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Moments of inertia

  1. Nov 10, 2003 #1
    In general, I=[inte]R2dm. First of all, what is R? R is the component of the position vector that is perpendicular to both the angular and linear velocity. Is this correct?

    EDIT: One more question. If we know the moments of interia of say two bodies, how do we find the moment of inertia of the system? Is it simply the sum of the moments?
    Last edited: Nov 10, 2003
  2. jcsd
  3. Nov 10, 2003 #2
    R is the distance from the axis of rotation to the mass element dm.

    No, the moment of inertia of a two-body system is not in general the sum of the individual moments of inertia, because those moments of inertia were calculated about two different axes.

    If the axes of the two bodies used to calculate their individual moments of inertia are parallel to the axis of two body system, you can do this: find the distance between those axes and the axis of the two-body system, and use the parallel-axis theorem to recalculate the moments of inertia of the two bodies about this common axis. Then you can sum them.

    If the individual axes aren't parallel to the two-body axis, then you pretty much just have to recalculate from scratch.
  4. Nov 10, 2003 #3


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    Moment of inertia is a scalar quantity (analagous to mass) and does not depend on the velocity (though moment of inertia is defined with repect to an axis of rotation, but I don't want to be pedantic.) Basically, r is the distance from the origin (the location of the axis you're calculating for) to dm, that infinitesimal piece of mass.

    No (though in many cases, it turns out that way). In general, you will have to use the parallel axis theorem to sum inertias.
  5. Nov 10, 2003 #4


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    Crud, that's twice in a row I answered an answered question; at least my post count will go up.
  6. Nov 10, 2003 #5
    A few seconds difference, and I would have answered an answered question. :smile:
  7. Nov 10, 2003 #6
    Re: Re: moments of inertia

    That's what I originally thought. But when we were doing a problem with torque and pulleys and such today in class, my professor thought differently. There was a mass hanging from a pulley such that it applied a force perpendicular to a radius of that pulley that was parallel to the ground. My professor told me that the moment of inertia of the mass about the axis of the pulley is given by R2M, where M is the mass of the object and R is the length of the radius of the pulley.
    For rigid bodies, each mass element moves in a circle about the axis. So linear velocity is of course perpendicular to any line drawn from the rotational axis. For a two body system such as the on just described, the velocity is not necessarily perpendicular to any given mass element.
  8. Nov 10, 2003 #7
    Is it really meaningful to speak about the moment of inertia of that mass about the axis of the pulley? What are you getting at?

    (I suppose at any given instant, if you're thinking of the mass/pulley as a single system, it acts as if the mass is located at the point where the string meets the horizontal radius of the pulley, doesn't it?)

    Edit: If you're thinking of the moment of inertia of the mass by itself, that would be constantly changing as r changes. How would you use that?
    Last edited: Nov 10, 2003
  9. Nov 10, 2003 #8
    Blasted 51000 byte maximum... I would have posted my notes for you to see.
    Anyway. He solves the problem by considering only the net torque on the system by external forces. If the mass of the object hanging from the pulley is M, then Mg is an external force (considering the mass and the pulley as a system). Other external forces cancel out (eg, mass pulley with tension of support). The net torque is given by t=(Mg)R, where R is the radius of the pulley.
    t=MgR=Ineta, where a is the angular acceleration
    Inet=1/2mR2 + MR2=1/2R2(m+2M), where m is the mass of the pulley
    So, a = 2MgR/(R2(m+2M))=2Mg/(R(m+2M)), which apparently agrees with the result obtained by considering internal forces.
  10. Nov 10, 2003 #9
    Yes, I see it does work out to exactly the same result both ways. So it is acting as if the hanging mass is attached to the perimeter of the pulley (which does seem to make sense, doesn't it?) (Edit: if you think of it "instantaneously").

    That relationship you wrote
    I=1/2mR^2 + MR^2 is just what you would calculate for a disk of mass m and radius R with an extra mass M attached at a single point on its perimeter.
    Last edited: Nov 10, 2003
  11. Nov 11, 2003 #10
    So we maintain the definition of I for rigid bodies. That's very good news.
    And it doesn't have anything to do with velocity... I guess.
  12. Nov 11, 2003 #11
    The falling mass's velocity is 0 relative to the point we're looking at.
  13. Nov 11, 2003 #12
    Which is very hard to comprehend, the more I think about it, since it's position keeps changing.
  14. Nov 11, 2003 #13
    I'm not sure what you mean by this. Isn't the mass accelerating?
  15. Nov 11, 2003 #14
    Yes, but at any given instant, that point on the perimeter of the pulley has the same linear velocity and acceleration as the falling mass.

    But at each successive instant we have to change our point of reference, so we're always looking at a point at the end of a horizontal radius of the pulley.

    It's just a weird way of looking at the problem, but it leads to a very neat solution.
  16. Nov 11, 2003 #15
    Ah, good point, the point is accelerating too and at the same rate, because
    I think that's right - I just learned this - it's all so new
  17. Nov 11, 2003 #16
    Looks right.

    Getting back to magnetism, which is even more baffling.......

    Have fun.
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