Two skaters, each of mass 80 kg , approach each other along parallel paths separ

Question

Two skaters, each of mass 80 kg, approach each other along parallel paths separated by 4.5 m. They have equal and opposite velocities of 2.0 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

here,

mass of skater , m = 80 kg

distance seprated , d = 4.5 m

velocity , v = 2 m/s

their angular speed , wi = v/(d/2)

w1 = 0.89 rad/s

the initial angular speed is 0.89 rad/s

when their distance is , d' = 1 m

let their angular speed be w'

uisng conservation of angular momentum

m * r^2 * w = m * r'^2 * w'

2.25^2 * 0.89 = 0.5^2 * w'

w' = 18.02 rad/s

the new angular speed is 18.02 rad/s

the ratio of the final kinetic energy to the initial , R = KEf /KEi

R = ( 0.5 * I' * w'^2 ) / ( 0.5 * I* w^2)

R = ( 0.5 * 2 * 80 * 0.5^2 * 18.02^2 ) / ( 0.5 * 2 * 80 * 2.25^2 * 0.89^2)

R = 20.25 :1

the RATIO of the final kinetic energy to the original kinetic energy is 20.25 : 1

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