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2015 Paper 2 Q9
D: 1600.0 B: 1484.0
circular motion rigid body energy conservation equilibrium equilateral triangle light rods rotation about fixed axis minimum speed for complete rotation

An equilateral triangle \(ABC\) is made of three light rods each of length \(a\). It is free to rotate in a vertical plane about a horizontal axis through \(A\). Particles of mass \(3m\) and \(5m\) are attached to \(B\) and \(C\) respectively. Initially, the system hangs in equilibrium with \(BC\) below \(A\).

  1. Show that, initially, the angle \(\theta\) that \(BC\) makes with the horizontal is given by \(\sin\theta = \frac17\).
  2. The triangle receives an impulse that imparts a speed \(v\) to the particle \(B\). Find the minimum speed \(v_0\) such that the system will perform complete rotations if \(v>v_0\).

Solution:

TikZ diagram
  1. The sine rule tells us: \begin{align*} && \frac{\frac58 a}{\sin(30^\circ + \theta)} &= \frac{a}{\sin(90^{\circ}-\theta)} \\ \Rightarrow &&\frac58 \cos \theta &= \frac12 \cos \theta+ \frac{\sqrt{3}}2 \sin \theta \\ \Rightarrow && \frac{1}{4\sqrt{3}} &= \tan \theta \\ \Rightarrow && \sin \theta &= \sqrt{\frac{1}{48+1}} = \frac17 \end{align*}
  2. \(\,\) \begin{align*} && \text{initial energy} &= \frac12(5m)v^2 + \frac12 (3m)v^2 - 3m \cdot g \cdot a \cos(30^{\circ}+\theta) -5m \cdot g \cdot a\cos(30^\circ - \theta) \\ &&&= 4m v^2 - amg(4\sqrt{3} \cos \theta + \sin \theta) \\ &&&= 4mv^2 - 7amg \\ && \text{energy at top} &= \frac12 m v_{top}^2 + 7amg \end{align*} We need this equation to be positive for all values of \(v_{top} \geq 0\), so \(4mv^2 \geq 14amg \Rightarrow v_0 = \sqrt{\frac{7ag}2}\)

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