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Moment of Inertia and Velocity

  1. Nov 27, 2005 #1
    I need to derive the formula for this "film projection wheel" (it just has four holes in the center). Ive looked all over google AND Hyperphysics but I havent found a site that shows how to derive the formulas.
    Please can someone help me (my phsyics teacher is a dick: hes been missing for 3/4 of the semester).:mad:
    [​IMG]
     
  2. jcsd
  3. Nov 27, 2005 #2

    lightgrav

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    Homework Helper

    You add Inertias when you have more than one mass, right?

    The film wheel is just a disk with 4 "negative mass disks" added to it, off-axis.
    (use parallel-axis : add the Inertia of the center-of-mass around the axis to the Inertia around the center-of-mass)
     
  4. Nov 27, 2005 #3
    the problem is i literally dont know where to begin. I understand subtracting the holes to get the moment. But i dont even know how to set-up the integral. I only know (integral)r(squared)dm, im lost as to where to go from there.
     
  5. Nov 27, 2005 #4

    lightgrav

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    You don't need to integrate. Most mortals look up Inertia formulas in a table.
    Inertia of a disk around its center-of-mass is ½ M R^2 (axis perp to surface).
     
  6. Nov 28, 2005 #5
    ok i found the MOI but what happens with the 4 holes? LIghtgrav, you said I should subtract the holes so would
    the total moment of inertia be:

    I= 1/2m(r)^2 - 4[1/2m(r)^2] but is it easy as subtracting the 4 holes, because
    they are not concentric?



    I was also told about the Parallel axis theorem but im not sure what its used for exactly; would the d in the parallel axis theorem be 0.25 m (from the picture I included)?
     
  7. Nov 28, 2005 #6

    lightgrav

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    As I see your diagram, the .25 m are the little (negative) disk radii.
    The Inertia of the center-of-mass around the axis "d" is "a", which
    points from the axis (the reel center) to the center of that (negative) disk.
    This is EXACTLY what the parallel axis is for. You'll need it again
    (the reel rotates around an axis at its edge, not at its center!)

    Careful with your symbols ... I = ½ M R^2 - 4 (½ m r^2 + m a^2).
    Can you figure out what "m" should be, if M = 10 kg ?
     
    Last edited: Nov 28, 2005
  8. Nov 28, 2005 #7
    Im confused now, because in an earlier post you said I could just subtract the four moments of inertia from the moment of inertia of the large circle. So is that correct or is the parallel axis theorem the correct formula to use? Also why is
    I = ½ M R^2 - 4 (½ m r^2 + m a^2) ?
    I assume your applying the Parallel Axis Theorem but how does this relate to the formula (im not questioning what you have, just why it is)?

    Also to answer you question, m would be 10kg-4[pi(little r)^2]
     
  9. Nov 29, 2005 #8
    So does this mean that saying I = I_cm + ml^2 is the proper moment of inertia for the reel rotating about is edge (I_cm is the moi that we derived and then subtracted the four "negative masses")?
    Basically, I'm bothered by not understanding what to do about the holes in the reel. And it's not even my problem :grumpy:
     
  10. Nov 29, 2005 #9

    Doc Al

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    Staff: Mentor

    Yes and yes. You need to find the moments of inertia of the holes about the axis of rotation (not just their centers); that requires using the Parallel Axis Theorem.

    For how the parallel axis theorem relates to this formula, see my comment above.

    Realize that in the formula provided by lightgrav:
    M is the mass of a solid disk of radius R (if the holes were filled)
    m is the mass of a disk of radius r with the same density as the solid disk​

    You are given that the mass of "the disk" is 10 Kg. I assume that this is the mass of the "film projection wheel", holes and all. You'll have to figure out the surface density of the disk and use that to calculate M and m.
     
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