Question

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly ${\mathbf{10}}^{\mathbf{14}}$times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was $\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{km}$(comparable to sun) its final radius is $\mathbf{16}\mathbf{}\mathbf{km}$. If the original star rotated once in 30 days, find the angular speed of the neutron star.

Short answer

Hence, the final angular speed is $4.6\times {10}^3\mathrm{rad}/\mathrm{s}$.

Step-by-step solution

Conservation of angular momentum.

If a system experiences no external torques from the environment, the total angular momentum is conserved.

$\Delta \overrightarrow{L}\to t=0$

Apply the law of conservation of angular momentum to a system whose moment of inertia changes gives:

$I_j{\omega }_j=l_f{\omega }_f=\text{constant}$

The given data in the question

Given that, initial radius of the original star is $R_j=7.0\times {10}^5\mathrm{km}$, and the final radius of the neutron star is $R_f=16\mathrm{km}$. The angular speed of the original star is

$$\begin{gathered} {\omega }_i=\left(\frac{1rev}{30days}\right)\left(\frac{2\pi rad}{1rev}\right)\left(\frac{1day}{24h}\right)\left(\frac{1h}{3600s}\right) \\ =2.424\times {10}^{-6}\mathrm{rad}/\mathrm{s} \end{gathered}$$

The star is modelled as a uniform, solid, rigid sphere both before and after the collapse.

The angular speed of the star.

Let the initial state of the system before the collapse and the final sate be after the collapse.

Consider there are no astronomical bodies are in the vicinity of the star, no forces or torques are exerted on the star. Thus, apply conservation of momentum principle from equation to the star between the initial and final states:

$$\begin{gathered} I_j{\omega }_i=l_f{\omega }_f \\ \frac{2}{5}M{R}_i^2{\omega }_i=\frac{2}{5}M{R}_f^2{\omega }_f \\ {R}_i^2{\omega }_i=R_f^2{\omega }_f \end{gathered}$$

Solve for ${\omega }_f$as follows:

$$\begin{gathered} {\omega }_f={\omega }_i{\left(\frac{R_i}{R_f}\right)}^2 \\ =2.424\times {10}^{-6}\left(\frac{7.0\times {10}^5}{16}\right) \\ =4.6\times {10}^3\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the final angular speed is $4.6\times {10}^3\mathrm{rad}/\mathrm{s}$.