Problems

A uniform disc with centre \(O\) and radius \(a\) is suspended from a point \(A\) on its circumference, so that it can swing freely about a horizontal axis \(L\) through \(A\). The plane of the disc is perpendicular to \(L\). A particle \(P\) is attached to a point on the circumference of the disc. The mass of the disc is \(M\) and the mass of the particle is \(m\). In equilibrium, the disc hangs with \(OP\) horizontal, and the angle between \(AO\) and the downward vertical through \(A\) is \(\beta\). Find \(\sin\beta\) in terms of \(M\) and \(m\) and show that \[ \frac{AP}{a} = \sqrt{\frac{2M}{M+m}} \,. \] The disc is rotated about \(L\) and then released. At later time \(t\), the angle between \(OP\) and the horizontal is \(\theta\); when \(P\) is higher than \(O\), \(\theta\) is positive and when \(P\) is lower than \(O\), \(\theta\) is negative. Show that \[ \tfrac12 I \dot\theta^2 + (1-\sin\beta)ma^2 \dot \theta^2 + (m+M)g a\cos\beta \, (1- \cos\theta) \] is constant during the motion, where \(I\) is the moment of inertia of the disc about \(L\). Given that \(m= \frac 32 M\) and that \(I=\frac32Ma^2\), show that the period of small oscillations is \[ 3\pi \sqrt{\frac {3a}{5g}} \,. \]

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Solution:

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First, notice that the centre of mass will lie directly below \(A\) and will be \(\frac{m}{M+m}\) of the way between \(O\) and \(P\). Therefore \(\sin \beta = \frac{m}{M+m}\). The cosine rule states that: \begin{align*} && AP^2 &= a^2 + a^2 - 2a^2 \cos \angle AOP \\ \Rightarrow && \frac{AP^2}{a^2} &= 2 - 2 \sin \beta \\ &&&= \frac{2M+2m-2m}{M+m} \\ &&&= \frac{2M}{M+m} \\ \Rightarrow && \frac{AP}{a} &= \sqrt{\frac{2M}{M+m}} \end{align*}

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Considering conservation of energy, we have: Rotational kinetic energy for the disc: \(\frac12 I \dot{\theta}^2\) Kinetic energy for the particle: \(\frac12 m (\dot{\theta} \sqrt{2-2\sin \beta} a)^2 = (1- \sin \beta)ma^2 \dot{\theta}^2\)

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GPE: The important thing is the vertical location of \(G\). The triangle \(OAG\) will still have angle \(\beta\) at \(A\). The vertical height below is: \(\cos \theta \cdot AG = \cos \theta a \cos \beta\). The distance from when \(\theta = 0\) will be \(a \cos \beta (1- \cos \theta)\) and so the GPE will be \((M+m)ga \cos \beta ( 1- \cos \theta)\) we can therefore say by conservation of energy: \[ \frac12 I \dot{\theta}^2 + (1- \sin \beta)ma^2 \dot{\theta}^2+(M+m)ga \cos \beta ( 1- \cos \theta) \] is constant. Suppose \(m = \frac32 M\) and \(I = \frac32 Ma^2\) then differentiating the constant wrt to \(\theta\) gives \(\sin \beta = \frac{m}{M+m} = \frac{3}{5}, \cos \beta = \frac45\) \begin{align*} && 0 &= \frac12 \frac32 M a^2 \cdot 2 \dot{\theta}\ddot{\theta} + (1- \sin \beta)\frac32M a^2 2 \dot{\theta}\ddot{\theta} + (M+\frac32M) ga \cos \beta \sin \theta \cdot \dot{\theta} \\ \Rightarrow && 0 &= \frac32 \ddot{\theta} + 3(1-\sin \beta) \ddot{\theta} + \frac{5}{2}\frac{g}{a} \cos \beta \sin \theta \\ &&&= (\frac32 + \frac65) \ddot{\theta} + \frac{2g}{a} \sin \theta \\ &&&= \frac{27}{10} \ddot{\theta} + \frac{2g}{a} \sin \theta \end{align*} If \(\theta\) is small, we can approximate this by: \(\frac{27}{10} \ddot{\theta} + \frac{2g}{a} \theta = 0\) which will have period \(\displaystyle 2 \pi \sqrt{\frac{27a}{10\cdot2 g}} = 2 \pi \sqrt{\frac{3a}{5g}}\) as required.

A uniform rod \(AB\) has mass \(M\) and length \(2a\). The point \(P\) lies on the rod a distance \(a-x\) from~\(A\). Show that the moment of inertia of the rod about an axis through \(P\) and perpendicular to the rod is \[ \tfrac13 M(a^2 +3x^2)\,. \] The rod is free to rotate, in a horizontal plane, about a fixed vertical axis through \(P\). Initially the rod is at rest. The end \(B\) is struck by a particle of mass \(m\) moving horizontally with speed \(u\) in a direction perpendicular to the rod. The coefficient of restitution between the rod and the particle is \(e\). Show that the angular velocity of the rod immediately after impact is \[ \frac{3mu(1+e)(a+x)}{M(a^2+3x^2) +3m(a+x)^2}\,. \] In the case \(m=2M\), find the value of \(x\) for which the angular velocity is greatest and show that this angular velocity is \(u(1+e)/a\,\).

Calculate the moment of inertia of a uniform thin circular hoop of mass \(m\) and radius \(a\) about an axis perpendicular to the plane of the hoop through a point on its circumference. The hoop, which is rough, rolls with speed \(v\) on a rough horizontal table straight towards the edge and rolls over the edge without initially losing contact with the edge. Show that the hoop will lose contact with the edge when it has rotated about the edge of the table through an angle \(\theta\), where \[ \cos\theta = \frac 12 +\frac {v^2}{2ag}. \] %Give the corresponding result for a smooth hoop and table.

A uniform right circular cone of mass \(m\) has base of radius \(a\) and perpendicular height \(h\) from base to apex. Show that its moment of inertia about its axis is \({3\over 10} ma^2\), and calculate its moment of inertia about an axis through its apex parallel to its base. \newline[{\em Any theorems used should be stated clearly.}] The cone is now suspended from its apex and allowed to perform small oscillations. Show that their period is $$ 2\pi\sqrt{ 4h^2 + a^2\over 5gh} \,. $$ \newline[{\em You may assume that the centre of mass of the cone is a distance \({3\over 4}h\) from its apex.}]

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.

The points \(O,A,B\) and \(C\) are the vertices of a uniform square lamina of mass \(M.\) The lamina can turn freely under gravity about a horizontal axis perpendicular to the plane of the lamina through \(O\). The sides of the lamina are of length \(2a.\) When the lamina is haning at rest with the diagonal \(OB\) vertically downwards it is struck at the midpoint of \(OC\) by a particle of mass \(6M\) moving horizontally in the plane of the lamina with speed \(V\). The particle adheres to the lamina. Find, in terms of \(a,M\) and \(g\), the value which \(V^{2}\) must exceed for the lamina and particle to make complete revolutions about the axis.

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Solution:

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Consider the moment of inertia of the lamina. The MoI about the centre of mass is \(\frac1{12}M((2a)^2 + (2a)^2) = \frac23Ma^2\). //el axis theorem, tells us the moment of inertia about \(O\) is \(I_O = I_G + Md^2_{OG} = \frac23Ma^2 + M2a^2 = \frac83Ma^2\) Moment of inertia of particle is \(6Ma^2\) Total moment of inertial is: \(\frac{26}{3}Ma^2\). Conservation of angular momentum states that \(6M \frac{\sqrt{2}}2Va = \frac{26}{3}Ma^2 \omega \Rightarrow \omega = \frac{9\sqrt{2}V}{26a}\) Consider the centre of mass (in the frame drawn) \begin{array}{c|c|c} \text{Shape} & \text{Mass} & \text{COM} \\ \hline \text{Square} & M & (0,-\sqrt{2}a) \\ \text{Particle} & 6M & (-\frac{\sqrt{2}}2a, -\frac{\sqrt{2}}{2}a) \\ \text{combined} & 7M & \left ( \frac{-3\sqrt{2}}{7} a, -\frac{4\sqrt{2}}{7}a \right) \end{array} The lamina/particle system will complete full circles if it still has positive angular velocity at the peak, ie: \begin{align*} && \underbrace{\frac12 I \omega^2}_{\text{initial rotational energy}} + mgh_{start} &\geq mgh_{top} \\ && \frac 12 \frac{26}{3} Ma^2 \frac{9^2 \cdot 2 V^2}{26^2 a^2} - (7M)g\frac{4\sqrt{2}}{7}a &\geq (7M)g\frac{5\sqrt{2}}{7}a \\ \Rightarrow && \frac{V^2 \cdot 27}{26} &\geq 9\sqrt{2}ga \\ \Rightarrow && V^2 & \geq \frac{26\sqrt{2}}{3}ga \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.