Recent content by shanklove

Ok I think I got it! I(initial)*Omega(initial)=I(final)*Omega(final) So I(initial)*(2pi/T(initial)) = I(final)*(2pi/T(final)) If the distribution is symmetric then kMR(initial)^2*(I(initial))=kMR(final)^2*(I(final)) So the kMR would cancel out on each side, so it would have...

Not following... From your link I= (2mr^2)/5 <----assuming you mean I= moment of inertia. But it would make more sense to me that I=r^2. So what do I do now with the rotational rate...? I'm so confused =X

It's just 2pi/T. So would it be 2pi/2592000?

We're doing oscillatory motion and rotational motion and angular momentum. Oh and I'm assuming that he was asking for the period of the planet, the period of the star wouldn't make sense.

Homework Statement A neutron star has a radius of 10,000 km, and takes a planet 30 days to complete one revolution around the star. When the star collapses, the new radius is 3 km. Find the new period of revolution of the newly formed neutron star. Homework Equations T=2(pi)/omega omega...