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Angular Mometum please help.

  1. Nov 16, 2005 #1
    Problem: A person stands, hands at his side, on a platform that is rotating at a rate of 1.30 rev/s. If he raises his arms to a horizontial position the speed of rotation decreases to 0.80 rev/s. (a) Why? (b)By what factor has the moment of inertia changed? Answer: (b) 1.6

    Okay, so I know part a which is because his rotational inertia increases.

    Someone please help explain where 1.6 is even coming from... I have yet to get a soild answer. :(
     
  2. jcsd
  3. Nov 16, 2005 #2

    Päällikkö

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    [tex]\vec{L} = I \vec{\omega}[/tex]
    L, angular momentum, is conserved. Can you take it from here?
     
    Last edited: Nov 16, 2005
  4. Nov 16, 2005 #3
    But how do I find I ? I know w is 1.30 - 0.80....
     
  5. Nov 17, 2005 #4

    Päällikkö

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    You don't.
    You have Ii and If = kIi (where k is a factor by which Ii must be multiplied to get If)

    I's should cancel out.
     
  6. Nov 17, 2005 #5

    FredGarvin

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    You do realize what conservation of linear momentum imples...

    [tex]\vec{L}_{initial} = \vec{L}_{final}[/tex]
     
  7. Nov 17, 2005 #6
    I am going to have to ask my instructor. I have no idea where the 1.6 is coming from. sorry.
     
  8. Nov 17, 2005 #7

    FredGarvin

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    Like I mentioned the previous time you brought this question up, the factor I get is 1.4. I think there is an error in the answer you were given.

    EDIT: After looking at sig figs, I ended up with the factor of 1.6.

    Since you're not grasping the conservation idea, how about this:

    [tex]{L}_1 = I_1 {\omega_1}[/tex] and [tex]{L}_2 = I_2 {\omega_2}[/tex]

    Now Since [tex]{L}_1 = {L}_2[/tex] then you can set the rest equal [tex] I_1 \omega_1 = I_2 \omega_2[/tex]

    To get the ratio you are looking for solve that for [tex]\frac{I_2}{I_1}[/tex]
     
    Last edited: Nov 17, 2005
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