Angular Momentum Question

Question

idku · Answer

An interesting phenomenon occurring in certain
pulsars is an event known as a "spin glitch", that is,
a quick change in the spin rate of the pulsar due to
a shift in mass location and a resulting rotational
change in inertia. Imagine a pulsar whose radius is
10.2 km and whose period of rotation is 25.037 ms.
The rotation period is observed to suddenly decrease
from 25.037 ms to 25.033 ms. If that decrease was
caused by a contraction of the star, by what amount
would the pulsar radius have had to shrink?

idku · Answer

So I am assuming the momentum is conserved, and thus,

$L_o=L_f$
$\omega_oI_o=\omega_fI_f$

I don't know what proportion that is for inertia, whether it is more like disk, ring, or whatever, so I will say $a$ for the coefficient (and it will cancel out regardless)

$\omega_o(aM_oR_o^{~~2})=\omega_f(aM_fR_f^{~~2})$
$\omega_o(M_oR_o^{~~2})=\omega_f(M_fR_f^{~~2})$

So, we would have to have two more unknowns that I can't determine and that do not cancel, SO, I therefore must assume the mass stays the same ($M_o=M_f$), and just the radius shrinks (so maybe the shape gets thicker but with smaller radius). So the mass will cancel out.

$\omega_o~R_o^{~~2}=\omega_f~R_f^{~~2}$
$ R_f^{~~2}=\frac{\omega_o}{\omega_f}~R_o^{~~2}$
$ R_f=\sqrt{\frac{\omega_o}{\omega_f}~R_o^{~~2}}$

So, therefore the difference in radii is,
$ R_o-R_f=R_o-\sqrt{\frac{\omega_f}{\omega_o}~R_f^{~~2}}$

idku · Answer

Ohh, my bad, the last equation should be,
$R_o-R_f=R_o-\sqrt{\frac{\omega_f}{\omega_o}~R_o^{~~2}}$

agent0smith · Answer

$$\Large I_1 \omega_1 = I_2 \omega_2 $$where I is for a sphere $$\large I = \frac{ 2 }{ 5 }m r^2$$
and w is the rotational velocity $$\large \omega = \frac{ 2 \pi }{ T }$$where T is the rotation period

idku · Answer

So, assuming the mass is constant (or else, I have too many unknowns), not only 2/5 (which I accounted for, with "a"), but also the masses cancel out, and so would the 2π, so I end up with,

$\displaystyle R_o-R_f=10200~\color{grey}{\rm m}-\sqrt{\frac{2\pi/T_f}{2\pi/T_o}~(10200~\color{grey}{\rm m})^2}$

$\displaystyle R_o-R_f=10200~\color{grey}{\rm m}-\sqrt{\frac{T_o}{T_f}~(10200~\color{grey}{\rm m})^2}$

$\displaystyle R_o-R_f=10200~\color{grey}{\rm m}-(10200~\color{grey}{\rm m})\sqrt{\frac{25.037~\color{grey}{\rm ms}}{25.033~\color{grey}{\rm ms}}}$

$\displaystyle R_o-R_f=10200~\color{grey}{\rm m}-(10200~\color{grey}{\rm m})\sqrt{\frac{25.037}{25.033}}$

idku · Answer

$\displaystyle R_o-R_f=0.81482665~\color{grey}{\rm m}$
$\displaystyle R_o-R_f=81.482665~\color{grey}{\rm cm}$

idku · Answer

That is my take on it ....

idku · Answer

Yes, it's correct !!