Question

(Edit: oops, this should be in college physics)

Answer 1

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 ]