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I Double Cone Moment of Inertia

  1. Nov 28, 2016 #1
    What is the correct formula for a double cone, is it 3/10 mr^2, or 3/5 mr^2..?
     
  2. jcsd
  3. Nov 28, 2016 #2

    berkeman

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    Welcome to the PF.

    Can you post your sources for those expressions, or show your work with the integral? :smile:
     
  4. Nov 28, 2016 #3
    Thanks for chiming in Berkeman, much appreciated. I'm not exactly sure what you're asking me, but my question is, what the correct formula for a double cone, is it: I = 3/10 x m x rr or I = 3/5 x m x rr..?
     
  5. Nov 28, 2016 #4

    berkeman

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    I understand what you are asking, but I'm trying to get you to help us help you to figure it out. The answer should be available via a Google search, and you offer two possible expressions which means you probably have done those searches. If you could post links to the two different results, we can probably help you to pick the right one.

    Otherwise, we can always just do the integration, but I'm lazy and would prefer that you do the integration yourself and post your work so I can check it. :smile:
     
  6. Nov 28, 2016 #5
    I believe the correct formula is 3/5 x m x rr. It would be similar to a sphere and hemisphere, whereas the hemisphere (1/2) is derived from the whole, with the sphere being a standard shape in physics. On the other hand, in the case of the double cone, the single cone is the standard shape, therefore, the formula for the cone has to double for the double cone, correct..?
     
  7. Nov 29, 2016 #6
  8. May 1, 2017 #7
    Hello everyone, I'm still trying to get a concrete answer to this puzzle. Is the MOI of a double cone 3/10 MR^2 or 3/5 MR^2..?
     
  9. May 1, 2017 #8

    jbriggs444

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    So you are talking about a double cone (e.g. a pair of ice cream cones tip to tip) rotating around the central axis of symmetry.

    From your searches, you have found a formula for the moment of inertia of a single cone: 3/10 MR2.

    If you have two identical cones, the moment of inertia should be twice as high as if you had only one, right?
    If you have two identical cones, the total mass will also be twice as much as if you had only one, right?

    Given this, there is no need to put an added factor of two into the formula. The factor of two is already present in the increased mass.
     
  10. May 1, 2017 #9
    Jbriggs, I appreciate your reply, but let's say the formula for the double cone was 3/5 MR^2. The single cone would be half the total mass, so the formula would also work, correct..? What I'm saying, is that the formula for the double cone is based starting from a single cone or 1/2 of that. Whereas the formula for a full circle is based on a full circle, not from the starting base of a semi circle. Because the single cone is a standard shape calculations are being concluded from that. Conversely, the shpere's MOI wasn't derived from a hemisphere, rather the hemisphere MOI was derived from the whole. Are you understanding my logic..?
     
  11. May 1, 2017 #10
    Here's another thought, let's call it the marble and the top theory. Wouldn't you agree that a spinning top is harder to move than a spinning sphere..? If you agree on that, then the MOI for a top must be higher than a sphere, correct..? Therefore, the MOI for a double cone ( basically a top) has to be higher than the MOI for a shpere. So, if a sphere's MOI is 2/5 MR^2, the double cone and/or a top has to be higher than that, which is why the correct formula should be 3/5 MR^2, not 3/10 MR^2. Your thoughts..?
     
  12. May 1, 2017 #11

    Nidum

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    Edit :

    I see from that linked document that you are talking about the base to base configuration .

    Confusion arose because 'double cone' has a conventional meaning in geometry and your usage of the words describes something different .

    Anyway - don't guess - do the sums .
     
    Last edited: May 1, 2017
  13. May 1, 2017 #12

    jbriggs444

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    Why would I agree to that? Per unit mass, the moments of inertia are 0.3 for the cone and 0.4 for the sphere.
     
  14. May 1, 2017 #13

    Nugatory

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    There's a bit of talking past one another going on here, because the answer is different for different axes of rotation. Just to be sure that we're talking about the same problem.

    Are the cones joined tip to tip or base to base? Is the axis of rotation a line that passes through both tips and the center of both bases, or is it perpendicular to that line? That's two questions with two answers for a total of four possibilities, and only two of the four have the same answer.
     
  15. May 1, 2017 #14
    That's if you believe the numbers we've been presented with. I'm contesting those numbers, with my arguments to back it up, along with a physics study report that was performed a while back in my post above dated Nov. 29th. Isn't it common sense knowledge that if you spin a marble and spin a top next to each other, if you blow at them, the marble will be blown off the table and the top will still be standing relatively close to the same position..? Or, if you blow the marble towards the top, upon impact, the marble will go flying, not the other way around..? Again, I don't believe the 3/10 MR^2 formula is correct, hence this discussion.
     
  16. May 1, 2017 #15
    The cones I'm referring to are joined base to base, with the axis of rotation spinning like a top.
     
  17. May 1, 2017 #16

    jbriggs444

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    The experiments you point to do not test moment of inertia.

    Further, any personal speculation you have that 3/10 MR^2 is incorrect is out of place on these forums.
     
  18. May 1, 2017 #17
    With all due respect, the resistance to angular acceleration, angular motion and/or a change in direction, is exactly what my experiments are pointing to. I'm having a tough time understanding why you'e demanding an end this discussion, when clearly my statements have validity..?
     
  19. May 1, 2017 #18
    Is it not possible for a mistake..? Are you suggesting that nothing can change, the earth is still flat or Einstein wasn't wrong on occasion..? Clearly, these forums do suggest Einsteins formulas are being put to the test and I'm quite sure, he has been wrong at least once in his life, correct..?
     
  20. May 1, 2017 #19

    Nugatory

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    That's not a test of the moment of inertia, it's a test of the surface area, surface velocity, and drag coefficient of the solid (with some more complicated second-order effects thrown in).
    It is possible, and a good way to check that possibility would be for you to evaluate the integral (as @berkeman suggested in post #4 above) and if you don't come up with the currently accepted results post your work. Either you've found a mistake that has gone undetected for more than three centuries or you've made a mistake yourself - and with the calculation posted people will be able to figure out which it is.
     
  21. May 1, 2017 #20

    berkeman

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    So one last chance -- post your work on the integrals showing the error, or this thread will be closed. We don't let newbie posters come here and show no work and waste the valuable time of our helpers. Please do this...
     
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