Answer:
The angular velocity is [tex]w_f = 1.531 \ rad/ s[/tex]
Explanation:
From the question we are told that
The mass of each astronauts is [tex]m = 50 \ kg[/tex]
The initial distance between the two astronauts [tex]d_i = 7 \ m[/tex]
Generally the radius is mathematically represented as [tex]r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m[/tex]
The initial angular velocity is [tex]w_1 = 0.5 \ rad /s[/tex]
The distance between the two astronauts after the rope is pulled is [tex]d_f = 4 \ m[/tex]
Generally the radius is mathematically represented as [tex]r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m[/tex]
Generally from the law of angular momentum conservation we have that
[tex]I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}[/tex]
Here [tex]I_{k_1 }[/tex] is the initial moment of inertia of the first astronauts which is equal to [tex]I_{p_1}[/tex] the initial moment of inertia of the second astronauts So
[tex]I_{k_1} = I_{p_1 } = m * r_i^2[/tex]
Also [tex]w_{k_1 }[/tex] is the initial angular velocity of the first astronauts which is equal to [tex]w_{p_1}[/tex] the initial angular velocity of the second astronauts So
[tex]w_{k_1} =w_{p_1 } = w_1[/tex]
Here [tex]I_{k_2 }[/tex] is the final moment of inertia of the first astronauts which is equal to [tex]I_{p_2}[/tex] the final moment of inertia of the second astronauts So
[tex]I_{k_2} = I_{p_2} = m * r_f^2[/tex]
Also [tex]w_{k_2 }[/tex] is the final angular velocity of the first astronauts which is equal to [tex]w_{p_2}[/tex] the final angular velocity of the second astronauts So
[tex]w_{k_2} =w_{p_2 } = w_2[/tex]
So
[tex]mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2[/tex]
=> [tex]2 mr_i^2 w_1 = 2 mr_f^2 w_2[/tex]
=> [tex]w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }[/tex]
=> [tex]w_f = \frac{3.5^2 * 0.5}{ 2^2 }[/tex]
=> [tex]w_f = 1.531 \ rad/ s[/tex]