Two skaters, each of mass 75 kg , approach each other along parallel paths separ

Question

Two skaters, each of mass 75 kg, approach each other along parallel paths separated by 12.1 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?

Calculate the ratio of the final kinetic energy to the original kinetic energy.

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

Explanation / Answer

V1 = V2 = V

M1 = M2 = M

L1 = M1*V1*(L/2)

L2 = M2*V2(L/2)

Li = L1+L2 = M*V*L

Lf = (M1*(L/2)^2) + M2*(L/2)^2) *W = M*L^2*W

Lf = Li

M*L^2*W = M*V*L

W = V/L = 2.2/12.1 = 0.181 rad/s

KEf/KEi = 0.5*M*L^2*W /M*V^2 = L^2W/2V^2 = (12.1^2*0.181)/(2*2.2^2) = 2.73

c) W' = V/L' = 2.2/1 = 2.2 rad/s

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