Respuesta :
Answer:
a) I = ([tex]\frac{M}{3}[/tex] + [tex]\frac{4m}{9}[/tex]) L² , b) w = (\frac{27 M}{18 m} + 2)⁻¹ Lv₀
Explanation:
a) The moment of inertia is a scalar that represents the inertia in circular motion, therefore it is an additive quantity.
The moment of inertia of a rod held at one end is
I₁ = 1/3 M L²
The moment of inertia of the mass at y = L
I₂ = m y²
The total inertia method
I = I₁ + I₂
I = \frac{1}{3} M L² + m (\frac{2}{3} L)²
I = ([tex]\frac{M}{3}[/tex] +[tex]\frac{4m}{9}[/tex] ) L²
b) The conservation of angular momentum, where the system is formed by the masses and the bar, in such a way that all the forces during the collision are internal.
Initial instant. Before the crash
L₀ = I₂ w₀
angular and linear velocity are related
w₀ = y v₀
w₀ = [tex]\frac{2}{3}[/tex]L v₀
L₀ = I₂ y v₀
Final moment. After the crash
[tex]L_{f}[/tex] = I w
how angular momentum is conserved
L₀ = L_{f}
I₂ y v₀ = I w
substitute
m ([tex]\frac{2L}{3}[/tex])² (\frac{2L}{3} v₀ = ([tex]\frac{M}{3}[/tex] +[tex]\frac{4m}{9}[/tex] ) L² w
[tex]\frac{6}{27}[/tex] m L³ v₀ = ([tex]\frac{M}{3}[/tex] +[tex]\frac{4m}{9}[/tex] ) L² w
[tex]\frac{6}{27}[/tex] m L v₀ = ([tex]\frac{M}{3}[/tex] +[tex]\frac{4m}{9}[/tex] ) w
L v₀ = [tex](\frac{27 M}{18 m} + 2)[/tex] w
w = (\frac{27 M}{18 m} + 2)⁻¹ Lv₀
