Respuesta :
Answer:
The angular speed of the neutron star is 3130.5 rad/s.
Explanation:
Given that,
Initial radius [tex]r_{1}=7\times10^{5}\ km[/tex]
Final radius [tex]r_{2}=18 km[/tex]
Density of a neutron [tex]\rho= 10^{14}[/tex]
Equal masses of two stars [tex]m_{1}=m_{2}[/tex]
Suppose, If the original star rotated once in 35 days, find the angular speed of the neutron star
Time period of original star T = 35 days = 3024000 s
We need to calculate the initial angular speed of original star
Using formula of angular star
[tex]\omega=\dfrac{2\pi}{T}[/tex]
Put the value into the formula
[tex]\omega_{1}=\dfrac{2\pi}{3024000}[/tex]
[tex]\omega_{1}=0.00207\times10^{-3}\ rad/s[/tex]
Let the initial moment of inertia of the star is
[tex]I_{1}=m_{1}r_{1}^2[/tex]
Final moment of inertia of the star is
[tex]I_{2}=m_{2}r_{2}^2[/tex]
From the conservation of angular momentum
[tex]I_{1}\omega_{1}=I_{2}\omega_{2}[/tex]
[tex]\omega_{2}=\dfrac{I_{1}\omega_{1}}{I_{2}}[/tex]
[tex]\omega_{2}=\dfrac{m_{1}r_{1}^2\omega_{1}}{m_{2}r_{2}^2}[/tex]
Put the value into the formula
[tex]\omega_{2}=\dfrac{(7.0\times10^{5})^2\times0.00207\times10^{-3}}{18^2}[/tex]
[tex]\omega_{2}=3130.5\ rad/s[/tex]
Hence, The angular speed of the neutron star is 3130.5 rad/s.
