A 40 kg sphere suspended by a thread of length l is oscillating in a vertical plane, the angular amplitudebeing${\theta }_0$ . What is the tenson in the thread when it makes an angle$\theta$ with the vertical during oscillations?If the thread can support a maximum tension of 80 kg-f, then what can be the maximum angularamplitude of oscillation of the sphere without breaking the rope

$\frac{1}{2}{\text{mu}}^2=\text{mg}l\left(1-\cos \theta \right)$

${\text{mu}}^{\text{2}}=2\text{mg}l\left(1-\cos {\theta }_{\theta }\right)$

$\text{for}\theta <{\theta }_0$                                           

Tension at $\theta \text{is}{\text{T}}_{\theta }=\text{mg}\cos \theta +\frac{{\text{mv}}_0^2}{l}$

${\text{v}}_{\theta }^2={\text{u}}^{\text{2}}-2\text{g}l\left(1-\cos \theta \right)$

${\text{mv}}_{\theta }^2={\text{mu}}^{\text{2}}-2\text{mg}l\left(1-\cos \theta \right)$

${\text{mv}}_{\theta }^2=2\text{mg}l\left(1-\cos {\theta }_0\right)-2\text{mg}l\left(1-\cos \theta \right)$

${\text{mv}}_{\theta }^2=2\text{mg}l-2\text{mg}l\cos {\theta }_0-2\text{mg}l+2\text{mg}l\cos \theta$

${\text{mv}}_{\theta }^2=2\text{mg}l\left(\cos \theta -{\theta }_0\right)$

$\therefore {\text{T}}_{\theta }=\text{mg}\cos \theta +\frac{{\text{mv}}_{\theta }^2}{l}$

$=\text{mg}\cos \theta +\frac{2\text{mg}l\left(\cos \theta -\cos {\theta }_0\right)}{l}$

${\text{T}}_{\theta }=3\text{mg}\cos \theta -2\text{mg}\cos {\theta }_0$

$\text{T}\cos \theta =\text{mg}$

$\text{80cos}\theta =\text{40}$

$\text{cos}\theta \text{=}\frac{1}{2}$

$\theta =60$