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A particle pendulum (thread length $l$) is at rest and is given an initial velocity such as, when the thread "gets loose" / "is not extended fully", the angle is $\frac{2\pi}{3}$ from the vertical neutral angle $0$.

I'm asked to determine the relationship between the kinetic energy of the $\frac{2\pi}{3}$ position and of the initial position.

I've introduced natural coordinates,

$$v = \dot{s}e_t$$ $$a = \ddot{s}e_t + \frac{\dot{s}^2}{l}e_n$$

I am assuming that the thread being lose means the force in $e_n$ direction is $0$ and I set $$mg\cos{\frac{2\pi}{3}} = m\frac{\dot{s}^2}{l} \iff \dot{s} = \sqrt{lg\cos{\frac{2\pi}{3}}} = \dot{s} = \sqrt{lg(-\frac{1}{2})}$$ which is nonsense (?). If I instead set $\cos{\frac{2\pi}{3}} = \frac{1}{2}$ I get the answer $\frac{2}{3}$ but it is supposed to be $\frac{1}{7}.$

Is my assumption wrong? If it is correct, how would I proceed?

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Your assumption is indeed wrong. Hint: the pendulum is undergoing circular motion up until the point when the thread is not fully extended.

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