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Center of Mass constant with near-correct attempt

  1. Jul 29, 2012 #1
    1. The problem statement, all variables and given/known data

    A wagon wheel is made entirely of wood. Its components consist of a rim, 16 spokes, and a hub. The rim has mass 5.1 kg, outer radius 0.90 m, and inner radius 0.86 m. The hub is a solid cylinder with mass 3.1 kg and radius 0.12 m. The spokes are thin rods of mass 1.1 kg that extend from the hub to the inner side of the rim. Determine the constant
    c = I/MR2 for this wagon wheel.

    Hi folks, this will be my last problem, promise.

    2. Relevant equations

    I for hub=> I for solid cylinder=> 1/2 M R^2
    I for wheel=> I for thick hollow cylinder or wheel (I THINK...) =1/2 M (R1^2+R2^2)
    Otherwise, I for hollow cylinder or hoop is MR^2
    I for thin rod is 1/12 M h^2. These are rotating about their end though, so by parallel axis theorem, their moment is : 1/12Mh^2+M(h/2)^2

    I(net)=cMR^2
    c=I/MR^2

    So there you have it.

    3. The attempt at a solution

    The correct answer is .356
    My answer is .344

    Here's what I did:

    For Hub:1/2(3.1)*.12^2=.002232
    For Wheel: 1/2(5.1)(.90^2+.86^2)=3.95148
    For thin rods: 16 *{(1/12)(1.1)(.74)^2 + (.74/2)^2(1.1)}=3.212586

    Then added them all to get 7.1832
    Then divided them by their combined mass (16*1.1+5.1+3.1)=25.8 and the entire radius^2
    so 7.1832/(25.8*.9^2)=.344

    I've tried with less sig figs, and it doesn't work. There's something off, but its small. I have a feeling a quick once-over might spot it. Many thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 29, 2012 #2
    I should note, I've given it many, many once-overs myself and can't find it. I don't think its alebraic, I think its conceptual, somewhere
     
  4. Jul 29, 2012 #3

    TSny

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    Hi jamesbiomed.

    Are you sure the spokes are rotating about an end? Aren't they rotating about the center of the wheel? But the spokes do not extend all the way to the center.
     
    Last edited: Jul 29, 2012
  5. Jul 29, 2012 #4
    Tsny, I think you're right. That makes a lot of sense, and changes h. Thanks for the help!
     
  6. Jul 29, 2012 #5
    Not h, but d distance during parallel axis theorem.
     
  7. Jul 29, 2012 #6
    Darn it, still off.. this time a little high.

    I added radius 12 to the distance in the P.a.th. Was that the correct adjustment?
     
  8. Jul 29, 2012 #7

    TSny

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    I think that's right. I get that d = .49 m for the parallel axis theorem.
     
  9. Jul 29, 2012 #8
    right, that's what I'm getting. Then I'm still using .74 for the h in 1/12mh^2.

    So even though that needed fixing, there was something else. I just reworked the whole thing start to finish and still got the same answer.

    For the R in 1/MR^2, would you say that's .9?

    It could be a glitch in the HW program, so I might give up soon
     
  10. Jul 29, 2012 #9
    Not to be pessimistic or anything, but that's all I can find
     
  11. Jul 29, 2012 #10

    TSny

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    I don't see anything wrong. I get an answer of .43 for c.
     
  12. Jul 29, 2012 #11
    That's what I get too. Not sure how they're getting their's. I feel good about the process, so I'll just run it by my prof in the morning. Thanks for the help.
     
  13. Jul 29, 2012 #12

    TSny

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    OK. Good luck with it.
     
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