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Angular momentum

  1. Dec 27, 2008 #1
    1st of all i apologise if my terminology is incorrect, as i am translating into english..

    on a spinning chair a man is sitting and holding 2 weights, of 10kg each, at a radius of 0.5m, the chair is turning at a frequency of 1Hz., the total moment of the man and the chair is I=2.5kgm2. how will the frequency change is the man moves the weights to a radius of 0.2m? what is the work done by the man in this case?

    what i did was, using conservation of momentum

    Iweights=mr2=20*0.52=5kgm2
    Iman+chair=2.5kgm2
    L=I[tex]\omega[/tex]=(5+2.5)(2[tex]\Pi[/tex]f)=15[tex]\Pi[/tex]

    conservation of momentum
    Iweights=mr2=20*0.22=0.8kgm2
    Iman+chair=2.5kgm2

    L=15[tex]\Pi[/tex]=(0.8+2.5)(2[tex]\Pi[/tex]f)

    f=25/11Hz

    is this correct,

    now to find the work done, can i say- work done is the change of energy and using that, say,

    Ei=Ek=0.5mv2
    vi=2[tex]\Pi[/tex]fi*Ri=[tex]\Pi[/tex]
    Ei=10[tex]\Pi[/tex]2j

    Ef=Ek=0.5mv2
    vf=2[tex]\Pi[/tex]ff*Rf=(10/11)[tex]\Pi[/tex]
    Ef=81.566j

    W=[tex]\Delta[/tex]E=0.5m(vf2-vi2)
    =10*(8.157-9.87)=-17.13j

    W=17.13j
     
  2. jcsd
  3. Dec 27, 2008 #2
    the correct answer (in my book) for the work done comes out around 188.4j so i'm way off.- what do i need to change. theres no answer as far as the frequency goes
     
  4. Dec 27, 2008 #3

    LowlyPion

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    Homework Helper

    The first part looks correct.

    But for the second wouldn't you want to consider the change in rotational kinetic energy?

    KE = ½Iω²

    ΔKE = ½I1ω1² - ½I2ω2²
     
  5. Dec 27, 2008 #4
    thanks a lot
     
  6. Jan 5, 2009 #5
    could you take a please look at this post for me, also momentum, thanks
     
  7. Jan 5, 2009 #6
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