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Collapsing Star

  1. Apr 8, 2008 #1
    The mass of a star is 1.4 × 10^31 kg and it performs one rotation in 26.3 day. Find its new period if the diameter suddenly shrinks to 0.55 times its present size. Assume a uniform mass distribution before and after.

    I don't know what I am doing wrong here:


    I(intial)=(2/5)(1.4 × 10^31 kg)(1)^2
    = 5.6 x 10^30

    I(final)=(2/5)(1.4 × 10^31 kg/.55)(.55)^2
    =9.317 x 10^29

    (5.6 x 10^30)(26.3 day) = (9.317 x 10^29)(w)

    w(final)=158.1 day

    This answer is wrong. Can someone tell me what I am doing wrong here?
    Thank You!
  2. jcsd
  3. Apr 8, 2008 #2
    Hi Hollywood!

    Now I'm not ever sure of anything I say or do, but I would try to equate the initial and final angular momentums L = m * v * r since they should be conserved.
    Where the tangential velocity v = (2 * pi * r )/ T

    So the initial Lo = (Mo * 2pi * Ro ^2)/To
    And the final Lf= (Mf * 2pi * Rf ^2)/Tf

    Since the mass of your star does not change (and hopefully, neither does '2*pi'), you can just write

    Tf * Ro ^2 = To * Rf ^2

    Since Ro = 1 and Rf = .55

    Tf = 26.3 * (.55^2) = 7.9...days

    Hope that's correct!
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