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Help with calculus physics

  1. Dec 1, 2007 #1
    I have to do this proof

    Use the spherical and cylindrical coordinates to calculate the moment of
    inertia of a uniform solid sphere and uniform solid cylinder.

    The problem is I am taking calc 2, and don't have knowledge of cylindrical / spherical coordinates since I haven't taken calc 3

    Any help is greatly appreciated
     
  2. jcsd
  3. Dec 1, 2007 #2

    Ouabache

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    Welcome to PF forums!! If you haven't looked around yet, I believe you will find many interesting & information discussions around the forum. There are also helpful folks who can aim you in a successful direction, when you have questions. Particularly on homework, it is a good idea to look at some of the examples in this thread to get an idea on how to construct a question.
    A poorly constructed question, will get little response.

    I don't know too much about proofs. But I have had calc3 and can get you started on understanding cylindrical & spherical coordinates.

    All three systems may be converted from one to another. The Cartesian system, with 3 dimensions [itex](x,y,z)[/itex]

    The cylindrical coordinate system, is like using polar coordinates in one plane [itex] (r,\theta) [/itex] and extending into the third dimension with displacement z. So a point in space may be specified by [itex](r,\theta,z)[/itex]

    With spherical coordiates: a point in space is given with two angular directions and one radial distance. The radial distance r is like you use for polar coordinates (distance from the origin), [itex]\theta[/itex] is the angular displacement taken in the horizontal plane, as you do with polar coordinates, and [itex]\phi[/itex] is the angular displacement from the z-axis (from straight up).
    So you have [itex](r,\theta,\phi)[/itex]

    For moments of inertia, if you don't already have this part explained, you might investigate the concept at hyperphysics website. They give examples of common moments of inertia.
     
    Last edited: Dec 1, 2007
  4. Dec 1, 2007 #3
    Thanks for the quick reply

    I guess it is not really a proof, but the question was worded vague so I really cannot improve it

    The moment of inertia of a solid sphere is
    [tex]2/5 MR^2[/tex]


    But how would I go at calculating it using the cylindrical coordinates or spherical coordinates?

    I been browsing about the cylindrical and spherical and I'm getting the concept, but I still don't know how to start the problem
     
  5. Dec 1, 2007 #4
    I think I found the derivation in spherical coordinates

    https://www.physicsforums.com/showthread.php?t=159293

    I gotta look for the cylindrical coordinates, and then I gotta study everything about the two systems :D


    Edit the cylinder would end in [tex]1/2 *MR^2[/tex]
     
    Last edited: Dec 1, 2007
  6. Dec 1, 2007 #5

    Ouabache

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    Good to see you are making some progress.

    I took a look at that thread you pointed to, do you understand their derivation? You can use a similar process to derive in cylindrical coordinates. Since you know the end solution, you have something to aim for. It is not a good idea to ask arunma for his derivation. This is your homework. Our policy is to help you solve a problem, but only after you show some effort.
     
  7. Dec 1, 2007 #6

    I just need the first part, I understand now the triple integration and the coordinate systems, I just don't seem to know where to start from

    eg

    to derive the moment of inertia of a solid sphere of uniform density and radius R.

    [tex] dI = r_{\perp}^{2} dm = r_{\perp}^{2} \rho dV [/tex]

    I need this part for the cylinder
    It seems the challenge for triple integrals is in setting them up, but any clues or even explanations of this part for the spherical proof is appreciated; so I can do the cylindrical .
     
    Last edited: Dec 1, 2007
  8. Dec 1, 2007 #7
    for the density function of a triple integral of the cylinder in cylindrical coordinates, what would be the limits of integration for the dz part?

    [tex]\int_0^{2\pi} \int_0^R \int_0^? (Kr) r \,dz\,dr\,d \theta[/tex]

    And would the density function be [tex]K\sqrt {x^2 + y^2} = Kr[/tex] ?
     
    Last edited: Dec 1, 2007
  9. Dec 2, 2007 #8

    cristo

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    What are you doing? What is K? You should use one coordinate system throughout your calculation.

    You have [itex]dI=r^2dm=r^2\rho dV[/itex] in cylindircal coordinates. Can you continue from here? You need to find the volume element and then integrate.
     
  10. Dec 2, 2007 #9
    So the density function for the cylinder is the same as for the sphere? [tex] r^2\rho dV[/tex]
    What would the limit of integration be for the dz part of the triple integral?

    Sorry if the questions seem trivial, I need more calc 3

    The K part was just a constant
     
  11. Dec 2, 2007 #10

    cristo

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    That's not the density function; that's the moment of inertia differential (dI).
    What do you think? What is the z coordinate measuring in the case of a cylinder?
     
  12. Dec 2, 2007 #11
    height, so it has to be from 0 to something, just don't know what the height should be to get the [tex]1/2 * MR^2[/tex]
     
  13. Dec 2, 2007 #12

    cristo

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    Well, it's whatever the height of the cylinder is. You can't just try and fudge things together to get the right answer; you need to sit down and do some work. Try and write down the integral you are trying to perform first.
     
  14. Dec 2, 2007 #13
    Sure, just getting used to putting in the tex code

    Btw I have no idea if I am even on the right track

    [tex]\int_0^{2\pi} \int_0^R \int_0^h \rho r^2 \,dz\,dr\,d \theta[/tex]


    [tex]= \rho * h \int_0^{2\pi} \int_0^R r^2 \,dr\,d \theta[/tex]

    [tex]= (\rho * h)/3 \int_0^{2\pi} r^3 \,d \theta[/tex]

    [tex]= 1/3 *\rho h {2\pi} r^3 [/tex]
     
  15. Dec 2, 2007 #14

    cristo

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    dV in cylindrical polars is [itex]rdrdzd\theta[/itex], so then the integral is[tex]\int\int\int\rho r^3 dz dr d\theta[/tex]. That's not gonna change much, so your solution is [tex]I=\frac{2\rho h\pi R^4}{4}[/tex]. Can you simplify this to the answer you quote?
     
  16. Dec 2, 2007 #15
    so
    [tex]\rho h \pi=M/R^2[/tex]?


    Those equations weren't even given to us :|
     
  17. Dec 2, 2007 #16

    cristo

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    Of course they were. You know that density=mass/volume, and you know that the volume of a cylinder is (pi r^2 h); or at least you should do.
     
  18. Dec 2, 2007 #17
    Yes to the volume part, but I have been to every class and this is the first time I even found rho in a problem
    I guess this would be a physics two problem?

    Thanks everyone in the thread for the input
     
    Last edited: Dec 2, 2007
  19. Dec 2, 2007 #18

    cristo

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    I don't know what is taught in each year over there (I'm from the UK) and thus "calc 3" and "physics 2" means nothing to me. I just presumed that since this is in the intro physics forum, that you'd know what the density is. If not, well, at least you've learned something!
     
  20. Dec 3, 2007 #19

    Ouabache

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    Thanks cristo for jumping in there. cotufa, i could only take you so far to help you understand the coordinate systems. I was hoping someone with stronger physics background than myself could assist in the derivation. That is the benefit of this forum. Many people can contribute and hopefully bring you to a successful result.
    cristo thanks for adding insight to this problem. Showing the relationship from first principles, makes it much clearer.

    cotufa, were you to choose which coordinate system to use depending on the solid body? It looked like you derived the moment of inertia for the solid cylinder using cylindrical coordinates only, and for the solid sphere, using spherical coordinates only.

    [itex]\rho[/itex] is explained in this reference, the "continuous mass density function". Good luck with the rest of your physics questions :smile:
     
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