Question

Two skaters, each of mass 40 kg, approach each other along parallel paths separated by 4.0 m. They have opposite velocities of 2.1 m/s each. One skater carries one endof 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 thatthe friction between skates and ice is negligible.


(a) What is the radius of the circle they now skate in?
2 meters

(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.

Answer 1

Without giving it ALL away, here goes! You will have to do the numbers.

a) they revolve around their common CM (center of mass), which is at the 3.8 m pole center.

b)The total angular momentum about their CM is the sum of their individual L's. L=(Sum mvr). Which, in turn, equals the moment of inertia times the angular speed. L=(I)(omega)

c)The KE = (1/2)( I )(omega)^2

d)Since no external forces or torques are acting, angular momentum is conserved. So since the moment of inertia is reduced, the angular speed is increased. ( L before)=(L after)

e)(KE after) =(1/2)( I after)(omega after)^2

f)internal energy of the skaters pulling on the pole changing the moment of inertia