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Parallel Axis Theorem homework

  1. Aug 2, 2005 #1


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    Inertia - Parallel Axis Theorem

    Two spheres with masses M and m are connected by a rigid rod of length L and negligible mass as in Figure 10.22 (M is on the left end and m is on the right end). For an axis perpendicular to the rod, show that the system has the minimum moment of intertia when the axis passes through the center of mass. (The center of mass is halfway between the center and the left end. And x is the distance from the left end to the center of mass.) Show that this moment of intertia is I = uL^2, where u = mM/(m + M).

    If you have Physics for Scientists and Engineers, by Serway, this is #22 on p. 324.

    I'm having trouble with the parallel axis theorem. The rod alone has
    I = 1/3ML^2 with the axis passing through its end. I'm not sure how the spheres change the formula for intertia. So I have tried to set it up as I would for a rod with no spheres, and the axis passing through the end. And using the parallel axis theorem, I've come up with this..
    I_tot = 1/3ML^2 + M(L-x)^2
    = 1/3ML^2 + M(L^2 - 2Lx + x^2)

    From here, I'm lost. I have no idea how that could lead to I = mML^2/(m + M), as the book is suggesting in the problem. Thanks in advance for any help!

    Last edited: Aug 3, 2005
  2. jcsd
  3. Aug 2, 2005 #2
    you must remeber that the parallel-axis theroem states that I = Icm + MD^2
  4. Aug 3, 2005 #3


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    D = (L - x) in this problem.
  5. Aug 4, 2005 #4


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    Homework Helper

    This problem is not about the moment of inertia of a rod. The massless rod is doing nothing more than holding the two spheres at a fixed separation L. Since the sphere radii are not given, you must assume all of the mass is to be considered located at the centers of the spheres.
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