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This was good practice for my upcoming physics exam too... so here we go.
You have two identical strings, which are 2.00m each.
First, find the angle the upper string (T1) makes with the upper half of the pole, which is:
cos θ = 1.50 / 2.00
θ = 41.41°
This also applies to the lower half, which will be referred to as (T2).
T1 and T2 combined result in the circular motion of the ball. The horizontal components are what causes the ball to move, so:
T1sinθ + T2sinθ = (mv^2)/r
The radius is part of the horizontal circular motion as well, so you must take the horizontal component. s = string
r = s(sinθ)
r = 2.00sin(41.41)
r = 1.32
Therefore:
T1sinθ + T2sinθ = (mv^2)/r
T1sin41.41 + T2sin41.41 = (3.50 * 4.95^2)/1.32
T1 + T2 = 65.0/(sin41.41)
T1 + T2 = 98.3 N
Now, analyze the vertical components. However, gravity also acts upon the object, so you must take that into consideration with the lower tension.
Components:
T1cosθ
T2cosθ
mg
T1cosθ = mg + T2cosθ
T1cos41.41 = (3.5 * 9.8) + T2cos41.41
T1 - T2 = 34.3 / (cos41.41)
T1 - T2 = 45.7 N
Thus:
T1 + T2 + (T1 - T2) = 98.3 + 45.7
2T1 = 144
T1 = 72 N
Plug the newly found T1 into T1 + T2 to get:
T1 + T2 = 98.3
72 + T2 = 98.3
T2 = 26.3 N
Answers:
T1 = 72.0 N
T2 = 26.3 N
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