Problems

A non-uniform rod \(AB\) has weight \(W\) and length \(3l\). When the rod is suspended horizontally in equilibrium by vertical strings attached to the ends \(A\) and \(B\), the tension in the string attached to \(A\) is \(T\). When instead the rod is held in equilibrium in a horizontal position by means of a smooth pivot at a distance \(l\) from \(A\) and a vertical string attached to \(B\), the tension in the string is \(T\). Show that \(5T = 2W\). When instead the end \(B\) of the rod rests on rough horizontal ground and the rod is held in equilibrium at an angle \(\theta\) to the horizontal by means of a string that is perpendicular to the rod and attached to \(A\), the tension in the string is \(\frac12 T\). Calculate \(\theta\) and find the smallest value of the coefficient of friction between the rod and the ground that will prevent slipping.

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Solution:

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Suppose the centre of mass of the rod is \(x\) away from \(A\). \begin{align*} \overset{\curvearrowleft}{B}: && (3l-x)W - 3lT &= 0 \\ \Rightarrow && x &= \frac{3l(W-T)}{W} \tag{1} \end{align*}

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In the second set up we have: \begin{align*} \overset{\curvearrowleft}{\text{pivot}}: && 2lT - (x-l)W &= 0 \\ \Rightarrow && x &= \frac{2lT + lW}{W} \tag{2} \\ \\ (1) \text{ & } (2): && 3l(W-T) &= l(2T+W) \\ \Rightarrow && 2W &= 5T \end{align*}

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\begin{align*} && x&= \frac{3l(W-T)}{W}\\ &&&= \frac{3l(W - \frac25 W)}{W} \\ &&&= \frac{9}{5}l\\ \overset{\curvearrowleft}{B}: && -\frac12 T (3l \sin \theta) + W \frac{6}{5}l \cos \theta &= 0 \\ \Rightarrow && \tan \theta &= \frac{4}{5} \frac{W}{T} \\ &&&= \frac45 \frac52 \\ &&&= 2 \\ \Rightarrow && \theta &= \tan^{-1} 2 \\ \\ \text{N2}(\uparrow): && R &= W \\ \text{N2}(\rightarrow): && F &= \frac12 T \\ \Rightarrow && F & \leq \mu R \\ \Rightarrow && \frac12 T &\leq \mu W \\ \Rightarrow && \mu &\geq \frac12 \frac{T}{W} = \frac12 \frac25 = \frac15 \end{align*}

A non-uniform rod \(AB\) of mass \(m\) is pivoted at one end \(A\) so that it can swing freely in a vertical plane. Its centre of mass is a distance \(d\) from \(A\) and its moment of inertia about any axis perpendicular to the rod through \(A\) is \(mk^{2}.\) A small ring of mass \(\alpha m\) is free to slide along the rod and the coefficient of friction between the ring and rod is \(\mu.\) The rod is initially held in a horizontal position with the ring a distance \(x\) from \(A\). If \(k^{2} > xd\), show that when the rod is released, the ring will start to slide when the rod makes an angle \(\theta\) with the downward vertical, where \[ \mu\tan\theta=\frac{3\alpha x^{2}+k^{2}+2xd}{k^{2}-xd}. \] Explain what will happen if (i) \(k^{2}=xd\) and (ii) \(k^{2} < xd\).

A thin non-uniform rod \(PQ\) of length \(2a\) has its centre of gravity a distance \(a+d\) from \(P\). It hangs (not vertically) in equilibrium suspended from a small smooth peg \(O\) by means of a light inextensible string of length \(2b\) which passes over the peg and is attached at its ends to \(P\) and \(Q\). Express \(OP\) and \(OQ\) in terms of \(a,b\) and \(d\). By considering the angle \(POQ\), or otherwise, show that \(d < a^{2}/b\).

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Solution:

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Resolving horizontally, it's clear that \(\angle POG = \angle GOQ\), in particular applying the sine rule: \begin{align*} && \sin \angle POG &= \frac{a+d}{2b-x} \sin \angle PGO \\ && \sin \angle GOP &= \frac{a-d}{x} \sin \angle OGQ \\ \Rightarrow && \frac{a+d}{2b-x} &= \frac{a-d}{x} \\ \Rightarrow && x(a+d) &= (2b-x)(a-d) \\ \Rightarrow && 2ax &= 2b(a-d) \\ \Rightarrow && x &= b - \frac{db}{a} \\ \Rightarrow && PO &= b+\frac{db}{a} \\ && OQ &= b - \frac{d}{a} \end{align*} Applying the cosine rule: \begin{align*} && \cos POQ &= \frac{(b + \frac{db}{a})^2 + (b - \frac{db}{a})^2 -4a^2}{2(b^2 - \frac{d^2b^2}{a^2})} \\ &&&= \frac{2b^2 + \frac{2d^2b^2}{a^2}-4a^2}{2(b^2 - \frac{d^2b^2}{a^2})} \\ &&&= \frac{2a^2b^2 + 2d^2b^2-4a^4}{2b^2(a^2 - d^2)} \\ &&&< 1 \\ \Leftrightarrow && 2a^2b^2 + 2d^2b^2-4a^4 &< 2b^2(a^2-d^2) \\ \Leftrightarrow && 2d^2b^2-4a^4 &< -2b^2d^2 \\ \Leftrightarrow && 4d^2b^2&< 4a^4 \\ \Leftrightarrow && d^2&< \frac{a^4}{b^2} \\ \Leftrightarrow && d&< \frac{a^2}{b} \\ \end{align*}

A body of mass \(m\) and centre of mass \(O\) is said to be dynamically equivalent to a system of particles of total mass \(m\) and centre of mass \(O\) if the moment of inertia of the system of particles is the same as the moment of inertia of the body, about any axis through \(O\). Show that this implies that the moment of inertia of the system of particles is the same as that of the body about any axis. Show that a uniform rod of length \(2a\) and mass \(m\) is dynamically equivalent to a suitable system of three particles, one at each end of the rod, and one at the midpoint. Use this result to deduce that a uniform rectangular lamina of mass \(M\) is dynamically equivalent to a system consisting of particles each of mass \(\frac{1}{36}M\) at the corners, particles each of mass \(\frac{1}{9}M\) at the midpoint of each side, and a particle of mass \(\frac{4}{9}M\) at the centre. Hence find the moment of inertia of a square lamina, of side \(2a\) and mass \(M,\) about one of its diagonals. The mass per unit length of a thin rod of mass \(m\) is proportional to the distance from one end of the rod, and a dynamically equivalent system consists of one particle at each end of the rod and one at the midpoint. Write down a set of equations which determines these masses, and show that, in fact, only two particles are required.