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Moment of Inertia

  1. Oct 25, 2005 #1
    I couldnt find the Moment of inertia of the next tube:

    moment.jpg

    hope you can help..thnks
     
  2. jcsd
  3. Oct 25, 2005 #2

    Tide

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    What, exactly, did you try to do?
     
  4. Oct 26, 2005 #3
    I am not a physics student yet, so I dont really know.
    By seeing diffrent exempls of moment of inertia I thoght it may be
    0.25M(r1^2-r2^2) + 0.25ML^2
    But this is only by intuation so I cant actually use it.
    please, help me if you can.
     
  5. Oct 26, 2005 #4

    Tide

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    You will not find the moment of inertia by just guessing. You need a place to start. Do you have a (mathematical) definition for moment of inertia? Have you had any calculus classes? What was the actual assignment?
     
  6. Oct 27, 2005 #5

    Mk

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  7. Oct 27, 2005 #6

    andrevdh

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    I do'nt think it is an assignment - he wants to build/construct something or work something out without having the necessary Physics background. so either we need to just give hime the answer or direct him to a tutorial.
    Chui, start by looking at http://wps.aw.com/aw_young_physics_11. Have a look at Part 1:Mechanics - 7.6 Rotational Inertia.
     
    Last edited: Oct 27, 2005
  8. Oct 27, 2005 #7
    andrevdh

    andrevdh rights! (thankyouverymuch)
    All I need is the formula, with no other choice, I can develpoe it by myself but it will take time.
    I want it for other calculation so I only need a formula.

    If anyone know,(or have a book which know) please write it.
    thanks.
     
  9. Oct 27, 2005 #8

    Tide

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    This is the homework section and we're not supposed to "give" the answer without the person showing some work. Otherwise, post in a different section! :)
     
  10. Oct 27, 2005 #9
    alright, I used Wikipedia, to learn some things with Integration,
    but it wasn't enought, I didn't understand too many things yet.
    however, I looked here:
    http://en.wikipedia.org/wiki/List_of_moments_of_inertia
    and learn the next things:
    1) an axis located at the center of a solid cylinder , its MomentumOfInertia should be this:
    (1/4)MR^2 + (1/12)ML^2 (number 3 on the list)
    2) a Rod with its axis on center its I should be:
    (1/12)ML^2 (one befort the last one on the list)
    3) a Rod with its axis on the end, like I need, its I should be:
    (1/3)ML^2 (the last one on the list)
    I put it all together with number 2 in the list to get this:
    (1/4)M(R1^2-R2^2) + (1/3)ML^2
    I still don't sure if I should put a minus or a plus between R1^2
    to R2^2
     
  11. Oct 27, 2005 #10

    Tide

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    It will be [itex]r_1^2 + r_2^2[/itex]. That part of the integration gives [itex]r_1^4-r_2^4[/itex] which factors into [itex](r_1^2-r_2^2)(r_1^2+r_2^2[/itex] and the first factor (minus sign) ultimately gets absorbed into M. I don't have time right now but I need to check on that factor of 1/4.
     
  12. Oct 27, 2005 #11
    i'm sorry I don't mean to change the subject, but how would I show that the Rotational Inertia would be 2/5MR^2 of a uniform sphere about any axis. I tried to first calculate the R.I. of a disk but I am not getting anywhere. . . please help if you can? I just became a memeber today and don't know how to post a new thread and this is the only thread relavent to what I need help with. . .
     
  13. Oct 28, 2005 #12

    Tide

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    bluejay,

    You need to integrate the square of the distance from the rotation axis. It looks like this:

    [tex]I = 2\pi \rho \int_0^R r^2 dr \int_0^\pi d\theta r^2 \sin^3 \theta[/tex]

    where the [itex]2\pi[/itex] if from integrating about the azimuth (based on symmetry). [itex]\rho = \frac {M}{4\pi R^3/3}[/itex] is the mass density. This works out to the desired result.
     
  14. Oct 28, 2005 #13

    andrevdh

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    Last edited: Oct 28, 2005
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