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Rotational Motion Problem - 8

  1. Dec 16, 2013 #1
    Hi friends,
    Please help me in solving this problem, I'll appreciate the help.

    The problem is as:

    https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-ash3/q71/s720x720/1503966_1461727630721009_1993563313_n.jpg

    Attempt -

    https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-prn2/q71/s720x720/1514999_1461727717387667_205197017_n.jpg
    https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn1/q71/s720x720/1497775_1461727790720993_14768769_n.jpg

    Thank you all in advance.
     
  2. jcsd
  3. Dec 16, 2013 #2

    haruspex

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    Your first method happens to work because the two displacements (from the centre of the rod to the end of the rod, then from the end of the rod to O) are at right angles. Adding the squares of those therefore gives the square of the distance from the centre of the rod to O.
    Your second one failed because you had the right angle in the wrong place, so subtracted the squares instead of adding.
     
  4. Dec 16, 2013 #3

    haruspex

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    Your first method happens to work because the two displacements (from the centre of the rod to the end of the rod, then from the end of the rod to O) are at right angles. Adding the squares of those therefore gives the square of the distance from the centre of the rod to O.
    Your second one failed because you had the right angle in the wrong place, so subtracted the squares instead of adding.
     
  5. Dec 17, 2013 #4
    Sorry haruspex I don't get it.
     
  6. Dec 17, 2013 #5

    haruspex

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    Label the left-hand end of the object A and the next corner B. So AC=CB and CBO is a right angle. But you drew the picture more as though BCO was a right angle, and this led you to the wrong equation for the distance CO.
     
  7. Dec 17, 2013 #6
    I got it, but what about this -
    https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-prn1/q71/s720x720/1525140_1462156974011408_1387363187_n.jpg
     
  8. Dec 17, 2013 #7

    haruspex

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    I'm not sure I understand your remaining difficulty.
    The MoI about C is mL2/12.
    By parallel axis theorem, MoI about B is mL2/12 + mL2/4 = mL2/3.
    Likewise, by parallel axis theorem, MoI about O is mL2/12 + m(CO2) = mL2/12 + m(CB2+BO2) (Pythagoras) = mL2/12 + mL2/4 + mL2 = 4mL2/3.
    It is not in general valid to apply the theorem to find the MoI about one point that's not the mass centre, then add the square of the displacement to a third point, but it works here because the two displacements, from C to B then from B to O, happen to be at right angles.
     
  9. Dec 17, 2013 #8
    The problem has been cleared.
     
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