Problems

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Solution:

\begin{align*} %\text{COE}: && \frac12 m u^2 - mga &= \frac12mv^2 + mga\cos \theta \\ \text{N2}(\swarrow): && R+mg\cos\theta &= \frac{m v^2}{a} \\ R = 0: && v^2 &= ag\cos \theta \\ \end{align*} So the particle will become a projectile moving tangent to the circle with \(v^2 = ag \cos \theta\). Therefore the velocity will be \(\displaystyle \sqrt{ag \cos \theta}\binom{-\cos \theta}{\sin \theta}\). We have: \begin{align*} && \mathbf{s} &= a\binom{\sin \theta}{\cos \theta}+\sqrt{ag \cos \theta}\binom{-\cos \theta}{\sin \theta} t + \frac12 \binom{0}{-g} t^2 \\ \Rightarrow && a^2 &= \mathbf{s} \cdot \mathbf{s} \\ &&&= a^2 + ag\cos \theta t^2 + \frac1{4} g^2t^4 -ag \cos \theta t^2 - \sqrt{ag \cos \theta} \sin \theta g t^3 \\ \Rightarrow && 0 &= \frac14 g t - \sqrt{ag \cos \theta} \sin \theta \\ \Rightarrow && t &= \frac{4\sqrt{a g \cos \theta} \sin \theta}{g} \end{align*} At this time, the vertical position will be: \begin{align*} && s_y &= a \cos \theta + \sqrt{ag \cos \theta} \sin \theta \frac{4\sqrt{a g \cos \theta} \sin \theta}{g} - \frac12 g \left ( \frac{4\sqrt{a g \cos \theta} \sin \theta}{g} \right)^2 \\ &&&= a \cos \theta + 4a\cos \theta \sin^2 \theta - 8a\cos \theta \sin^2 \theta \\ &&&= a \cos \theta - 4 a \cos \theta \sin^2 \theta \\ &&&= a \cos \theta (1-4 \sin^2 \theta) \\ \underbrace{\Rightarrow}_{s_y < 0} && 0 &> 1 - 4 \sin^2 \theta \\ \Rightarrow && \sin\theta &> \frac12 \\ \Rightarrow && \theta & > \frac{\pi}{6} \end{align*}