Center of Mass: Definition & Formulas

[AudioFlux]

What does it mean for a body (or many bodies) to have the centre of mass at a certain point?

How does the formula for finding the centre of mass find it?

How can you describe the centre of mass with respect to the other particles of an object?

 

[BruceW]

The centre of mass (in one dimension) of a collection of objects is:
$$\frac{ x_1 m_1 + x_2 m_2 + x_3 m_3 + ... }{ m_1 + m_2 + m_3 + ... }$$
And for 3 dimensions, just replace x with the position vector.
(where the numbers 1,2,3 represent each object in the collection you are considering)

 

[BruceW]

$$\frac{ \sum_n \ {\vec{r}}_n m_n }{ \sum_n \ m_n }$$
is a more mathematical way of saying it.
And the physical meaning is 'the average location of all the masses' (where the greater masses are more important in the sum)

 

[AudioFlux]

The centre of mass (in one dimension) of a collection of objects is:
$$\frac{ x_1 m_1 + x_2 m_2 + x_3 m_3 + ... }{ m_1 + m_2 + m_3 + ... }$$
And for 3 dimensions, just replace x with the position vector.

thanks, I know the formula. my question is, how does this formula give the centre of mass?

 

[AudioFlux]

$$\frac{ \sum_n \ {\vec{r}}_n m_n }{ \sum_n \ m_n }$$
is a more mathematical way of saying it.
And the physical meaning is 'the average location of all the masses' (where the greater masses are more important in the sum)

so, if there are two masses m1 and m2, the mass at their centre of mass is (m1 + m2)/2?

 

[BruceW]

The formula is the definition of the centre of mass.
So the phrase 'centre of mass' simply means 'this formula'.
What the formula does is gives an 'weighted average' of the average position of all the bodies in the sense that the bodies of greater mass contribute more strongly to the average.
The reason it is divided by the sum of the masses is so that the weighting is done correctly.

 

[BruceW]

so, if there are two masses m1 and m2, the mass at their centre of mass is (m1 + m2)/2?

The centre of mass for two masses m1, m2 at x1, x2 will be at (m1x1 + m2x2)/(m1+m2)
and if you choose the origin to be right in the middle of the two masses, then the centre of mass is at: x(m2-m1)/(m1+m2)
(this is because x1 = -x and x2 = x)
(assuming the first mass is to the left of the origin)

 

[BruceW]

And the total mass of a collection of objects is simply the sum of the masses, so in this case, the total mass is m1 + m2, which is at the position x(m2-m1)/(m1+m2)

 

[BruceW]

To be clear: 'centre of mass' gives the average location of a collection of masses (it has nothing to do with what the total mass is)

 

[AudioFlux]

suppose a lever has masses m1 and m2 on its either sides (where m1 is not equal to m2).
why doesn't the fulcrum lie in the centre (you said the centre of mass is the average location of all the masses)?

 

[BruceW]

Yes, if the masses are not equal, the fulcrum will not lie in the centre.
The centre of mass is the weighted average of the locations of the masses, so if one of the objects is more massive, the centre of mass will be closer to the more massive object.

 

[BruceW]

In other words, when calculating the centre of mass, greater importance is given to more massive objects.

 

[AudioFlux]

is there a derivation for the formula, or is it common sense?

 

[BruceW]

On the wikipedia page for 'centre of mass', they show a derivation that relies on the fact that when there are no external forces, the centre of mass of a collection of bodies will travel in uniform motion. (i.e. they derive it from Newton's laws).
But the concept of a centre of mass doesn't rely on Newton's laws. Think of the concept 'centre of mass' as a definition, which can then be used to simplify physical problems.

 

[I_am_learning]

The formula can be derived if we first decide on what we mean by center of mass.
But, if we define center-of-mass as being the point given by that formula, then ofcourse there is no derivation.
I think, we mostly do the former. The center of mass is defined as the point where the force of gravity can be supposed to apply so that the net external effect is the same.
What This means is that suppose you have a pretty big irregular rock. You let it drop. The force of gravity applies to every small masses on it. But since the body is rigid they all add up to provide a net effect.
Instead of thinking of gravity being applied to every portion, you can do calculation, by assuming the gravity is being applied at a specific point on the stone. You can't arbitrarily select this point, because applying force on this point should create same net Force and moment on the body as the gravity did. Based on this criteria, the center of gravity (mass) can be worked out and this gives the given formula.

 

[BruceW]

iamlearning has got the right idea. The formula defines the centre of mass. Then this concept can be used in certain physical situations.

 

[sophiecentaur]

You need to remember that the CM is not necessarily the point to which objects will be attracted. The objects need to be far enough away from the 'object' to be considered to be at infinity.
If you take a dumbell shape (e.g. the Earth and Moon), when you are nearer to either of the large masses, that is the mass to which you will be attracted. You stand ON the Moon's surface and are not attracted to the CM of the two bodies, which is somewhere below the surface of Earth. CM is a mathematically derived point which may not have any actual mass at the location.

 

[TobyC]

Because of Newton's 3rd law that every action has an equal and opposite reaction, the laws of physics for any given system have a special property, and that is that you can define a special point in space which moves as if it were a particle with a mass equal to the total mass of the system, and a force acting on it equal to the sum of all forces on the system. This means for example that this special point for a closed system always stays in the same place.

The above result can be derived relatively easily from Newton's 3rd law, and it is incredibly useful for solving problems because it can simplify the analysis of a complicated system, and the special point it defines is called the 'centre of mass', and as you derive the above result you'll find that its definition is equal to the formula already posted in this thread.

For example, if you're floating in space and you wave your arms around to try and move, it won't work, because you are a closed system and you can't change the position of your centre of mass. If however you throw a brick in some direction, although the centre of mass of you and the brick doesn't move, because the brick is shooting off, you have to shoot off in the other direction, and that does give a way of moving in space (that's how rockets do it).

Another important application of the centre of mass is in rigid bodies, because the motion of a rigid body can be greatly simplified by treating it as a particle located at its centre of mass, and then you simply sum all the forces on the body and put them on the particle instead. The only thing you have left to worry about is the torque about the centre of mass.

The fact that gravity can be thought of as a single force acting on the centre of mass of a rigid body is a consequence of the fact that in a uniform gravitational field (like we roughly have at the surface of the Earth), the gravitational force on a body will exert no torque on it. The reason I say that is that the centre of mass is not only special because of gravity, it is a useful concept in its own right.

 

[AudioFlux]

For example, if you're floating in space and you wave your arms around to try and move, it won't work, because you are a closed system and you can't change the position of your centre of mass. If however you throw a brick in some direction, although the centre of mass of you and the brick doesn't move, because the brick is shooting off, you have to shoot off in the other direction, and that does give a way of moving in space (that's how rockets do it).

nice! so, if you take a closed system like me and the brick, the centre of mass will always be conserved no matter how fast I throw it? This is like conservation of momentum,
m1 x v1 = m2 x v2

 

[AudioFlux]

You need to remember that the CM is not necessarily the point to which objects will be attracted. The objects need to be far enough away from the 'object' to be considered to be at infinity.
If you take a dumbell shape (e.g. the Earth and Moon), when you are nearer to either of the large masses, that is the mass to which you will be attracted. You stand ON the Moon's surface and are not attracted to the CM of the two bodies, which is somewhere below the surface of Earth. CM is a mathematically derived point which may not have any actual mass at the location.

If you take the centre of mass of the Earth and moon separately, then that's where I get attracted if I stand on their respective surface

 

[sophiecentaur]

If you take the centre of mass of the Earth and moon separately, then that's where I get attracted if I stand on their respective surface

Of course, but only if you are talking approximately. The other body always has an effect on the actual direction of the gravitational pull when standing on the other. i.e. if you are standing on ' one side' of the Moon, you would fall not vertically towards its centre but slightly towards the Earth, because there is a measurable extra force. That exta force acounts for the tides on Earth, for example.
My point is that you wouldn't fall towards the cm of the pair unless you were a long way away - far enough to ignore the separation of the two masses.

 

[AudioFlux]

nice! so, if you take a closed system like me and the brick, the centre of mass will always be conserved no matter how fast I throw it? This is like conservation of momentum,
m1 x v1 = m2 x v2

A small derivation-verification (if m1 x v1 = m2 x v2 is correct):

m1 x v1 = m2 x v2

we know, velocity(v) = distance(x) / time
hence, velocity is directly proportional to distance traveled.

so, m1 * x1 = m2 * x2(refer to the diagram)

using the already given formula,

x1 = m2 (x1 + x2) / (m1 + m2)

simplify this equation, you'll get m1 * x1 = m2 * x2
voila!

 

[TobyC]

A small derivation-verification (if m1 x v1 = m2 x v2 is correct):

m1 x v1 = m2 x v2

we know, velocity(v) = distance(x) / time
hence, velocity is directly proportional to distance traveled.

so, m1 * x1 = m2 * x2


(refer to the diagram)

using the already given formula,

x1 = m2 (x1 + x2) / (m1 + m2)

solve this equation, you'll get m1 * x1 = m2 * x2
voila!

Excellent yes, conservation of momentum can also be thought of as a consequence of Newton's 3rd law. Conservation of momentum and conservation of centre of mass are closely related. You have just proved that the for a closed system containing two bodies, the centre of mass must always stay in the same place because of conservation of momentum. If you extend that derivation for n bodies then you will have proved it for any closed system!