Problems

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Solution:

\begin{align*} \text{N2}(\nwarrow): && T - mg \sin \theta &= m \left ( \frac{v^2}{r}\right) \\ &&&= \frac{m v^2}{a} \\ \text{COE}:&& \underbrace{0}_{\text{assume initial GPE level is }0} &= \frac12 m v^2 - mga\sin \theta \\ \Rightarrow && v^2 &= 2ag \sin \theta \\ \Rightarrow && T &= \frac{m}{a} \cdot 2 ag \sin \theta + mg \sin \theta \\ &&&= 3mg \sin\theta \end{align*} Considering the force on the edge of the table will be: \begin{align*} && \mathbf{R} &= \binom{-T}{0} + \binom{T \cos \theta}{-T \sin \theta} \\ &&&= \binom{T(1-\cos \theta)}{-T \sin \theta} \\ &&&= 3mg \sin \theta \binom{1-\cos \theta}{-\sin \theta} \\ \Rightarrow && |\mathbf{R}| &= 3mg \sin \theta \sqrt{(1-\cos \theta)^2 + \sin ^2 \theta} \\ &&&= 3mg \sin \theta \sqrt{2 - 2 \cos \theta} \\ &&&= 3mg \sin \theta\sqrt{4 \sin^2 \tfrac{\theta} {2}} \\ &&&= 6mg \sin \theta |\sin \tfrac{\theta} {2} | \\ s = \sin \tfrac \theta2:&&&= 12mg s^2 \sqrt{1-s^2} \end{align*} We can maximise \(V = x\sqrt{1-x}\) by differentiating: \begin{align*} && \frac{\d V}{\d x} &= \sqrt{1-x} - \frac{x}{2\sqrt{1-x}} \\ &&&= \sqrt{1-x} \left ( 1 - \frac{x}{2-2x}\right) \\ &&&= \sqrt{1-x} \frac{2-3x}{2-2x} \\ \Rightarrow && x &= \frac23 \end{align*} Therefore the maximum for will be: \begin{align*} |\mathbf{R}| &= 12 mg \frac 23 \sqrt{\frac13} \\ &= 8mg/\sqrt{3} \end{align*} as required.