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

Question

Two skaters, each of mass 80 kg, approach each other along parallel paths separated by 3.9 m. They have equal and opposite velocities of 1.4 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?

I figured out the first one.
7.18×10-1 rad/s

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.

Can anyone help me get started on the second part?

Explanation / Answer

Let w represent angular velocity. 2 m v r = I w = 2 m r^2 w conservation of angular momentum w = v/r = 1.4 / 1.95 = .718 = w1 initial angular speed I1 w1 = I2 w2 conservation of angular momentum I1 = 2 m r1^2 = 2 * 80 * 1.95^2 = 608 kg-m^2 I2 = 2 m r2^2 = 2 * 80 * .5^2 = 40 kg-m^2 w2 = I1 * w1 / I2 = 608 * .718 / 40 = 10.9 / s KEf / KEi = 1/2 I2 w2^2 / (1/2 I1 w1^2) = I2 w2^2 / (I1 w1^2) KEf / KEi = = (40/608) * (10.9/.718)^2 = 15.2

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