Answer:
v = √[ gL (3 + 2√3) / 2 ]
Explanation:
Draw a free body diagram of the ball. There are three forces acting on the ball:
Weight force mg pulling down
Tension force T₁ pulling left
Tension force T₂ pulling at angle θ
d = L/2, so as you found, θ = 30° and r = √3/2 L.
Sum the forces in the y direction:
∑F = ma
T₂ sin 30° − mg = 0
T₂ = 2mg
Sum the forces in the radial direction:
∑F = ma
T₂ cos 30° + T₁ = m v² / r
√3/2 T₂ + T₁ = m v² / (√3/2 L)
3/4 T₂ + √3/2 T₁ = m v² / L
3T₂ + 2√3 T₁ = 4m v² / L
v² = L (3T₂ + 2√3 T₁) / (4m)
v = √[ L (3T₂ + 2√3 T₁) / (4m) ]
Substitute 2mg for T₂:
v = √[ L (6mg + 2√3 T₁) / (4m) ]
If the tensions are equal, then T₁ = 2mg.
v = √[ L (6mg + 4√3 mg) / (4m) ]
v = √[ L (3g + 2√3 g) / 2 ]
v = √[ gL (3 + 2√3) / 2 ]
