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Moment Of Inertia (calc vs physics different answer)

  1. Jul 25, 2012 #1
    I'm trying to understand Moment Of Inertia using integration.

    But it seems the calculus definition of physics definition are different.
    (I'm trying to apply my math skill to physics)

    example of apparent contradiction:

    it says the moment of inertia of "hoop about symmetric axis" is


    but here:
    page 21 of the book:
    http://www.eng.auburn.edu/~marghitu/MECH2110/C_4.pdf [Broken]
    says the moment of inertia of a circle centered at (0,0) about the z-axis is

    Pi r^4 /4

    does this mean the mass of a circle is "Pi r^2/4"

    my goal: be able to calculate the moment of inertia of a, x-y planar figure about the z-axis, given the parametric (or polar) equation of the curve.

    please help me solve the apparent contradiction between the calculus and physics definition of moment of inertia
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 25, 2012 #2
    The mass of the disk is the area times the density, ie. pi*r^2*density. This mass appears explicitly in the first expression, whereas in the second expression the author has substituted m for pi*r^2, implicitly assuming that the density is equal to 1.

    With this assumption, both expressions are correct (and equal).
  4. Jul 25, 2012 #3
    The first object is a thin hoop with no thickness, all the mass resides at a radius, r from the axis of rotation.

    The second object is is a flywheel in which the mass is distributed in a range from R1 to R2, so not all the mass is the same distance away from the axis of rotation.

    Good luck with those substitutions and geometry. I find that some of the objects that have symmetry can be quite a pain to find I. For myself, I accomplish doing one then look back a couple of years later and forget how I did it... so I redo it all over again.
    Last edited by a moderator: May 6, 2017
  5. Jul 25, 2012 #4
    thanks for the replies that helps, but i found another answer some where else:

    I found an equation here http://www.math24.net/physical-applications-of-line-integrals.html
    that looks like what I want, but that equation says the same circle around the z axis would have a moment of inertia of
    x[t_] := R Cos[t];
    y[t_] := R Sin[t];

    Integrate[ (x[t]^2 + y[t]^2)*Sqrt[D[x[t], t]^2 + D[y[t], t]^2] , {t, 0, 2 Pi}]

    0 2 pi(x[t]^2 + y[t]^2)*√( x'[t]^2 + y'[t]^2) dt=
    0 2 pi (cos(t)^2 + sin(t)^2)*√( sin(t)^2 + cos(t)^2) dt=
    2 Pi R ^3

    please help me solve the apparent contradiction between the calculus and physics
  6. Jul 25, 2012 #5
    I don't see how that correlates at all. You didn't quote an expression for I_z, you quoted the parabolic form of a circle! I've already explained why your two original expressions are actually equal, but I don't see where this one comes from, not even after searching your link.
  7. Jul 25, 2012 #6
    I'm thinking x= R cos(t), y= R sin(t) is a circle when t goes from 0 to 2Pi.

    at all times the distance from axis is r, and so r^2 = x^2 + y^2

    I thin this is the formula:
    ∫i(x[t]^2 + y[t]^2)*√( x'[t]^2 + y'[t]^2) dt

    because above it says:

    "The moments of inertia about the x-axis, y-axis and z-axis are given by the formulas"

    IZ = ∫C ( x2 + y2 ) rho(x,y,z) ds

    from how its done above(on the web page) I get that :
    ds = Sqrt[(dx/dt)2+(dy/dt)2] dt

    x = r cos[t] -> dx = -r sin[t] dt ; y = r sin[t] -> dy = r cos[t] dt;
    rho = 1;

    IZ = ∫C (r cos[t]2 + r sin[t]2 ) * Sqrt[(dx/dt)2+(dy/dt)2] dt

    IZ = ∫C (r cos[t]2 + r sin[t]2 ) * Sqrt[(-r sin[t] dt)2+(r cos[t] dt)2] dt

    IZ = ∫C (r cos[t]2 + r sin[t]2 ) * Sqrt[(-r sin[t] )2+(r cos[t])2] dt

    IZ = ∫C (r 2 ) * Sqrt[(-r sin[t] )2+(r cos[t])2] dt

    IZ = ∫C (r 2 ) * Sqrt[r2] dt

    IZ = ∫C (r 3 ) * dt

    t from 0 to 2pi

    IZ = 2 pi R^3
    Last edited: Jul 25, 2012
  8. Jul 25, 2012 #7
    This seems to be a different problem, where you're finding the moment of inertia of a wire, not a disk- is this what you want?
  9. Jul 25, 2012 #8
    I'm trying to answer: http://ocw.mit.edu/courses/physics/8-03-physics-iii-vibrations-and-waves-fall-2004/assignments/

    problem set 1 # 1.6

    there is an answer provided, but he assumes that the rod is the same as two pendulums.

    I'm trying to solve it directly using a line integral as the path of the rod.

    Specifically I want to know how the equation

    IZ = ∫C ( x2 + y2 ) rho(x,y,z) ds

    can arrive at the moment of inertia.

    i have the equation r = - R*sin(t), {t, -pi/6 to +pi/6}
    as the physical pendulum.

    I think it is:

    Integrate[MR^2 Sin[t]^2, {t, -Pi/6, Pi/6}],
    But i get an answer nowhere near the one in the solution provided.
  10. Jul 25, 2012 #9


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    Gold Member

    I suggest you consider an element distance s around the arc from the apex, the angle this subtends at the centre of curvature, and express the distance of the element from the pivot in terms of that angle.
  11. Jul 26, 2012 #10
    your circle has an equation

    r = 2Rsin(t)

    (flipping y=-y without loss of generality) furthermore,

    r2 = 4R2sin2(t)
    dr/dt = 2Rcos(t)
    (dr/dt)2 = 4R2cos2(t)

    your line element in polar coordinates is

    ds2 = dr2 + r2dt2


    (ds/dt)2 = (dr/dt)2 + r2 = 4R2cos(t) + 4R2sin2(t) = 4R2

    ds/dt = 2R
    ds = 2Rdt

    so we have a fairly simple line element

    the moment of inertia due to this line element is the distance from the origin squared times the mass of the line element. If we say the material has a density ρ (which we can work out from the given information) then the mass of the element is

    dm = ρds

    and the moment of inertia of the element is

    dI = r2ρds = 4R2sin2(t)*ρ*2Rdt = 8ρR3sin2(t)dt

    now we need to integrate this over the limits, which can be found in the helpful diagram provided: the right hand section is in the range t = [0, pi/6] and the left hand section is in the range [5pi/6, pi], but we don't need to use two separate integrals because sin(x) = sin(pi-x). The two integrals will therefore be equal and we can simply do the first one and multiply it by two:

    I = 2∫dI (from 0 to pi/6)

    substitute the expression for dI in,

    I = 16ρR3∫sin2(t) dt

    and thanks to wolfram alpha the result is:

    I = (2/3)ρR3(2pi - 3*sqrt(3))

    the total length is (2/3)*piR and the mass is m so the density of the material is

    ρ = (3m)/(2*pi*R)

    so the final value for I is

    I = mR2(2 - 3*sqrt(3)/pi)

    You can check the plausability of your answer by checking the units- moment of inertial must be in mass*length^2 units, so if your result is a nondimensional scalar times mR^3 then it's wrong.
    Last edited: Jul 26, 2012
  12. Jul 26, 2012 #11
    Thanks a lot. I need some time to digest this. I'll let you all know how it goes. Thanks again
  13. Jul 26, 2012 #12
    Mikey w, thank you so much. Your example makes sense. i now see the apparent contradiction between math and physics is that math examples take the density as 1, but in the case ofa Wire, density equal m over the length of the wire, then the answers are consistent. Thank you
  14. Jul 27, 2012 #13
    No problem- I learnt a few things trying to figure it out as well!
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