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Sagot :
Answer:
1) T = 649.86 s, 2) L₀ = L_f, [tex]\frac{K_o}{K_f}[/tex] = 4.8
Explanation:
1) As the system of the two bodies is isolated, its angular momentum is conserved
initial instant. r₀ = 155 m, T₀= 385.3 s
L₀ = I₀ w₀
final instant. r = 119.35 m
L_f = I w
L₀ = L_f
I₀ w₀ = I w
w = [tex]\frac{I_o}{I} \ w_o[/tex]
let's consider each object as punctual
I = m r²
at angle velocity and period are related
w = 2pi / T
we substitute
[tex]\frac{2\pi }{T} = \frac{m r^2}{m _o^2 } \ \frac{2\pi }{T_o}[/tex]
[tex]\frac{1}{T} = ( \frac{r}{r_o} )^2 \ \frac{1}{T_o}[/tex]
T = [tex](\frac{r_o}{r} )^2 \ T_o[/tex]
let's calculate
T = [tex]( \frac{155}{119.35} )^2 \ 385.3[/tex]
T = 649.86 s
2) The angular momentum is conserved because the system is isolated.
Let's look for kinetic energy
K_total = 2 K = 2 (½ I w²)
K_total = I 4π² / T²
K_total = 2m r² 4 π² / T²
for r = 155 m
K₀ = 8π² m r₀² / T₀²
for r = 119.35 m
K_f = 8π² m r² / T²
the relationship is
[tex]\frac{K_o}{K_f} = ( \frac{r_o \ T}{ r \ \ T_o} )^2[/tex]
[tex]\frac{K_o}{K} = ( \frac{ 155 \ \ \ 649.86}{ 119.35 \ 385.3})^2[/tex]
[tex]\frac{K_o}{K_f}[/tex] = 4.8
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