How Does Angular Speed Change When Skaters Connect by a Pole?

[Punchlinegirl]

Two skaters, each of mass 85 kg, approach each other along parallel paths separated by 3.5 m. They have equal and opposite velocities of 1.6 m/s. The first skater carries one end of a long pole with negligible mass, and the second skater grabs the other end of it as she passes. Assume frictionless ice. Describe quantitatively the motion of the skaters after they have become connected by the pole. What is their angular speed?

I know that to solve this I need to use conservation of angular momentum, but I'm unsure what the moment of inertia is. Would it just be .5MR^2?

 

[Galileo]

Just add the moments of inertia of the two skaters with respect to the axis of rotation (center of pole). The moment of inertia of one person (treated as a point particle) is mr^2.

 

[Punchlinegirl]

Ok, I used:
L init= L final
Iw= Iw
mr^2 = (mr^2 + mr^2) w
I converted the linear velocity to angular velocity, v=wr
subsituting gives
85(3.5^2)(.457)= 2(85)(3.5^2) w
and dividing gives w= .229 rad/s, which isn't right.
Can someone tell me what I did wrong?

 

[asrodan]

Since the parameters for both skaters are the exactly the same both sides of the angular momentum equation will be exactly the same as well if you write them in terms of the same quantities.

I.e. Iw = Iw (2mr^2)w = (2mr^2)w

try using the fact that L = (m+m)vr = 2mvr along with Iw