Problems

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, \(\ell>2a\)?

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Solution:

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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.

A thin beam is fixed at a height \(2a\) above a horizontal plane. A uniform straight rod \(ACB\) of length \(9a\) and mass \(m\) is supported by the beam at \(C\). Initially, the rod is held so that it is horizontal and perpendicular to the beam. The distance \(AC\) is \(3a\), and the coefficient of friction between the beam and the rod is \(\mu\). The rod is now released. Find the minimum value of \(\mu\) for which \(B\) strikes the horizontal plane before slipping takes place at \(C\).

In this question, all gravitational forces are to be neglected. A rigid frame is constructed from 12 equal uniform rods, each of length \(a\) and mass \(m,\) forming the edges of a cube. Three of the edges are \(OA,OB\) and \(OC,\) and the vertices opposite \(O,A,B\) and \(C\) are \(O',A',B'\) and \(C'\) respectively. Forces act along the lines as follows, in the directions indicated by the order of the letters: \begin{alignat*}{3} 2mg\mbox{ along }OA, & \qquad & mg\mbox{ along }AC', & \qquad & \sqrt{2}mg\mbox{ along }O'A,\\ \sqrt{2}mg\mbox{ along }OA', & & 2mg\mbox{ along }C'B, & & mg\mbox{ along }A'C. \end{alignat*}

1. The frame is freely pivoted at \(O\). Show that the direction of the line about which it will start to rotate is $\begin{pmatrix}1\\ 1\\ 2 \end{pmatrix}$ with respect to axes along \(OA\), \(OB\) and \(OC\) respectively.
2. Show that the moment of inertia of the rod \(OA\) about the axis \(OO'\) is \(2ma^2/9\) and about a parallel axis through its mid-point is \(ma^2/18\). Hence find the moment of inertia of \(B'C\) about \(OO'\) and show that the moment of inertia of the frame about \(OO'\) is \(14ma^2/3\). If the frame is freely pivoted about the line \(OO'\) and the forces continue to act along the specified lines, find the initial angular acceleration of the frame.