Two skaters, each of mass 60 kg , approach each other along parallel paths separ

Question

Two skaters, each of mass 60 kg, approach each other along parallel paths separated by 6.7 m. They have equal and opposite velocities of 2.2 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 the figure. Assume frictionless ice. Describe quantitatively the motion of the skaters after they have become connected by the pole. What is their angular speed?

By pulling on the pole, the skaters reduce their separation to 1.0 m. What is their angular speed then?

Calculate the ration of the final kinetic energy to the original kinetic energy?

Explanation / Answer

m1=m2=m=60kg, r1=r2=r=6.7/2 = 3.35m , v1=v2=v=2.2m/s

Before pulling on the pole,

Angular velocity =w=v/r = 2.2/3.35 = 0.66rad/s

Total moment of inertia = ITotal = m1*r1^2+m1*r1^2 = 2*m*r^2 = 2* 60*3.35^2 = 1346.7 kg*m^2

Total KE= ½*I*w^2 = 1/2*1346.7*0.66^2 = 293.3 J

After pulling on the pole,
By law of conservation of angular momentum,

Iiwi=Ifwf

wf= (Iiwi)/If

If = 2*m*r^2 = 2*60*0.5^2= 30kg*m/s

wf= (Iiwi)/If = (1346.7*0.66)/30 = 29.6 rad/s

KE = ½*I*w^2 = ½*30*29.6^2 = 13.14 kJ

KEf/KEi = 13142 /293.3 = 44.8

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