In the figure below two skaters each of mass 35 kg approach each other long para
ID: 1646294 • Letter: I
Question
In the figure below two skaters each of mass 35 kg approach each other long parallel paths separated by 2.4 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 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? What is the kinetic energy of the two-cater system? J (d) Next, the skaters pull along the pole until they are separated by 0.8 m. What is their angular speed then? rad/s (e) Calculate the kinetic energy of the system box..Explanation / Answer
(a)
radius = d/2 = 24/2 = 1.2 m
(b)
w= v/r = 1.9/1.2 = 1.583 rad/s
(c)
The total inertia is
I_total = 2 mr^2 = 2(35) ( 1.2)^2 = 100.8 kg m^2
KE= 1/2 I w^2 = 1/2 * ( 100.8) (1.583)^2 = 126.29 J
(d)
I_total = 2 mr^2 = 2( 35) ( 0.4)^2 = 11.2 J
from the law of conservation of angualr momentum
Ii wi = ifwf
wf = Ii wi/ If
=100.8(1.583)/11.2
=14.247 J
(e)
final kinetic energy is
Kf= 1/2 I w^2 = 1/2 * ( 11.2)( 14.247)^2 = 1136.6 J