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Mass concentrated at a single point?

  1. Apr 20, 2009 #1
    I'm trying to understand this concept. For example if there is a wooden sphere and we said that all of its mass is concentrated at a single point, can I think of this as the mass of the entire sphere being contained in one of its atoms?
     
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  3. Apr 20, 2009 #2

    russ_watters

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    Staff: Mentor

    Welcome to PF.

    No need for atoms here - this is just a useful mathematical device for solving problems. You don't have to try to visualize it relating to reality.
     
  4. Apr 21, 2009 #3

    Andy Resnick

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    Minor quibble- it's not a mathematical device, it's a *conceptual* device. Replacing an extended object with a point object located at the center of mass, having the same mass as the original object is a conceptual simplification.
     
  5. Apr 21, 2009 #4
    Any compound system is described with its center of inertia coordinates (they do not correspond to a real material point-like heavy particle) and with the relative coordinates.
    It is not a replacement of objects, it is a variable change.

    When you describe your wooden sphere with only its center of inertia coordinates, you know what a "simplification" you do - you just omit the relative coordinate equations.

    Bob.
     
  6. Apr 21, 2009 #5
    Let's say we want to build up a shape of mass (a sphere, for example), i.e. by repeatedly placing identical masses of [itex]dm[/itex] until we've made our shape and keeping track of their centre of mass (which is a [itex]dm\mathbf{r}[/itex]. If we have two particles we get

    [itex]\frac{dm\mathbf{r_1}+dm\mathbf{r_2}}{2dm}[/itex]

    if we have three we have

    [itex]\frac{dm\mathbf{r_1}+dm\mathbf{r_2}+dm\mathbf{r_3}}{3}[/itex]

    and so on. All we're finding is the average here. Now let's say the 4th mass, as an example, lies at exactly the same place as where the second mass placed was (i.e. there is more mass at that position) then our sum becomes

    [itex]\frac{dm\mathbf{r_1}+dm\mathbf{r_2}+dm\mathbf{r_3}+dm\mathbf{r_4}}{4}[/itex]

    but [itex]r_4=r_2[/itex], therefore, we can say

    [itex]\frac{dm\mathbf{r_1}+2dm\mathbf{r_2}+dm\mathbf{r_3}}{4}[/itex].

    Hopefully from here you can see how we arrive at the center of mass concept
     
  7. Apr 21, 2009 #6
    Now, the reason we do this is because it ends up simplifying things conceptually but keeping the identical mathematics. It kinds puts the many-body aspect (the fact that we're dealing with a ton of different masses) under the rug, so to speak. To demonstrate, let's take two time derivative of our center of mass. and since our masses are constant in time the derivative only acts on our [itex]\frac{d^2\mathbf{r_i}}{dt^2}=\frac{\mathbf{v_i}}[/itex] so our center of mass equation becomes:

    [itex]\frac{d^2\mathbf{R_{CM}}{dt^2}=\mathbf{A_{CM}}=\frac{\Sigma_i m_i \mathbf{a_i}{M}[/itex]

    Now if we times this by our total mass ([itex]M=\Sigma_i m_i[/itex]) we get:

    [itex]M\mathbf{A_{CM}}=M\frac{\Sigma_i m_i \mathbf{a_i}}{M}=\Sigma_i m_i \mathbf{a_i}[/itex]

    which we see is exactly the equation for the total net force of a system of particles [itex]\Sigma_i m_i \mathbf{a_i} = \Sigma_i \mathbf{F_i} = \mathbf{F_{total}}[/itex]

    so we see that by treating the system as a single giant particle of mass M and position [itex]\mathbf{R_{CM}}[/itex] we get the exact same result as if we'd just treated each particle individually. It's just a lot easier.
     
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