Period of a Star Please help

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Period of a Star...Please help!!!

Homework Statement


The mass of a star is 1.250×1031 kg and it performs one rotation in 36.30 day. Find its new period (in days) if the diameter suddenly shrinks to 0.590 times its present size. Assume a uniform mass distribution before and after.

I don't know where to begin!!!
 

Answers and Replies

  • #2
hage567
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Think about conservation of angular momentum. What do you know about that?
 
  • #3
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Torque(net) * Change in T = Change in L
Ii * Wi = If * Wf
 
  • #4
hage567
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Ii * Wi = If * Wf
OK, so what do you think you might do next? What is I for a solid sphere?
 
  • #5
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2/5MR^2 but how do we get R?
1 rotation/36.3days * pi = W
 
  • #6
hage567
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You don't need to know the radius. In the question, you are given the final radius in terms of the first, so it will cancel out.

1 rotation/36.3days * pi = W
Be careful. This isn't quite right. If you are going to convert this, remember there are 2*pi radians in one revolution.
 
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Right thats what I thought, but I don't know how to convert rotations to revolutions....
 
  • #8
hage567
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1 rotation = 1 revolution. They mean the same thing. There are 2*pi radians per rotation (or revolution).
 
  • #9
hage567
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You could leave Wi in terms of rotation/day for this particular problem since all of your other units will cancel out. Of course, if you're unsure how to do it, it's good practice to give it a try.
 
  • #10
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Ii * Wi = If * Wf
2/5MR^2 * 1rot/36.3 days = If * Wf
I am not sure how we figure out If?
 
  • #11
hage567
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If will be the same equation where the only thing that changes is R since the star becomes smaller in diameter. You are given information in the question that allows you to determine how much smaller it becomes

Find its new period (in days) if the diameter suddenly shrinks to 0.590 times its present size.
 
  • #12
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Ii * Wi = If * Wf
2/5MR^2 * 1rot/36.3 days = (2/5MR^2*0.590) * Wf
 
  • #13
hage567
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Ii * Wi = If * Wf
2/5MR^2 * 1rot/36.3 days = (2/5MR^2*0.590) * Wf
This needs to be squared as well, since it is part of the radius term.
 

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