Physics Lesson 6.2.2 - Centre of Mass of a System Composed by Two or More Distant Objects

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Welcome to our Physics lesson on Centre of Mass of a System Composed by Two or More Distant Objects, this is the second 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 System Composed by Two or More Distant Objects

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.

Example 2

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.

Solution 2

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:

x1 = 1 m
x2 = 1 m + 6 m = 7 m
m1 = 4 kg
m2 = 2 kg
xC = ?
Applying the equation (1), we obtain

xC = x1 × m1 + x2 × m2/m1 + m2
= 1 × 4 + 7 × 2/4 + 2
= 18/6 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:

x1 = 4 m
x2 = 4 m + 6 m = 10 m
m1 = 4 kg
m2 = 2 kg
xC = ?

Applying again the equation (1) we obtain

xC = x1 × m1 + x2 × m2/m1 + m2
= 4 × 4 + 10 × 2/4 + 2
= 36/6 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).

You have reached the end of Physics lesson 6.2.2 Centre of Mass of a System Composed by Two or More Distant Objects. 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

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6.2Determining the Centre of Mass in Objects and Systems of Objects
Lesson IDPhysics Lesson TitleLessonVideo
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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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