Basic Center Of Mass

I've recently begun learning about how to calculate the COM of objects/particles. So, I've acquired the knowledge that calculating the COM for a finite number of particles is quite simple and is given by the following: $$X_{\mathrm{com}} = \frac{x_1m_1 + x_2m_2}{m_1+m_2}=\displaystyle\frac{\sum_{i=1}^{i=2} x_im_i}{M}$$

However, I'm struggling to believe this. If we wanted to calculate the COM for two particles, the COM will be roughly in the middle or nearer the greater mass. Newton's second law, $\vec{F} = m\vec{a}$ assumes we apply forces at the COM. However, this doesn't make intuitive sense. If I placed two particles on my bedroom floor (assuming my floor obeys the law of inertia) and applied a force at the COM which would be some point on my floor, I'm convinced that this system would not accelerate.

Second Question:

When we say a object must have uniform density in order to easily find the COM, what do we mean by uniform density? Does this mean every particle forming the particle must have the same density?

• @QuirkyTurtle98 Thanks, couldn't figure out how to do that :p – dsfsdsdfsdfsdf Jun 11 '17 at 12:18
• @sammy gerbil could you help me again? :) – dsfsdsdfsdfsdf Jun 11 '17 at 12:28

What I believe you're missing is the fact that "system of particles" does not mean "rigid-body".

If you try to apply a force to the COM of a stick (which, if uniformly dense, is in the middle) it starts behaving as $F=ma$ predicts: even if you are pushing on a little portion of it, the whole body moves because Tensions between the particles it is composed of occur. These tensions propagate through the body at the speed of sound and get all the particles to start moving "together", as a system.

This doesn't happen in your example 'cause there is no "internal force" that gets your two particles to act like a rigid body.

If you don't push the COM then Torques and rotation start predicting the behavior, of course.

Newton's second law, $\vec F=m\vec a$ assumes we apply forces at the COM.

It does not assumes that. In a Newtonian system of particles, each of the external forces acting on the system is assumed to be applied to some individual particle rather than to the center of mass. However, the dynamical behavior of the center of mass of the system of particles obeys

$$M_\text{tot}\, {\ddot{\vec x}}_{\text{com}} = \vec F_\text{ext,tot} = \sum_i \vec F_{\text{ext};i} = \sum_i \sum_j \vec F_{\text{ext};i,j} \tag{1}$$

Working from left to right, the first equality ($M_\text{tot}\, {\ddot{\vec x}}_{\text{com}} = \vec F_\text{ext,tot}$) looks exactly like Newton's second law. It says that the center of mass of the system accelerates exactly the same as would a point mass whose mass is the total mass of the system of particles and is acted up by the net external force acting on the system. This does not however mean you truly do apply the forces at the center of mass. This is an as-if expression. This distinction becomes very important when looking at the rotational behavior of the system of particles.

The next equality ($\vec F_\text{ext,tot} = \sum_i \vec F_{\text{ext};i}$) says the total external force on the system of particles is the sum of the total external force on each of the individual particles. So what is the total external force on an individual particle? That's what the final equality ($\sum_i \vec F_{\text{ext};i} = \sum_i \sum_j \vec F_{\text{ext};i,j}$) tells us: The total external force on an individual particle is the sum of the external forces acting on that specific particle. Forces are additive, vectorially.

Whether all of this is meaningful depends on whether the system boundary is meaningful. Imagine a rock on the Moon and a chunk of ice on Pluto. One could denote these two objects as comprising a system of particles. This is a rather meaningless system of particles. While equation (1) does faithfully describes the time evolution of the center of mass of this system, it's not very meaningful. There are plenty of applications where a partitioning into system versus external does make sense, and it's these applications where equation (1) is very helpful.

The Center-of-Mass refers to a system that is bound together in some way, such as a thin steel rod lying along the x-axis (as a simple example). The steel is made up of different atoms, but is of uniform density the throughout the object (assuming it has the same radius along the whole length of course). In this case you can mathematically divide it up into any number of pieces and, using the equation given, find the x coordinate of its center of mass. Be sure to use the distance to the center of each piece when you multiply x*m In more advanced math, you'll be performing integrations instead of summations, and the objects will not be of uniform density.

protected by Qmechanic♦Jun 11 '17 at 13:04

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