# Physics Lesson 6.2.1 - Centre of Mass of a Regular Non-Homogenous Object

Welcome to our Physics lesson on Centre of Mass of a Regular Non-Homogenous Object, this is the first lesson of our suite of physics lessons covering the topic of Determining the Centre of Mass in Objects and Systems of Objects, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

## Centre of Mass of a Regular Non-Homogenous Object

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 C1 and C2 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 x1 and x2 the values for the x-coordinates of each individual centre of mass. For example, if the lengths of each individual bar are L1 and L2 respectively, we have for the coordinates of their centre of gravity,

x1 = L1/2

and

x2 = L1 + L2/2

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 m1 and m2 are the masses of each individual bar respectively, the centre of mass xC of this system is calculated by the formula

### Equation 1

xC = x1 × m1 + x2 × m2/m1 + m2

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.

### Example 1

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 L1 : L2 : L3 = 1 : 2 : 3. Also, the total mass of the bar is 18 kg and the ratio of pieces in respect to mass is m1 : m2 : m3 = 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.

### Solution 1

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 = L1 + L2 + L3 = 12 m

Also (as a ratio),

L1 ∶ L2 ∶ L3 = 1∶2∶3

Hence, we can write

L1 = 1 × x, L2 = 2 × x and L3 = 3 × x

Therefore,

1 × x + 2 × x + 3 × x = 12 m
6 × x = 12 m
x = 2m

Thus,

L1 = 1 × 2m = 2m
L2 = 2 × 2m = 4m
L3 = 3 × 2m = 6m

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

x1 = L1/2 = 2m/2 = 1m
x2 = L1 + L2/2 = 2m + 4m/2 = 4m
x3 = L1 + L2 + L3/2 = 2m + 4m + 6m/2 = 9m

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

M = M1 + M2 + M3 = 18 kg

Also,

m1 ∶ m2 ∶m3 = 3∶2∶4

Hence, we can write

m1 = 3 × x, m2 = 2 × x and m3 = 4 × x

Therefore,

3 × x + 2 × x + 4 × x = 18 kg
9 × x = 18 kg
x = 2 kg

Thus,

m1 = 3 × 2 kg = 6 kg
m2 = 2 × 2 kg = 4 kg
m3 = 4 × 2 m = 8 kg

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

xC = x1 × m1 + x2 × m2 + x3 × m3/m1 + m2 + m3
= 1 × 6 + 4 × 4 + 9 × 8/6 + 4 + 8
= 94/18 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.

### What does this result mean?

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 xC = 5.2 m, it will lean aside. Look at the figures below:

You have reached the end of Physics lesson 6.2.1 Centre of Mass of a Regular Non-Homogenous Object. There are 4 lessons in this physics tutorial covering Determining the Centre of Mass in Objects and Systems of Objects, you can access all the lessons from this tutorial below.

## More Determining the Centre of Mass in Objects and Systems of Objects Lessons and Learning Resources

Centre of Mass and Linear Momentum Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
6.2Determining the Centre of Mass in Objects and Systems of Objects
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
6.2.1Centre of Mass of a Regular Non-Homogenous Object
6.2.2Centre of Mass of a System Composed by Two or More Distant Objects
6.2.3Centre of Mass of Three Collinear Objects
6.2.4Centre of Mass of Three non-Collinear Objects

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