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zachzach
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Homework Statement
Beginning with
[tex] \frac{d^2 r}{dt^2} = -G\frac{M_{r}}{r^2} [/tex]
, 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
[tex] r_{f} = \frac{{{\omega_{0}}^{2} {r_{0}}^{4}}} {{2 G M_{r}}} [/tex]
where [tex] M_{r} [/tex] is the interior mass. Assume the initial radial velocity of the cloud is zero and that [tex] r_{f} \ll r_{0} [/tex]
Hint: [tex] \frac{d^2 r}{dt^2} = v_{r} \frac{dv_{r}}{dr} [/tex]
Homework Equations
Adding the centripetal term:[tex] \frac{d^2 r}{dt^2} = r{\omega}^2 - G\frac{M_{r}}{r^2} [/tex]
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 :[tex] L_{i} = \frac{1}{2}M_{r}{{r_{0}^{2}}{\omega}_{0} [/tex] [tex] L_{f} = \frac{1}{2}M_{r}r_{f}^{2}{\omega}_{f} [/tex][tex] L_{i} = L_{f} [/tex]
[tex] {\omega}_{f} = \frac{r_{0}^{2}}{r_{f}^{2}}{\omega}_{0} [/tex]
When the cloud stops collapsing [tex] v_{r} = 0 [/tex]
so from the equation given:[tex] r_{f}\omega_{f}^2 = G\frac{M_{r}}{r_{f}^{2}}
[/tex]
Plugging in for [tex] {\omega_{f}} [/tex][tex] 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} [/tex]
which 2 times too large.
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