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Mistake on wikipedia (Moment of inertia of a quadrant)

  1. Oct 4, 2012 #1
  2. jcsd
  3. Oct 4, 2012 #2
    Would you like to display your reasoning?
     
  4. Oct 4, 2012 #3

    vanhees71

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    Well, let's check :-). The quartercircle is described by the parametrization in polar coordinates
    [tex]\begin{pmatrix}
    x \\ y
    \end{pmatrix}=
    \rho \begin{pmatrix}
    \cos \varphi \\ \sin \varphi
    \end{pmatrix}, \quad \rho \in [0,r], \quad \varphi \in [0,\pi/2].
    [/tex]
    The moment of inertia thus is
    [tex]
    I=\int_F \mathrm{d} F y^2 = \int_0^r \mathrm{d} \rho \int_0^{\pi/2} \mathrm{d} \varphi \; \rho^3 \sin^2 \varphi.
    [/tex]
    The integral over [itex]\rho[/itex] is straight forward:
    [tex]
    I=\frac{r^4}{4} \int_0^{\pi/2} \mathrm{d} \varphi \; \sin^2 \varphi \qquad (1).
    [/tex]
    To evaluate the integral over the angle, note that
    [tex]\cos(2 \varphi)=\cos^2 \varphi-\sin^2 \varphi=1-2 \sin^2 \varphi[/tex]
    and thus
    [tex]\sin^2 \varphi=\frac{1-\cos(2 \varphi)}{2}.[/tex]
    Integrating this over [itex]\varphi[/itex] gives
    [tex]
    \int_0^{\pi/2} \mathrm{d} \varphi \; \sin^2 \varphi=\left . \frac{\varphi}{2}-\frac{\sin(2 \varphi)}{4} \right |_{\varphi=0}^{\pi/2}=\frac{\pi}{4}.
    [/tex]
    Plugging this into (1) gives
    [tex]
    I=\frac{\pi r^4}{16},
    [/tex]
    which is what's claimed on Wikipedia.
     
  5. Oct 4, 2012 #4
    Are you sure you didn't simply look at the wrong entry in a table?

    Your figure would be correct for the quadrant product of inertia or the half circle moment of inertia.
     
  6. Oct 4, 2012 #5
    The moment of inertia of a disc of radius "r" and mass "M" = (Mr^2)/2

    A quadrant is one fourth the mass of that, thus the moment of inertia would be :

    (Mr^2)/8

    @vanhees71
    Would u please define the variables you used in ur proof :

    d, F, φ, ρ
     
  7. Oct 4, 2012 #6
    Why do you say that?
     
  8. Oct 4, 2012 #7
    You do realise that the moment of inertia for a disc you propose is the polar moment of inertia (ie the moment about the Z axis).

    Wikipedia and Vanhees are talking about the diametral moment of inertia in the plane of the disk, the moment of inertia about the X and Y axes for instance.

    Edit

    I am happy to discuss this, but you don't seem to want to?
     
    Last edited: Oct 4, 2012
  9. Oct 4, 2012 #8

    vanhees71

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    First of all, perhaps I'm misreading the Wikipedia, but I thought this thread is about the entry of the table labeled in the first column as

    "a filled quarter circle with radius r entirely in the 1st quadrant of the Cartesian coordinate system"

    Further, I assumed that the axis of rotation under consideration is the line with the arrow as depicted in the second column, i.e., my [itex]x-[/itex]-Axis.

    I just use standard poloar coordinates [itex](\rho,\varphi)[/itex] with [itex]\rho[/itex] the distance of the point under consideration from the origin and [itex]\varphi[/itex] the angle measured from the [itex]x[/itex]-axis of a Cartesian coordinate system. Then the Cartesian coordinates of a point are given by
    [tex]
    x=\rho \cos \varphi, \quad y=\rho \sin \varphi.
    [/tex]
    The area element of a parallelogram spanned by coordinate lines then obviously is
    [tex]
    \mathrm{d} F=\mathrm{d} \rho \; \mathrm{d} \varphi \rho.
    [/tex]
    This you also can derive formally (i.e., without basic geometry) using the Jacobian:
    [tex]
    \mathrm{d} x \; \mathrm{d} y = \mathrm{d} \rho \; \mathrm{d} \varphi \; \mathrm{det} \frac{\partial(x,y)}{\partial(\rho,\varphi)}.
    [/tex]
    The Jacobian matrix is straight-forwardly calculated by taking the derivatives
    [tex]
    J=\frac{\partial(x,y)}{\partial(\rho,\varphi)}=
    \begin{pmatrix}
    \cos \varphi & -\rho \sin \varphi \\
    \sin \varphi & \rho \cos \varphi
    \end{pmatrix}
    \; \Rightarrow \; \mathrm{det} J=\rho.
    [/tex]
     
  10. Oct 4, 2012 #9
    Vanhees, there's nothing wrong your with your analysis, it's just that the OP hasn't read the first entry in his own first reference (Wikepedia table of Moments of Inertia) properly.

    There it clearly states

    Which is Ixx

    and the first entry clearly states

    I0 = ∏r4 / 4 not as in his post#5 : I think we can forgive the mix up between second and fourth powers as a typo.

    So the OP has obtained his Moment from another source and it is the Polar moment

    J = Ixx + Iyy = ∏r4 / 4 +∏r4 / 4 = ∏r4 / 2
     
    Last edited: Oct 4, 2012
  11. Oct 4, 2012 #10
    I am sorry for the confusion, i forgot to mention the axis of rotation.

    I am assuming one side of the quadrant as the axis of rotation .
    @vanhees71 , so the axis of rotation is the x axis (as shown in the wikipedia link here :

    http://en.wikipedia.org/wiki/File:Ar...rtercircle.svg [Broken]




    If what the wikipedia page says that the moment of inertia is about the Center of Mass of the Quarter Circle, then i can safely say that wikipedia answer is correct.

    However the issue remains , my axis of rotation is the x axis



    @Studiot , i believe that u are applying the Perpendicular Axes Theorum for calculating the moment of inertia ( http://en.wikipedia.org/wiki/Perpendicular_axis_theorem )

    J = Ixx + Iyy = ∏r^4 / 4 +∏r^4 / 4 = ∏r^4 / 2

    Where Ixx and Iyy are both in the plane of the Quarter Circle (and represent the X and Y axes Respectively)

    I Assume that what u referred to as "J" is the Z axis perpendicular to the plane of the Quarter Circle.

    I still dont quite get how you proved it.

    Is there any other way (other than polar coordinates) to prove it ?
     
    Last edited by a moderator: May 6, 2017
  12. Oct 4, 2012 #11
    You should pay attention to both the Maths and the English, success in both subjects comes as a result of attention to detail.

    No, J is not an axis.

    J is the moment of inertia of the disk about the Z axis and is often required in torsion, torque or rotating machinery calculations.
    It is called the polar moment of inertia.
    Polar coordinates have nothing directly to do with J the polar moment of inertia.

    Yes the Z axis is perpendicular to the plane of the quarter circle.

    No I didn't prove anything, merely stated some results.

    In particular the value (formula) you stated in post#5 for the moment of inertia is for J, not the moment of inertia about any axis in the plane of the quarter circle.

    Yes you can use rectangular coordinates and do the integration that way

    Since X2 + Y2 = r2 for any radius r

    Y2 = r2 - X2

    Ixx = ∫Y2dA (The definition of the moment of I about the x axis)

    This is a double integral you need to substitute for Y into and integrate it from x= 0 to x=R and Y=0 to y=R. You will find the rectangular integral tricky which is why polar coordinates are used.
     
  13. Oct 4, 2012 #12

    vanhees71

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    No! At the Wikipedia page they give the moment of inertia wrt. an axis in the plane of the body. It's given by the axis with an arrow drawn in the second column of the table. What I've calculated was the quarter-circle disk with the axis of rotation identical to one of its straight edges.

    Wikipedia also gives the result for a cicular disk with the axis of rotation in the plane of the disk going through the center of mass. As is said there, they use Steiner's Law. So for the second case we only need to calculate the center of mass of the quarter-circle disk. With the same coordinates used in my previous posting it's given by
    [tex]
    \vec{x}_{\text{cm}}=\frac{4}{\pi r^2} \int_0^r \mathrm{d} \rho \int_0^{\pi/2} \mathrm d \varphi \; \rho^2 \begin{pmatrix}
    \cos \varphi \\ sin \varphi
    \end{pmatrix}.
    [/tex]
    The angle integrals give
    [tex]
    \int_0^{\pi/2} \mathrm{d} \varphi \; \cos \varphi=1, \quad \int_0^{\pi/2} \mathrm{d} \varphi \; \sin \varphi=1.[/tex]
    This gives
    [tex]
    \vec{x}_{\text{cm}}=\frac{4 r}{3\pi} \begin{pmatrix}
    1 \\ 1
    \end{pmatrix}.
    [/tex]
    Steiner's Law then tells us that
    [tex]I_{xx}=\frac{\pi r^4}{16}=I_{\text{cm}}+\frac{\pi}{4} r^2 y_{\text{cm}}^2=I_{\text{cm}}+\frac{4 r^4}{9 \pi}[/tex]
    Solving the equation for the area momentum of inertia around the cm axis, finally gives
    [tex]
    I_{\text{cm}}=r^4 \left (\frac{\pi}{16}-\frac{4}{9 \pi} \right ),
    [/tex]
    again in accordance with Wikipedia :-).
     
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