A star of mass twice the solar mass and radius $106km$ rotates about its axis with an angular speed of $10^{‑6}$ rad per sec. What is the angular speed of the star when it collapses (due to inward gravitational forces) to a radius of $10^4km$? Solar mass = $1.99 \times 10^{23}kg$.
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According to the law of conservation of angular momentum, $\text{I}_1\omega_1=\text{I}_2\omega_2$ Since sun is a asphere, so $\text{I}_1=\frac{2}{5}\text{MR}^2_1, R_1$ = Radius of sun M.I. of star $\text{I}_2\frac{2}{5}2\text{M}\text{R}^2_2,R_2$ = radius of star
$\therefore\frac{3}{5}\text{MR}^2_1\omega_1=\frac{4}{5}\text{MR}^2_2\omega_2$
$\text{or }\omega_2=\frac{\text{R}^2_1}{2\text{R}^2_2}\omega_1$ Since $R_1 = 10^6km; R_2 = 10^4km$; $\omega_1=10^{-6}\text{/sec}$
$\therefore\omega_2=\frac{(10^6)^2}{2(10^4)^2}\times10^{-6}$
$=5\times10^{-3}\text{rad/sec.}$
$\therefore\frac{3}{5}\text{MR}^2_1\omega_1=\frac{4}{5}\text{MR}^2_2\omega_2$
$\text{or }\omega_2=\frac{\text{R}^2_1}{2\text{R}^2_2}\omega_1$ Since $R_1 = 10^6km; R_2 = 10^4km$; $\omega_1=10^{-6}\text{/sec}$
$\therefore\omega_2=\frac{(10^6)^2}{2(10^4)^2}\times10^{-6}$
$=5\times10^{-3}\text{rad/sec.}$
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