# Homework Help: Angular Speed of a star collapse

1. Nov 9, 2007

### Heat

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

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 9.0×10^5 km (comparable to our sun); its final radius is 16 km.

If the original star rotated once in 31 days, find the angular speed of the neutron star.

3. The attempt at a solution

Rotation of Original Star: (31 days/rev)(86400s/1day)(1rev/2pi) = 288770.73 rad/s

My first attempt was proportions

[(288770.72 rad/s) / (9.0x10^5 km)] = [(w)/(16km)]

w = 5.13x10^10

2. Nov 9, 2007

### Bill Foster

Conservation of angular momentum

3. Nov 9, 2007

### Heat

but all that I am given is that the density is 10^14 greater than the before scenario.

4. Nov 9, 2007

### Bill Foster

Density is mass divided by volume.

The original density is $$\rho_0=\frac{m_0}{V_0}$$

The new density is $$\rho_1=\frac{m_1}{V_1}=10^{14}\rho_0=10^{14}\frac{m_0}{V_0}$$

5. Nov 9, 2007

### Bill Foster

Also, $$I_0\omega_0=I_1\omega_1$$

Rearranging:

$$\omega_1=\frac{I_0\omega_0}{I_1}$$

What is $$I$$ for a sphere?

$$I=\frac{2}{5} m r^2$$

Now let's combine the equations:

$$\frac{m_0 r_0^2\omega_0}{m_1 r_1^2}=\omega_1$$

Now to write mass in terms of density:

$$m=\rho V$$

And you know the following:

$$\rho_1=\frac{m_1}{V_1}=10^{14}\rho_0=10^{14}\frac{ m_0}{V_0}$$

Rearranging:

$$\frac{m_0}{m_1}=10^{-14}\frac{V_0}{V_1}$$

Now that can be substituted in the other equation, giving:

$$10^{-14}\omega_0\frac{V_0 r_0^2}{V_1 r_1^2}=\omega_1$$

6. Nov 9, 2007

### aq1q

Last edited: Nov 9, 2007
7. Nov 9, 2007

### Bill Foster

And the volume of a sphere is $$\frac{4}{3}\pi r^3$$

$$V_0=\frac{4}{3}\pi r_0^3$$

$$V_1=\frac{4}{3}\pi r_1^3$$

$$\frac{V_0}{V_1}=\frac{r_0^3}{r_1^3}$$

8. Nov 9, 2007

### Bill Foster

$$10^{-14}\omega_0\frac{r_0^5}{r_1^5}=\omega_1$$

9. Nov 9, 2007

### aq1q

ahh exactly, nice.. and if u read above, i explain how to solve for Wo. Wo=(1rev/31days)(1day/86400s)(2PIrad/1rev)

10. Nov 9, 2007

### Heat

wo = 2.35x10^-6

10^-14 ( .00000235)[(9.0x10^5)^5/(16)^5] = 13233.68

In two sig figs: 1.3 x 10 ^ 4

which is incorrect :O

11. Nov 9, 2007

### hage567

It says the density increases. That does not (necessarily) mean the mass changes. I think you can assume that the mass is the same before and after the collapse.

12. Nov 9, 2007

### aq1q

the thing is, there is a good possibility that the mass will change.. because the density AND the radius changes... i=initial f=final
Pi=Mass(i)/Volume(i) Pf= (10^14)Pi .... now if we were to assume that masses are the same.. then lets call mass=x
Pi=X/(4pi/3(9.0×10^5 km)^3) Pf= (X/(4pi/3(16)^3)) we know that Pf/Pi=(10^14)
(X/(4pi/3(16)^3)) / (X/(4pi/3(9.0×10^5 km)^3)) must = to 10^14
after doing the math (the X's cancel out)
it comes out to be 1.8x10^14.. so masses must change

13. Nov 10, 2007

### Heat

so masses do change, but it seems that Bill Foster already took that into account and simplified it to post #8 equation.

yet I went wrong somewhere. :(

14. Nov 10, 2007

### hage567

I'm just saying if the "mass changing approach" doesn't give the right answer, maybe the simpler approach is what the question is really asking for. I think the 10^14 may just be extra information that is not needed to solve the problem.

You have $$I_o \omega_o = I_f \omega_f$$

where

$$I_o = \frac{2}{5}MR_o^2$$
$$I_f = \frac{2}{5}MR_f^2$$ assume the total mass stays constant.

You have $$\omega_o$$

So put everything together, the mass cancels and you find $$\omega_f$$

I could be wrong, of course. But did you at least try it this way to see what answer you get?

15. Nov 10, 2007

### aq1q

Maybe.. but its unlikely.. because the question wouldn't go into so much details about density.

Last edited: Nov 10, 2007
16. Nov 10, 2007

### aq1q

if Heat gets the right answer using the "no mass change approach" then..well... I would say that the book/question is "technically" wrong.

However, there is some chance that Heat plugged the numbers incorrectly.. or something.. im not sure.. lets see if Heat gets the right answer.

17. Nov 22, 2010

### mclame22

I know this is a very old post, but using hage567's method works. The mass does not change, and using his expression gives the correct answer.