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Answer :
Answer:
w = 1.14 rad / s
Explanation:
This is an angular momentum exercise. Let's define a system formed by the three bodies, the platform, the bananas and the monkey, in such a way that the torques during the collision have been internal and the angular momentum is preserved.
Initial instant. The platform alone
L₀ = I w₀
Final moment. When the bananas are on the shelf
we approximate the bananas as a point load and the distance is indicated
x = 0.45m
L_f = (m x² + I ) w₁
angular momentum is conserved
L₀ = L_f
I w₀ = (m x² + I) w₁
w₁ = [tex]\frac{I}{m x^2 + I} \ w_o[/tex]
Let's repeat for the platform with the bananas and the monkey is the one that falls for x₂ = 1.73 m
initial instant. The platform and bananas alone
L₀ = I₁ w₁
I₁ = (m x² + I)
final instant. After the crash
L_f = I w
L_f = (I₁ + M x₂²) w
the moment is preserved
L₀ = L_f
(m x² + I) w₁ = ((m x² + I) + M x₂²) w
(m x² + I) w₁ = (I + m x² + M x₂²) w
we substitute
w = [tex]\frac{m x^2 +I}{I + m x^2 + M x_2^2} \ \frac{I}{m x^2 + I} \ w_o[/tex]
w = [tex]\frac{I}{I + m x^2 + M x_2^2} \ w_o[/tex]
the moment of inertia of a circular disk is
I = ½ m_p x₂²
we substitute
w = [tex]\frac{ \frac{1}{2} m_p x_2^2 }{ \frac{1}{2} m_p x_2^2 + M x_2^2 + m x^2} \ \ w_o[/tex]
let's calculate
w =[tex]\frac{ \frac{1}{2} \ 97.3 \ 1.73^2 }{ \frac{1}{2} \ 97.3 \ 1.73^2 + 21.9 \ 1.73^2 + 9.67 \ 0.45^2 } \ \ 1.67[/tex]
w = [tex]\frac{145.60 }{145.60 \ + 65.54 \ + 1.958} \ \ 1.67[/tex]
w = 1.14 rad / s
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