Conservation of Momentum; finding the angular speed of a Neutron star

  • #1
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1

Homework Statement


Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly 10^{14} times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 6.0×10^5km (comparable to our sun); its final radius is 16km

If the original star rotated once in 33 days, find the angular speed of the neutron star.
Express your answer using two significant figures and in rad/s

Ri = initial radius
Rf = final radius

Homework Equations


Conservation of Momentum

I = 2/5MR^2

L = Iw


The Attempt at a Solution



I first converted the 1 rev/ 33 days to rad/s

1/33 rev/ day *(2pi)*(1 day/24 hours)*(1hour/60 min)*(1 min/60 sec)
w = 0.000002 rad/s

I = (2/5)*(M)*(Ri^2)*(0.000002) = (2/5)*(M)*(Rf^2)wf

The masses (M) and the 2/5 cancel

3.6x1017 = 2.56x108*wf

wf = 1.40625x109

It says it's wrong. The only thing that I can think of that is wrong is that I did the angular velocity (w) conversion wrong.
 
Last edited:

Answers and Replies

  • #2
TSny
Homework Helper
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I first converted the 1 rev/ 33 days to rad/s

1/33 rev/ day *(2pi)*(1 day/24 hours)*(1hour/60 min)*(1 min/60 sec)
w = 0.000002 rad/s
Hello,
Note that you've only kept 1 significant figure in w. How many sig figs should you have? Also, you might want to use scientific notation for this very small number.

I = (2/5)*(M)*(Ri^2)*(0.000002) = (2/5)*(M)*(Rf^2)wf
I'm sure you meant to type L rather than I at the beginning of the equation.

The masses (M) and the 2/5 cancel

3.6x1017 = 2.56x108*wf
Did you forget to multiply by wi on the left side?
 
  • #3
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I think you're right. So I get:
(6x10^8)^2 * 0.000002 = (16000)^2 * wf

wf = 2812.5 rad/s

But my answer is still wrong
 
  • #4
TSny
Homework Helper
Gold Member
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Look's like your method is correct. But I really do think you should include more significant figures in your expression for wi. The number .000002 has only one significant figure. The problem gives data with two significant figures. What do you get for wi expressed to two significant figures? How does this affect your final answer?
 
  • #5
249
1
You are correct. I ran the calculation again using the exact value my calculator gave me. wf = 3098.95.

My calculator is a TI-89 and I was used the approximate function to get the answer, which gave me 0.000002. The real value for wi is pi/1425600.
Thank you for your help.
 

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