Respuesta :
Answer:
The angular speed of the neutron star is [tex]3976.978\ rads^{-1}[/tex]
Solution:
As per the question:
Initial radius of the star, r = [tex]7.0\times 10^{5}\ km = 7\times 10^{8}\ m[/tex]
Final Radius, r' = 16 km
Period of rotation of the star, T = 35 days = [tex]35\times 24\times 3600 = 3024000\ s[/tex]
Now,
To calculate the angular speed of the neutron star:
We use the principle of conservation of angular momentum:
[tex]I\omega = I'\omega'[/tex]
where
I = Moment of inertia
[tex]\omega = \frac{2\pi}{T}[/tex]
[tex](7\times 10^{8})^{2}\times \frac{2\pi}{T} = (16000)^{2}\omega'[/tex]
[tex]\omega = 3976.978\ rads^{-1}[/tex]
The angular speed of the neutron star is; 4871.29 rad/s
What is the angular speed?
We are given;
Initial Radius of Original Star; R₁ = 7 * 10⁵ km
Final Radius of neutron star; R₂ = 16 km
Angular speed of original star for once in 35 days; ω₁ = 2.545 * 10⁻⁶ rad/s
Applying the conservation of momentum principle, we have;
I₁ω₁ = I₂ω₂
where;
I₁ = ²/₅MR₁²
I₂ = ²/₅MR₂²
Thus;
²/₅MR₁²ω₁ = ²/₅MR₂²ω₂
This reduces to R₁²ω₁ = R₂²ω₂
ω₂ = (R₁/R₂)²
ω₂ = (7 * 10⁵/16)² * 2.545 * 10⁻⁶
ω₂ = 4871.29 rad/s
Read more about Angular Speed at; https://brainly.com/question/6860269
