Here is an example problem:

A woman of a certain mass stands on the left side of a canoe in water. She walks to the other side. How far does the canoe go?

With my understanding of centre of mass, the centre of mass calculated initially will be slightly to the left of the centre of the canoe. When she walks to the other side, the centre of mass will shift once more but this time to the right of the centre of the canoe.

I should then calculate the initial centre of mass and equate that to the final centre of mass to solve for x, the position that the canoe moved in.

My question is, why is the centre of mass conserved like that (as if it's like conservation of energy where initial equals final)?


3 Answers 3


It's not a "conservation of center of mass"$^*$. You just know that the center of mass will not move. This is because we are assuming that the water does not supply any horizontal force to the canoe, so we know that the net external force acting on the woman-canoe system is $0$. This means that $$\mathbf F_\text{ext}=\frac{\text d\mathbf p_\text {total}}{\text dt}=M\frac{\text d\mathbf x_\text {COM}}{\text dt}=0$$

Since the center of mass was not moving before the woman starts walking, this means that it cannot move while the woman is walking also.

$^*$Typically we reserve the word "conserved" for quantities that can be gained or lost by the system, such as energy or momentum. It also suggests that there could be a "flow", i.e. for a closed system of two bodies, they can exchange energy while the total energy of the system is conserved.

Systems don't really "gain" or "lose" center of mass, so it's odd to say that this is "conserved". Additionally, it's not like one object can give "center of mass" to another object. I suppose you could argue in this scenario that center of mass is "conserved", but I think it's somewhat of an odd term for this specific quantity.


The woman and canoe are an internal system. If the water and air are considered frictionless, then the woman and canoe can move back and forth about their COM. As the woman moves, her action causes a reaction on the canoe, moving it the opposite direction, so the COM does not move relative to an external observer standing on the bank. It would take an external force to move the COM.


(This started as a comment, but got too long. Maybe someone will find this reasoning helpful.)

The thing is to understand that you can express any rigid system as: $$\sum {m_i\cdot \vec{r_i}} = \vec{r_c}\sum {m_i} = {M_c}\vec{r_c} = \vec {M_c}$$ Where $({m_i}, \vec{r_i})$ are constituent material points.

(The only special thing about center of mass is that origin point is chosen that $\vec{r_c} = 0$ but it doesn't change any conclusions.)

This has a very nice property of being additive. So you can split your system into any sub-sums of sub-systems, so you can combine material points into larger rigid subsystems. E.g:

$$\vec {M_c} = M_{woman} \cdot \vec r_{woman} + M_{canoe} \cdot \vec r_{canoe}$$

Also, if you apply some force to it, because of additiveness of derivatives and intergrals you can write: $$\vec{F_{net}} \cdot \vec {M_c} = \sum {m_i\cdot \vec{r_i} \cdot \vec{F_i}}$$

Above means that if you have a combination of some "small forces" acting on your constituants (or sub-systems) you can write it down as some net force acting on center of mass. Now, if you assume there is no net force for the whole system described by $\vec {M_c}$, then by definition it can't move. That is very definition of force. Non-zero force causes displacement. This works both ways. $\Delta \vec {M_c} = 0$ means there was no force, no force means center of mass stayed the same.

Conversely for work you get: $$W = \vec{F_{net}} \cdot \Delta\vec {M_c} = \sum {m_i\cdot \Delta\vec{r_i} \cdot \vec{F_i}}$$

To move center of mass you would need to spend internal energy or have an external net force to do some work on the system.

But sub-systems can move! $\vec{F_i}$ do not have to be 0.

So after application and work done by force you'll have: $$\vec {M_c} = \vec {M'_c} = M_{woman} \cdot \vec r'_{woman} + M_{canoe} \cdot \vec r'_{canoe}$$

You could even say, that woman did some work on the canoe... and canoe on the woman with the opposite sign, but energy of their whole system did not change.

The beauty of it is that you absolutely don't care how many exact forces worked on which parts of the system. You can wrap it up as some net force working on center of mass, and because of additiveness you can split it into subs systems. So as long as the woman and canoe look exactly the same in both pictures, it doesn't even matter what shapes they took in between.

You couldn't reverse argument a bit and say that since $$M_{woman} \cdot \vec r_{woman}+ M_{canoe} \cdot \vec r_{canoe} - (M_{woman} \cdot \vec r'_{woman}+ M_{canoe} \cdot \vec r'_{canoe}) = 0$$ i.e. their center of mass did not move (even though their configuration changed) you conclude there was no external force working on them.


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