint of the side opposite that corner intersect.

The center of mass of the system will lie along the line joining the centers of mass of the triangle and square as shown here.

When we added mass to the lamina, the center of mass of the system moved toward this added mass. However, here we have taken away mass or added “negative mass,” so the center of mass will move away from the removed mass.

It is worth noting that while we say that this removed lamina has “negative mass,” it is physically impossible for a body to have a negative mass. All we are doing is modeling the removal of mass as a “negative mass.”

Let us now look at an example of how the negative mass method can be used.

Example 1: Finding the Center of Mass of a Rectangular Lamina with a Rectangle Removed

Find the coordinates of the center of mass of the following figure, which is drawn on a grid of unit squares.

Answer

To determine the position of the center of mass of the figure, we are obliged to assume that it is uniform. Assuming uniformity, the center of mass of the figure can be determined without calculation by modeling it as a positive mass lamina, with a lamina removed from it. We can then model the removed lamina as a lamina with “negative mass.” The negative mass lamina must be uniform as the positive mass lamina is uniform.

The following figure shows a diagrammatic determination of the position of the center of mass of the positive mass and negative mass laminas, confirming that they coincide.

The coordinates of the center of mass of either lamina, and hence of the figure, are 92,4.

It is useful to note that the position of the center of mass of the figure would still be at this point if the negative mass lamina instead had positive mass, provided the lamina was uniform.

Now let us look at an example of using the negative mass method in which the centers of mass of the positive mass and negative mass laminas are at different positions.

Example 2: Finding the Center of Mass of an Arrow-Shaped Lamina

The diagram shows a uniform lamina 𝐴𝐵𝐶 from which a triangle 𝐺𝐵𝐶 has been cut out. 𝐴𝐵𝐶 was an equilateral triangle with a side length of 93 cm and a center of mass 𝐺. Find the coordinates of the new center of mass of the resulting lamina. Round your answer to two decimal places if necessary.

Answer

Determining the center of mass of the arrow-shaped lamina can be achieved by modeling it as a system consisting of a positive mass lamina corresponding to the triangle 𝐴𝐵𝐶 with the triangle 𝐺𝐶𝐵 removed, which we can be modeled as a “negative mass” lamina. Both laminas are uniform and so their centers of mass are at their geometric centers.

The first thing we can notice about the lamina 𝐴𝐵𝐺𝐶 is that it has a vertical line of symmetry. This means that the 𝑥-coordinate of its center of mass will be on this line. The line of symmetry will be halfway along the base of 𝐴𝐵𝐶, so the 𝑥-coordinate of the center of mass will be 932=46.5.

The geometric center of an equilateral triangle is the point where each of the medians of the triangle intersect, which is one-third of the length of a median from the midpoint of a side of the triangle. The following figure shows the center of mass of the positive mass lamina at 𝑦 and the center of mass of the “negative mass” lamina at 𝑦.

In order to determine the lengths 𝑦 and 𝑦, we first need to know the height of the triangle 𝐴𝐵𝐶, which we can call 𝑎.

Since 𝐴𝐵𝐶 is an equilateral triangle, all of its angles are 60. The vertical line in the diagram bisects the angle at the top of the triangle, so we know that 𝜃=30. Using the trigonometric ratio tanopp.adj.(𝜃)=, we have tan(30)=𝑎.

This can be rearranged to 𝑎=932(30)=9332.tancm

Now we are able to find the center of mass of 𝐴𝐵𝐶 in the 𝑦-direction, since the geometric center of a triangle is 13 of the way along the line connecting the middle of the base and the opposite corner. Hence, we can say that 𝑦=𝑎3=9336.cm

Now, 𝐺 is at the geometric center of 𝐴𝐵𝐶, so the geometric center of 𝐺𝐵𝐶 in the 𝑦-direction can be found to be 𝑦=𝑦3=93318.cm

The question tells us that the lamina is uniform, which means it has an even density across the lamina. Since we are using the negative mass method, this also means that we can also say that the “negative mass” lamina, 𝐺𝐵𝐶, is uniform and that its density is equal but opposite in sign to the density of 𝐴𝐵𝐶. What this means is that the masses of the laminas are proportional to their areas. Hence, we can calculate the relative mass of 𝐴𝐵𝐶 to be 𝑚=12×93×9332=9334.

And the relative mass of 𝐺𝐵𝐶, remembering that it will be a “negative mass,” is given by 𝑚=12×93×9336=93312.

Using these two, we can find the relative mass of the lamina 𝐴𝐵𝐺𝐶 to be 𝑚=933493312=9336.

Using this information, we can form a table consisting of the relative masses of the different laminas and the 𝑦-coordinates of their centers of mass.

Lamina𝐴𝐵𝐶𝐺𝐵𝐶𝐴𝐵𝐺𝐶
Relative Mass9334933129336
𝑦-Coordinate933693318𝑦

Since lamina 𝐴𝐵𝐺𝐶 is made by combining laminas 𝐴𝐵𝐶 and 𝐺𝐵𝐶, we can form the following equation from this table: 9334×933693312×93318=9336×𝑦.

Now, all we need to do is rearrange this equation to solve for 𝑦. 𝑦=××==933493336=6233.cm

We have now found the 𝑥- and 𝑦-coordinates of the center of mass. All we need to do is round the 𝑦-coordinate to 2 decimal places. In doing this, we reach our solution that the center of mass of the lamina is (46.5,35.80).

Now let us look at an example where laminas of different shapes are combined.

Example 3: Finding the Center of Mass of a Lamina with a Circle Removed

A square-shaped uniform lamina 𝐴𝐵𝐶𝐷 has a side length of 28 cm. A circular disk of radius 7 cm was cut out of the lamina such that its center was at a distance of 17 cm from both 𝐴𝐵 and 𝐵𝐶. Determine the coordinates of the center of mass of the resulting lamina. Take 𝜋=227.

Answer

The following figure shows the positions of the centers of mass of the positive mass square lamina and the negative mass circular lamina.

The value of 𝑥 is given by 𝑥=14+1417=11.

The coordinates of the center of mass of the square lamina are (14,14), and the coordinates of the center of mass of the circular lamina are (11,17).

The masses of the laminas are proportional to their areas.

The mass of the square lamina is given by 𝑚=28=784.

The mass of the circular lamina is given by 𝑚=𝜋𝑟.

Using the values for 𝜋 and 𝑟, we have that 𝑚=227×7=154.

At this stage, we could form a table consisting of the masses of the laminas and their centers of mass and use this to solve the question, but we will be using an alternative method. This is using the formula to find the 𝑥- and 𝑦-coordinates of the center of mass.

The formula for the 𝑥-coordinate is 𝑥=𝑚𝑥+𝑚𝑥𝑚+𝑚.

Substituting in the values we have just found, we obtain 𝑥=784×14154×11784154.

We can simplify this to give 𝑥=22115.

Similarly, the 𝑦-coordinate of center of mass of the system is given by 𝑦=784×14154×17784154=19915.

Hence, we have found the center of mass of the system to be 22115,19915.

Adding the position of the center of mass to the system, we can see that it lies on the line that intersects the center of mass of the square and circular laminas. The center of mass of the system is displaced from the center of mass of the square in the direction that points away from the center of mass of the circular lamina (which has a negative mass).

Let us look at an example in which the negative mass method could be used.

Example 4: Finding the Center of Mass of a Lamina with Additional Weights Added

A uniform square-shaped lamina 𝐴𝐵𝐶𝐷 of side length 222 cm has a mass of one kilogram. The midpoints of 𝐴𝐷,𝐴𝐵, and 𝐵𝐶 are denoted by 𝑇, 𝑁, and 𝐾 respectively. The corners 𝑇𝐴𝑁 and 𝑁𝐵𝐾 were folded over so that they lie flat on the surface of the lamina. Bodies of masses 365 g and 294 g were attached to the points 𝑇 and 𝐾 respectively. Find the coordinates of the center of mass of the system rounding your answer to two decimal places if necessary.

Answer

This question can be solved by modeling the system as consisting only of positive mass bodies. In this case, the positive mass values would be those of the rectangle 𝐶𝐷𝑇𝐾, the triangle 𝐾𝑇𝑁, and the added masses at 𝑇 and 𝐾.

Using the negative mass method, the system is modeled as containing positive mass bodies consisting of the added masses at 𝑇 and 𝐾, the square 𝐴𝐵𝐶𝐷, and the triangle 𝐾𝑇𝑁, and also containing “negative mass” bodies consisting of triangles 𝑇𝐴𝑁 and 𝐾𝐵𝑁. We will be using the negative mass method in order to reach a solution.

First, let us determine the 𝑥-coordinate of the center of mass of the system.

The added masses both have 𝑥-coordinates of 111.

The square 𝐴𝐵𝐶𝐷 is uniform and has a side length of 222 cm, so the position of its center of mass is its geometric center, which has an 𝑥-coordinate of 111.

The square lamina has a mass of 1 kilogram, which equals 1‎ ‎000 grams.

The center of mass of the triangle 𝐾𝑇𝑁 has an 𝑥-coordinate equal to the length of 𝐷𝑇 plus one-third of the length of 𝐴𝑇, which is given by 111+1113=148.

The mass of triangle 𝐾𝑇𝑁 is one-quarter of the mass of the square lamina, which is 250 grams.

The center of mass of the negative mass triangles has an 𝑥-coordinate equal to the length of 𝐶𝐵 minus one-third of the length of 𝐾𝐵, which is given by 2221113=185.

The total mass of the negative mass triangles is minus one-quarter of the mass of the square lamina, which is 250 grams. This is necessarily equal to the magnitude of the mass of triangle 𝐾𝑇𝑁, as the total mass of the system is the mass of the square lamina and the added masses, the triangular laminas only representing the mass of the part of the square lamina being redistributed.

Let us first summarize the data we have found in a table.

Lamina or Mass𝐴𝐵𝐶𝐷Mass at 𝑇Mass at 𝐾𝐾𝑇𝑁𝑇𝐴𝑁𝐾𝐵𝑁
Relative Mass1‎ ‎000365294250125125
𝑥-Coordinate111111111148185185

In order to find the 𝑥-coordinate of the system, we can multiply the mass of each body in the system by the 𝑥-coordinate of the relative body’s center of mass and then divide by the sum of the masses. Doing this give us 𝑥=1000×111+365×111+294×111+250×148125×185125×1851000+294+365+250125125, which simplifies to 𝑥=1748991659.

To two decimal places, this is 105.42.

The 𝑦-coordinate of the center of mass of the square lamina is 111, which is also the value for the 𝑦-coordinate of the center of mass of the positive mass triangular lamina.

The 𝑦-coordinates of the centers of mass of the “negative mass” triangular laminas are distributed symmetrically about 𝑦=111, so the center of mass of both of the “negative mass” laminas is also 111.

The 294-gram added mass has a 𝑦-coordinate of 0, while the 365-gram added mass has a 𝑦-coordinate of 222.

Let us summarize these values in a table.

Lamina or Mass𝐴𝐵𝐶𝐷Mass at 𝑇Mass at 𝐾𝐾𝑇𝑁𝑇𝐴𝑁 and 𝐾𝐵𝑁
Relative Mass1‎ ‎000365294250250
𝑦-Coordinate1112220111111

Taking these values into account, we see that 𝑦=1000×111+365×222+294×0+250×111250×1111659, which simplifies to 𝑦=1920301659.

To two decimal places, this is 115.75.

The coordinates of the system are therefore (105.42,115.75).

Let us now consider another similar example.

Example 5: Solving an Applied Problem Involving Center of Mass

A uniform lamina is in the form of a rectangle 𝐴𝐵𝐶𝐷, where 𝐴𝐵=56cm and 𝐵𝐶=35cm. Two points 𝐸 and 𝐹 are on 𝐴𝐵 such that 𝐴𝐸=𝐵𝐹=14cm. The triangle 𝑀𝐸𝐹, where 𝑀 is the center of the rectangle, is cut out of the lamina. Find the coordinate of the center of mass of the resulting lamina. Given that the lamina was freely suspended from 𝐷, find the tangent of the angle that 𝐷𝐴 makes to the vertical, tan𝜃, when the lamina is hanging in its equilibrium position.

Answer

The following figure shows how the lamina can be modeled using only positive mass laminas.

In this case, it is simpler to use the negative mass method as doing so requires only two rather than five laminas.

The following figure shows the positions of the centers of mass of a positive mass rectangle lamina and a negative mass triangular lamina that correspond to the compound lamina.

The length 𝑥 is given by 𝑥=2(2814)=28.cm

The coordinates of the center of mass of the rectangular lamina are 28,352.

The 𝑦-coordinate of the center of mass of the triangular lamina is given by 353=1756.

The coordinates of the center of mass of the triangular lamina are 28,1756.

The mass of the rectangular lamina is proportional to its area. Therefore, we can say that the relative mass of the rectangular lamina is given by 𝑚=35×56=1960.

The relative mass of the triangular lamina is given by 𝑚=282×352=245.

We are now ready to find the 𝑥-coordinate of the center of mass of the system. Substituting the values into the formula, we obtain 𝑥=1960×28245×281960245=28.

The 𝑦-coordinate of the center of mass of the system is given by 𝑦=1960×245×1960245=205800428756×1715.

This fraction can be simplified to 𝑦=956.

The coordinates of the center of mass of the system are, therefore, 28,956.

The 𝑦-coordinate of the center of mass of the system is displaced away from the center of mass of the rectangular lamina by 53.

Suspending the compound lamina from point 𝐷 will result in the lamina rotating clockwise until the line of action of the tension in the string that suspends the lamina intersects the center of mass of the lamina, as shown in the following figure.

The angle 𝜃 from the vertical from which the lamina is suspended can be seen from the figure to have a tangent given by tan𝜃=28=16895.

Let us now summarize what we learned in this explainer.

Key Points

  • We can find the center of mass of a system of bodies by creating a table consisting of the mass and the coordinates of the center of mass of each body and then forming and solving an equation using this table.
  • The position vector of the center of mass of a system of 𝑛 bodies in space is given by 𝑟=𝑚𝑟𝑚, where 𝑚,,𝑚 are the masses of the bodies and 𝑟,,𝑟 are the position vectors of these bodies relative to the origin of the system.
  • In a 2-dimensional system defined using Cartesian coordinates, we can define the position vector of the center of mass as 𝑟=𝑥,𝑦.
  • We often find the coordinates of the center of mass of a system by considering each of the orthogonal coordinates separately and using the equations 𝑥=𝑚𝑥𝑚,𝑦=𝑚𝑦𝑚.
  • A uniform compound lamina can be modeled as a system of laminas, where a gap in a lamina with a positive mass corresponds to a lamina with a “negative mass.”
  • The center of mass of a system containing a positive-mass lamina (𝐴) and a negative-mass lamina (𝐵) lies on the line that intersects the centers of mass of 𝐴 and 𝐵. The center of mass of the system is displaced from the center of mass of 𝐴 in the direction that points away from the center of mass of 𝐵 (the negative-mass lamina).

Download the Nagwa Classes App

Attend sessions, chat with your teacher and class, and access class-specific questions. Download Nagwa Classes app today!

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.