How Does Shortening a Rope Affect Astronauts' Angular Momentum?

[haydn]

Homework Statement

Two astronauts, each having a mass M, are connected by a rope of length d having negligible mass. They are isolated in space, orbiting their center of mass at speeds v. (Use M, d, and v as appropriate in your equations for each of the following questions.)

By pulling on the rope, one of the astronauts shortens the distance between them to d/3.
(c) What is the new angular momentum of the system?

Homework Equations

L = mvrsin$$\theta$$

L = I$$\omega$$

The Attempt at a Solution

I calculated the angular momentum when the two astronauts were a distance d between each other and got the correct answer, L = Mvd

For the new angular momentum, I thought I would do the same thing just replacing r in the first equation listed above:

distance between astronaut and pivot point = (1/6)d

L (per astronaut) = Mv(1/6)d
2L = (1/3)Mvd

The website I'm using is telling me it's wrong though... can anyone help?P.S. in the picture it shows the pivot point being the center of the rope connecting the astronauts.

 

[Delphi51]

Angular momentum is conserved, so it is still mvd where v and d are from the original situation. It is also m*v2*d2, where the new v2 is larger than the old v and the new d2 is smaller than the old d.

 

[nlightner]

Your approach is correct, but there is a small error in your calculation. When you shorten the distance between the astronauts to d/3, the distance between each astronaut and the pivot point is not (1/6)d, but (2/3)d. This is because the distance between the astronauts is now (1/3)d, so each astronaut is (1/6)d away from the pivot point.

Therefore, the new angular momentum for each astronaut would be:

L (per astronaut) = Mv(2/3)d

And the total new angular momentum for the system would be:

2L = (4/3)Mvd

I hope this helps!