1. Oct 21, 2007

### BuBbLeS01

1. The problem statement, all variables and given/known data
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!!!

2. Oct 21, 2007

### hage567

Think about conservation of angular momentum. What do you know about that?

3. Oct 21, 2007

### BuBbLeS01

Torque(net) * Change in T = Change in L
Ii * Wi = If * Wf

4. Oct 21, 2007

### hage567

OK, so what do you think you might do next? What is I for a solid sphere?

5. Oct 21, 2007

### BuBbLeS01

2/5MR^2 but how do we get R?
1 rotation/36.3days * pi = W

6. Oct 21, 2007

### hage567

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.

Be careful. This isn't quite right. If you are going to convert this, remember there are 2*pi radians in one revolution.

7. Oct 21, 2007

### BuBbLeS01

Right thats what I thought, but I don't know how to convert rotations to revolutions....

8. Oct 21, 2007

### hage567

1 rotation = 1 revolution. They mean the same thing. There are 2*pi radians per rotation (or revolution).

9. Oct 21, 2007

### hage567

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. Oct 21, 2007

### BuBbLeS01

Ii * Wi = If * Wf
2/5MR^2 * 1rot/36.3 days = If * Wf
I am not sure how we figure out If?

11. Oct 21, 2007

### hage567

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

12. Oct 21, 2007

### BuBbLeS01

Ii * Wi = If * Wf
2/5MR^2 * 1rot/36.3 days = (2/5MR^2*0.590) * Wf

13. Oct 21, 2007

### hage567

This needs to be squared as well, since it is part of the radius term.