Two skaters, each of mass 45 kg, approach each other along parallel paths separa
ID: 3161673 • Letter: T
Question
Two skaters, each of mass 45 kg, approach each other along parallel paths separated by 3.2 m. They have equal and opposite velocities of 2.1 m/s. The first skater carries one end of a long pole with negligible mass, and the second skater grabs the other end of it as she passes. See Fig. 12-42 (from HRW). Assume frictionless ice.
(a) Let the system include the skaters' bodies and the pole. By assuming that the ice is frictionless, we are really making which of the following assumptions
the angular momentum of the system is conserved
the kinetic energy of the system is conserved
there are no unbalanced external torques acting on the system
(b) After they become connected by the pole, what is the distance from each skater to the axis of rotation? m
The angular velocity rad/s
(c) By pulling on the pole, the skaters reduce their separation to 0.8 m. What is their angular speed then?
rad/s
(d) Calculate the kinetic energy of the system in (b) and (c).
J (energy for system a)
J (energy for system b)
(e) What caused the kinetic energy to change from (b) to (c)?
friction and air resistance
chemical potential energy was converted to kinetic energy
the skaters' hands did work when they moved toward the center.
Explanation / Answer
part a )
the angular momentum of the system is conserved
part b )
total linear momentum is zero . thus their center of mass remains fixed and they execute circular motion
center of mass = middle of long stick
r = 1.6 m
w = v/r = 2.1/1.6 = 1.3125 rad/s
part c )
here angular momentum is conserved
Li = Lf
L = Iw
I = mr^2
initially r = 1.6 m
Ii = 2*45 * 1.6^2 = 230.4 kg-m^2
If = r = 0.4m = 14.4 kg-m^2
Ii*wi =If*wf
wf = 230.4 * 1.3125 / 14.4
wf = 21 rad/s
part d )
KE = Iw^2/2
initially
w = 1.3125
KEi = 1/2 * 230.4 * 1.3125^2
KEi = 198.45 J
KEf = 1/2*If*wf^2
KEf = 1/2 * 14.4 * 21^2
KEf = 3175.2 J
part e )
the skaters' hands did work when they moved toward the center.
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