Texts two astronauts each of mass m are connected by a rope...

Texts: Two astronauts, each of mass 𝑚, are connected by a rope of length 𝑑 (assume the rope is massless). They are isolated in space and are each spinning in a circle with the axis of rotation at the center of the rope. a) At some time, the astronauts both pull on the rope, so that the distance between them becomes less than 𝑑. Is the angular momentum of the system (the two astronauts) conserved during this pull? b) Compare the moment of inertia of the system about the axis before the pull and after the pull. Choose one answer from below and clearly explain how you arrive at your answer! i. The moment of inertia is larger after the pull than before. ii. The moment of inertia is the same before and after the pull. iii. The moment of inertia is smaller after the pull than before. c) Compare the angular speed of the astronauts before the pull and after the pull. Choose one answer from below and clearly explain how you arrive at your answer! i. The astronauts are rotating faster after the pull than before. ii. The astronauts are rotating with the same angular speed after the pull as before. iii. The astronauts are rotating slower after the pull than before.

Transcript

00:01 In this question, it has been given there two astronauts in deep space which is almost isolated.

00:15 That is no other force is present in this space.

00:20 So, and the mass of each astronaut is 99 .5 kg and they are connected by a rope of length 10 meter and they are revolving or rotating in a circle such that at the linear speed of the individual astronaut is 4 .9 meter per second.

01:02 Now in the question it has been given that at the midpoint of the rope they are rotating counterclockwise about this point.

01:18 So about this point we can calculate, let's say point o, we can calculate the angular momentum of the system.

01:25 Now angular momentum of system can be calculated by the formula sigma m1, v1, r1, where m1 is the mass of the individual particle or individual body, sorry, and v1 is the linear velocity and r1 is the radius.

01:52 Of rotation for this question this will be the r1 and let's say this is the mass m2 v2 r2 so this will be r2 so angular momentum can be calculated that is mass 99 .5 then 4 .9 into radius that is equals to 5 5 meter plus the angular momentum of the second astronaut that is that is also 99 .5 into 4 .9 into 5 so some of this will be equals to that is angular momentum of system will be equals to 4875 .5 kg meter square per second so this will be the answer for part a.

03:02 Now in part b it has been asking that rotational energy of this system.

03:09 So rotational energy will be equals to half i omega 1 square plus half i2 omega 2 square.

03:27 Now here i1 can be written as m1r1 square where i2.

03:36 R1 is the radius of rotation that is these are the two individual astronauts then r1 will be from point 1 to this length and this one will be r2 so now this is multiplied by omega 1 square now omega 1 can be written as v1 square divided by r1 square so similarly this can we can also write half into m2 r2 square into v2 square by r2 square.

04:14 So solving this, we will get half m1 v1 square plus half m2 v2 square.

04:23 So this can be written as the formula of energy.

04:28 Now in this question, the m1 equals to m2, that is the mass of each individual astronaut is 99 .5 kg.

04:35 So putting the formula, sorry the value, we will get half into 99 .5 into b1 squared that is 4 .9 square multiplied by 2 because values are same.

04:53 So we will get energy equals to 2388 .99 or one can say 2389 jule.

05:05 This is the kinetic energy or sorry rotational energy of this system.

05:12 So this is the answer for part b...

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