Angular Momentum of Collapsing Cloud

[zachzach]

Homework Statement

Beginning with


$$\frac{d^2 r}{dt^2} = -G\frac{M_{r}}{r^2}$$
, adding a centripetal acceleration term, and using conservation of angular momentum, show that the collapse of a cloud will stop in a plane perpendicular to its axis of rotation when the radius reaches
$$r_{f} = \frac{{{\omega_{0}}^{2} {r_{0}}^{4}}} {{2 G M_{r}}}$$

where $$M_{r}$$ is the interior mass. Assume the initial radial velocity of the cloud is zero and that $$r_{f} \ll r_{0}$$

Hint: $$\frac{d^2 r}{dt^2} = v_{r} \frac{dv_{r}}{dr}$$

Homework Equations

Adding the centripetal term:$$\frac{d^2 r}{dt^2} = r{\omega}^2 - G\frac{M_{r}}{r^2}$$

The Attempt at a Solution

Since we are only concerned with the plane perpendicular to the axis of rotation, the mass concerned will be a thin disk the whole time so :

$$L_{i} = \frac{1}{2}M_{r}{{r_{0}^{2}}{\omega}_{0}$$ $$L_{f} = \frac{1}{2}M_{r}r_{f}^{2}{\omega}_{f}$$$$L_{i} = L_{f}$$

$${\omega}_{f} = \frac{r_{0}^{2}}{r_{f}^{2}}{\omega}_{0}$$

When the cloud stops collapsing $$v_{r} = 0$$

so from the equation given:$$r_{f}\omega_{f}^2 = G\frac{M_{r}}{r_{f}^{2}}$$

Plugging in for $${\omega_{f}}$$$$r_{f}\left[\frac{r_{0}^{4}}{r_{f}^{4}}\omega_{0}^{2}\right] = \frac{GM}{r_{f}^{2}}\Rightarrow r_{f} = \frac{r_0^{4}{\omega_{0}^{2}}}{GM}$$

which 2 times too large.

 

[zachzach]

Any help? PLEEEASE.

 

[fzero]

You found a condition such that

$$v_r \frac{dv_r}{dr} =0,$$

you didn't show that $$v_r=0$$ at this radius because it could be (and is) the case that $$dv_r/dr$$ vanishes there instead. What you should do is use angular momentum conservation to find the $$r$$ dependence of $$\omega$$. Then you can integrate the equation of motion to determine $$v_r(r)$$. Determining when that vanishes will yield the expected result. You will want to use the suggested approximation to simplify finding the root.

 

[zachzach]

That makes perfect sense. Thank you very much I have arrived at the correct answer.