Chapter 33, Problem 36GP

To determine

The angular velocity of a neutron star.

Answer to Problem 36GP

The angular velocity of a neutron star is $11.9\times {10}^3\text{rad}/\text{s}$ .

Explanation of Solution

Given:

The time period of rotation is $T_1=1month$ .

The initial radius of the star is $R_1=7\times {10}^8\text{m}$ .

The final radius of the star is $R_2=10\text{km}=10\times {10}^3\text{m}$ .

Formula used:

The expression of principle of conservation of angular momentum is,

  $\begin{array}{l}{I}_1{\omega }_1=I_2{\omega }_2\\ \left(\frac{2}{5}M{R}_1^2\right)\left(\frac{2\pi }{T_1}\right)=\frac{2}{5}M{R}_2^2{\omega }_2\\ {\omega }_2=\frac{2\pi }{T_1}\times {\left(\frac{R_1}{R_2}\right)}^2\end{array}$

Here, $T_1$ is the period of rotation, ${\omega }_1$ is the initial angular velocity and ${\omega }_2$ is the final angular velocity.

Calculation:

The final angular velocity is,

  $\begin{array}{c}{\omega }_2=\left(\frac{2\pi }{T_1}\right){\left(\frac{R_1}{R_2}\right)}^2\\ =\left(\frac{2\pi }{1\text{month}\times \frac{30\text{days}}{1\text{month}}\times \frac{24\text{h}}{30\text{days}}\times \frac{3600\text{s}}{1\text{h}}}\right){\left(\frac{7\times {10}^8}{10000}\right)}^2\\ =\left(\frac{2\pi }{30\times 24\times 3600}\right){\left(\frac{7\times {10}^8}{10000}\right)}^2\\ =11.9\times {10}^3\text{rad}/\text{s}\end{array}$

Conclusion:

The angular velocity of a neutron star is $11.9\times {10}^3\text{rad}/\text{s}$ .