Answer to Question #110783 in Classical Mechanics for Jujin

Question

A star rotates with a period of 30 days about an axis through its center. The period is the time interval required for a point on the star’s equator to make one complete revolution around the axis of rotation. After the star undergoes a supernova explosion, the stellar core, which had a radius of 1.0 x 10^4 km, collapses into a neutron star of radius 3.0 km. The period of rotation of the neutron star is 0.23 s. Calculate the work done by the 'supernova' forces per kg of star mass. (MJ/kg = 10^6 J/kg)

Answer 1

The initial kinetic energy of the core was

$\frac{1}{2}I_c\omega_c^2=\frac{1}{2}\bigg(\frac{2}{5}mr_c^2\bigg)\frac{4\pi^2}{T^2}=\frac{4}{5}m\bigg(\frac{\pi r_c}{T_c}\bigg)^2$

By analogy, the kinetic energy of the neutron star will be

$\frac{1}{2}I_n\omega_n^2=\frac{4}{5}m\bigg(\frac{\pi r_n}{T_n}\bigg)^2$

The work done by supernova per kilogram of mass is

$\frac{W}{m}=\frac{\Delta KE}{m}=\frac{4\pi^2}{5}\bigg[\bigg(\frac{r_n}{T_n}\bigg)^2-\bigg(\frac{r_c}{T_c}\bigg)^2\bigg]=\\ \space\\ =1340\text{ MJ/kg}.$