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

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

Step-by-step 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 = (d_i)/(2) = (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 = (d_f)/(2) = (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 = (2 * m * r_i^2 w_1)/(2 * m * r_f^2 )`

=>
`w_f = (3.5^2 * 0.5)/( 2^2 )`

=>
`w_f = 1.531 \ rad/ s`