Answer:
Ring  v² = gh
solid wheel (cylinder)  v² = 4/3 gh
Explanation:
Let's use conservation of energy to find the speed of the wheels at the bottom of the hill.
starting point. Point before starting movement
      Em₀ = mgh
final point. At the bottom of the hill
      Em_f = K = ½ m v² + ½ I w²
energy is conserved
      Emo = Em_f
      mgh = ½ m v² + ½ I w²
angular and linear velocity are related
      v = w r
we substitute
     mgh = ½ m v² + ½ I v² / r²
     mgh = [tex]\frac{1}{2} \ ( m+ \frac{I}{r^2} ) \ v^2[/tex]½ (m + I / r²) v²
     v² = [tex]\frac{2mgh}{m+ \frac{I}{r^2} }[/tex]
the moments of inertia are tabulated
Ring
   I = mr²
     v² = 2 m g h / (m + m)
     v² = gh
solid wheel (cylinder)
   I = ½ m r²
    v² = 2m gh / (m + m / 2)
    v² = 4/3 gh
We can see that due to the difference in the moment of inertia of each body it is different, the solid wheel has more speed when it reaches the lower part of the ramp
