In the figure below, two skaters, each of mass 60 kg, approach each other along
ID: 1302917 • Letter: I
Question
In the figure below, two skaters, each of mass 60 kg, approach each other along parallel paths separated by 2.6 m. They have opposite velocities of 1.9 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?
m
(b) What is the angular speed of the skaters?
rad/s
(c) What is the kinetic energy of the two-skater system?
J
(d) Next, the skaters pull along the pole until they are separated by 0.6 m. What is their angular speed then?
rad/s
(e) Calculate the kinetic energy of the system now.
J
(d) What provided the energy for the increased kinetic energy?frictiongravitycentripetal forceinternal energy of the skaterscentrifugal force
Explanation / Answer
a.)r=1.3 m
b.)w=v/r=1.9/1.3= 1.462 rad/s
c.)I=2mr^2=2*60*(1.3)^2=202.8 kg.m^2
KE=0.5*I*w^2=0.5*202.8*1.462^2= 216.6 J
d.)Iiwi=Ifwf=202.8*1.462
rf=0.3 m
If=2*60*(0.3)^2=10.8 kg.m^2
So wf=202.8*1.462/10.8= 27.45 rad/s
e.)Kf=0.5*If*wf^2=0.5*10.8*(27.45)^2= 4069.84 J
f.)We account for the large increase in kinetic energy (part (e) minus part (c)) by noting that the
skaters do a great deal of work (converting their internal energy into mechanical energy) as they
pull themselves closer