Respuesta :
Answer:
a) SPHERE must be the winner
b) BOX is most fastest
Explanation:
For this exercise we must use conservation of energy
starting point. Highest part of the plane
Em₀ = U = m g h
final point. Lowest part of the plane
Em_f = K = ½ m v² + ½ I w²
Note that as the objects roll, the kinetic energy of rotation is included. Energy is conserved
Em₀ = Em_f
mg h = ½ m v² + ½ I w²
the linear and rotational variables are related
v = w r
w = v / r
the moment of inertia of the bodies is tabulated
ring (empty can) I = m r²
cylinder I = ½ m r²
sphere (marbles) I = 2/5 m r²
mgh = ½ m v² + ½ I v²/r² = ½ mv² (+ I / mr²)
2gh / (1 + I / mr²) = v² (1)
Let's analyze the value of I / mr2
can I / mr² = mr² / mr² = 1
cylinder I / mr² = ½ mr² / mr² = ½
sphere I / mr² = 2/5 mr² / mr² = 2/5
we substitute in equation 1
v = [tex]\sqrt{ \frac{2gh}{ (1+ \frac{I}{m r^2}) } }[/tex]
can v = √(2gh / 2) = √ gh
v = √gh
cylinder v= [tex]\sqrt{ \frac{2gh}{ \frac{3}{2} } }[/tex] = √(4gh/3)
v = 1,155 √gh
sphere v = [tex]\sqrt{ \frac{2gh}{ \frac{7}{5} } }[/tex] = √(10gh/7)
v = 1.20 √gh
therefore the object with the highest speed is the one that takes less time, consequently the SPHERE must be the cattle.
b) for a box on a frictionless surface, there is no rotational kinetic energy
mgh = ½ m v²
v = √2gh
v = 1.41 √(gh)
When comparing with the latter, this would be the one that arrives first
