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Collapsing Star

  • Thread starter Hollywood
  • Start date
5
0
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:

Iw(intial)=Iw(final)

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!
 

Answers and Replies

1
0
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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