Answer in Mechanics | Relativity for Valentine Agun #108074
For a single rotating body, the law of conservation of momentum must be fulfilled. We will assume (which is a rough approximation) that the mass of the star is distributed evenly over its volume. Then the moment of inertia of the ball is written as “I=frac{2}{5}mR^2”, where m, R – its mass and radius [1]. the net angular momentum of a ball around a principal axis is “L=Iw”. When our Sun is normal star
(1) “L_o=frac{2}{5}M_{bigodot} R^2 cdot w_o”
After it collapses into a white dwarf, losing about half its mass in the process and winding up with a radius 1.0% (=0.01) of its existing radius the angular momentum will be
(2) “L_1=frac{2}{5}frac{M_{bigodot}}{2} (0.01cdot R)^2 cdot w_1”
According the law of conservation angular momentum (the lost mass carries away no angular momentum) we have “L_1=L_o” . Thus for new angular velocity “w_1” we have
(3) “frac{2}{5}frac{M_{bigodot}}{2} (0.01cdot R)^2 cdot w_1=frac{2}{5}M_{bigodot} R^2 cdot w_o\nw_1=2 cdot 100^2 w_o=2cdot 10^4 w_o”
“w_o=frac{2pi}{30 day}=frac{2pi}{30cdot 24cdot 60cdot 60 s}=2pi cdot 3.9cdot 10^{-7} s^{-1}” Finally for new rotation we get
“w_1=2pi cdot 7.8cdot 10^{-3} s^{-1}” Thus our Sun’s white dwarf will rotate with period “T=frac{1}{7.8cdot 10^{-3}}s=128s”
The KE we can find by the equation
“E_k=frac{Iw^2}{2}=frac{Lw}{2}” Initial KE is “E_o=frac {L_ow_o}{2}”. The final KE is “E_1=frac {L_1w_1}{2}”. We can find a relationship
(4) “frac {E_1}{E_o}=frac {L_1 w_1}{L_ow_o}=frac {w_1}{w_o}=2cdot 10^4”
Where we taken into account the law of conservation angular momentum and (3). In our assumptions, the gravitational potential energy of the Sun’s mass is completely converted into the kinetic energy of rotation of its remainder.
Answer: The Sun’s new rotation period will be about 128 s. The KE of white dwarf will exceed the initial value for the Sun by “2cdot 10^4” times.
[1] https://en.wikipedia.org/wiki/Moment_of_inertia
