Problems

1. A horizontal disc of radius \(r\) rotates about a vertical axis through its centre with angular speed \(\omega\). One end of a light rod is fixed by a smooth hinge to the edge of the disc so that it can rotate freely in a vertical plane through the centre of the disc. A particle \(P\) of mass \(m\) is attached to the rod at a distance \(d\) from the hinge. The rod makes a constant angle \(\alpha\) with the upward vertical, as shown in the diagram, and \(d\sin\alpha < r\).
   

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   By considering moments about the hinge for the (light) rod, show that the force exerted on the rod by \(P\) is parallel to the rod. Show also that \[ r\cot\alpha = a + d \cos\alpha \,, \] where \(a = \dfrac {g \;} {\omega^2}\,\). State clearly the direction of the force exerted by the hinge on the rod, and find an expression for its magnitude in terms of \(m\), \(g\) and \(\alpha\).
2. The disc and rod rotate as in part (i), but two particles (instead of \(P\)) are attached to the rod. The masses of the particles are \(m_1\) and \(m_2\) and they are attached to the rod at distances \(d_1\) and \(d_2\) from the hinge, respectively. The rod makes a constant angle \(\beta\) with the upward vertical and \(d_1\sin\beta < d_2\sin\beta < r\). Show that \(\beta\) satisfies an equation of the form \[ r\cot\beta = a+ b \cos\beta \,, \] where \(b\) should be expressed in terms of \(d_1\), \(d_2\), \(m_1\) and \(m_2\).

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1. In the case \(W> T\sin(\alpha+\beta)\), show that the block will remain at rest provided \[ W\sin\lambda \ge T\cos(\alpha+\beta- \lambda)\,, \] where \(\lambda\) is the acute angle such that \(\tan\lambda = \mu\).
2. In the case \(W=T\tan\phi\), where \(2\phi =\alpha+\beta\), show that the block will start to move in a direction that makes an angle \(\phi\) with the horizontal.

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Solution:

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1. Assuming the block is at rest we must have: \begin{align*} \text{N2}(\uparrow): && 0 &= T \sin \beta\cos \alpha + T \cos \beta \sin \alpha +R -W \\ \Rightarrow && W &> T \sin \beta\cos \alpha + T \cos \beta \sin \alpha \\ &&&= T\sin(\alpha+\beta) \\ \Rightarrow && R &= W-T\sin(\alpha+\beta)\\ \\ \text{N2}(\rightarrow): && 0 &= T \cos \beta \cos \alpha - T \sin \beta \sin \alpha - F \\ \Rightarrow && T \cos(\alpha+\beta) &= F \\ &&&\leq \mu (W-T\sin(\alpha+\beta)) \\ \Rightarrow && W \sin \lambda &\geq T \cos (\alpha+\beta)\cos \lambda +T \sin (\alpha+\beta) \sin \lambda \\ &&&= T\cos(\alpha+\beta-\lambda) \end{align*}
2. If \(W = T\tan \phi\) where \(2\phi = \alpha + \beta\) then \begin{align*} \text{N2}(\uparrow): && ma_y &= T\sin(\alpha+\beta) - W \\ &&&= T \sin(\alpha+\beta) - T \tan \left ( \frac{\alpha+\beta}{2} \right ) \\ &&&= T \tan \left ( \frac{\alpha+\beta}{2} \right ) \left ( 2 \cos^2 \left ( \frac{\alpha+\beta}{2} \right ) -1\right) \\ &&&= T \tan \phi \cos \left ( \alpha+\beta\right ) \tag{notice this is positive so \(R=F=0\)} \\ \text{N2}(\rightarrow): && ma_x &= T \cos(\alpha+\beta) \\ \Rightarrow && \frac{a_y}{a_x} &= \tan \phi \end{align*} Therefore we are accelerating at an angle \(\phi\) to the horizontal

Two particles, \(A\) and \(B\), of masses \(m\) and \(2m\), respectively, are placed on a line of greatest slope, \(\ell\), of a rough inclined plane which makes an angle of \(30^{\circ}\) with the horizontal. The coefficient of friction between \(A\) and the plane is \(\frac16\sqrt{3}\) and the coefficient of friction between \(B\) and the plane is \(\frac13 \sqrt{3}\). The particles are at rest with \(B\) higher up \(\ell\) than \(A\) and are connected by a light inextensible string which is taut. A force \(P\) is applied to \(B\).

1. Show that the least magnitude of \(P\) for which the two particles move upwards along \(\ell\) is \(\frac{11}8 \sqrt{3}\, mg\) and give, in this case, the direction in which \(P\) acts.
2. Find the least magnitude of \(P\) for which the particles do not slip downwards along~\(\ell\).

A light smoothly jointed planar framework in the form of a regular hexagon \(ABCDEF\) is suspended smoothly from \(A\) and a weight 1kg is suspended from \(C\). The framework is kept rigid by three light rods \(BD\), \(BE\) and \(BF\). What is the direction and magnitude of the supporting force which must be exerted on the framework at \(A\)? Indicate on a labelled diagram which rods are in thrust (compression) and which are in tension. Find the magnitude of the force in \(BE\).

The point \(A\) is vertically above the point \(B\). A light inextensible string, with a smooth ring \(P\) of mass \(m\) threaded onto it, has its ends attached at \(A\) and \(B\). The plane \(APB\) rotates about \(AB\) with constant angular velocity \(\omega\) so that \(P\) describes a horizontal circle of radius \(r\) and the string is taut. The angle \(BAP\) has value \(\theta\) and the angle \(ABP\) has value \(\phi\). Show that \[\tan\frac{\phi-\theta}{2}=\frac{g}{r\omega^{2}}.\] Find the tension in the string in terms of \(m\), \(g\), \(r\), \(\omega\) and \(\sin\frac{1}{2}(\theta+\phi)\). Deduce from your results that if \(r\omega^2\) is small compared with \(g\), then the tension is approximately \(\frac{mg}{2}\)

Show Solution

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Solution: None \begin{multicols}{2}

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\columnbreak \begin{align*} N2(\uparrow): && T \cos \theta - T \cos \phi - mg &= 0 \\ N2(\rightarrow): && T \sin \theta + T \sin \phi &= m r \omega^2 \\ \\ && T \cos \theta - T \cos \phi &= mg \tag{\(*\)}\\ && T \sin \theta + T \sin \phi &= m r \omega^2 \tag{{\(**\)}} \end{align*} \end{multicols} Dividing \((*)\) by \((**)\) we obtain: \begin{align*} \frac{g}{r\omega^2} &= \frac{\cos \theta - \cos \phi}{\sin \theta + \sin \phi} \\ &= \frac{2 \sin \left ( \frac{\theta + \phi}2 \right )\sin \left (\frac{\phi - \theta}2 \right )}{2 \sin \left ( \frac{\theta + \phi}2 \right )\cos \left (\frac{\phi - \theta}2 \right )} \\ &= \tan \left ( \frac{\phi - \theta}2 \right ) \end{align*} as required. Squaring and adding \((*)\) and \((**)\) we obtain: \begin{align*} && m^2(g^2 + r^2 \omega^4) &= T^2(2 + \sin \theta \sin \phi - \cos \theta \cos \phi) \\ && &= T^2(2 - 2\cos (\theta + \phi)) \\ && &= T^2(2 - 2(1 - 2 \sin^2 \left ( \frac{\theta + \phi}2 \right ) )) \\ && &= T^2(4 \sin^2 \left ( \frac{\theta + \phi}2 \right )) \\ \Rightarrow && T &= \frac{m\sqrt{g^2 + r^2 \omega^4}}{2 \sin \left ( \frac{\theta + \phi}2 \right )} \\ \Rightarrow && T &= \frac{mg\sqrt{1 + \frac{r^2 \omega^4}{g^2}}}{2 \sin \left ( \frac{\theta + \phi}2 \right )} \end{align*} If \(r \omega^2 \ll g\) then \(\tan \l \frac{\phi - \theta}2 \r\) is very large, so \(\phi - \theta \approx \pi\) and so \(\phi + \theta \approx \pi\). We can then say that \[ T \approx \frac{mg}{2}\]

The diagram shows a crude step-ladder constructed by smoothly hinging-together two light ladders \(AB\) and \(AC,\) each of length \(l,\) at \(A\). A uniform rod of wood, of mass \(m\), is pin-jointed to \(X\) on \(AB\) and to \(Y\) on \(AC\), where \(AX=\frac{3}{4}l=AY.\) The angle \(\angle XAY\) is \(2\theta.\) \noindent

A uniform ladder of mass \(M\) rests with its upper end against a smooth vertical wall, and with its lower end on a rough slope which rises upwards towards the wall and makes an angle of \(\phi\) with the horizontal. The acute angle between the ladder and the wall is \(\theta\). If the ladder is in equilibrium, show that \(N\) and \(F\), the normal reaction and frictional force at the foot of the ladder are given by \[ N=Mg\left(\cos\phi-\frac{\tan\theta\sin\phi}{2}\right), \] \[ F=Mg\left(\sin\phi+\frac{\tan\theta\cos\phi}{2}\right). \] If the coefficient of friction between the ladder and the slope is \(2\), and \(\phi=45^{\circ}\), what is the largest value of \(\theta\) for which the ladder can rest in equilibrium?