Answer:
N₁ = 393.96 N and N = 197.96 N
Explanation:
In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop
Lowest point, we write Newton's second law n for the y-axis
N -W = m a
where the acceleration is ccentripeta
a = v² / r
N = W + m v² / r
N = mg + mv² / r
we can use energy to find the speed at the bottom of the circle
starting point. Highest point where the ball is released
Em₀ = U = m g h
lowest point. Stop curl down
= K = ½ m v²
Emo = Em_{f}
m g h = ½ m v²
v² = 2 gh
we substitute
N = m (g + 2gh / r)
N = mg (1 + 2h / r)
let's calculate
N₁ = 5 9.8 (1 + 2 17.6 / 5)
N₁ = 393.96 N
headed up
we repeat the calculation in the longest part of the loop
-N -W = - m v₂² / r
N = m v₂² / r - W
N = m (v₂²/r - g)
we seek speed with the conservation of energy
Em₀ = U = m g h
final point. Top of circle with height 2r
= K + U = ½ m v₂² + mg (2r)
Em₀ = Em_{f}
mgh = ½ m v₂² + 2mgr
v₂² = 2 g (h-2r)
we substitute
N = m (2g (h-2r) / r - g)
N = mg (2 (h-r) / r 1) = mg (2h/r -2 -1)
N = mg (2h/r - 3)
N = 5 9.8 (2 17.6 / 5 -3)
N = 197.96 N
Directed down
