How Does a Star's Rotation Period Change If Its Diameter Shrinks?

[Punchlinegirl]

The mass of a star is 1.570 x 10^31 kg and it performs one rotation in 27.70 days. Find its new period (in days) if the diameter suddenly shrinks to 0.550 times its present size. Assume a uniform mass distribution before and after.
I used conservation of momentum
I \omega_o = I \omega_f
I said that the initial diameter was 1 since it wasn't given. This would mean the initial radius was .5.
I found the initial angular velocity by doing 2 \pi rad / 27.70 day, divided by 24 hr, and then divided by 60 min, and divided by 60 s to get 2.63 x 10^-6 rad/s.
So L_o = (1.570 x 10^{31})(.5^2)(2.63x 10^{-6}) = 1.03 x 10^{25}
The diameter is then shrunk by .550. This means the new radius is .275.
So L_f= (1.570 x 10^{31})(.275^2) \omega = 1.19 x 10^{30} \omega
Solving for omega gave me 8.66 x 10^ {-6} \frac {rad} {s} .
My answer has to be in days, and I'm not really sure how to convert or if I even did this right. Please help!

 

[mathphys]

The mass of a star is 1.570 x 10^31 kg and it performs one rotation in 27.70 days. Find its new period (in days) if the diameter suddenly shrinks to 0.550 times its present size. Assume a uniform mass distribution before and after.
I used conservation of momentum
I \omega_o = I \omega_f
I said that the initial diameter was 1 since it wasn't given. This would mean the initial radius was .5.
I found the initial angular velocity by doing 2 \pi rad / 27.70 day, divided by 24 hr, and then divided by 60 min, and divided by 60 s to get 2.63 x 10^-6 rad/s.
So L_o = (1.570 x 10^31)(.5^2)(2.63x 10^-6) = 1.03 x 10^25
The diameter is then shrunk by .550. This means the new radius is .275.
So L_f= (1.570 x 10^31)(.275^2) \omega=1.19 x 10^30 \omega
Solving for omega gave me 8.66 x 10^ -6 rad/s .
My answer has to be in days, and I'm not really sure how to convert or if I even did this right. Please help!

First do not give any value to the radius, just call it r.
yes , use L=I_0w_0=I_1w_1 (*)

Then use the momentum of inertia for a sphere I=(2/5) m r^2.



don't convert to seconds there's no need (I_0, I_1 do not depend on time) and calculate w_0. Use rad/days as units. Find w_1 from (*), and from it the new rotation period.

:)

 

[Punchlinegirl]

ok i used (2/5)MR^2 \omega_o = (2/5)MR^2 \omega_ f
I found \omega_o to be 174 by doing 27.70 x 2\pi
Dividing both sides by the moment of inertia would cancel out them out meaning that the final angular velocity= 174 days
Is that right? Wouldn't the .550 need to come in somewhere?

 

[mezarashi]

Yes, the 0.55 comes in:

\frac{2}{5}MR^2\omega_o = \frac{2}{5}M(0.55R)^2\omega_f<br />

Converting radians to days: (radians/s) = 2pi/T. T = 2pi/(radians/s). convert seconds into days.

 

[mathphys]

I_0 is not equal to I_1 because the radius of the sphere is different in each case.
In one case we have a radius given by R, and in the other one we have a smaller radius: 0.550R.

You have the correct expression for w0 in your first post w0=2pi rad/T_0=2pi rad /27.7 days.

 

[Punchlinegirl]

I got it. Thank you

 

[lightgrav]

I hope you're not too ticked off that your \omega_f was correct in your first post.
All you needed to do was the reciprocal units conversion that you did first!