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Parallel-Axis Theorem Clarification

  1. Apr 3, 2008 #1
    Hey guys, we are currently studying chapter 10 in the Fundamentals of Physics 8th. Ed. Halliday & Resnick Textbook and I am having a little problem understanding the Parallel-Axis Theorem. I understand that the Moment of Inertia (Rotational Inertia) is derived from the Kinetic Energy Formula for Rotational Motion of a rigid body. However, the Parallel-Axis Theorem allows us to formulate the Moment of Inertia of a particular rigid body about an axis parallel to the axis of the center of mass. What I do not understand is why the formula for the Parallel-Axis Theorem has Icm + MD^2??? Why do you include the Icm and not just take the differential mass element at some distance "r" from the axis of rotation and calculate it that way - is this way more complicated? The way I interpreted the Icm part is that it represents the differential mass element of the rigid body right at some distance "r" from the new axis of rotation right? I totally understand the whole Moment of Inertia is I = integral r^2 dm which is the same thing as the MD^2 portion of the parallel-axis theorem right? I guess I am just trying to figure out why there is that plus in the middle of the formula and what it all means? Could someone please help clarify this for me because its just confusing to me. Thanks.
  2. jcsd
  3. Apr 4, 2008 #2
    When a body rotates about an axis parallel to its centroidal axis, not only do you need to move that mass around the parallel axis (which can be considered a point mass at a distance r from the rotation axis), but it also needs to rotate about its own centroidal axis, which means you need to add the centroidal moment term.
  4. Apr 4, 2008 #3
    All you need is this two formulas:
    1. Integral[alfa*f+beta*g]=alfa*Integral[f]+beta*Integral[g]
    2. Integral[(x-<x>)*dm]=0 (definition of center of mass <x>)

    Below I used x' for (x-<x>) and y' for (y-<y>):

    Integral[r^2*dm]=Int[x^2+y^2*dm]=Int[( (x'+<x>)^2+(y'+<y>)^2 )*dm]=
  5. Apr 11, 2008 #4
    Could you clarify this a little better? Is the x the distance along the x axis to the particle? If so, then you are subtracting out the location on the x axis of the CM, <x>, right? Which is basically the distance along the x axis from the CM to the particle.

    Now comes my confusion, why does the integral of {(the distance from the CM to the particle)*(mass of the particle)} equal zero. Why not int(x'*dm) = (x' as constant) x'*m?

    Is the definition of center of mass supposed to tip me off that we can think of it as having no mass there (x') and all being concentrated at the CM? Therefore int(x'*dm) = x'*m, but the m is zero everywhere except x'=0, which of course makes the expression zero everywhere.

    This just feels a little fishy to me, at one point (when calculating the moment of inertia) the Center of Mass is not brought into the reasoning (except to find the axis to measure from). For instance when we calculate the moment of inertia as a sum of all the infinitely small masses (dm) multiplied by the square of their respective distances (r), in other symbols integral(r^2 * dm), we do not think of the mass concentrated at the CM. Of course it would be a trivial calculation if we considered all of the mass to be at the origin of the calculation, the integral would be zero.

    I feel like I am missing something so I am just wanting to get the specifics of the interpretation correct. Thanks.
    Last edited: Apr 11, 2008
  6. Apr 12, 2008 #5
    The reason for usefullness of average values is easier evaluation of the sum or integral with which that average was defined.
    Center of mass is defined as:

    <x>=Integral[x*dm]/m (m=Integral[dm]=total mass)

    This means we can use <x> to evaluate the integral:


    The equation I used follows:


    In general it is not correct to replace a variable in an integral by it's average value. This is possible only if the average was defined with that integral.
  7. Apr 12, 2008 #6
    Ah, I see it now. Once you take the integral of the entire object, by definition of the center of mass, every m*x' will inevitably have an equal but opposite -(m*x') that then cancels everything to zero. I'm with you on that now, I appreciate the help.
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