A hoop is rotating at 12 rad/s on a horizontal plane about an a vertical axis that passes through its edge when a second body with mass one quarter of the disc's is placed safely on it just opposite of the
pivot. What is the percentage change of the kinetic energy of the hoop?
A. 20%
B. 80%
C. 36%
D. 64%​

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

        =

        \frac{K_f}{K_o} = (6.67 / 12)²

        \frac{K_f}{K_o} = 0.309

        = 31%

the correct answer is C