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Calculation of Moments of Inertia

  1. Nov 30, 2004 #1

    Bri

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    A uniform thin solid door has height 2.20 m, width .870 m, and mass 23.0 kg. Find its moment of inertia for rotation on its hinges. Is any piece of data unnecessary?

    So far, I don't understand how to calculate moments of inertia for things like this at all. I can do a system of particles, but when it comes to any ridgid objects, such as this door or rods or cylinders, I don't get it.
    So basically I have no idea where to even start with this.
     
  2. jcsd
  3. Nov 30, 2004 #2
    Ok. Here is the moment of inertia equation for a, thin rectangular sheet, axis along one edge. (This is the door, hinge on one side)

    I = (1/3)(mass)(length)^2

    Length is the Width that you are given in the equation.

    You should be able to get it from there.
     
  4. Nov 30, 2004 #3
    My book is open to the page with your answer this very second:) Mine has it different than the preceding post though, if you're curious, says the MOI of a rectangular thin plate of length l and width w is (1/12)(mass)(l^2+w^2) if that helps at all

    Also, my book handily lists how to find the MOI of a variety of objects like that, so I could only guess yours might too
     
  5. Nov 30, 2004 #4
    Can you calculate

    [tex]I = \int_V r^2 \rho dV[/tex]

    ?

    If you can't, you'll have to either learn how to or look up the formula in a book.

    --J
     
  6. Nov 30, 2004 #5

    Doc Al

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    Staff: Mentor

    That's the rotational inertia of a thin plate rotated about an axis perpendicular to the plate and through the center of mass. That's not what's needed here.

    For this problem, Nonok gave the correct formula.
     
  7. Nov 30, 2004 #6

    Doc Al

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    Staff: Mentor

    Bri, to calculate moments of inertia for solid objects, you need to integrate. Start here for some examples: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mig
     
  8. Nov 30, 2004 #7
  9. Nov 30, 2004 #8
    What about if the density varies with the radius? inversely I mean. what would you integrate then... hypothetically speaking?
     
  10. Nov 30, 2004 #9

    Doc Al

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    Staff: Mentor

    The setup of the integral would be exactly the same. But solving it would be harder. :smile:
     
  11. Nov 30, 2004 #10

    Bri

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    First of all, thanks to everyone for your input.

    I can do integration, but that equation there doesn't really make any sense to me...
    What is the V? Is it supposed to be a definite integral from V to... something?
    Other than the V thing, wouldn't it be [tex]r^2 \rho V[/tex]? (r and [tex]\rho[/tex] are constants, right?)

    On the site Doc Al linked to (http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mig), they kind of lose me on the integration for the cylinder and the sphere...
    On the first page it says to integrate [tex]r^2 dm[/tex] over the mass, from 0 to M. Then they don't do that for the cylinder and sphere. I don't really understand what they do, though.
     
    Last edited: Nov 30, 2004
  12. Nov 30, 2004 #11
    [itex]\rho[/itex] is not necessarily a constant. It could just as well depend on position.

    V is the volume of whatever you're trying to find the moment of inertia for. The V under the integral is just a notation saying you have to choose your limits of integration such that you integrate over the entire volume. The dV is just notation saying that this is a volume integral, and that dV will be replaced by the volume differentials appropriate for your coordinates. For instance, dV for cartesian coordinates is just dx dy dz. For spherical coordinates, we have [itex]r^2 \sin{\theta} dr d\theta d\phi[/itex].

    Refer to a calculus textbook for more.

    As for your question about dm, [itex]dm = \rho dV[/itex]. The mass density times a bit of volume gives you the bit of mass.

    --J
     
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