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Moment of inertia

  1. Jul 16, 2011 #1
    1. The problem statement, all variables and given/known data
    part c and part d
    attachment.php?attachmentid=37226&stc=1&d=1310800089.jpg

    And the ans.
    a.)140
    b.)130/21
    c.)6.92M
    d.)150M/7

    2. Relevant equations


    attachment.php?attachmentid=37225&stc=1&d=1310800089.jpg
    3. The attempt at a solution
    attachment.php?attachmentid=37227&stc=1&d=1310800092.jpg

    It seems the ans is wrong, I am not sure.
    Can you help me?
     

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  2. jcsd
  3. Jul 16, 2011 #2
    the density you used is defined from the lighter end and you have to find I about heavier end.

    you may take the dx element at distance x from heavy end and then use (10-x) in the equation of density
     
  4. Jul 16, 2011 #3

    gneill

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    Staff: Mentor

    The parallel axis theorem works fine for objects with constant density. This object doesn't have constant density -- the density varies from end to end. Therefore there will be a light end and a heavy end. If you grab the rod by the light end and swing the heavy end around, you'll find a lot more angular inertia than if you grab the rod by the heavy end and swing the light end around.

    So, see if you can set up your integral to directly find the moment of inertia about the heavy end.
     
    Last edited: Jul 16, 2011
  5. Jul 16, 2011 #4
    Is it a rule that I must start from heavy end?

    Why it is wrong to start from light end?
     
  6. Jul 16, 2011 #5

    gneill

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    It's not a rule, and it's not wrong to start at the light end. You may start anywhere you wish! But note that the problem is asking for the moment of inertia about the heavy end. Therefore, in the the calculation of the moment of inertia, the radial distances of the mass elements must be with respect to the heavy end. And be sure that the density at that distance from the heavy end is the correct density for that position along the rod.
     
  7. Jul 16, 2011 #6
    Oh! Sorry for my carelessness, I didn't read the question carefully.
    I have solved part d, but part c is still very complicated.
    attachment.php?attachmentid=37230&stc=1&d=1310822855.jpg

    The result I obtained is so ugly that I don't want to write it out.
    But I have check it and it is wrong.

    So what's wrong with my work this time?
     

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  8. Jul 16, 2011 #7

    gneill

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    Suppose you break the integral into two. The first runs from x=0 to x= xcm, the second from xcm to 10. The density function remains as 2x + 4 for any value of x, so leave it alone. What is the distance from xcm to the mass element in each case?
     
  9. Jul 16, 2011 #8
    I am so sorry that I don't understand.
    Can you explain it more clearly?:shy:
     
  10. Jul 16, 2011 #9

    gneill

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    attachment.php?attachmentid=37235&stc=1&d=1310826902.gif

    x is the position along the rod from the LHS. For the region to the left of the center of mass, the distance from the center of mass to x is xcm - x. To the right of the center of mass, the distance is x - xcm. The density at point x remains [itex] \rho (x) = 2 x + 4[/itex], since here x is always with respect to the light end of the rod.
     

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    Last edited: Jul 16, 2011
  11. Jul 16, 2011 #10
    Do you mean this?
    I have just finished half of the integral.

    attachment.php?attachmentid=37236&stc=1&d=1310828027.jpg
     

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  12. Jul 16, 2011 #11
    Oh! It works and I get the answer!

    But can you explain what's wrong with my method?
    Why I can't set the center of mass as the orgin?
     
  13. Jul 16, 2011 #12
    Work in #6 turns out to be correct, I just type it wrongly in my calculator.

    By the way, how to generate the picture in #9?
    What software are you using?
     
  14. Jul 16, 2011 #13

    gneill

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    You can set the center of mass at the origin if you translate the density equation accordingly for each section of the rod. Personally I tend to find it easier to leave the density function alone and work out the required distances as functions of the same x.
     
  15. Jul 16, 2011 #14

    gneill

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    I use Visio to draw the pictures, then paste it into MS Paint to save it as a .gif type file.
     
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