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Rotational Inertia Problem

  1. Oct 28, 2013 #1
    1. The problem statement, all variables and given/known data

    in Attachment

    2. Relevant equations

    I = .5*mr^2

    3. The attempt at a solution

    I have no idea what is going on. This is a pre section quiz and I guessed on one and got it right. I dont know what is meant by the equations not giving I. I googled and noticed I has various equations. I plugged in the one I though most pertinent but it was marked wrong. Someone assist me please!
     

    Attached Files:

  2. jcsd
  3. Oct 28, 2013 #2
    The moment of inertia, I should be .5*MR^2, not .5*mR^2. See if that helps.
     
  4. Oct 28, 2013 #3
    I don't see how. I tried calculating with both m and M but I was wrong both times
     
  5. Oct 28, 2013 #4

    haruspex

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    The wheel is not stated to be a uniform disc. That is why I is given as an input variable. It cannot be obtained from a formula. But the question setter seems to have forgotten to provide the numerical value of I.
     
  6. Oct 28, 2013 #5
    Are you sure? Lol
    Not to insult you but googling it has the same question solved with I not given. And I can figure it out for the life of me :/
     
  7. Oct 28, 2013 #6
    Since you "guessed" the answer for a correctly, you can substitute that value into eq. 3 and with a little alfebra get T = mg - ma. Using that, you get T = 4.524 N. Plugging I=.5*MR^2 into eq. 4 should give you the same result.
     
  8. Oct 28, 2013 #7
    I got T=4.6N,a=4m/s2,α(angular acceleration)=16.25rad/s2

    I used the equations

    mg-T=ma

    TR=Iα

    a=Rα

    Solve these equations you will get the answer
     
  9. Oct 28, 2013 #8

    haruspex

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    There are 6 variables there: m, T, R, a, I, α. You have three equations. Therefore you need the values of three variables. You are told the numerical values of m and R, that's all. You must have used a fourth equation.
     
  10. Oct 28, 2013 #9

    haruspex

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    Quite sure. Note that equation (4) has I in the final formula before evaluation, implying you are supposed to substitute a number for I. If you were expected to obtain I from M and R there would be an equation for that somewhere in the text.
    If you think you have found somewhere on the net that solves this on the information given, pls post the link.
     
  11. Oct 28, 2013 #10
    we also have value of I
    I have assumed it to be a thin disc.
     
    Last edited: Oct 28, 2013
  12. Oct 28, 2013 #11

    haruspex

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    You are not given that information, nor a formula. Instead, I is given symbolically as an input and also appears in the final equations. This strongly suggests it is not supposed to be a uniform disc, but the numerical value was to be given.
    If you read back through the posts you'll see that xtrubambinoxpr tried using your assumption (though once with the wrong mass) and the answer was rejected.
     
  13. Oct 29, 2013 #12
    But the acceleration looks right
     
  14. Oct 29, 2013 #13

    haruspex

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    In what way does it 'look right'? I don't see the answer supplied anywhere in the thread.
     
  15. Oct 29, 2013 #14
    See in the attachment he has got acceleration right
     
  16. Oct 29, 2013 #15

    haruspex

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    Ok, I see. It could be rounding error, i.e. 4 is accepted as close enough.
    xtrubambinoxpr, it accepted a=4. Since α = a/R, if a is right then it should be easy to get α right. Similarly, T = mg-ma. What did you enter for α and T?
     
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