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 π‘šοƒ with coordinates (π‘₯,𝑦), the π‘₯- and 𝑦-coordinates of the center of gravity, denoted COG and COG, respectively, are given by COGCOGο—οƒοƒοƒο˜οƒοƒοƒ=βˆ‘π‘šπ‘₯βˆ‘π‘š,=βˆ‘π‘šπ‘¦βˆ‘π‘š.

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

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

Answer

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 𝐴𝐡, while the masses of magnitudes 5 kg and 1 kg are concentrated at 𝐴 and 𝐡 respectively. Using centimetres as the length unit and placing point 𝐴 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 𝐴 and 𝐡, which is given by 0+42=2. So, the mass of the uniform rod is concentrated at the midpoint. Using the coordinates of all three masses, we construct the following table:

Position024
Mass5 kg4 kg1 kg

We recall that if a system of particles has mass π‘šοƒ at position π‘₯, then the position of its center of gravity, denoted COG, is given by COG=βˆ‘π‘šπ‘₯βˆ‘π‘š.

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: ο„šπ‘šπ‘₯=0Γ—5+2Γ—4+4Γ—1=12β‹….kgcm

The denominator of COG is the total mass of the system, which is obtained by summing over the second row of our table. Hence, ο„šπ‘š=5+4+1=10.kg

Substituting these values into the formula for COG, COG=βˆ‘π‘šπ‘₯βˆ‘π‘š=1210=1.2.

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

Since point 𝐴 is at the origin and the length unit is centimetres, the distance between point 𝐴 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 π‘₯-coordinate of the center of gravity, denoted COG. Let the uniform rods in a wireframe have lengths 𝐿. Denoting the linear density of the uniform rod 𝜌, the mass π‘šοƒ of a uniform rod is π‘š=πΏΓ—πœŒ.

Substituting this expression into the formula for COG leads to COG=βˆ‘π‘šπ‘₯βˆ‘π‘š=βˆ‘πΏπœŒπ‘₯βˆ‘πΏπœŒ=πœŒο€Ήβˆ‘πΏπ‘₯ο…πœŒο€Ήβˆ‘πΏο…=βˆ‘πΏπ‘₯βˆ‘πΏ.

We note that the resulting expression is independent of the linear density 𝜌. It is clear that the 𝑦-coordinate, COG, will also be independent of 𝜌. 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 𝐴𝐡𝐢𝐷 of length 10 cm where 𝐴𝐡=𝐡𝐢=2𝐢𝐷=4cm. Find the coordinates of the center of gravity of the wire with respect to axes 𝐡𝐴 and οƒͺ𝐡𝐢.

Answer

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: massofkgmassofkgmassofkg𝐴𝐡=4,𝐡𝐢=4,𝐢𝐷=4Γ—12=2.

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: 𝐴=(0,4),𝐡=(0,0),𝐢=(4,0),𝐷=(4,βˆ’2).

The coordinates of the midpoint of each rod are the averages of the coordinates of the respective endpoints: midpointofmidpointofmidpointof𝐴𝐡=ο€Ό0,0+42=(0,2),𝐡𝐢=ο€Ό0+42,0=(2,0),𝐢𝐷=ο€Ό4,0βˆ’22=(4,βˆ’1).

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

Rods𝐴𝐡𝐡𝐢𝐢𝐷
π‘₯-Coordinate024
𝑦-Coordinate20βˆ’1
Mass4 kg4 kg2 kg

We recall that if a system of particles has mass π‘šοƒ at position (π‘₯,𝑦), then the π‘₯- and 𝑦-coordinates of its center of gravity, denoted COG and COG, respectively, are given by COGCOGο—οƒοƒοƒο˜οƒοƒοƒ=βˆ‘π‘šπ‘₯βˆ‘π‘š,=βˆ‘π‘šπ‘¦βˆ‘π‘š.

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 ο„šπ‘šπ‘₯=0Γ—4+2Γ—4+4Γ—2=16β‹…,ο„šπ‘šπ‘¦=2Γ—4+0Γ—4+(βˆ’1)Γ—2=6β‹….kgcmkgcm

The denominator ο„šπ‘šοƒ in both π‘₯- and 𝑦-coordinates is the total mass of the system, which is obtained by summing over the last row of our table. Hence, ο„šπ‘š=4+4+2=10.kg

Substituting these values into the formula for COG, COGCOGο—οƒοƒοƒο˜οƒοƒοƒ=βˆ‘π‘šπ‘₯βˆ‘π‘š=1610=85,=βˆ‘π‘šπ‘¦βˆ‘π‘š=610=35.

Hence, the coordinates of the center of gravity are ο€Ό85,35.

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 Sheet

A thin steel frame is in the form of a trapezoid 𝐴𝐡𝐢𝐷 in which 𝐴𝐷=22cm, 𝐢𝐷=33cm, 𝐡𝐢=66cm, and π‘šβˆ πΆ=π‘šβˆ π·=90∘. The frame is located in a Cartesian plane such that 𝐴 is at the origin and 𝐷 is on the π‘₯-axis. The section 𝐴𝐷 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.

Answer

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 𝐴𝐷 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 𝐴𝐡, 𝐡𝐢, and 𝐢𝐷 to be 1 kg/cm, which means that the density of rod 𝐴𝐷 is 2 kg/cm. This means that the mass of rod 𝐴𝐷 is twice its length, while the masses of the other three rods are equal to their lengths: massofkgmassofkgmassofkg𝐴𝐷=2Γ—22=44,𝐡𝐢=66,𝐢𝐷=33.

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

Applying the Pythagorean theorem on this right triangle, we obtain the length of 𝐴𝐡: 𝐴𝐡=√44+33=55.cm

Hence, the mass of rod 𝐴𝐡 is 55 kg.

Using centimetres as the length unit, we can identify the coordinates of each vertex from the diagram: 𝐴=(0,0),𝐡=(βˆ’44,33),𝐢=(22,33),𝐷=(22,0).

We can compute the coordinates of the midpoint of each rod by taking the average of the coordinates of the endpoints: midpointofmidpointofmidpointofmidpointof𝐴𝐡=ο€Ό0βˆ’442,0+332=ο€Όβˆ’22,332,𝐡𝐢=ο€Όβˆ’44+222,33=(βˆ’11,33),𝐢𝐷=ο€Ό22,33+02=ο€Ό22,332,𝐷𝐴=ο€Ό22+02,0=(11,0).

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

Rods𝐴𝐡𝐡𝐢𝐢𝐷𝐷𝐴
π‘₯-Coordinateβˆ’22βˆ’112211
𝑦-Coordinate332333320
Mass55 kg66 kg33 kg44 kg

We recall that if a system of particles has mass π‘šοƒ at position (π‘₯,𝑦), then the π‘₯- and 𝑦-coordinates of its center of gravity, denoted COG and COG, respectively, are given by COGCOGο—οƒοƒοƒο˜οƒοƒοƒ=βˆ‘π‘šπ‘₯βˆ‘π‘š,=βˆ‘π‘šπ‘¦βˆ‘π‘š.

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 ο„šπ‘šπ‘₯=(βˆ’22)Γ—55+(βˆ’11)Γ—66+22Γ—33+11Γ—44=βˆ’726β‹…,ο„šπ‘šπ‘¦=332Γ—55+33Γ—66+332Γ—33+0Γ—44=3630β‹….kgcmkgcm

The denominator ο„šπ‘šοƒ in both the π‘₯- and 𝑦-coordinates is the total mass of the system, which is obtained by summing over the last row of our table. Hence, we obtain ο„šπ‘š=55+66+33+44=198.kg

Substituting these values into the formula for COG, COGCOGο—οƒοƒοƒο˜οƒοƒοƒ=βˆ‘π‘šπ‘₯βˆ‘π‘š=βˆ’726198=βˆ’113,=βˆ‘π‘šπ‘¦βˆ‘π‘š=3630198=553.

Hence, the coordinates of the center of gravity are ο€Όβˆ’113,553.

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

Answer

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 𝐴𝐡, 𝐡𝐢, and 𝐢𝐷. 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: 𝐴=(0,49),𝐡=(0,0),𝐢=(36,0),𝐷=(36,21).

Then, the coordinates of the midpoints of each rod are the average of the coordinates of its endpoints: midpointofmidpointofmidpointof𝐴𝐡=ο€Ό0,49+02=ο€Ό0,492,𝐡𝐢=ο€Ό0+362,0=(18,0),𝐢𝐷=ο€Ό36,0+212=ο€Ό36,212.

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

Rods𝐴𝐡𝐡𝐢𝐢𝐷
π‘₯-Coordinate01836
𝑦-Coordinate4920212
Mass49 kg36 kg21 kg

We recall that if a system of particles has mass π‘šοƒ at position (π‘₯,𝑦), then the π‘₯- and 𝑦-coordinates of its center of gravity, denoted COG and COG, respectively, are given by COGCOGο—οƒοƒοƒο˜οƒοƒοƒ=βˆ‘π‘šπ‘₯βˆ‘π‘š,=βˆ‘π‘šπ‘¦βˆ‘π‘š.

Using our table above, we can multiply each coordinate with the mass in the same column and sum over the products to obtain ο„šπ‘šπ‘₯=0Γ—49+18Γ—36+36Γ—21=1404β‹…,ο„šπ‘šπ‘¦=492Γ—49+0Γ—36+212Γ—21=1421β‹….kgcmkgcm

The denominator ο„šπ‘šοƒ in both π‘₯- and 𝑦-coordinates is the total mass of the system, which is obtained by summing over the last row of our table. Hence, we obtain ο„šπ‘š=49+36+21=106.kg

Substituting these values into the formula for COG, COGCOGο—οƒοƒοƒο˜οƒοƒοƒ=βˆ‘π‘šπ‘₯βˆ‘π‘š=1404106,=βˆ‘π‘šπ‘¦βˆ‘π‘š=1421106.

Hence, the coordinates of the center of gravity are ο€Ό1404106,1421106. The line connecting the suspended vertex 𝐴(0,49) and ο€Ό1404106,1421106 forms the vertical in the equilibrium position. Now, we are ready to find the angle of inclination between the vertical and 𝐴𝐡.

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

Both side lengths π‘Ž and 𝑏 can be obtained from the coordinates of point 𝐴 and the center of gravity: π‘Ž=49βˆ’1421106=35.745…,𝑏=1404106=13.245….

By applying right triangle trigonometry, we obtain the angle of inclination πœƒ=13.245…35.753…=20.4111….tan∘

Since 1=60β€²βˆ˜, we can convert the decimal portion of the degree to the nearest minute by 0.4111…×60=24.667…′, which is 25β€² to the nearest minute. Hence, the measure of the inclination angle from 𝐴𝐡 to the vertical when the body is at its equilibrium is 2025β€²βˆ˜.

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

Answer

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 𝐴 at equilibrium.

We can consider the bent rod as a system of two uniform rods that are joined at the point 𝐡. Each uniform rod, 𝐴𝐡 and 𝐡𝐢, has length 462=23cm. 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 𝑀 kg and the distance between the two centers of gravity to be 𝐿 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 π‘šοƒ at position π‘₯, then the position of its center of gravity, denoted COG, is given by COG=βˆ‘π‘šπ‘₯βˆ‘π‘š.

In this system, the masses of both particles are 𝑀 kg and the coordinates of the masses are 0 and 𝐿. Hence, COG=βˆ‘π‘šπ‘₯βˆ‘π‘š=𝑀×0+𝑀×𝐿𝑀+𝑀=𝑀𝐿2𝑀=𝐿2.

The coordinate of the center of gravity is 𝐿2, 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 wire and the wire is suspended from point 𝐴, the line connecting point 𝐴 and the center of gravity must be the vertical in the equilibrium position. Since we are given that 𝐡𝐢 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 𝐴 and the center of gravity is given by the length of 𝐴𝐸. To find this length, we will use similar and congruent triangles. To achieve this, we first add perpendicular feet from point 𝐷 to the vertical 𝐴𝐹 and the horizontal 𝐡𝐢 and label the new points of intersection.

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

We begin by noticing that the right triangles 𝐴𝐻𝐷 and 𝐴𝐹𝐡 are similar, which means the proportion of their sides is constant. Since 𝐴𝐡 is double 𝐴𝐷, we must have that 𝐴𝐻=12𝐴𝐹.

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

To find the length 𝐡𝐹, we need to notice that points 𝐼 and 𝐹 trisect the line 𝐡𝐺. We can see this by observing that triangles 𝐴𝐻𝐷 and 𝐷𝐼𝐡 are congruent, giving 𝐷𝐻=𝐡𝐼. Also, since 𝐷𝐻 and 𝐼𝐹 are opposite sides of rectangle 𝐷𝐻𝐹𝐼, 𝐷𝐻=𝐼𝐹. Finally, since triangles 𝐷𝐻𝐸 and 𝐺𝐹𝐸 are congruent, the lengths 𝐷𝐻 and 𝐹𝐺 are equal. Hence, we have obtained 𝐷𝐻=𝐡𝐼,𝐷𝐻=𝐼𝐹,𝐷𝐻=𝐹𝐺.

This lead to the fact that 𝐡𝐼=𝐼𝐹=𝐹𝐺, which means that 𝐼 and 𝐹 trisect the line 𝐡𝐺. Since 𝐺 is the midpoint of the line 𝐡𝐢 and 𝐡𝐢=23 cm, we know that 𝐡𝐺=232: 𝐡𝐹=𝐡𝐼+𝐼𝐹=13𝐡𝐺+13𝐡𝐺=23𝐡𝐺=23Γ—232=233.

Then, applying the Pythagorean theorem, we obtain 𝐴𝐹=βˆšπ΄π΅βˆ’π΅πΉ=ο„Ÿ23βˆ’ο€Ό233=46√23.

This leads to 𝐴𝐻=12𝐴𝐹=23√23.

Next, let us find the length 𝐻𝐸. By the congruence of triangles 𝐷𝐻𝐸 and 𝐺𝐹𝐸, length 𝐻𝐸 is equal to length 𝐸𝐹. On the other hand, triangle 𝐺𝐸𝐹 is similar to triangle 𝐺𝐷𝐼, so the ratios of corresponding sides must be constant. Since the side length 𝐺𝐷 is double the length 𝐺𝐸, length 𝐷𝐼 must be twice 𝐸𝐹. Hence, 𝐻𝐸=12𝐷𝐼.

We can find length 𝐷𝐼 by applying the Pythagorean theorem to the triangle 𝐷𝐼𝐡. Hypotenuse 𝐷𝐡 is half of length 𝐴𝐡. Then, 𝐷𝐡=232cm. Also, we know that the length of base 𝐡𝐼 is a third of the length of 𝐡𝐺. So, 𝐡𝐼=236. Applying the Pythagorean theorem, 𝐷𝐼=βˆšπ·π΅βˆ’π΅πΌ=ο„Ÿο€Ό232οˆβˆ’ο€Ό236=23√23.

Substituting this above, we obtain 𝐻𝐸=12𝐷𝐼=23√26.

Finally, adding up the two lengths, we obtain length 𝐴𝐸: 𝐴𝐸=𝐴𝐻+𝐻𝐸=23√23+23√26=23√22.

Therefore, the distance from point 𝐴 to the center of gravity 𝐸 is 23√22 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.

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