Lesson Explainer: Center of Gravity of Uniform Rods | Nagwa Lesson Explainer: Center of Gravity of Uniform Rods | Nagwa

# Lesson Explainer: Center of Gravity of Uniform Rods Mathematics

In this explainer, we will learn how to find the center of gravity of uniform rods.

In a uniform mass field, the center of gravity is the the unique point from which the object’s weight force acts. In other words, we can effectively assume that the object’s mass is concentrated at the center of gravity as if it were a particle. If we support a rigid body at its center of gravity, then it will be perfectly balanced at this single point.

In a rigid body with constant density, the center of gravity is located at the geometric center of the body. A uniform rod is a linear object with a constant linear density, and its center of gravity is at its midpoint.

As we can see in the diagram above, a uniform rod would be perfectly balanced at its midpoint.

### Definition: Center of Gravity of a Uniform Rod

The center of gravity of a uniform rod is at its midpoint.

Hence, we can assume that the mass of a uniform rod is concentrated at the center of gravity, which is located at its midpoint. In other words, we can treat the mass of a uniform rod as a particle located at its midpoint. When dealing with a system of masses involving a uniform rod, we can find the center of gravity of the system by treating the uniform rod as a particle and finding the center of gravity of the resulting system of particles.

We recall the formula for finding the center of gravity for a system of particles in a coordinate system.

### Theorem: Center of Gravity for a System of Particles

The center of gravity of a system of particles is the average position of the particle weighted according to its mass. In other words, given mass with coordinates , the - and -coordinates of the center of gravity, denoted and , respectively, are given by

In the first example, we will find the center of gravity of a uniform rod when masses are added at both ends of the rod.

### Example 1: Finding the Center of Gravity of a Uniform Rod with Added Masses

is a uniform rod of length 4 cm and mass 4 kg. A mass of magnitude 5 kg is fixed at and another mass of magnitude 1 kg is fixed at . Find the distance from the center of gravity of the system to .

Since a uniform rod has a constant density, the center of gravity for the uniform rod is located at its midpoint. We can assume that the mass of the uniform rod, 4 kg, is concentrated at the midpoint of , while the masses of magnitudes 5 kg and 1 kg are concentrated at and respectively. Using centimetres as the length unit and placing point at the origin, we can draw a coordinate line to indicate the positions of these masses.

The coordinate of the midpoint is the average of the coordinates for and , which is given by . So, the mass of the uniform rod is concentrated at the midpoint. Using the coordinates of all three masses, we construct the following table:

 Position Mass 0 2 4 5 kg 4 kg 1 kg

We recall that if a system of particles has mass at position , then the position of its center of gravity, denoted COG, is given by

In other words, the center of gravity is the average of each particle’s position weighted with its mass. Using our table constructed above, we can multiply through the column and sum over the products to obtain the numerator of the fraction:

The denominator of COG is the total mass of the system, which is obtained by summing over the second row of our table. Hence,

Substituting these values into the formula for COG,

Hence, the coordinate of the center of gravity is 1.2.

Since point is at the origin and the length unit is centimetres, the distance between point and the center of gravity is 1.2 cm.

In the previous example, we considered a system of masses consisting of a uniform rod with two masses at both ends. We were able to identify the center of gravity for this system by considering the mass of the uniform rod to be concentrated at its midpoint.

We can use the same strategy when we have a system of multiple uniform rods. When multiple uniform rods are connected together, they form a rigid body called a wireframe. We can find the center of gravity of a wireframe by treating the mass of each rod to be concentrated at its midpoint. This leads to a system of particles whose center of gravity coincides with that of the wireframe.

If each rod in the wireframe has the same density, then the value of this density does not influence the center of gravity. We will demonstrate this for the -coordinate of the center of gravity, denoted . Let the uniform rods in a wireframe have lengths . Denoting the linear density of the uniform rod , the mass of a uniform rod is

Substituting this expression into the formula for leads to

We note that the resulting expression is independent of the linear density . It is clear that the -coordinate, , will also be independent of . For this reason, the density, or exact weight of uniform rods, is often not specified in examples. To simplify computations, we can set the linear density to be 1 so that the mass of a uniform rod is given by its length.

In the next example, we will find the center of gravity of a system of uniform rods, particularly a Z-shaped wireframe, placed in a Cartesian coordinate plane.

### Example 2: Finding the Coordinates of the Center of Gravity of a Uniform Z-Shaped Wire

The figure shows a uniform wire of length 10 cm where . Find the coordinates of the center of gravity of the wire with respect to axes and .

Recall that the center of gravity of a uniform rod is located at its midpoint. The given system of masses is composed of three uniform rods. Since we can identify the center of gravity for each rod at its midpoint, we can effectively treat this as a system of three particles whose mass is concentrated at the three centers of gravity.

Since the uniform rod has a constant mass distribution, the mass of each rod is proportional to its length. The value of density does not influence the location of the center of gravity, so we can set the linear density to be 1 kg/cm. Then, the mass of each rod is given by its length:

Since the center of gravity of each rod is at its midpoint, we can think of these masses to be effectively concentrated at their respective midpoints. In other words, we can think of this wireframe as a system of particles located at the midpoints of each rod as in the diagram below.

Using centimetres as the length unit in the coordinate system, we can compute the coordinates of each point:

The coordinates of the midpoint of each rod are the averages of the coordinates of the respective endpoints:

Hence, we can construct a table of the coordinates and the masses.

Rods 𝑥-Coordinate 𝑦-Coordinate 𝐴𝐵 𝐵𝐶 𝐶𝐷 0 2 4 2 0 −1 4 kg 4 kg 2 kg

We recall that if a system of particles has mass at position , then the - and -coordinates of its center of gravity, denoted and , respectively, are given by

In other words, each coordinate of the center of gravity is the average of each particle’s coordinate weighted with its mass. Using our table above, we can multiply each coordinate with the mass in the same column and sum over the products to obtain

The denominator in both - and -coordinates is the total mass of the system, which is obtained by summing over the last row of our table. Hence,

Substituting these values into the formula for COG,

Hence, the coordinates of the center of gravity are .

In the next example, we will find the center of gravity of a system of uniform rods on a coordinate plane where the densities of the uniform rods are different.

### Example 3: Determining the Coordinates of the Center of Gravity of a Nonuniform Trapezoid Frame

A thin steel frame is in the form of a trapezoid in which , , , and . The frame is located in a Cartesian plane such that is at the origin and is on the -axis. The section is made of a metal whose density is twice that of the metal used in the remaining part of the frame. Determine the coordinates of the center of gravity of the frame.

We recall that the center of gravity of each uniform rod is at its midpoint. We can find the center of gravity of the given system by treating the mass of each rod to be concentrated at its midpoints. We begin by identifying the mass of each uniform rod.

We note that the density and the mass of the uniform rods are not specified in this example. However, we are given that the density of rod is twice that of the other three rods. Since the center of gravity in a system of particles is given by the relative position of each particle with respect to its masses, we know that the exact value of the density does not affect the center of gravity. Hence, let us define the linear density of the three rods , , and to be 1 kg/cm, which means that the density of rod is 2 kg/cm. This means that the mass of rod is twice its length, while the masses of the other three rods are equal to their lengths:

To find the mass of rod , we need to compute its length. We draw a right triangle with as the hypotenuse.

Applying the Pythagorean theorem on this right triangle, we obtain the length of :

Hence, the mass of rod is 55 kg.

Using centimetres as the length unit, we can identify the coordinates of each vertex from the diagram:

We can compute the coordinates of the midpoint of each rod by taking the average of the coordinates of the endpoints:

Using what we have obtained so far, let us construct a table of the coordinates and the masses.

Rods 𝑥-Coordinate 𝑦-Coordinate Mass 𝐴𝐵 𝐵𝐶 𝐶𝐷 𝐷𝐴 −22 −11 22 11 332 33 332 0 55 kg 66 kg 33 kg 44 kg

We recall that if a system of particles has mass at position , then the - and -coordinates of its center of gravity, denoted and , respectively, are given by

In other words, each coordinate of the center of gravity is the average of each particle’s coordinate weighted with its mass. Using our table above, we can multiply each coordinate with the mass in the same column and sum over the products to obtain

The denominator in both the - and -coordinates is the total mass of the system, which is obtained by summing over the last row of our table. Hence, we obtain

Substituting these values into the formula for COG,

Hence, the coordinates of the center of gravity are .

In the previous two examples, we found the coordinates of the center of gravity of a system of uniform rods, or a wireframe. Recall that, in a uniform gravitational field, the weight of a rigid body is a single force that acts at its center of gravity. Since a wireframe is a rigid body, its weight is a downward force acting at its center of gravity.

Now, suppose that we suspend a wireframe from one of its vertices. When this system comes to an equilibrium, the wireframe will not rotate about the suspended vertex. Since the weight of the wireframe is a single force acting on its center of gravity, the equilibrium is only possible when the center of gravity is vertically under the suspended point.

### Theorem: Vertical Line in a Wireframe Suspended at a Vertex

When a wireframe is suspended at a vertex, the equilibrium position occurs when the center of gravity is directly under the suspended vertex.

In other words, we can obtain the vertical line of the system under equilibrium by connecting the suspended vertex and the center of gravity. We can always identify the vertical using this approach as long as the center of gravity is not at the suspended vertex. We cannot use this method when the center of gravity coincides with the suspended vertex, since the rigid body suspended at its center of gravity is at equilibrium at any given position. In other words, if the suspended vertex is at the center of gravity, then the vertical direction is not well defined since any direction is a possible vertical direction at equilibrium.

In our next example, we will find the center of gravity of a U-shaped wireframe and use it to find the angle between the vertical and one of the rods, the angle of inclination, in its equilibrium position.

### Example 4: Finding the Angle between a Uniform U-Shaped Wire and the Vertical When It Is Freely Suspended

The figure shows a uniform wire . It has been bent at and to form right-angles. The wire was freely suspended from . Find the measure of the inclination angle from to the vertical when the body is hanging in its equilibrium position. Round your answer to the nearest minute.

We recall that, in a rigid body under a uniform gravitational field, the weight is a force acting at the center of gravity of the rigid body. If a rigid body is suspended at a point, the equilibrium position is possible when the center of gravity is directly under the suspended point. Hence, we can identify the vertical at equilibrium by connecting the suspended point and the center of gravity of the rigid body.

Since the given wire is a rigid body, we begin by identifying its center of gravity. We can consider this wire as a system of three uniform rods , , and . To identify the center of gravity, we assume that the mass of each rod is concentrated at its center of gravity, which is at its midpoint. Then, we can find the center of gravity of the resulting three-particle system.

Recall that if a system of uniform rods has a constant density, the value of the density does not affect the center of gravity. By defining the linear density of the wire to be 1 kg/cm, we can assume that the mass of each rod is equal to its lengths in kilograms. We place the wire in a coordinate system and label the masses and the centers of gravity of the uniform rods.

Using centimetres as the length unit, we can identify the coordinates of each point:

Then, the coordinates of the midpoints of each rod are the average of the coordinates of its endpoints:

Let us organize this information into a table of the coordinates and the masses.

Rods 𝑥-Coordinate 𝑦-Coordinate 𝐴𝐵 𝐵𝐶 𝐶𝐷 0 18 36 492 0 212 49 kg 36 kg 21 kg

We recall that if a system of particles has mass at position , then the - and -coordinates of its center of gravity, denoted and , respectively, are given by

Using our table above, we can multiply each coordinate with the mass in the same column and sum over the products to obtain

The denominator in both - and -coordinates is the total mass of the system, which is obtained by summing over the last row of our table. Hence, we obtain

Substituting these values into the formula for COG,

Hence, the coordinates of the center of gravity are . The line connecting the suspended vertex and forms the vertical in the equilibrium position. Now, we are ready to find the angle of inclination between the vertical and .

To find the angle of inclination, denoted , we draw a right triangle by adding a perpendicular foot from the center of gravity to the -axis.

Both side lengths and can be obtained from the coordinates of point and the center of gravity:

By applying right triangle trigonometry, we obtain the angle of inclination

Since , we can convert the decimal portion of the degree to the nearest minute by which is to the nearest minute. Hence, the measure of the inclination angle from to the vertical when the body is at its equilibrium is .

In our final example. we will find the center of gravity of a wireframe when we are given information about the position at equilibrium.

### Example 5: Finding the Center of Gravity of a Freely Suspended Bent Rod in Equilibrium

A uniform rod of length 46 cm was bent at its midpoint then suspended freely from . Given that is horizontal when the rod is hanging in its equilibrium position, determine the distance between the center of gravity of the rod and .

We recall that, in a uniform gravitational field, the center of gravity is the point at which the weight of a rigid body acts. Since the bent rod is a rigid body, the center of gravity must lie directly under the suspended point at equilibrium.

We can consider the bent rod as a system of two uniform rods that are joined at the point . Each uniform rod, and , has length . Since the lengths of both uniform rods are the same, their masses must also be the same. Since the density of a uniform rod is constant, the center of gravity of a uniform rod is at its midpoint. Thus, we can consider the masses of the two rods to be concentrated at their midpoints.

This leads to a system of two particles. Since both rods have equal mass, the two particles have equal mass. We want to find the center of gravity of this system. Let us consider the mass of each rod to be kg and the distance between the two centers of gravity to be cm. We can draw a coordinate line through these two points with one of them at the origin.

We recall that if a system of particles has mass at position , then the position of its center of gravity, denoted COG, is given by

In this system, the masses of both particles are kg and the coordinates of the masses are 0 and . Hence,

The coordinate of the center of gravity is , which is the midpoint of the line segment connecting the two centers of gravity. We add the center of gravity of the system to our diagram.

Since the green point in the diagram above is the center of gravity of the bent rod and the rod is suspended from point , the line connecting point and the center of gravity must be the vertical in the equilibrium position. Since we are given that is horizontal at equilibrium, these two lines must intersect perpendicularly. Let us add the vertical to the diagram and label all points of intersection.

From the diagram above, the distance between and the center of gravity is given by the length of . To find this length, we will use similar and congruent triangles. To achieve this, we first add perpendicular feet from point to the vertical and the horizontal and label the new points of intersection.

We are looking for the length of , which is the sum of the lengths of and .

We begin by noticing that the right triangles and are similar, which means the proportion of their sides is constant. Since is double , we must have that

We can obtain the length by applying the Pythagorean theorem to the right triangle . We know that the hypotenuse of this right triangle is 23 cm, so we need to find the length of its base .

To find the length , we need to notice that points and trisect the line . We can see this by observing that triangles and are congruent, giving . Also, since and are opposite sides of rectangle , . Finally, since triangles and are congruent, the lengths and are equal. Hence, we have obtained

This lead to the fact that , which means that and trisect the line . Since is the midpoint of the line and cm, we know that :

Then, applying the Pythagorean theorem, we obtain

Next, let us find the length . By the congruence of triangles and , length is equal to length . On the other hand, triangle is similar to triangle , so the ratios of corresponding sides must be constant. Since the side length is double the length , length must be twice . Hence,

We can find length by applying the Pythagorean theorem to the triangle . Hypotenuse is half of length . Then, . Also, we know that the length of base is a third of the length of . So, . Applying the Pythagorean theorem,

Substituting this above, we obtain

Finally, adding up the two lengths, we obtain length :

Therefore, the distance from point to the center of gravity is cm.

Let us finish by recapping a few important concepts from this explainer.

### Key Points

• The center of gravity of a uniform rod is at its midpoint.
• To find the center of gravity of a system of masses composed of uniform rods and particles, we can consider the mass of each uniform rod to be concentrated at its midpoint.
• A rigid body comprised of multiple uniform rods is called a wireframe. When a wireframe is suspended at a vertex, the center of gravity of the wireframe is directly under the suspended vertex at equilibrium. We can find the vertical at equilibrium by connecting the suspended point and the center of gravity, provided that they are not the same.