Two skaters, each of mass 55 kg , approach each other along parallel paths separ
ID: 1700785 • Letter: T
Question
Two skaters, each of mass 55 kg, approach each other along parallel paths separated by 11.9 m. They have equal and opposite velocities of 1.3 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.
What is their angular speed? = .218rad/s
Calculate the ratio of the final kinetic energy to the original kinetic energy.
Explanation / Answer
Mass of each skater, m = 55 kg Velocity of each, v = 1.3 m/s Radius of circular path, r = 11.9 / 2 = 5.95 m Angular momentum of each, L = m v r = 425.43 kg m^2 s^-1 Moment of inertia of each, I = m r^2 = 55 * 5.95^2 = 1947.13 kg m^2 Angular momentum of both, L' = 2L = 850.86 kg m^2 s^-1 Moment of inertia of both, I' = 2 I = 3894.3 kg m^2 (a) L' = I' w Angular speed, w = 850.86 / 3894.3 = 0.218 rad/s (b) Radius of circular path, r = 1 / 2 = 0.5 m Moment of inertia of each, I1= m r^2 = 55 * 0.5^2 = 13.75 kg m^2 Moment of inertia of both, I1' = 2 I = 27.5 kg m^2 According to law of conservation of angular momentum, L ( after ) = L ( before ) I1' * w' = 850.86 27.5 * w' = 850.86 w' = 30.94 rad/s (c) Ratio of final to initial kinetic energy = (1/2) I1' w'^2 / (1/2) I' w^2 = 27.5 * 30.94^2 / [ 3894.3 * 0.218^2 ] = 142.2
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