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Sagot :
Answer:
a) w = 0.4726 rad / s, b) w_{f} = 4.56 rad / s
Explanation:
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
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