Question

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center of mass with an angular velocity of 0.5 rad/s. One of the astronauts then pulls on the rope shortening the distance between the two astronauts to 4 m. What was the averageangular speed exerted by the astronaut on the rope?

Answer 1

Answer:

The angular  velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

     The mass of each astronauts is  $m = 50 \ kg$

      The initial  distance between the two  astronauts  $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

      The initial  angular velocity is  $w_1 = 0.5 \ rad /s$

       The  distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

           $I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts  So

      $I_{k_1} = I_{p_1 } = m * r_i^2$

Also   $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts  So

      $w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts  So

      $I_{k_2} = I_{p_2} = m * r_f^2$

Also   $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the  final angular velocity of the second astronauts  So

      $w_{k_2} =w_{p_2 } = w_2$

So

      $mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2$

=>   $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=>   $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

=>    $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

=>   $w_f = 1.531 \ rad/ s$