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2017 Paper 3 Q10
D: 1700.0 B: 1484.0
moments of inertia rigid body hinge friction angular acceleration energy conservation normal reaction slipping condition rod with particle rotational dynamics

A uniform rod \(PQ\) of mass \(m\) and length \(3a\) is freely hinged at \(P\). The rod is held horizontally and a particle of mass \(m\) is placed on top of the rod at a distance~\(\ell\) from \(P\), where \(\ell <2a\). The coefficient of friction between the rod and the particle is \(\mu\). The rod is then released. Show that, while the particle does not slip along the rod, \[ (3a^2+\ell^2)\dot \theta^2 = g(3a+2\ell)\sin\theta \,, \] where \(\theta\) is the angle through which the rod has turned, and the dot denotes the time derivative. Hence, or otherwise, find an expression for \(\ddot \theta\) and show that the normal reaction of the rod on the particle is non-zero when~\(\theta\) is acute. Show further that, when the particle is on the point of slipping, \[ \tan\theta = \frac{\mu a (2a-\ell)}{2(\ell^2 + a\ell +a^2)} \,. \] What happens at the moment the rod is released if, instead, \(\ell>2a\)?

Solution:

TikZ diagram
By energy considerations, the initial energy is \(0\).
Inital\@ \(\theta\)
Rotational KE of rod\(0\)\(\frac{1}{2}I\dot{\theta}^2 = \frac{1}{2} \frac{1}{3} m (3a)^2 \dot{\theta}^2 = \frac32 m a^2 \dot{\theta}^2\)
KE of particle\(0\)\(\frac12 m \ell^2\dot{\theta}^2\)
GPE of rod\(0\)\(-\frac{3}{2}mga \sin \theta\)
GPE of particle\(0\)\(-mg \ell \sin \theta\)
Total\(0\)\(\frac12m \l \l 3a^2 + \ell^2\r \dot{\theta}^2 - \l 3a + 2\ell \r g \sin \theta \r\)
Therefore: \begin{align*} && \l 3a^2 + \ell^2\r \dot{\theta}^2 &= \l 3a + 2\ell \r g \sin \theta \\ \Rightarrow && \l 3a^2 + \ell^2\r 2\dot{\theta} \ddot{\theta} &= \l 3a + 2\ell \r g \cos\theta \dot{\theta} \tag{\(\frac{\d}{\d t}\)} \\ \Rightarrow && 2\l 3a^2 + \ell^2\r \ddot{\theta} &= \l 3a + 2\ell \r g \cos\theta \\ \Rightarrow && \ddot{\theta} &= \boxed{\frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta} \\ \end{align*} \begin{align*} \text{N}2(\perp PQ): && mg \cos \theta - R &= m \ell \ddot{\theta} \\ && R &= mg \cos \theta - m \ell \l \frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta \r \\ && &= mg\cos \theta \l 1 - \ell \frac{3a + 2\ell }{2(3a^2 + \ell^2)} \r \\ && &= mg \cos \theta \l \frac{6a^2 + 2\ell^2 - 3a\ell - 2\ell^2}{2(3a^2 + \ell^2)} \r \\ && &= mg \cos \theta \l \frac{3a(2a - \ell)}{2(3a^2 + \ell^2)} \r > 0 \tag{since \(2a > \ell\)} \end{align*} At limiting equilibrium, \(F = \mu R\). \begin{align*} \text{N}2(\parallel PQ): && \mu R - mg \sin \theta &= m \ell \dot{\theta}^2 \\ \Rightarrow && \mu mg \cos \theta \l \frac{3a(2a - \ell)}{2(3a^2 + \ell^2)} \r - mg \sin \theta &= m \ell \frac{(3a+2\ell)}{(3a^2+\ell^2)} g \sin \theta \\ \Rightarrow && \mu \l 3a(2a - \ell) \r - \l 2(3a^2 + \ell^2) \r \tan \theta &= 2\ell (3a+2\ell) \tan \theta \\ \Rightarrow && \mu \l 3a(2a - \ell) \r &= \l 6a\ell + 6a^2 + 6\ell^2 \r \tan \theta \\ \Rightarrow && \tan \theta &= \boxed{\frac{\mu a(2a-\ell)}{2(a^2 + a\ell + \ell^2)}} \end{align*} If \(\ell > 2a\), then the initial reaction force will be \(0\), ie the particle will have no contact with the rod. In other words, the rod will rotate faster than the particle will free-fall and the particle immediately loses contact with the rod.

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