two astronauts angular momentum unkown r

diego1404

Two astronauts, each having a mass M are connected by a length of rope of length d have a negligible mass. They are isolated in space, orbiting their center of mass at an angular speed of ω0. By pulling on the rope, one of the astronauts shortens the total distance between them to 0.668d. Treat the astronauts as point particles (in terms of their moments of inertia).

a) What is the final angular speed of the astronauts as a fraction/multiple of ω0 ? (E.g. If you find that the final angular speed is half the initial angular speed enter 0.5.) Use angular momentum conservation.
b) What work does the astronaut do to shorten the rope as a multiple/fraction of the quantity Md2ω02 (which has dimensions of energy)?


essentially i ended up having the equation

initial---->Iw = Iw <----final
2mr^2w = 2mr^2 w

Wf = (R/Rf)^2 * W

as you can see the problem comes when you realize that you dont know what the initial distance is and that almost all the rotational dynamics needs distance. so what can i substitute for the initial distance

tiny-tim

hi diego1404!
call the initial distance d, and the final distance 0.668d

diego1404

every single time....... thank you. i have to remind myself to read things properly