Respuesta :
Answer:
a) h=3.16 m, b) Â v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
      Em₀ = U = mg h
final point. Lowest point
      [tex]Em_{f}[/tex] = K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
      K = ½ m [tex]v_{cm }^{2}[/tex] + ½ [tex]I_{cm}[/tex] w²
angular and linear speed are related
      v = w r
      w = v / r
      K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
      Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
       Em₀ = Em_{f}
       mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)     (1)
       h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
       I_{cm} = ⅔ m r²
we substitute
      h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
      h = ½ v_{cm }^{2}/g   5/3
       h = 5/6 v_{cm }^{2} / g
     Â
let's calculate
      h = 5/6 6.1 2 / 9.8
      h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
     v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
       I_{cm} = ½ m r²
we substitute
       v_{cm } = √ [2gh / (1 + ½)]
       v_{cm } = √(4/3 gh)
let's calculate
       v_{cm } = √ (4/3 9.8 3.16)
       v_{cm }^ = 6.43 m / s
