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Period of a star / momentum

  1. Nov 24, 2005 #1
    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
    [tex] I \omega_o = I \omega_f [/tex]
    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 [tex] 2 \pi rad / 27.70 day [/tex], 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 [tex] L_o = (1.570 x 10^{31})(.5^2)(2.63x 10^{-6})[/tex] = 1.03 x 10^{25}
    The diameter is then shrunk by .550. This means the new radius is .275.
    So [tex] L_f= (1.570 x 10^{31})(.275^2) \omega = 1.19 x 10^{30} \omega [/tex]
    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!
    Last edited by a moderator: Nov 25, 2005
  2. jcsd
  3. Nov 24, 2005 #2
    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.

    Last edited: Nov 24, 2005
  4. Nov 24, 2005 #3
    ok i used [tex] (2/5)MR^2 \omega_o = (2/5)MR^2 \omega_ f [/tex]
    I found [tex] \omega_o [/tex] to be 174 by doing 27.70 x 2[tex]\pi [/tex]
    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?
  5. Nov 24, 2005 #4


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    Homework Helper

    Yes, the 0.55 comes in:

    [tex]\frac{2}{5}MR^2\omega_o = \frac{2}{5}M(0.55R)^2\omega_f

    Converting radians to days: (radians/s) = 2pi/T. T = 2pi/(radians/s). convert seconds into days.
  6. Nov 25, 2005 #5
    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.
    Last edited: Nov 25, 2005
  7. Nov 25, 2005 #6
    I got it. Thank you
  8. Nov 25, 2005 #7


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    Homework Helper

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