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Moment of inertia ?s

  1. Feb 20, 2005 #1
    Moment of inertia time in physics, hard as hell concept, but heres the question:

    You swing a small, heavy weight on the end of a light stick (the stick is light enough you can neglect its moment of inertia, and you can consider the weight as a point mass). Remember that the moment of inertia is a measure of how an object resists angular acceleration. How would the following actions effect the moment of inertia of what you are swinging

    A) doubling mass of weight
    B) adding an equal weight to the first at the center of the stick
    C) adding a weight that is 4 times the original mass to the center
    D) using a stick that isnt light, but weighs as much as the original weight on the end

    so, im using the equation I = mr^2, for A i got that if you double the mass, the moment of inertia is doubled as well. for B i believe that if you add a weight to the center of the stick, it shouldn't really affect it because it is not a distance away from the axis of rotation, same thing with C. For D i put that the moment of inertia increases because you get mass stick * r^2 + mass weight * r^2.

    Is what im saying here right? I think A and D are right but im quite unsure of B and C, any help would be appreciated
  2. jcsd
  3. Feb 20, 2005 #2


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    For A) your answer is correct.

    For B) and C) : the implication from the question is that you're grabbing the stick by one end and swinging the mass from the other end. If you add the mass to the center of the stick, you're adding it midway between the original mass and the axis of rotation. Can you figure out the correct expressions now ?

    For D) you need to know that for more complex masses than point masses situated at a certain distance from an axis, the expression for the moment of inertia is not simply [itex]mr^2[/itex]. For a uniform thin rod, the moment of inertia about the end is [itex]I = \frac{1}{3}mr^2[/itex], where r is the length of the rod (and the radius of the rotation). (To prove that you can use simple calculus). Now figure out the total moment of inertia of the rod-mass combination.
  4. Feb 20, 2005 #3
    Just some extra insight on that last idea:

    The same fundemental principle applies: you are still multiplying each point mass by the square of its distance from the axis. The difference for an object that is not itself a point mass is that now you have a collection of infinitely small point masses of infinite number. We express the moment of inertia (the infinite sum of all the moments of inertia of these theoretical point masses that make up the object) as a continuous distribution using the integration over the mass:


    This general expression is then used to develop the specific equations for moments of inertia of different objects.
  5. Feb 21, 2005 #4
    thank you both for your help
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