Answer:
2.38 m/s, 4.31 m/s, lower
Explanation:
a)
Initial energy = final energy
½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²
Since the ball is rolling without slipping, ω = v / r.
For a hollow sphere, I = ⅔ m r².
½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²
½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²
⅚ m v₀² = mgh + ⅚ m v₁²
⅚ v₀² = gh + ⅚ v₁²
v₀² = 1.2gh + v₁²
v₁ = √(v₀² − 1.2gh)
Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:
v₁ = √((4.03)² − 1.2 (9.80) (0.900))
v₁ ≈ 2.38 m/s
At the top of the loop, the sum of the forces in the radial direction is:
∑F = ma
W + N = m v² / R
N = m v² / R - mg
N = m (v² / R - g)
Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:
N = m ((2.38)² / 0.450 - 9.80)
N = 2.77m
N ≥ 0, so the ball stays on the track.
b)
Initial energy = final energy
Borrowing from part a):
v₂ = √(v₀² − 1.2gh)
This time, h = -0.200 m:
v₂ = √((4.03)² − 1.2 (9.80) (-0.200))
v₂ ≈ 4.31 m/s
c)
Without the rotational energy:
½ m v₀² = mgh + ½ m v₁²
½ v₀² = gh + ½ v₁²
v₀² = 2gh + v₁²
v₁ = √(v₀² - 2gh)
This is less than v₁ we calculated earlier.
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