Physics
19.07.2020 03:54
127
163
6
Solved by an expert

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope.

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?
Show Answers
Jperez6541
Jperez6541
5,0(75 marks)

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

       

tianna08
tianna08
5,0(57 marks)

please find attached pdf

Explanation:

Popular Questions about the subject: Physics

What is the major source of energy for humans...
Physics
24.09.2022 20:07
What is one situation where potential energy might be useful...
Physics
10.12.2020 04:55
How long does it take for the speed of light and sound to travel around...
Physics
14.08.2020 20:22
What occurs when water freezes?...
Physics
18.07.2021 21:40

New questions by subject

Who want to trade on rocket league got to be on ps4 though...
Business
06.07.2020 11:08
Which of the following is equivalent to –(–5.25) ? 5 5.25 –5 –5.25 please...
Mathematics
09.11.2022 03:22
Please help me on dis...
Mathematics
27.05.2022 23:26
Do all of the chemistry shown below...
Chemistry
01.10.2021 08:51
Please help ill give brainliest...
English
08.07.2022 06:26
The range of which function includes -4? Oy5 Y = AX-5 y= √x+5 5 Y = 4X+5...
Mathematics
20.06.2022 01:55
CAN SOMEONE PLEASEE PLEASEE SOLVE THIS PROBLEM ? PLEASEEEE...
Mathematics
15.01.2020 17:28
Please help! (look at picture) Poem about A fence is......
English
14.06.2022 04:09
#
#
#
#
# #

We expand our knowledge with many expert answers