(hrw8c11p51) Two skaters, each of mass 80 kg , approach each other along paralle

Question

(hrw8c11p51) Two skaters, each of mass 80 kg, approach each other along parallel paths separated by 11.7 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 ratio of the final kinetic energy to the original kinetic energy.

Explanation / Answer

Given: mass of each skater 'M' = 80 Kg; Length of rod (their seperation) = 11.7 m; and

speed of each skater 'v' = 2.2 m/s

Now as soon as they gets joined with the rod they will start to rotating about the center of mass of the system which in this case would be the center of the rod as their net translational speed would become zero.

Moreover net angular momentum of the system about the center of mass was:

L = 2*m*v*(L/2)

Hence the angular speed from angular momentum conservation would be:

L = I*0 where 'I0' = 2*m*(L/2)2 is the moment of inertia of the system about the center of mass   

Hence 0 = L/I0 = (2*m*v*L)/(m*L2) = 2*v/L = 2*2.2/11.7 = 0.376 rad/s

As the new seperation becomes Ln = 1 m so the new moment of inertia of the system becomes: I = 2*m*(Ln/2)2

Hence the new angular speed from momentum conservation becomes:

   = 2*v/Ln  = 2*2.2/1 = 4.4 rad/s

Ratio of final kinetic energy to initial kinetic energy Ef/Ei   = (0.5*I0*02)/(0.5*I*2)

Ef/Ei   = (L2*02)/(Ln2*2) = (11.7*11.7*0.376*0.376)/(1*1*4.4*4.4) = 0.999

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