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Polar moment of inertia in a rod?

  1. Jan 6, 2015 #1
    How is the polar moment of inertia in a rod calculated?

    Thanks.
     
  2. jcsd
  3. Jan 6, 2015 #2

    BvU

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    No idea what it is. This link something ?
     
  4. Jan 6, 2015 #3
    Yeah, I've had a look at that. Formula doesn't quite make sense to me. I know what moment of inertia is but the 'polar' bit has thrown me off. Thanks anyway.
     
  5. Jan 7, 2015 #4

    BvU

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    Since there is no context whatsoever in the original posting, it's difficult to assist any further here.
     
  6. Jan 7, 2015 #5
    Here's the problem., 4. (b) (i) is the one I'm looking at.

    H7jim.jpg
     
  7. Jan 7, 2015 #6

    BvU

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    Well, at least now I can understand your confusion. However, it doesn't literally ask for the polar moment of inertia (a.k.a area moment of inertia), but for the polar mass moment of inertia. I see that used for the rotational moment of inertia, so with the context of b (ii) and b(iii) that seems the most logical choice.
     
  8. Jan 7, 2015 #7
    What is the formula for rotational moment of inertia?
     
  9. Jan 7, 2015 #8

    BvU

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    Can't imagine you haven't seen it before ! What did you use for 4a ?

    ## I=\int dI = {\displaystyle \int_0^M r^2 \; dm}##
     
  10. Jan 7, 2015 #9
    I'm on 4. (a) (ii) currently. Where should it be used in (a)?

    Ok, thanks, I have seen it but haven't used it previously.
     
  11. Jan 7, 2015 #10
    I've found this... http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html#irod3

    Looking at the bottom set of equations, so dm = (M/L) x dr...

    So essentially, the equation is r^2 x (M/L) x dr...

    Using these number that would be 0.075^2 x (20/2) x 0.15 x 0.075...

    Is that correct?
     
  12. Jan 7, 2015 #11

    BvU

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    Tempting indeed (*), but I think this time the rod is turning around its length axis. I.e. it's a solid cylinder !

    And you will need a moment of inertia in 4a (ii) as well: you don't just have to accelerate the cart, but also the drum.
    However, in 4a (ii) they give you a radius of gyration, so you use that to calculate the moment of inertia..


    (*) well, not really. did you really think dr stands for diameter x radius (because I see a 0.15 x 0.075, and then a few dots, indicating justified doubt, I hope) ?
     
  13. Jan 7, 2015 #12
    Ok, I am familiar with the equation for mass moment of inertia for a cylinder, which fits with this.

    So are you saying the equation you gave me wouldn't work with this? I find it odd they say it's a rod and expect a cylinder equation to be used.

    And yes I knew it wasn't the correct answer as I have the solutions and it didn't match...I see that dr is used in the integration equation now.

    So what is the formula?
    dm = elemental mass =
    (M/L) x dr so the whole equation is r^2 x (M/L) x dr
    And then?
     
  14. Jan 7, 2015 #13

    BvU

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    It's a drilling rod. Look here
     
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