Two skaters, each of mass 51.2 kg, approach each other along parallel paths separated by 2.92 m. They have equal and op- posite velocities of 1.38 m/s. The first skater carries a long light pole 2.92 m long, and the second skater grabs the end of it as he passes; see Fig. 10-30. Assume frictionless ice. (a) Describe quantitatively the motion of the skaters after they are connected by the pole. (b) By pulling on the pole, the skaters reduce their separation to 0.940 m. Find their angular speed then.

a) the two skaters have an angular momentum, for which we will form a system formed by the two of them, in this case when they collide the forces are internal and the moment is conserved.

Initial instant. When the skaters approach, just before touching

L₀ = r p = r m v

in this case the result is the same if we take the reference system at the midpoint or on one of the skaters. We place the reference system on one of the skaters

Final moment. After being joined by the pole

[tex]L_{f}[/tex] = I w

as the forces are internal, the ngualr moment is conserved

L₀ = L_{f}

r m v = I w

suppose we approximate the skaters as particles

I = m r²

we substitute

r m v = m r² w

w = v /r

we calculate

w = 1.38 /2.92

w = 0.4726 rad / s

What happens is that when the two skaters lower the pole, their velocity acts as a torque, creating a rotational movement with angular velocity w.

b) When the skaters who are at r₀ = 2.92 m approach [tex]r_{f}[/tex] = 0.940 m as the system is isolated, the angular momentum is conserved

initial instant. r₀ = 2.92

L₀ = I w

final instant r_{f} = 0.940 m

L_{f} = I_{f} w_{f}

L₀ = L_{f}

I w = I_{f} w_{f}

let's approximate skaters as particles

I = m r²

we substitute

m r² w = m r_{f}² wf

[tex]w_{f}[/tex] = [tex]\frac{r^{2} }{r_{f}^{2} } w[/tex]

we calculate

w_{f} = [tex]\frac{2.92^{2} }{ 0.940^{2} }[/tex] 0.4726

w_{f} = 4.56 rad / s

we see that the angular velocity increases