66If we could find the speed at the angle θ, then from the free body diagram it is easy to see that
T-mg cosθ=mv2/l
from where the tension T in the thread can be calculated easily.
From energy conservation, we get
mgl(cosθ - cosθ0)=mv2/2
which gives mv2 as 2mgl(cosθ - cosθ0). Hence,
T=mgcosθ + 2mg(cosθ - cosθ0)=mg(3cosθ - 2cosθ0)
From this expression for T, we see that the maximum value of T is obtained for θ = 0. And
Tmax=mg(3 - 2cosθ0)
In order that the thread does not break
Tmax ≤ Tbreaking = T0 (say)
from where we get
mg(3 - 2cosθ0) ≤ T0
i.e.
cosθ0 ≥ 32 - T02mg
Since, cos is a decreasing function of its argument, we finally get
θ0 ≤ cos-1 (32 - T02mg) = θ0,max
Plugging in the values, we get θ0,max = 60°
