A certain star approximated as a uniform sphere with radius

A certain star, approximated as a uniform sphere with radius 8 Times 10^8m, has a rotational period of about 30 days (~ 2.592 Times 10^6s.) If this star were to collapse inward to form a red dwarf with approximately the same radius as the earth, R_e = 6.38 Times 10^6m, find the new rotational period. The moment of inertia of a uniform sphere is 2/5 MR^2. Very briefly explain how you can calculate this answer and why.

Solution

Initial Moment of Inertia

I_initial=(2/5)M(8*10⁸)²

Final Moment of inertia

I_final=(2/5)M*(6.38*10⁶)²

Initial rotational speed

W_initial=2pi/T =2pi/(2.592*10⁶) =2.424*10^-6 rad/s

By Conservation of angular momentum

I_initialW_initial=I_finalW_final

(2/5)M(8*10⁸)²*(2.424*10^-6)=((2/5)M*(6.38*10⁶)²*W_final

W_final = 0.038 rad/s