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 \(m_2\).

Two particles of mass \(M\) and \(m\) \((M>m)\) are attached to the ends of a light rod of length \(2l.\) The rod is fixed at its midpoint to a point \(O\) on a horizontal axle so that the rod can swing freely about \(O\) in a vertical plane normal to the axle. The axle rotates about a vertical axis through \(O\) at a constant angular speed \(\omega\) such that the rod makes a constant angle \(\alpha\) \((0<\alpha<\frac{1}{2}\pi)\) with the vertical. Show that \[ \omega^{2}=\left(\frac{M-m}{M+m}\right)\frac{g}{l\cos\alpha}. \] Show also that the force of reaction of the rod on the axle is inclined at an angle \[ \tan^{-1}\left[\left(\frac{M-m}{M+m}\right)^{2}\tan\alpha\right] \] with the downward vertical.

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Solution:

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The accelerations of \(M\) and \(m\) are \(l \sin \alpha \omega^2\) and \(-l \sin \alpha \omega^2\) so the forces \(R_M\) and \(R_m\) are \(M\binom{l \sin \alpha \omega^2}{g}, \,m \binom{-l \sin \alpha \omega^2}{g}\). Since the axle is rotating freely, the moment about \(O\) for the rod must be \(O\). The moment for \(M\) will be \(M\binom{l \sin \alpha\omega^2}{g} \cdot \binom{-l\cos \alpha}{l \sin \alpha} = lM\sin\alpha (g - l \cos \alpha\omega^2)\). The moment for \(m\) will be \(m \binom{-l \sin \alpha\omega^2}{g} \cdot \binom{-l\cos \alpha\omega^2}{l \sin \alpha} = lm \sin \alpha(g+l \cos \alpha\omega^2)\) Therefore \begin{align*} && lM\sin\alpha (g - l \cos \alpha\omega^2) &= lm \sin \alpha(g+l \cos \alpha\omega^2) \\ && M(g - l \cos \alpha \omega^2) &= m(g + l \cos \alpha \omega^2 ) \\ \Rightarrow && g(M-m) &= l \cos \alpha (M+m) \omega^2 \\ \Rightarrow && \omega^2 &= \left (\frac{M-m}{M+m} \right) \frac{g}{l \cos \alpha} \end{align*} as required. The total force on the rod is \(\mathbf{0}\) so the reaction force must be \(M\binom{l \sin \alpha \omega^2}{g}+ \,m \binom{-l \sin \alpha \omega^2}{g} = \binom{l \sin \alpha \omega^2 (M-m)}{(M+m)g}\) Therefore the angle this makes with downward vertical is: \begin{align*} \theta &= \tan^{-1} \left ( \frac{l \sin \alpha \omega^2 (M-m)}{(M+m)g} \right) \\ &= \tan^{-1} \left ( \frac{l \sin \alpha (M-m)}{(M+m)g} \omega^2\right) \\ &= \tan^{-1} \left ( \frac{l \sin \alpha (M-m)}{(M+m)g} \left (\frac{M-m}{M+m} \right) \frac{g}{l \cos \alpha}\right) \\ &= \tan^{-1}\left[\left(\frac{M-m}{M+m}\right)^{2}\tan\alpha\right] \end{align*} as required.