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 sphere of radius \(R\) and uniform density \(\rho_{\text{s}}\) is floating in a large tank of liquid of uniform density \(\rho\). Given that the centre of the sphere is a distance \(x\) above the level of the liquid, where \(x < R\), show that the volume of liquid displaced is \[ \frac \pi 3 (2R^3-3R^2x +x^3)\,. \] The sphere is acted upon by two forces only: its weight and an upward force equal in magnitude to the weight of the liquid it has displaced. Show that \[ 4 R^3\rho_{\text{s}} (g+\ddot x) = (2R^3 -3R^2x +x^3)\rho g\,. \] Given that the sphere is in equilibrium when \(x=\frac12 R\), find \(\rho_{\text{s}}\) in terms of \(\rho\). Find, in terms of \(R\) and \(g\), the period of small oscillations about this equilibrium position.

A particle \(P\) of mass \(m\) is attached to points \(A\) and \(B\), where \(A\) is a distance \(9a\) vertically above \(B\), by elastic strings, each of which has modulus of elasticity \(6mg\). The string \(AP\) has natural length \(6a\) and the string \(BP\) has natural length \(2a\). Let \(x\) be the distance \(AP\). The system is released from rest with \(P\) on the vertical line \(AB\) and \(x = 6a\). Show that the acceleration \(\ddot{x}\) of \(P\) is \(\ds{4g \over a}(7a - x)\) for \(6a < x < 7a\) and \(\ds{g \over a}(7a - x)\) for \(7a < x < 9a\,\). Find the time taken for the particle to reach \(B\).

A particle is attached to a point \(P\) of an unstretched light uniform spring \(AB\) of modulus of elasticity \(\lambda\) in such a way that \(AP\) has length \(a\) and \(PB\) has length \(b\). The ends \(A\) and \(B\) of the spring are now fixed to points in a vertical line a distance \(l\) apart, The particle oscillates along this line. Show that the motion is simple harmonic. Show also that the period is the same whatever the value of \(l\) and whichever end of the string is uppermost.

Two small spheres \(A\) and \(B\) of equal mass \(m\) are suspended in contact by two light inextensible strings of equal length so that the strings are vertical and the line of centres is horizontal. The coefficient of restitution between the spheres is \(e\). The sphere \(A\) is drawn aside through a very small distance in the plane of the strings and allowed to fall back and collide with the other sphere \(B\), its speed on impact being \(u\). Explain briefly why the succeeding collisions will all occur at the lowest point. (Hint: Consider the periods of the two pendulums involved.) Show that the speed of sphere \(A\) immediately after the second impact is \(\frac{1}{2}u(1+e^{2})\) and find the speed, then, of sphere \(B\).

Consider a simple pendulum of length \(l\) and angular displacement \(\theta\), which is {\bf not} assumed to be small. Show that $$ {1\over 2}l \left({\d\theta\over \d t}\right)^2 = g(\cos\theta -\cos\gamma)\,, $$ where \(\gamma\) is the maximum value of \(\theta\). Show also that the period \(P\) is given by $$ P= 2 \sqrt{l\over g} \int_0^\gamma \left( \sin^2(\gamma/2)-\sin^2(\theta/2) \right)^{-{1\over 2}} \,\d\theta \,. $$ By using the substitution \(\sin(\theta/2)=\sin(\gamma/2) \sin\phi\), and then finding an approximate expression for the integrand using the binomial expansion, show that for small values of \(\gamma\) the period is approximately $$ 2\pi \sqrt{l\over g} \left(1+{\gamma^2\over 16}\right) \,. $$

A smooth circular wire of radius \(a\) is held fixed in a vertical plane with light elastic strings of natural length \(a\) and modulus \(\lambda\) attached to the upper and lower extremities, \(A\) and \(C\) respectively, of the vertical diameter. The other ends of the two strings are attached to a small ring \(B\) which is free to slide on the wire. Show that, while both strings remain taut, the equation for the motion of the ring is $$2ma \ddot\theta=\lambda(\cos\theta-\sin\theta)-mg\sin\theta,$$ where \(\theta\) is the angle \( \angle{CAB}\). Initially the system is at rest in equilibrium with \(\sin\theta=\frac{3}{5}\). Deduce that \(5\lambda=24mg\). The ring is now displaced slightly. Show that, in the ensuing motion, it will oscillate with period approximately $$10\pi\sqrt{a\over91g}\,.$$

The points \(A,B,C,D\) and \(E\) lie on a thin smooth horizontal table and are equally spaced on a circle with centre \(O\) and radius \(a\). At each of these points there is a small smooth hole in the table. Five elastic strings are threaded through the holes, one end of each beging attached at \(O\) under the table and the other end of each being attached to a particle \(P\) of mass \(m\) on top of the table. Each of the string has natural length \(a\) and modulus of elasticity \(\lambda.\) If \(P\) is displaced from \(O\) to any point \(F\) on the table and released from rest, show that \(P\) moves with simple harmonic motion of period \(T\), where \[ T=2\pi\sqrt{\frac{am}{5\lambda}}. \] The string \(PAO\) is replaced by one of natural length \(a\) and modulus \(k\lambda.\) \(P\) is displaced along \(OA\) from its equilibrium position and released. Show that \(P\) still moves in a straight line with simple harmonic motion, and, given that the period is \(T/2,\) find \(k\).

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Solution:

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The extension of \(OAP\) is \(|AP|\) and so the tension \(T_a = \frac{\lambda}{a} |AP|\). To simplify calculations, let \(A = a, B = a \omega, C = a \omega^2, \cdots\) where \(\omega = e^{2 \pi i/5}\) and let \(P = z\). then we can calculate the force as: \begin{align*} &&\sum_{p}T_p \mathbf{n}_{z \to p} &= \sum_{p} \frac{\lambda}{a} |z-p| \frac{p-z}{|p-z|} \\ &&&= \frac{\lambda}{a} \sum_{p} ( p - z) \\ &&&= -\frac{5\lambda}{a}z \end{align*} Therefore the force has magnitude \(\frac{5 \lambda}{a} |OP|\) directly towards the origin. Therefore if we set up our coordinate axis such that \(OP\) is the \(x\) axis, the particle will remain on the \(x\) axis and will move under the equation: \[ m \ddot{x} + \frac{5 \lambda}{a} x = 0 \] But then we can say that \(P\) moves under SHM with period \(\displaystyle 2 \pi \sqrt{\frac{am}{5 \lambda}}\) as required. Now suppose that \(PAO\) has been replaced with the string of modulus \(k \lambda\) but that \(P\) is along \(OA\). \begin{align*} F &= \frac{\lambda}{a}\left ( (a \omega - z) + (a \omega^2 - z)+ (a \omega^3 -z)+ (a \omega^4 - z) + k(a -z) \right) \\ &= \frac{\lambda}{a}(-a - 4z+ka -kz) \\ &= \frac{\lambda}{a}((k-1)a-(k+4)z) \end{align*} Notice that if \(z\) is real, this expression is also real, so all forces are acting along \(OA\). Therefore the particle will remain on the line \(OA\). We can also notice that the particle will move under the differential equation \[ m \ddot{x} + \frac{(k+4) \lambda}{a}x = \lambda(k-1) \] Therefore it will move with SHM about a point slightly displaced from the origin. The period will be: \(\displaystyle 2 \pi \sqrt{\frac{ma}{(k+4)\lambda}}\) which is equal to \(T/2\) if \((k+4) = 20 \Rightarrow k = 16\)

A thin uniform elastic band of mass \(m,\) length \(l\) and modulus of elasticity \(\lambda\) is pushed on to a smooth circular cone of vertex angle \(2\alpha,\) in such a way that all elements of the band are the same distance from the vertex. It is then released from rest. Let \(x(t)\) be the length of the band at time \(t\) after release, and let \(t_{0}\) be the time at which the band becomes slack. Assuming that a small element of the band which subtends an angle \(\delta\theta\) at the axis of the cone experiences a force, due to the tension \(T\) in the band, of magnitude \(T\delta\theta\) directed towards the axis, and ignoring the effects of gravity, show that \[ \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+\frac{4\pi^{2}\lambda}{ml}(x-l)\sin^{2}\alpha=0,\qquad(0< t< t_{0}). \] Find the value of \(t_{0}.\)

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Solution:

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\begin{align*} \text{N2}(\nwarrow): && T\delta \theta \sin \alpha &= -m\frac{\delta \theta}{2\pi} \ddot{d} \end{align*} Notice that \(r = d \sin \alpha\) and \(x = 2 \pi r\), so \(x = 2\pi d \sin \alpha\) and \(\ddot{x} = 2\pi \sin \alpha \ddot{d} \Rightarrow \ddot{d} = \ddot{x} \frac{1}{2 \pi \sin \alpha}\) Notice also that \(T = \frac{\lambda}{l}(x-l)\) so. \begin{align*} && \frac{m}{4 \pi^2 \sin\alpha} \ddot{x} &= -\frac{\lambda}{l}(x-l) \sin\alpha \\ \Rightarrow && \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+\frac{4\pi^{2}\lambda}{ml}(x-l)\sin^{2}\alpha&=0 \end{align*} The solution to the differential equation we have is: \begin{align*} && x(t) &= A \sin \left (\sqrt{\frac{4 \pi^2 \lambda}{ml}\sin^2 \alpha} \cdot t \right) + B \sin \left (\sqrt{\frac{4 \pi^2 \lambda}{ml}\sin^2 \alpha} \cdot t \right) + l \\ &&&= A \sin \left (2 \pi \sin \alpha\sqrt{\frac{ \lambda}{ml}} \cdot t \right) +B \sin \left (2 \pi \sin \alpha\sqrt{\frac{ \lambda}{ml}} \cdot t \right) + l\\ && \dot{x}(0) = 0 \\ \Rightarrow && B &= 0 \\ && x(t) &= (x(0)-l) \sin \left (2 \pi \sin \alpha\sqrt{\frac{ \lambda}{ml}} \cdot t \right) + l \\ && x(t_0) &= l \\ \Rightarrow && t_0 &= \frac{1}{4\sin \alpha} \sqrt{\frac{ml}{\lambda}} \end{align*}