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Homework Help: Moment Of Inertia Of Non-uniform Rod

  1. Feb 8, 2015 #1

    I am trying to find the moment of inertia of a uniform rod, that has a mass added to it at some position along it's length, which is equal to the mass of the rod itself, and the axis of rotation is at one end.

    1. The problem statement, all variables and given/known data

    A uniform, [itex]\mathrm{1.00m}[/itex] stick hangs from a horizontal axis at one end and oscillates as a physical pendulum with period [itex]T_{0}[/itex]. A small object of mass equal to that of the stick can be clamped to the stick at a distance [itex]y[/itex] below the axis. The system then has a period [itex]T[/itex].

    Find the ratio [itex]\frac{T}{T_{0}}[/itex]

    2. Relevant equations

    3. The attempt at a solution

    I know that the moment of inertia of a uniform rod with the axis about one end is equal to;

    [itex]I = \frac{1}{3}ML^{2}[/itex]

    The period of a physical pendulum is given by;

    [itex]T = 2 \pi \sqrt{\frac{I}{mgd}}[/itex] where d is the distance from the pivot to the center of gravity. So, for the initial case, we have a period of

    [itex]T_{0} = 2 \pi \sqrt{\frac{\frac{1}{3}ML^{2}}{mg\frac{L}{2}}} = 2 \pi \sqrt{\frac{2L}{3g}}[/itex]

    I am stuck here though, because I don't know how to find a moment of inertia for the second case. I understand how to calculate moments of inertia of non-uniform rods using integration, but I don't have a function for the linear density...

    Thanks for any help you can give!
  2. jcsd
  3. Feb 8, 2015 #2

    Doc Al

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    Staff: Mentor

    What's the second case? Where the mass is stuck to the rod? What's the moment of inertia of a point mass? Just add the separate moments of inertia to find the total.
  4. Feb 8, 2015 #3
    Thanks, I was making that far more complicated in my head than it needed to be.

    How do I describe the center of mass of such an object? I need to find an expression that describes the center of mass as a function of y, the masses distance along the rod.
  5. Feb 8, 2015 #4


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    Homework Helper

    You treat it as a point mass i.e. the mass itself is dimensionless so it has no radius and so on to consider. So you use your basic definition of I for a point mass.

    What will change is your center of gravity of the new system.
  6. Feb 8, 2015 #5

    Doc Al

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    Staff: Mentor

    You can find the center of mass of each separately (which is trivial, of course). Then treat them as two point masses for the purpose of find the center of mass of the system.
  7. Feb 8, 2015 #6
    So I have my uniform rod, and I say that the left end is at 0 and the right is at L. I treat the rod as a point mass at L/2. I then consider another point mass at position y, where y is between 0 and L somewhere along the length of the rod.

    So the center of mass of the system is [itex]CM = \frac{M\frac{L}{2} + My}{2M}[/itex]
  8. Feb 8, 2015 #7

    Doc Al

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    Staff: Mentor

  9. Feb 8, 2015 #8
    Thank you.
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