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Moment of inertia | Integral form

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Starting from the sum: I=Ʃ mαα2 and replacing it by the appropriate integral, find the moment of inertia of a uniform thin square with side length 2*b, lying in the x-y plane, rotated about the x-axis. Calculate its moment of inertia.


    2. Relevant equations
    The integral form turns to I=∫ρ(r)*r2dV


    3. The attempt at a solution
    I rotated it around the z-axis, giving me a cylinder with radius 2b and height 2b. I know the volume of this cylinder is 2*∏*r2*h. And that ρ=mass/volume. I need help in setting up and evaluating the integral because what I got seems wrong.

    I used cylindrical coordinates, claiming that:
    θ goes from 0 to 2∏
    r goes from 0 to 2b
    z goes from 0 to 2b

    and my volume element dV=r*dr*dz*dθ. I end up with mass*b2, and I do not believe this is correct. Help please?
     
  2. jcsd
  3. Jan 29, 2013 #2

    Dick

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    You have the integral wrong because you have the concept wrong. You don't want to integrate over the cylinder, you want to integrate over the square in the x-y plane. Since they say the square is thin you can ignore the z coordinate and make it a 2d integral. The m_a should be replaced by the surface mass density of the square. The ρ should be replaced by the distance of a point in the square to the x-axis. And you don't want to do polar coordinates, cartesian will do fine. Try thinking about it that way for a bit.
     
  4. Jan 29, 2013 #3
    The problem actually has one more line that I neglected to add, it goes: rotated about the x-axis and passing through its center. I have no idea what they mean by "passing through its center".

    I am very confused why this is a 2d problem, the problem tells me to rotate the square, so that makes me think cylinders and volumes. Is there a reason why this is 2d? Am I missing something?
     
  5. Jan 29, 2013 #4

    Dick

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    Moment of inertia is usually used for calculating rotational motion around an axis. The moment of inertia depends on the axis you rotate around. So the phrase "find the moment of inertia of the square rotated around the x-axis" doesn't mean "find the moment of inertia of the volume generated by rotating the square". It means find the moment of inertia of the square when it's rotated around the given axis.
     
  6. Jan 29, 2013 #5
    oh my god im so stupid! I was stuck in calc 2 mode for some awful reason!

    Ok I will give that a shot and hopefully get it, thank you very much Dick!
     
  7. Jan 29, 2013 #6
    So here is what I got: ∫(from 0 to 2b)M/A*x2dx

    This turns out to be 2/3*M*b.

    The answer in the back of the book says it should be 2/3*M*b2. What did I do wrong here?
     
  8. Jan 29, 2013 #7

    Dick

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    A couple of things. For one, the axis goes through the center of the square. You should by integrating from -b to b. For another, it should be a double integral dx*dy. For a third thing, I seem to keep getting a different answer from the book answer. What do you get?
     
    Last edited: Jan 30, 2013
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