Problems

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Solution:

While the particle is not sliding, we can consider the whole system. Considering the moment of inertia about the end, we have: \begin{align*} I &= \frac13 \cdot 3m \cdot (2a)^2 + m a^2 \\ &= 5ma^2 \end{align*} Taking the level of the pivot as the \(0\) GPE level, the initial energy is \(0\). The energy once it has rotated through an angle \(\theta\) is: \begin{align*} && 0 &= \text{rotational ke} + \text{gpe} \\ &&&= \frac12 I \dot{\theta}^2 - 4mg \sin \theta \\ &&&= \frac12 5am \dot{\theta}^2 -4mg \sin \theta \\ \Rightarrow && 5a\dot{\theta}^2 &= 8g \sin \theta \end{align*} as required. We also have \(5a \ddot{\theta} = 4g \cos \theta\) The acceleration towards the pivot required to maintain circular motion is \(m \frac{v^2}{r} = m a \dot{\theta}^2\). When we are on the point of sliding: \begin{align*} \text{N2}(\nearrow): && R - mg\cos \theta &= -ma \ddot{\theta} \\ \Rightarrow && R &= mg \cos \theta - ma \frac{4mg \cos \theta}{5a} \\ &&&= \frac15mg \cos \theta \end{align*} Therefore we must have: \begin{align*} \text{N2}(\nwarrow):&&\mu R - mg \sin \theta &= ma \dot{\theta}^2 \\ && \frac12 \cdot \frac 15 mg \cos \theta &= m \frac{13}5 g \sin \theta \\ \Rightarrow && \tan \theta &= \frac{1}{26} \end{align*}