Lesson Explainer: Center of Mass of Laminas

In this explainer, we will learn how to find the position of the center of mass (sometimes called the center of gravity) of a lamina composed of standard shapes in 2D.

The weight of a body is a force, and a force acts at a point. For a body that is a particle, we can say that the weight of the body acts at the position of the body.

Consider however a rigid, horizontal rod. The rod can be modeled as a system of 𝑛 particles, each separated by a distance 𝑑, as shown by the following figure.

We can imagine comparing a single particle, P, to the system of 𝑛 particles. The mass of P is equal to the sum of the masses of the 𝑛 particles. Because P has no length, it cannot rotate, so any force that acts on it can only produce translational acceleration.

If a net force ⃑𝐹 acts on the system of 𝑛 particles at a particular point, π‘₯, that is unique to the system, each particle in the system is accelerated due to ⃑𝐹 equivalently to the acceleration of P due to ⃑𝐹.

If ⃑𝐹 acts at π‘₯, the system is given only translational acceleration due to ⃑𝐹. The system does not rotate due to ⃑𝐹.

When the acceleration of P and that of the system are equal, point π‘₯ is that of the center of mass of the system.

Let us define the center of mass of a one-dimensional system of particles.

Definition: The Center of Mass of a One-Dimensional System of Particles

The position of π‘₯, the center of mass of a one-dimensional system of particles, is given by π‘₯=βˆ‘π‘šπ‘₯𝑛,οŠοƒοŠ²οŠ§οƒοƒ where π‘šοƒ is the mass of the particle of index 𝑖 and π‘₯ is the distance from the origin of a coordinate system of the particle of index 𝑖.

The approximation of a system of 𝑛 particles can be made more realistic by allowing π‘šοƒ to tend to zero and, hence, 𝑛 to tend to infinity. This requires replacing the summation with the following integral: π‘₯=∫π‘₯π‘šπ‘›.d

In the case of a uniform rod, π‘₯ is given by π‘₯=∫π‘₯β‹…π‘₯𝑛=1𝐿×π‘₯2=𝐿2,οŒ«οŠ¦οŠοŒ«οŠ¨ο—οŠ²οŒ«ο—οŠ²οŠ¦d where 𝐿 is the length of the rod and π‘₯ is, therefore, at the midpoint of the rod.

A force acting on the center of mass of a rod acts equivalently to a set of forces acting in the same direction at every point along the rod. The same substitution of one force for a set of forces can be considered for the force of the weight of the rod.

A rod resting on a surface exerts forces at every point of contact between the rod and the surface, but the weight of a rod can be modeled as a single force that acts at the rod’s center of mass.

If a rod is in contact with a surface, the surface will produce a reaction force, ⃑𝑅, in the opposite direction to the weight of the rod, οƒŸπ‘Š, where οƒŸπ‘Š will act along a line passing through the center of mass. In order for οƒŸπ‘Š and ⃑𝑅 to act to make the body be in equilibrium, the center of mass of the rod must be vertically above a part of the surface that provides ⃑𝑅, as shown in the following figure.

The position of the center of mass of the rod relative to the surface determines whether or not the weight will produce a rotational effect.

In this explainer, only bodies of negligible thickness are considered. These bodies are called laminas. A lamina is two-dimensional.

For a lamina, the center of mass can be defined by a 2D coordinate in a given coordinate system. The origin of the coordinate system used to define the center of mass of a lamina can, for convenience, be one of the vertices of the lamina, assuming the lamina has any vertices.

Let us look at an example of determining the position of the center of mass of a lamina.

Example 1: Finding the Center of Mass of a Lamina in the Shape of a Right Triangle

In the figure shown, find the position of the center of mass of the uniform triangular lamina 𝐴𝐡𝐢, considering 𝐴 to be the origin point.

Answer

For uniform laminas, the center of mass of the body is the geometric center of the body. The lamina is regular, so its center of mass is its geometric center.

The geometric center of a right triangle is the point of intersection of two lines where for each line one end intersects a vertex of the triangle and the other end intersects the midpoint of the line opposite that vertex. These lines are the medians of the triangle.

The following figure shows two lines that intersect at the point (π‘₯,𝑦), the geometric center of 𝐴𝐡𝐢, called its centroid. The coordinates of the intercepts with the π‘₯- and 𝑦-axes of these lines are included in the figure.

The values of π‘₯ and 𝑦 can be determined algebraically by finding the values of π‘₯ and 𝑦 where the equation of line 𝐸𝐢 equals the equation of line 𝐡𝐷.

The equation of 𝐡𝐷 is 𝑦=π‘Žο€Όο€Ό0βˆ’52βˆ’0π‘₯+5οˆπ‘¦=π‘Žο€Όβˆ’52π‘₯+5.

The equation of 𝐸𝐢 is 𝑦=π‘Žο‚ο‚0βˆ’4βˆ’0π‘₯+52οŽπ‘¦=π‘Žο€Όβˆ’58π‘₯+52.

These equations are equal, where π‘₯ and 𝑦 are the coordinates of the centroid of 𝐴𝐡𝐢: π‘Žο€Όβˆ’52π‘₯+5=π‘Žο€Όβˆ’58π‘₯+52.

There is a common factor of π‘Ž that can be eliminated to give βˆ’52π‘₯+5=βˆ’58π‘₯+52.

This can be rearranged as follows to determine π‘₯: βˆ’52π‘₯βˆ’βˆ’58π‘₯=52βˆ’5ο€Όβˆ’52βˆ’βˆ’58π‘₯=52βˆ’5ο€Όβˆ’208+58π‘₯=βˆ’52βˆ’158π‘₯=βˆ’5215π‘₯=402π‘₯=43.

Substituting this value of π‘₯ into the equation of 𝐡𝐷 gives 𝑦=π‘Žο€Όο€Όβˆ’52Γ—43+5οˆπ‘¦=π‘Žο€Όβˆ’206+5οˆπ‘¦=π‘Žο€Όβˆ’206+306οˆπ‘¦=106π‘Ž=5π‘Ž3.

To obtain the π‘₯-coordinate of the point π‘Ž(π‘₯,𝑦), the value of π‘₯ must be multiplied by π‘Ž, so π‘₯=4π‘Ž3.

Therefore, the coordinates of the center of mass of 𝐴𝐡𝐢 are given by ο€Ό4π‘Ž3,5π‘Ž3.

The method of finding the centroid by intersection of medians could have been simplified greatly as 𝐴𝐡𝐢 is a triangle, and we know that centroid divides the median in the ratio 2∢1. This means that the position of the centroid is necessarily 13 of the length of its median, and so the 𝑦-coordinate value is necessarily 13Γ—5π‘Ž. As the length of the side of 𝐴𝐡𝐢 in the π‘₯-direction is 45 the length of the side in the 𝑦-direction, the π‘₯-coordinate value is necessarily 13Γ—4π‘Ž.

When we are locating the center of mass of a uniform lamina, we need to identify the geometric center of the lamina. This is because when we are considering a uniform lamina, the two are equal. Let us consider why this is the case.

Consider the following uniform elliptical lamina with focus points at 𝐴 and 𝐡.

The horizontal line of symmetry has been added, which passes through both 𝐴 and 𝐡. Now, if we were to split the ellipse along this line of symmetry, we would end up with two congruent shapes. Since this is a uniform lamina, the two shapes would also have equal mass. Let us now consider the centers of mass of these two new laminas. Since the laminas are congruent and uniform, the centers of mass will be in the same place on both of them.

Note that since the ellipse is a regular shape and this is a uniform lamina, it is logical that the centers of mass of the two shapes will lie on the vertical line of symmetry of the ellipse. In the diagram, we can see the centers of mass, 𝐷 and 𝐸, and that the line segment 𝐷𝐸 is perpendicular to ⃖⃗𝐴𝐡.

We know that the mass of each of these laminas is the same, so the center of mass of the ellipse will lie at the midpoint of 𝐷𝐸, which will be at the point where it intersects ⃖⃗𝐴𝐡. Hence, the center of mass of the ellipse is at the intersection of the lines of symmetry.

By looking at this ellipse, we have seen a property of uniform laminas.

Property: The Center of Mass of a Symmetrical Lamina

If a geometrical axis of symmetry of a uniform lamina exists, then the center of mass lies along that line of symmetry.

Similarly, if a uniform lamina has more than one geometrical line of symmetry, then the center of mass will lie at the intersection of these lines of symmetry.

The point at which the lines of symmetry of the lamina intersect can also be called the geometric center. Hence, the center of mass lies at the same point as the geometric center of a uniform lamina.

We will only be focusing on laminas in this lesson, but the same concept can be extended to 3D shapes with uniform density. If they have any planes of symmetry, then the center of mass will lie somewhere in that plane. If they have two planes of symmetry, we can narrow it down to the line where the two planes intersect, and if it has three or more planes of symmetry, then the center of mass will be at the point where all three planes intersect.

In the next example, we will see how we can find the center of mass of a system involving a lamina and some masses.

Example 2: Finding the Center of Mass of a Lamina with Additional Masses Added

A uniform lamina in the form of a square 𝐴𝐡𝐢𝐷 of side length 28 cm has a mass of 54 grams. Masses of 10, 8, 4, and 8 grams are fixed at 𝐴, 𝐡, 𝐢, and 𝐷 respectively. Find the coordinates of the center of mass of the system.

Answer

What has been presented in the question can be considered as a system consisting of five particles. The four suspended masses are effectively particles, and the center of mass of the lamina is also effectively a particle. Because the lamina is a uniform square, the center of mass of the lamina is at the center of the square.

The system of particles is a two-dimensional system rather than a one-dimensional system. It is, however, possible to model the two-dimensional system as two one-dimensional systems, in the π‘₯- and 𝑦-directions respectively.

The one-dimensional system for the π‘₯-direction can be represented by the following table:

Total
Mass108548484
Distance from 𝐴𝐡00142828
Mass Γ— Distance007562241121β€Žβ€‰β€Ž092

The center of mass of the system in the π‘₯-direction is given by π‘₯=109284=13.cm

Finding the position of the center of mass of the system in the π‘₯-direction is equivalent to using the following formula: π‘₯=π‘šπ‘₯+π‘šπ‘₯+π‘šπ‘₯π‘š+π‘š+π‘š, where the values of π‘šοŠ§, π‘šοŠ¨, and π‘šοŠ© are found by summing the values of π‘š that are found at a given π‘₯-value. Therefore, π‘š=10+8=18,π‘š=54,π‘š=8+4=12.

The square has sides of length 28 cm, and the mass of the square acts at the midpoint of these lines. The value of π‘₯, taking 𝐴 as being at π‘₯=0, is then given by π‘₯=(18Γ—0)+(54Γ—14)+(12Γ—28)18+54+12=109284=13.cm

The same method can be applied in the 𝑦-direction to give the position of the center of mass in the 𝑦-direction, and it so happens that the distribution of masses around the system of particles gives the same values for π‘šοŠ§, π‘šοŠ¨, and π‘šοŠ© and values for π‘¦οŠ§, π‘¦οŠ¨, and π‘¦οŠ© equal to π‘₯, π‘₯, and π‘₯, respectively, as shown in the following figure.

Therefore, the value of 𝑦 is also 13 cm.

The position of the center of mass of the system is (13,13).cm

The center of mass of the system is very close to the center of mass of the lamina. This is to be expected, firstly, as most of the mass of the system is due to the lamina, secondly, because the other masses are geometrically distributed symmetrically around the center of the lamina, and, thirdly, because the range of the values of the other masses is small compared to the mass of the lamina.

Let us look at an example of a compound lamina.

Example 3: Finding the Center of Mass of a Lamina Composed of a Rectangle and a Triangle

A uniform square lamina 𝐴𝐡𝐢𝐷 has a side length 𝑙. Another uniform lamina 𝐡𝐢𝐸 of the same density, shaped as an isosceles triangle, is attached to the square such that 𝐸 lies outside the square and 𝐡𝐸=𝐢𝐸. Given that the square’s side length is 53 times the triangle’s height, find the center of mass of the system.

Answer

The centers of mass of the square and the triangle that the compound lamina consists of are both readily determinable.

The center of mass of the square is the center of the square, so it has the coordinates 𝑙2,𝑙2, taking 𝐷 as the origin of a two-dimensional coordinate system.

The π‘₯-position of the center of mass of the triangle is at a point 13 of the length of the median of the triangle from its base; the coordinates of this point are 𝑙+ο€Ό13οˆο€½3𝑙5,𝑙2=ο€½6𝑙5,𝑙2.

The lamina is uniform, so the masses of the square and the triangle are proportional to their relative areas.

The area of the square is 𝑙×𝑙=π‘™οŠ¨ and the area of the triangle is 𝑙×𝑙2=310𝑙, and so the mass of the triangle is 310 the mass of the square.

The centers of mass of the square and the triangle form a one-dimensional system for the π‘₯-direction that can be represented by the following table:

Total
Mass13101310
Distance from 𝐴𝐷𝑙26𝑙5
Mass Γ— Distance𝑙218𝑙5043𝑙50

The center of mass of the system in the π‘₯-direction is given by π‘₯==13=4365𝑙.οŠͺοŠͺ

This gives the coordinates of the center of mass of the two-dimensional system as ο€½43𝑙65,𝑙2.

Now, let us look at an example where a uniform lamina is folded.

Example 4: Finding the Coordinates of the Center of Gravity of a Nonuniform Rectangular Lamina

A uniform rectangular lamina has a length of 63 cm and a width of 59 cm. It is divided into three equal-sized rectangles along its length; the last of these rectangles has been folded over so that it lies flat on the middle rectangle as show in the figure. Find the coordinates of the center of gravity of the lamina in this form.

Answer

The lamina is divided into three equal-sized rectangles. The center of mass of each rectangle is at the geometric center of each of the rectangles. Each rectangle has a mass π‘š.

After the lamina is folded, the area bounded by the dashed line has the same coordinates as the shaded area in the figure. As the center of mass of each rectangle is at the geometric center of each of the rectangles, the positions of the centers of mass of the two rectangles are shown in the following figure, including their distances from 𝐴 in the π‘₯-direction.

The shaded area consists of two stacked rectangles, so the mass of this area is double the mass of the unshaded area.

The one-dimensional system for the π‘₯-direction can be represented by the following table:

Total
Mass123
Distance from 𝐴𝐷2123ο€Ό212
Mass Γ— Distance2126ο€Ό2127ο€Ό212

The π‘₯-coordinate of the center of mass of the system in the π‘₯-direction is given by π‘₯=73=1476=492.

The result can be confirmed using the following formula: π‘₯=π‘šπ‘₯+π‘šπ‘₯π‘š+π‘š.

Substituting the values for π‘š and π‘₯ gives π‘₯=π‘š+2π‘šο€»3×3π‘š.

The factor of π‘š appears in both the numerator and the denominator, so it can be eliminated to give π‘₯=+2ο€»33π‘₯=+63=1476=492.

Folding the lamina does not change the distribution of its mass in the 𝑦-direction, so the center of mass of the lamina in the 𝑦-direction is simply the midpoint of the 59 cm length of the lamina in the 𝑦-direction.

The coordinates of the center of mass of the lamina are, therefore, ο€Ό492,592.

The weight of a body acts at the center of mass; therefore, a body that is acted on by its weight and by an applied force can only be in equilibrium if the applied force and the weight have the same line of action.

Consider a uniform square lamina that is suspended at equilibrium from a light string at a point 𝑃, as shown in the following figure.

The vertically upward force on the body due to the string tension has the same line of action as the weight of the lamina.

If the lamina was nonuniform, suspending it as shown would only establish that the line of action of the applied force passed through the center of mass of the lamina, not the position of the center of mass. The position of the center can be determined by suspension only by suspending the lamina from two points and finding the intersections of the lines of action of the applied forces that act at the suspension points, as shown in the following figure.

Let us look now at an example where a compound lamina is suspended from a point.

Example 5: Finding the Angle between a Straight Line and the Vertical Line in a Uniform Lamina Given That the Lamina Is Hung Freely from a Point

Find, to the nearest degree, the angle that the line π‘ˆπ‘‡ makes with the vertical if the given uniform lamina is hung freely from 𝑄.

Answer

Rather than suspending the lamina from a point to determine the position of its center of mass, the question involves determining the position of the center of mass of the lamina in order to determine the angle a side of the lamina will make from the vertical when it is suspended from a point.

The position of the center of mass of the lamina can be determined by first determining the positions of the centers of mass of the rectangular parts of the lamina. The center of mass of each rectangular part of the lamina is the geometric center of that rectangle.

Before making any calculations, it is possible to approximately locate the center of mass of the lamina by a scale drawing method, as shown in the following figure.

The scale drawing indicates that the center of mass of the lamina has an π‘₯-direction position very near to that of the point from which the lamina is to be suspended.

Modeling the lamina as two rectangles, the positions of the centers of mass of the rectangles, π‘šοŠ§ and π‘šοŠ¨, are shown by the following figure.

The coordinates of π‘šοŠ§ are ο€Ό4,152, and the coordinates of π‘šοŠ¨ are ο€Ό13,92.

As the rectangles are uniform, their relative masses are proportional to their relative areas; therefore, π‘šπ‘š=15Γ—810Γ—9=43.

From this, it can also be seen that π‘š=43π‘š, and so π‘š+π‘š=73π‘š.

The one-dimensional system for the π‘₯-direction can be represented by the following table:

Total
Mass43173
Distance from π‘ˆπ‘ƒ413
Mass Γ— Distance16313553

The π‘₯-position of the center of mass of the lamina is given by π‘₯==557.

The result can be confirmed using the following formula: π‘₯=π‘šπ‘₯+π‘šπ‘₯π‘š+π‘š.

Substituting known values gives π‘₯=ο€»π‘šΓ—4+(π‘šΓ—13)π‘š.οŠͺ

The factor of π‘šοŠ¨ appears in both the numerator and the denominator and can be eliminated to give π‘₯=ο€»Γ—4+13.οŠͺ

This can be rearranged as follows: π‘₯=3Γ—ο€»+137=557.

Following an equivalent procedure to find the 𝑦-position of the center of mass of the lamina gives 𝑦=3Γ—ο€»10+7=8714.

The angle between the center of mass and the line of action of the force that acts to suspend the lamina is shown in the following figure. The angle is very small.

When the lamina is suspended, it will rotate counterclockwise through this angle about its suspension point.

The vertical distance from suspension point to center of mass is given by 𝑑=15βˆ’8714=12314, and the horizontal distance from suspension point to center of mass is given by 𝑑=8βˆ’557=17.

Therefore, angle πœƒ is given by tan(πœƒ)==14861,οŠͺ giving a value for πœƒ of πœƒ=055β€².∘

The line π‘ˆπ‘‡ will rotate through the same angle. As π‘ˆπ‘‡ is initially horizontal, its initial angle to the vertical is 90∘. Decreasing this angle by 055β€²βˆ˜ results in a final angle, rounded to the nearest degree, of 89∘.

Key Points

  • If a net force ⃑𝐹 acts on a system of 𝑛 particles at the center of mass of the system, each particle in the system is accelerated due to ⃑𝐹 equivalently to the acceleration by ⃑𝐹 of a single particle that has a mass equal to the mass of the system.
  • The weight of a body acts at its center of mass.
  • For a uniform lamina, the position of the center of mass is the geometric center of the lamina.
  • The position of the center of mass of a one-dimensional system of particles is given by π‘₯=βˆ‘π‘šπ‘₯𝑛,οŠοƒοŠ²οŠ§οƒοƒ where π‘šοƒ is the mass of the particle of index 𝑖 and π‘₯ is the distance from the origin of a coordinate system of the particle of index 𝑖.
  • A two-dimensional system of particles can be treated as a pair of one-dimensional systems.
  • The center of mass of a lamina is the point that intersects the lines of action of multiple forces that each individually act to suspend the lamina at equilibrium.

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