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Period of a star

  • Thread starter lew44
  • Start date
  • #1
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Homework Statement




The mass of a star is 1.370×1031 kg and it performs one rotation in 38.10 day. Find its new period (in days) if the diameter suddenly shrinks to 0.330 times its present size. Assume a uniform mass distribution before and after.

Homework Equations


Ii * Wi = If * Wf
2/5MR^2 * 1rot/38.1 days = (2/5MR^2*0.330^2) * Wf


The Attempt at a Solution



I am not sure what cancels out

I thought it would be something like

1rot/38.1days = .330^2 * Wf

solve for Wf

But that didn't work. .241s was not correct
 

Answers and Replies

  • #2
ehild
Homework Helper
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Take care of the units. rot/day is not second.

ehild
 
  • #3
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I still don't understand because if I change the bottom number to seconds I still have rot/sec and that is not seconds. Plus it asks for the period in days so doesn't that factor in?
 
  • #4
ehild
Homework Helper
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1,854
wi and wf have the same units.

ehild
 
  • #5
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it asks for the new period in days! I am not sure what to do with this problem
 
  • #6
7
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I am getting like 2198.249 as an answer.
 
  • #7
ehild
Homework Helper
15,478
1,854
You replaced wi by 1rot/39.1days. Then you should do the same with wf: wf=1rot/xdays.

Rewrite your equation

1rot/38.1days = .330^2 * Wf

as

1rot/38.1days = 0.330^2* (1rot/x days)--->1/38.1 =0.330^2/x.

Solve for x.

ehild
 
  • #8
7
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thank you, i think i just didn't understand that you needed the x on the other side of the equation and that is what was giving me such a large answer
 

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