In the figure below, two skaters, each of mass 40 kg, approach each other along
ID: 1560359 • Letter: I
Question
In the figure below, two skaters, each of mass 40 kg, approach each other along parallel paths separated by 2.5 m. They have opposite velocities of 2.2 m/s each. One skater carries one end of a long pole with negligible mass, and the other skater grabs the other end as she passes. The skaters then rotate around the center of the pole. Assume that the friction skates and ice is negligible. (a) What is the radius of the circle? (b) What is the angular speed of the skaters? (c) What is the kinetic energy of the two-skater system? Next, the skaters pull along the pole until they are separated by 1.2 m. (d) What then is their angular speed? (e) What then is the kinetic energy of the system? (f) What provided the energy for the increased kinetic energy? (Select all that apply.)Explanation / Answer
a) radius of the circle r = d / 2 = 2.5 / 2 = 1.25 m
b) Angular velocity w = v / r = 2.2 / 1.25 = 1.76 rad/s
c) Total rotational inertia I = 2 * m r2 = 2 * 40 * 1.252 = 125 kg.m2
Total kinetic energy = 1/2 I w2 = 1/2 * 125 * 1.762 = 193.6 J
d) If = 2 * 40 * (1.2/2)2 = 28.8 kg.m2
wf = I1 wi / If = (125 * 1.76) / (28.8)
= 7.64 rad/s
e) kf = 1/2 * 28.8 * 7.642= 840 J