1. The problem statement, all variables and given/known data An ice-skater, with a mass of 60.0 kg, glides in a circle of radius 1.4 m with a tangential speed of 6.0 m/s. A second skater, with a mass of 50.0 kg, glides on the same circular path with a tangential speed of 5.0 m/s. At an instant of time, both skaters grab the ends of a lightweight, rigid set of rods, set at 90 degrees to each other, that can freely rotate about a pole, fixed in place on the ice. A) If each rod is 1.4 m long what is the tangential speed of the skaters after they grab the rods? B) What is the direction of the angular momentum of the before and after the skaters "collide" with the rods? 2. Relevant equations A)Vtang=r(omega) m1Vi1+m2Vi2=(m1+m2)Vf B)L=I(omega) 3. The attempt at a solution A) Vf=(m1mVi1+m2Vi2)/(m1+m2)=[60kg(6.0 m/s)+50kg(5.0m/s)]/(60kg+50kg)=5.54m/s I didn't include the radius in my calculations because it remains the same for both skaters before and after the collision with the rods. B)For the second part: I looked up the rotational inertia of a rod rotating at an axis located at its end: 1/3ML^2. Because the problem says lightweight rods, I think I need to set it equal to zero, but then I=0 and the angular acceleration would equal zero. After the skaters collide with the rods: omega is Vtang/r= (5.54m/s)/1.4m=3.95 1/3ML^2 for the first skater- 1/3(60kg)(1.4m)^2*(omega)= 39.2 kg m*3.95/sec= 154 kg m/s for the second skater- 1/3(50kg)(1.4m)^2*(omega)=32.67 kg m*3.95/sec=129 kg m/s 154+129 kg m/s= 283 kg m/s Is this right? Any help would be very much appreciated.