The problem is just too easy, with both of them having the same mass. So the center of rotation is at the midpoint, and radius of rotation is half length.
Itotal=2Ieach= 2*mr^2
after reducting the length, then
I total=2m*(r-.27r)^2
conservation of momentum
initial=final
2mr^2*wi=2mr^2(.73)^2 wf^2
solve for wf
I get wf=wi(.73) in my head. check that
work? find the change of KE in the before and after rotational energies
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 ù0. By pulling on the rope, one of the astronauts shortens the total distance between them to 0.54d. 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 ù0? (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 Md^2ù0^2 (which has dimensions of energy)? (Hint: Calculate the change in rotational kinetic energy of the system instead, since you have no idea how the astronaut actually carried out the shortening.)
2 answers
It's actually wf=wi(1/.54^2)=wi(3.43)
I can't seem to get the second part though.
I can't seem to get the second part though.
1 answer