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 rough ring of radius \(a\) is fixed so that it lies in a plane inclined at an angle \(\alpha\) to the horizontal. A uniform heavy rod of length \(b(>a)\) has one end smoothly pivoted at the centre of the ring, so that the rod is free to move in any direction. It rests on the circumference of the ring, making an angle \(\theta\) with the radius to the highest point on the circumference. Find the relation between \(\alpha,\theta\) and the coefficient of friction, \(\mu,\) which must hold when the rod is in limiting equilibrium.

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Solution:

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It is important to define clear coordinate axes, so let the \(x\)-axis point up the line of greatest slope of the ring. The \(z\)-axis perpendicular to the ring, and the \(y\)-axis perpendicular to both of these. Our method is going to be to take moments about \(O\) to avoid worrying about the force at the pivot. There are \(3\) forces we need to worry about:

• The mass of the rod
• The reaction where it meets the ring
• The friction at the ring

In our coordinate frame, the reaction will act in the \(z\)-direction, \(\displaystyle \begin{pmatrix} 0 \\ 0 \\ R \end{pmatrix}\), the friction force will act in the \(x-y\) plane: \(\displaystyle \begin{pmatrix} \mu R \sin \theta \\ -\mu R \cos \theta \\ 0 \end{pmatrix}\). We don't know the mass, but we know it will be acting "vertically", so \(\cos \alpha\) of it will act in the \(z\)-axis and \(\sin \alpha\) will act in the \(y\)-axis, ie it will act parallel to \(\displaystyle \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}\). When taking moments, we need to consider \(\mathbf{r}\) the direction of the rod. This will be \(\displaystyle \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0 \end{pmatrix}\). The moment of the weight will all be parallel to \(\mathbf{r} \times \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}\). Similarly the moments of the contact forces will be \(\mathbf{r} \times \begin{pmatrix} \mu R \sin \theta \\ -\mu R \cos \theta \\ R \end{pmatrix}\). Since these moments sum to \(\mathbf{0}\) as we are in equilibrium, these vectors must be parallel. Therefore it is sufficient to check the vector triple product, \begin{align*} && 0 &= \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0 \end{pmatrix} \cdot \left ( \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix} \times \begin{pmatrix} \mu \sin \theta \\ -\mu \cos \theta \\ 1 \end{pmatrix} \right ) \\ &&&= \cos \theta (\mu \cos \theta \cos \alpha)-\sin \theta (\sin \alpha - \mu \sin \theta \cos \alpha) \\ &&&= \mu((\sin^2 \theta+\cos^2 \theta) \cos \alpha) -\sin \theta \sin \alpha \\ \Rightarrow && \mu &= \tan \alpha \sin \theta \end{align*}