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In this Physics tutorial, you will learn:

- How to determine the centre of mass position for a non-homogenous object
- The same for a system of collinear distant objects
- How to find the centre of mass of a system composed by three or more objects using coordinates

In the previous tutorial, "Centre of Mass. Types of Equilibrium", we discussed in general about centre of mass for a single object and on how to determine it experimentally. The only case in which we can find mathematically the centre of mass making use only of the information provided in the previous tutorial, is when the object is regular and homogenous.

Now, we will explain how to determine mathematically the centre of mass in irregular objects. Furthermore, we will also explain how to calculate the centre of mass in systems composed by more than one object, whether attached or detached.

The simplest example of a regular non-homogenous object is a long and thin bar composed by two different materials, which are attached longwise as shown below.

Each individual bar has its own centre of mass if taken separately. These centres of mass are denoted by C_{1} and C_{2} respectively. Since the bars are both homogenous and regularly shaped (they are either prisms or cylinders, depending on their base shape), the centre of gravity for each bar is located at middle of their respective length.

First of all, we must choose a reference point or origin (as discussed in the Kinematics section). For convenience, we can chose the origin at the left-end of the first bar, which corresponds to the left-end of the entire system. Since the bars are very thin, we need only the x-dimension for calculating the centre of mass of the whole system. Therefore, we can write as x_{1} and x_{2} the values for the x-coordinates of each individual centre of mass. For example, if the lengths of each individual bar are L_{1} and L_{2} respectively, we have for the coordinates of their centre of gravity,

x_{1} = *L*_{1}*/**2*

and

x_{2} = L_{1} + *L*_{2}*/**2*

Look at the figure below to create a clearer idea in this regard.

To determine the centre of mass C of the system, we must also consider the masses of each individual bar (since they are made from different materials, we don't expect to have equal masses although the lengths may be equal). Thus, if m_{1} and m_{2} are the masses of each individual bar respectively, the centre of mass x_{C} of this system is calculated by the formula

The above Equation is widely used in other field of science as well. You may encounter it (only as a structure but with different quantities obviously) in mathematics, finance, etc. This Equation can be extended for more than two materials as well.

A long and thin bar is made from three different materials. The total length of the bar is 12 m and the ratio of its pieces in respect to length is L_{1} : L_{2} : L_{3} = 1 : 2 : 3. Also, the total mass of the bar is 18 kg and the ratio of pieces in respect to mass is m_{1} : m_{2} : m_{3} = 3 : 2 : 4.

If the order from left to right is 1 - 2 - 3, calculate the centre of gravity from the left end of the bar.

First, we must draw a figure to have a better idea on the situation described in this problem.

Now, let's determine the values for lengths and masses of each piece.

We have:

L = L_{1} + L_{2} + L_{3} = 12 m

Also (as a ratio),

L_{1} ∶ L_{2} ∶ L_{3} = 1∶2∶3

Hence, we can write

L_{1} = 1 × x, L_{2} = 2 × x and L_{3} = 3 × x

Therefore,

1 × x + 2 × x + 3 × x = 12 m

6 × x = 12 m

x = 2m

6 × x = 12 m

x = 2m

Thus,

L_{1} = 1 × 2m = 2m

L_{2} = 2 × 2m = 4m

L_{3} = 3 × 2m = 6m

L

L

Thus, for the x-coordinates of each centre of mass, we have

x_{1} = *L*_{1}*/**2* = *2m**/**2* = 1m

x_{2} = L_{1} + *L*_{2}*/**2* = 2m + *4m**/**2* = 4m

x_{3} = L_{1} + L_{2} + *L*_{3}*/**2* = 2m + 4m + *6m**/**2* = 9m

x

x

The same procedure is used to calculate the masses of each piece. We have

M = M_{1} + M_{2} + M_{3} = 18 kg

Also,

m_{1} ∶ m_{2} ∶m_{3} = 3∶2∶4

Hence, we can write

m_{1} = 3 × x, m_{2} = 2 × x and m_{3} = 4 × x

Therefore,

3 × x + 2 × x + 4 × x = 18 kg

9 × x = 18 kg

x = 2 kg

9 × x = 18 kg

x = 2 kg

Thus,

m_{1} = 3 × 2 kg = 6 kg

m_{2} = 2 × 2 kg = 4 kg

m_{3} = 4 × 2 m = 8 kg

m

m

Now, let's use the equation (1) to calculate the centre of mass of this system. We have:

x_{C} = *x*_{1} × m_{1} + x_{2} × m_{2} + x_{3} × m_{3}*/**m*_{1} + m_{2} + m_{3}

=*1 × 6 + 4 × 4 + 9 × 8**/**6 + 4 + 8*

=*94**/**18 m*

≈ 5.2 m

=

=

≈ 5.2 m

This means the centre of gravity of this system is at the second bar, 0.8 m on the left of its end (or the beginning of the third bar), as shown in the figure below.

The result obtained means that if we hang or put the object on a pivot, it will stay in equilibrium only if the string or the pivot are at 5.2 m from the left side of the composite bar. If we hang or put it at whatever point other than x_{C} = 5.2 m, it will lean aside. Look at the figures below:

If there is a system composed by two or more objects, which have a certain distance between them, we can use a similar procedure to the one described in the previous paragraph for calculating its centre of mass if masses and positions (coordinates) of each object are known. Even when the objects are irregularly shaped, we can ignore their shape and concentrate only on their respective centres of mass.

Despite the objects are distant, we consider them as a single system of objects connected through very light sticks whose masses are neglected.

In the following example, we will show that it is not important where you take the origin when calculating the centre of mass of a system of objects as long as you make the calculations correctly.

Two objects A and B of masses 4 kg and 2 kg respectively, are 6 m away from each other. Calculate the centre of gravity of the system.

We will take two different points of reference to show that this is not relevant in determining the centre of gravity of the system.

**1.** Let's take the reference point 1 m on the left of the first object as shown in the figure.

The clues for this configuration are:

x_{1} = 1 m

x_{2} = 1 m + 6 m = 7 m

m_{1} = 4 kg

m_{2} = 2 kg

x_{C} = ?

Applying the equation (1), we obtain x

m

m

x

x_{C} = *x*_{1} × m_{1} + x_{2} × m_{2}*/**m*_{1} + m_{2}

=*1 × 4 + 7 × 2**/**4 + 2*

=*18**/**6 m*

= 3 m

=

=

= 3 m

This result means the centre of gravity of the system is 2m on the right of the first object (3m - 1m = 2m) and 4m on the left of the second object (7m - 3m = 4m). Look at the figure:

**2.** If we take the origin at another position, for example 4m on the left of the first object, we obtain the following figure:

The clues for this new configuration are:

x_{1} = 4 m

x_{2} = 4 m + 6 m = 10 m

m_{1} = 4 kg

m_{2} = 2 kg

x_{C} = ?

x

m

m

x

Applying again the equation (1) we obtain

x_{C} = *x*_{1} × m_{1} + x_{2} × m_{2}*/**m*_{1} + m_{2}

=*4 × 4 + 10 × 2**/**4 + 2*

=*36**/**6 m*

= 6 m

=

=

= 6 m

This result means the centre of gravity of the system is again 2m on the right of the first object (6m - 4m = 2m) and 4m on the left of the second object (10m - 6m = 4m). Look at the figure:

Therefore, we get the same result, no matter where we choose to take the reference point (origin).

When we have three collinear objects, the centre of gravity of the system is determined at the same way as for two distant objects described above. It is better to choose the reference point on the left of the first object, so that all three objects are on its right and as such, we can take all x-coordinates for the individual centres of mass as positive. Then, the equation (1) written for three objects shown below, is used.

x_{C} = *x*_{1} × m_{1} + x_{2} × m_{2} + x_{3} × m_{3}*/**m*_{1} + m_{2} + m_{3}

Three objects are placed at the same line as shown below.

What is the position of the centre of mass of this system? All distances are calculated from the origin.

Using the equation (1) for three objects,

x_{C} = *x*_{1} × m_{1} + x_{2} × m_{2} + x_{3} × m_{3}*/**m*_{1} + m_{2} + m_{3}

we obtain after the substitutions,

x_{C} = *3 × 2 + 6 × 8 + 1 × 14**/**2 + 6 + 1*

=*68**/**9 m*

= 7.56 m

=

= 7.56 m

Therefore, the centre of mass of the system is slightly on the left of the second object.

In this case, we must consider a triangle in which objects' centres of gravity act as vertices as shown in the figure.

Again, we think the sides of this triangle as very light sticks with negligible weight. They are drawn in the figure only for calculation purpose. Also, we can think the inner part of the triangle as made by a very thin and light layer, such as a transparent vinyl sheet or anything similar.

It is obvious that the centre of gravity of the system is somewhere inside the internal area of triangle. It acts as an equilibrium point of the system, i.e. if we put the system on a pivot, which is exactly below the centre of mass, it stays in equilibrium as shown below.

We must consider two dimensions in order to work out the centre of mass if the system. Therefore, the equation (1) is applied for both x and y coordinates. This means the centre of mass of the system is written in the format (x_{C}, y_{C}). Hence, we write

x_{C} = *x*_{1} × m_{1} + x_{2} × m_{2} + x_{3} × m_{3}*/**m*_{1} + m_{2} + m_{3}

and

y_{C} = *y*_{1} × m_{1} + y_{2} × m_{2} + y_{3} × m_{3}*/**m*_{1} + m_{2} + m_{3}

A comprehensive approach in this regard involves dealing with some mathematical concepts such as the equation of a line that passes through two known points, the angles of a triangle when the side lengths are known, etc. These concepts are not difficult; a student learns all them in high school math. However, dealing with such concepts goes beyond the scope of this tutorial. Here we will discuss only situations involving objects whose centre of mass have known coordinates, so that we can apply directly the equation (1) for two dimensions.

Let's illustrate this point through an example.

Three objects are connected with very light sticks to form a single system. The first object has a mass of 4 kg and it is at (0m, 3m). The second object has a mass of 1 kg and it is at (-1m, 0m), and the third object has a mass of 2 kg and it is at (5m, 4m). Where is the centre of mass of the system?

In this example, it is better to draw a coordinate system and show all objects visually.

Now, let's write the clues. We have:

m_{1} = 4 kg

m_{2} = 1 kg

m_{3} = 2 kg

x_{1} = 0 m

x_{2} = -1 m

x_{3} = 5 m

y_{1} = 3 m

y_{2} = 0 m

y_{3} = 4 m

x_{C} = ?

y_{C} = ?

m

m

x

x

x

y

y

y

x

y

Applying the equation (1) for the x and y-direction separately, we obtain

x_{C} = *x*_{1} × m_{1} + x_{2} × m_{2} + x_{3} × m_{3}*/**m*_{1} + m_{2} + m_{3}

=*0 × 4 + (-1) × 1 + 5 × 2**/**4 + 1 + 2*

=*-1 + 10**/**7*

=*9**/**7 m*

≈ 1.29 m

=

=

=

≈ 1.29 m

and

y_{C} = *y*_{1} × m_{1} + y_{2} × m_{2} + y_{3} × m_{3}*/**m*_{1} + m_{2} + m_{3}

=*3 × 4 + 0 × 1 + 4 × 2**/**4 + 1 + 2*

=*12 + 8**/**7*

=*20**/**7*

≈ 2.86 m

=

=

=

≈ 2.86 m

Therefore, the centre of gravity of the system is at (1.29 m, 2.86 m). Look at the figure:

Even for three dimensions, we use the same method. The only difference is that we have another dimension, the z-dimension. Therefore, we must apply three times the equation (1).

Centre of mass of a regular non-homogenous object is determined by taking each homogenous part as a separate object and then applying the equation

x_{C} = *x*_{1} × m_{1} + x_{2} × m_{2} + …*/**m*_{1} + m_{2} + …

y_{C} = *y*_{1} × m_{1} + y_{2} × m_{2} + …*/**m*_{1} + m_{2} + …

z_{C} = *z*_{1} × m_{1} + z_{2} × m_{2} + …*/**m*_{1} + m_{2} + …

y

z

where x_{1}, x_{2}, y_{1}, y_{2}, z1, z2 etc., are the coordinates of centres of mass for each single regular part of the object or system and m_{1}, m_{2}, etc are their respective masses.

In one dimension (in a straight line), we use only the x-coordinates, in two dimensions (in a plane) we use both x and y-coordinates and in three dimensions (in space) we use all three coordinates.

If there is a system composed by two or more distant objects, we can use a similar procedure for calculating its centre of mass if the individual masses and positions (coordinates) of each object are known. Even when the objects are irregularly shaped, we can ignore their shape and concentrate only on their respective centres of mass.

Despite the objects are distant, we consider them as a single system of objects connected through very light sticks whose masses are neglected.

It is not important where you take the origin when calculating the centre of mass of a system of objects as long as you make the calculations correctly.

**1)** Two long and thin bars made of different material are stuck longwise as shown in the figure.

The golden bar is 4 m long and it weighs 3 kg, while the brownish bar is 3 m long and it weighs 1 kg. Where is the centre of gravity of the system if measured from the left end?

- 3.75 m
- 2.14 m
- 1.25 m
- 1.86 m

**Correct Answer: A**

**2)** Two objects of masses 5kg and 3 kg respectively are placed at 4 m away from each other and they are connected with a massless stick as shown below.

What is the position of the centre of gravity of this system from the 5 kg object?

- 2.5 m on the right of the 5 kg object
- 1.5 m on the right of the 5 kg object
- 2.7 m on the left of the 5 kg object
- 1.7 m on the left of the 5 kg object

**Correct Answer: B**

**3)** Three objects of masses m_{1} = 8 kg, m_{2} = 3 kg and m_{3} = 1 kg respectively are placed at the positions shown in the figure.

What are the coordinates of the centre of gravity of the system if the objects are connected with massless sticks?

- x
_{C}= 1 m, y_{C}= 0 m - x
_{C}= 0.75 m, y_{C}= - 0.083 m - x
_{C}= 0.75 m, y_{C}= - 1 m - x
_{C}= 0 m, y_{C}= 1 m

**Correct Answer: B**

We hope you found this Physics tutorial "Determining the Centre of Mass in Objects and Systems of Objects" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of "Linear Momentum Conditions Of Equilibrium" with our Physics tutorial on Newton's Second Law for a System of Particles.

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