There are more than one center of mass for a n-particles system

[ManishR]

R_COM is a vector from an inertial frame to the point called the center of mass. The system behaves as if all the mass is concentrated at the center of mass and all the external forces act at that point. so

$$\frac{d^{2}}{dt^{2}}\overrightarrow{R}{}_{COM}=\frac{\sum m_{i}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{i}}}{\sum m_{i}}$$ ...[1]

or

$$\frac{d^{2}}{dt^{2}}(\overrightarrow{R}{}_{COM}+\overrightarrow{k})=\frac{\sum m_{i}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{i}}}{\sum m_{i}}$$

so there are more than one center of mass for a system.

but if center of mass is

$$\overrightarrow{R}{}_{COM}=\frac{\sum m_{i}\overrightarrow{r_{i}}}{\sum m_{i}}$$ ...[2]

or there is unique center of mass for a system.

whats wrong here ? how did we reach from [1] to [2] ?

 

[mathman]

There is a unique center of mass. If you add a constant vector and then take a derivative, there is no difference in the result, since the derivative of a constant is zero.

 

[JohnSimpson]

I think it's more instructive to start at [2] and go to [1]. You can think of [2] as a definition.

 

[ManishR]

There is a unique center of mass. If you add a constant vector and then take a derivative, there is no difference in the result, since the derivative of a constant is zero.

yes, but it means COM is not unique.

let say R_COM is the position vector of COM from inertial frame then from the same inertial frame a position vector = (R_COM vector + K vector) also satisfy the equation [1].
or
mathematically, differentiation of something is actually a function, so the "something" has more than one value not unique value.

I think it's more instructive to start at [2] and go to [1]. You can think of [2] as a definition.

the idea behind COM is that, it is a point such that all mass of system is concentrated into that point and the external forces that were acting on each particle of the system, act on that COM. so
F=ma
F = sum of all external forces
m = mass of system
a = acceleration of COM from an inertial frame

so, the position of R_COM is given by equation [1].

Equation [1] is what we have to start with, not equation [2]

however its equation [2] which tells us that there is unique COM.

 

[Doc Al]

the idea behind COM is that, it is a point such that all mass of system is concentrated into that point and the external forces that were acting on each particle of the system, act on that COM. so
F=ma
F = sum of all external forces
m = mass of system
a = acceleration of COM from an inertial frame

so, the position of R_COM is given by equation [1].

I'd say that was incorrect. Equation [1] is not the defining equation for COM. If all you care about is acceleration, you can always add a constant to the COM and get the same acceleration. So what?

Your equation [2] is the definition of COM.

 

[ManishR]

I'd say that was incorrect. Equation [1] is not the defining equation for COM. If all you care about is acceleration, you can always add a constant to the COM and get the same acceleration. So what?

Your equation [2] is the definition of COM.

yes that's the only explanation, but why ?

look jpg which is pg 116 of An Introduction to Mechanics by Daniel Kleppner, Robert J. Kolenkow

every point according to equation [1] will act as if all mass is concentrated into that point.
but it seems like in all those points we prefer a point which is defined by equation [2]
but i don't understand why ? its wrong too because equation [2] does not even account for external force. if there is no external force, equation [2] still says COM is unique but it should not be unique.

 

[Doc Al]

every point according to equation [1] will act as if all mass is concentrated into that point.

No. Equation [1] just deals with translational acceleration. It is insufficient to define a unique center of mass.

Say you wanted to find the balance point of a non-uniform beam. The COM (defined by equation [2]) answers that question. Adding an arbitrary constant won't work.

 

[ManishR]

every point according to equation [1] will act as if all mass is concentrated into that point.

No.

consider a system and if there is a particle whose mass is equal to the mass of system and the force on it is equal to sum of all external force then the position vector R of that point is

$$\frac{d^{2}}{dt^{2}}\overrightarrow{R}=\frac{\sum m_{i}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{i}}}{\sum m_{i}}$$

Equation [1] just deals with translational acceleration.
No.

since m_i is a point size particle so rotation of the point on its own axis is not possible. however the rotation of system on its own axis is possible and it is defined by equation [1]. so equation [1] deals with any kind of motion.

Say you wanted to find the balance point of a non-uniform beam. The COM (defined by equation [2]) answers that question. Adding an arbitrary constant won't work.
No.

thats different thing. COM is not a point which balance a body.