Two skaters, each of mass 50 kg, approach each other along parallel paths separated by 4.0 m. They have opposite velocities of 1.7 m/s each. One skater carries one end of a long pole with negligible mass, and the second skater grabs the other end of it as she passes. The skaters then rotate around the center of the pole. Assume that the friction between skates and ice is negligible.
(a) What is the radius of the circle they now skate in?
(b) What is the angular speed of the skaters?
(c) What is the kinetic energy of the two-skater system?
(d) Next, the skaters pull along the pole until they are separated by 0.7 m. What is their angular speed then?
(e) Calculate the kinetic energy of the system now.
The Attempt at a Solution
I have figured out parts (a), (b), and (c). For part (d), I used the same process as in part (b). In (d), I assumed that the new radius was .35 m, and the velocity was still 1.7. So, 1.7/.35 should equal the new angular speed, but that answer does not work. In part (e), I still used what I think is the new radius, .35 m, and put the values into the Kinetic energy equation: 0.5*(50*(.35)^2+50*(.35)^2)*(1.7/.35)^2 = 0.5*I*w^2. This answer does not work either, but I have to be missing something in part (d) to begin with, and I have to get that first anyway to solve (e). Does anyone see what I'm forgetting here? Thanks.