Respuesta :
Answer:
the correct answer is C
Explanation:
For this exercise we can use the conservation of angular momentum, let's define a system formed by the disk and the other body, so that the torques during the collision have been internal and the moment is conserved.
Initial instant. Before placing the second body
L₀ = I₁ w₁
Final moment. Right after placing the body
L_f = (I₁ + I₂) w
the moment of inertia of the ring with respect to an axis passing through its center is tabulated
I_{cm} = m r²
we use the parallel axes theorem to find the moment of inertia about an axis passing through one end
I₁ = I_{cm} + m d²
where d is the distance from the center of mass to the new axis
d = r / 2
we substitute
I₁ = mr² + m (r/2)²
I₁ = 5/4 m r²
The second body does not specify a specific shape, so we can assume it to be punctual
I₂ = m’ (2r)²
I₂ = 4 m’ r²
in the exercise indicate that
m ’= m / 4
I₂ = m r²
how the moment of inertia is conserved
L₀ = L_f
I₁ w₁ = (I₁ + I₂) w
5/4 m r² w₁ = (5/4 m r² + m r²) w
5/4 m r² w₁ =9/4 m r² w
w = 5/9 w₁
we calculate
w = 5/9 12
w = 6.67 rad / s
having the initial and final angular velocities we can find the kinetic energy of the hoop
K₀ = ½ I₁ w₁²
K_f = ½ (I₁) w2
the energy ratio is
[tex]\frac{K_f}{K_o}[/tex] = [tex]\frac{w^2}{w_1^2}[/tex]
\frac{K_f}{K_o} = (6.67 / 12)²
\frac{K_f}{K_o} = 0.309
[tex]\frac{K_f}{K_o} %[/tex] = 31%
the correct answer is C
