two astronauts in free space each have mass M and are connected by a rope of neg

Question

two astronauts in free space each have mass M and are connected by a rope of negligable mass and length d. they each have a tangential speed of v and are orbiting their center of mass. calculate:

a. the magnitude of their total angular moment

b. their total energy

by pulling on the rope, one of the astronauts shortens the active rope between them to d/2. now calculate:

c. new magnitude of their angular momentum

d. what are the astronauts  new tangential speeds

e. their total energy

f. how much work was done by the astronaut who shortened the rope

Explanation / Answer

Given that the mass of astronaut is= m
separation between the eachastronaut is = d
speed of each astronautis = v
----------------------------------------------------------------
The separation between the astronaut and center of rotationis r = d/2
(a) the angular momentum of the system is L1 = 2*m*v*r = m*v*d
(b) rotational kinetic energy is K1 =(1/2)I?2
=(1/2)[ 2mr2 ][ v / r]2
= mv2
If the new separation between the persons is d/2 then r = d/4
(c) Since there is no external torque on thesystem so the angular momentum is conserved
L1 =L2
L2 = L1
(d) But the angular momentum of the system is L2 =2*m*v*r
v = L2 / 2*m*r ( where r = d/4)
(e) The new rotational energy is K2 =(1/2)I?2
= (1/2)[ 2mr2 ][ v / r]2
= mv2
( f ) work done W = change in kinetic energy
= K1 - K2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.