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Help with proving the parallel axis theorem

  1. Jun 27, 2009 #1
    Hi, this is my first post here, so please forgive any errors in my post. :)

    I'm recently thinking about switching from my major of computer science to physics, and have been brushing up on the first few semesters of physics I had taken a few years ago. I'm currently in the section on rotational motion and moments of inertia, and was looking at the parallel axis theorem and the proof they provided. I had a question about one of the parts of the proof, so I'll list what was written in my book (see attached image for reference):

    View attachment 19484

    Suppose that an object rotates in the xy-plane about the z-axis. The coordinates of the center of mass are [tex] x_{CM}, y_{CM} [/tex]. Let the mass element dm have coordinates x, y. Because this element is a distance [tex]r = \sqrt{x^2 + y^2}[/tex] from the z-axis, the moment of inertia about the z-axis is:

    [tex]I = \int r^2 dm = \int \left(x^2 + y^2\right) dm[/tex]

    However, we can relate the coordinates x, y of the mass element dm to the coordinates of this same element located in a coordinate system having the object's center of mass as its origin. If the coordinates of the center of mass are [tex]x_{CM}, y_{CM}[/tex] in the original coordinate system centered on O, then from the attached figure we see that the relationships between the unprimed and primed coordinates are [tex]x = x' + x_{CM}[/tex] and [tex]y = y' + y_{CM}[/tex] Therefore,

    [tex]I = \int \left[\left(x' + x_{CM}\right)^2 + \left(y' + y_{CM}\right)^2\right] dm[/tex]
    [tex]I = \int \left[\left(x'\right)^2 + \left(y'\right)^2\right] \,dm + 2x_{CM} \int x' \,dm + 2y_{CM} \int y' \,dm + \left({x_{CM}}^2 + {y_{CM}}^2\right) \int \,dm[/tex]

    The first integral is, by definition, the moment of inertia about an axis that is parallel to the z-axis and passes through the center of mass. The second two integrals are zero because, by definition of the center of mass,

    [tex]\int x' dm = \int y' dm = 0[/tex]


    The last integral is simply [tex]MD^2[/tex] because [tex]D^2 = {x_{CM}}^2 + {y_{CM}}^2[/tex] and

    [tex]\int dm = M[/tex]

    Therefore, we conclude that

    [tex]I = I_{CM} + MD^2[/tex]

    Now, I must admit that all of this proof totally makes sense, with the exception of the bold text "The second two integrals are zero because, by definition of the center of mass" followed by the integrals of [tex]x' dm[/tex] and [tex]y' dm[/tex] being equal to zero. This part stumps me - why are the equal to zero? I'm not sure what they mean by "by definition of the center of mass" and how it makes those integrals equal to zero. Could someone please explain why those two integrals are equal to zero? Thank you in advance! :)

    -Keith
     
  2. jcsd
  3. Jun 27, 2009 #2

    G01

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    Recall that the center of mass coordinates are given by:

    [tex]X_{CM}=\frac{1}{M}\int x dm[/tex]

    and


    [tex]Y_{CM}=\frac{1}{M}\int y dm[/tex]

    Thus, the center of mass coordinates in the primes reference frame (the ref. frame with the CM as the origin) are:

    [tex]X'_{CM}=\frac{1}{M}\int x' dm[/tex]

    and


    [tex]Y'_{CM}=\frac{1}{M}\int y' dm[/tex]

    So, what are the CM coordinates in the frame where the CM is the origin? What does this mean, then, about the integrals you are asking about?
     
  4. Jun 27, 2009 #3
    Thanks for your reply, G01. In the frame where the CM is the origin, the CM coordinates are (0, 0). I'm guessing what you mean is that:

    [tex]2x \int x' dm = 2(0) \int x' dm = 0[/tex]
    [tex]2y \int y' dm = 2(0) \int y' dm = 0[/tex]

    If I am on the right track, how do we know that we are treating the center of mass as the origin there? I'm sorry, I'm still a little confused - I guess I'm still trying to put it all together. :/
     
  5. Jun 28, 2009 #4

    G01

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    No that is not what I mean.

    X_cm and Y_cm are the coordinates of the CM in the first frame, where the CM is not at the origin. So, they can't be zero.

    Think about it again.

    I know the notation is tricky but you have to keep all variables straight in order to understand this proof. Remember which variables are from which reference frame and you should be fine. Unprimed coordinates are from the first reference frame. Primed coordinates are coordinates from the second reference frame where the CM is at the origin:

    The CM x coordinate in the second frame with the CM at the origin is:

    [tex]X'_{cm}=\frac{1}{M}\int x' dm = 0[/tex]

    X'_cm is equal to zero because, in the second reference frame, the CM is at the origin, meaning X'_cm ,the CM x coordinate in the second reference frame, is equal to zero.

    Now, what does the expression above for X'_cm tell you about the integral:

    [tex]\int x' dm[/tex]

    HINT: Solve for the integral. Similar logic shows the second integral with y' to be zero as well.
     
    Last edited: Jun 28, 2009
  6. Jun 28, 2009 #5
    Okay, I think I get it, but I wanted to run my reasoning by you real quick to make sure that I'm okay. :) (This will probably be a rehash of what you've been explaining, but it'll probably make me feel better and more sure of myself.)

    [tex]$\int x' dm = 0$[/tex] because if we look at the definition of the center of mass for the x-axis using the second reference frame (where the center of mass is the origin), then we get [tex]$\tfrac{1}{M} \int x' dm = 0$[/tex] This is because finding the center of mass when the origin is centered on it will just produce the origin again. Since [tex]$\tfrac{1}{M}$[/tex] is just a constant multiplier, then the integral by itself must be equal to zero, therefore the expression in the proof will be equal to zero as well.

    Is my reasoning okay? I think I didn't really understand what integrating x' and y' meant, and how to relate it to the center of mass equation. Please, just let me know if my logic is okay, and thank you very much for your help! :)
     
  7. Jun 28, 2009 #6

    G01

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    Yes, I think you got it now. :smile:
     
  8. Feb 21, 2010 #7
    Can you repost the attached image? I want to work out this proof and the link is broken. :\
     
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