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How to calculate moment of inertia of this irregular system?

  1. Jun 13, 2016 #1
    • Homework posted in wrong forum, so no template
    hello all, I am currently doing a lab investigating the relationship between the moment of inertia and speed.
    I have attached some pictures of my 'wheel' that is going to be rolled down a ramp. FYI i will be using a photogate to measure the time it takes the black strip to pass through. Initially, I thought the magnets were creating the 'hollow' part of the disc, but apparently i am wrong, and I am wondering how to calculate the moment of inertia of the magnet.

    I am desperate for help. As my teacher has not yet taught anything on this specific topic.
     

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  2. jcsd
  3. Jun 15, 2016 #2

    BiGyElLoWhAt

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    Can you come up with a piece wise density function? So the density is this from r = 2cm to r =4 cm, then the magnets have a density of that and they're centered at 5 cm along the x with radius 1cm or something like that.

    If you can, then all you have to do is run the integral and sum them.

    Worth noting: You only need to calculate the moment of inertia about the axis that you're planning on having rotation about. So you only need to worry about radial density, and you can assume all the mass lies in a plane.
     
  4. Jun 15, 2016 #3
    Hello, thank you for your suggestion,
    i actually went on the internet and tried to derive the moment of inertia of the solid wheels, which is 1/2 m r^2 .
    for the moment of inertia for the 8 magnets, I could either assume they are point mass, and use this formula I=mr^2. however, as i move the magnets towards the center of rotation, the error will increase.
    thus, i went online again and found this formula mass of magnets( distance from front of magnet to center of wheel^2+2distance from back of magnet to center^2)
    After i found the moment of inertia of the system. I will sub it into the equation mgh= 1/2mv^2 (translation)+ 1/2 I(omega)^2
     
  5. Jun 15, 2016 #4

    BiGyElLoWhAt

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    If your circle is small, I wouldn't assume that they are point masses. But I suppose it is a lab, so experimental error haha
     
  6. Jun 15, 2016 #5
    But for some reason, I cant get like a reasonable value with these formulas.
    MgH=v^2(m/2+Isystem/2r^2)
    where M is the mass of the whole wheel system 0.369kg.
    g is 9.8
    H is the initial height on the ramp which is 0.105m
    V is measured with a photogate ,0.10m/0.11746s= 0.85135m/s
    radius of the circle is 8cm.
    I of the system is (mr^2/2)+(mr^2/2)+8(mass of magnets (0.019kg(0.53m^2+2(0.078m^2))
    where the mass of bgi wheel is 0.142kg and small wheel 0.074kg.
     
  7. Jun 15, 2016 #6

    BiGyElLoWhAt

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    What I think is happening (maybe this is significant, and maybe it isn't), you're neglecting the holes in the board. Can you calculate what the 2 wheels would weigh if they were just the wheels, with no holes or magnets? Also, you have a giant hole in the center. Your 1/2mr^2 is not valid.
    Calculate the correct moment for the 2 discs. Sum them. Then add the magnets.
     
  8. Jun 15, 2016 #7
    y
    um

    um maybe the photo isnt too clear, but the wheel is actually a clear plastic, i used black tape so the photogate can sense it. in other words, the wheel is not a hollow cylinder.
     
  9. Jun 15, 2016 #8

    BiGyElLoWhAt

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    OK, so is it solid all the way through? Even where the magnets are placed? I was thinking that there was a hole in the center.
     
  10. Jun 15, 2016 #9

    BiGyElLoWhAt

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  11. Jun 15, 2016 #10
    yes the plastic disc is solid.
     
  12. Jun 15, 2016 #11
    sorry
    can you explain this theorem to me as I have not yet learn the moment of inertia at school and i dont really understand this part
     
  13. Jun 15, 2016 #12

    BiGyElLoWhAt

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    So then ##I_{sys} = I_{disc\ 1} + I_{disc\ 2} + I_{mags}##
    And I_mags is not at zero, so you need to shift it.
     
  14. Jun 15, 2016 #13
    I think i will read the parallet axis theorem again. tahksn
     
  15. Jun 15, 2016 #14

    BiGyElLoWhAt

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    So basically, it tells you how to move the axis. You have 2 places where there are magnets, and they're off center. So you need to compensate for that. You know the moment of inertia of a disc, so shift it, and add it to the original moment. It's a rather simple formula.
    You have ##I_{d}= I_{cm} + md^2##
     
  16. Jun 15, 2016 #15
    but i thought the magnets and the wheel share the rotation axis
     
  17. Jun 15, 2016 #16

    BiGyElLoWhAt

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    They do rotate about the same axis, but the axis the magnets rotate about is not it's center of mass. You know the moment of inertia for a disc about it's center of mass ##I_{cm}##, and you have a formula that allows you to shift that axis by a distance d that uses the formula that you already know.
     
  18. Jun 15, 2016 #17

    BiGyElLoWhAt

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    You should replace this part of the formula with the parallel axis theorem. which should not include the distance to the edges, only to the center of the magnets.
     
  19. Jun 15, 2016 #18
     
  20. Jun 15, 2016 #19

    BiGyElLoWhAt

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    I am not assuming that the magnets are a point mass.
    It is not mr^2. It is ##I_{cm} + mr^2## for the magnets. The I_cm takes care of the fact that they are not point masses.
     
  21. Jun 15, 2016 #20
    i think i understand the parallel axis theorem now. Thank you
    i have another question thought, how do i find the initial moment of inertia of the magnets?
     
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