What exactly does centre of mass mean? [closed]

Question

The definition of centre of mass is the point at which the entire mass of a body can be said to be concentrated. Now, there are some bodies which have their centre of mass outside of the region that they occupy. This is tantamount to saying that the mass of that body is concentrated outside of that body, which doesn't make any sense to me. So, what exactly am I getting wrong here?

Answer 1

The center of mass (or, center-of-mass position) is defined so that we can apply Newton's Second Law, which is an equation for a single point particle, to a collection of particles instead.

We can construct an equation that governs the dynamics of a composite object of many particles, just by summing their Newton Second Laws: \begin{align} \vec{F}_{{\rm net}, n} &= m_n \, \vec{a}_n \quad (n = 1, \ldots, N)\\ & \rightarrow \sum_n \vec{F}_{{\rm net}, n} = \sum_n \, m_n \, \vec{a}_n \end{align} This equation can be made more useful by recognizing a few things:

• The net force on object $n$ can be broken into $$ \vec{F}_{{\rm net}, n} = \vec{F}_{{\rm ext}, n} + \sum_m \vec{F}_{{\rm int}, m \; {\rm on} \; n} $$ meaning that it consists of the vector sum of the external forces on the object, and the forces exerted on it by the other $N-1$ particles ("internal" to the system of $N$ particles).
• In the above sum of $\vec{F}_{{\rm net}, n}$ over all particles, all internal forces will cancel out, by Newton's Third Law: $\vec{F}_{ m \; {\rm on} \; n} = - \vec{F}_{ n \; {\rm on} \; m}$. So we can write the sum as: $$ \sum_n \vec{F}_{{\rm net}, n} = \sum_n \vec{F}_{{\rm ext}, n} = \vec{F}_{\rm ext} $$ where we mean the sum of all external forces on the system in the last equality.
• On the right-hand side of Newton's Second Law for the composite system, we can make the following transformations: \begin{align} \sum_n \, m_n \, \vec{a}_n &= \sum_n \, m_n \, \frac{{\rm d} \vec{v}_n}{{\rm d}t} = \sum_n \, \frac{{\rm d} \left( m_n \vec{v}_n\right)}{{\rm d}t} = \sum_n \, M_{\rm tot} \, \frac{{\rm d} \left( m_n \vec{v}_n/M_{\rm tot} \right)}{{\rm d}t} \\ & = M_{\rm tot} \frac{\rm d}{{\rm d}t} \left(\sum_n \, \frac{m_n \vec{v}_n}{M_{\rm tot}}\right) \end{align} where $M_{\rm tot} = \sum m_n$ is the total mass of the system. We call the quantity $$ \vec{v}_{\rm com} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2 + \ldots + m_N\vec{v}_N}{M_{\rm tot}} $$ the center-of-mass velocity of the system.
• The center-of-mass position is then given by $\vec{r}_{\rm com}$, where: $$ \vec{v}_{\rm com} = \frac{{\rm d}}{{\rm d}t} \vec{r}_{\rm com} = \frac{{\rm d}}{{\rm d}t}\left( \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + \ldots + m_N\vec{r}_N}{M_{\rm tot}}\right) $$

So, we see that the set of Newton's Second Law equations for individual particles leads to the very nice expression: $$ \vec{F}_{\rm ext} = M_{\rm tot} \frac{{\rm d}^2 \vec{r}_{\rm com}}{{\rm d} t^2} $$ showing that the dynamics of a collection of particles (rigid body, cloud, ...) behaves just like a single particle with $M_{\rm tot}$ located at the center-of-mass. Which is exactly what we assume in introductory physics when applying the particle version of Newton II to macroscopic objects.

Answer 2

"The definition of centre of mass is the point at which the entire mass of a body can be said to be concentrated."

That definition is incorrect. The definition of the centre of mass is the average position of the mass distribution.

For some purposes, a body made up of lots of particles linked together by internal forces can be treated as if it was a single particle of the same total mass concentrated at the centre of mass location. The average effect of a force on the collection of particles works out the same as the force's effect would be on the average of the particles. It's a handy shortcut for calculations. But it doesn't work for all properties.

As you say, the centre of mass can be outside the body of an object. For example, the centre of mass of a uniform circular ring is at the centre of the ring, which is empty space. But suppose we consider the question of which way an object will hang (still, not swinging) when it is suspended from a point. Hold the ring loosely at a point on its circumference and let it settle. It will rotate until the centre of mass is directly below the suspension point - exactly like it would if the entire mass of the ring was concentrated in a particle at that point.

You can attach weights to various parts of the ring to move the centre of mass around. The rule is exactly the same. You can find the centre of mass of any body by suspending it from different points, letting it hang freely, and drawing the vertical line down through the suspension point. They will all meet at the position of the centre of mass.

For the purposes of deciding in what orientation a body loosely suspended from a point will hang, the whole of the weighted ring acts as if it was a single particle at the centre of mass position, even though there is no mass there. It's just the average position of the mass.

Answer 3

It's a mathematical concept that is useful for calculations. It's the point of space that moves in the way a point-particle with the same mass as the object would move. For a rigid body, this is very useful as you can account for every other point in the body just knowing where the center of mass is as the distance is fixed.