Problems

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A rough circular cylinder of mass \(M\) and radius \(a\) rests on a rough horizontal plane. The curved surface of the cylinder is in contact with a smooth rail, parallel to the axis of the cylinder, which touches the cylinder at a height \(a/2\) above the plane. Initially the cylinder is held at rest. A particle of mass \(m\) rests in equilibrium on the cylinder, and the normal reaction of the cylinder on the particle makes an angle of \(\theta\) with the upward vertical. The particle is on the same side of the centre of the cylinder as the rail. The coefficient of friction between the cylinder and the particle and between the cylinder and the plane are both \(\mu\). Obtain the condition on \(\theta\) for the particle to rest in equilibrium. Show that, if the cylinder is released, equilibrium of both particle and cylinder is possible provided another inequality involving \(\mu\) and \(\theta\) (which should be found explicitly) is satisfied. Determine the largest possible value of \(\theta\) for equilibrium, if \(m=7M\) and \(\mu=0.75\).

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Solution:

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\begin{align*} \text{N2}(\nwarrow): && R -mg \cos \theta &= 0 \\ \text{N2}(\rightarrow): && -R \sin \theta + F \cos \theta &= 0 \\ \\ \Rightarrow && F &= \tan \theta R \\ \\ && F & \leq \mu R \\ \Rightarrow && \tan \theta R &\leq \mu R \\ \Rightarrow && \tan \theta &\leq \mu \end{align*} (Notice also \(F = mg \sin \theta\)) Once everything is released, we have the following situation. (Red forces act on the cylinder, blue forces on the particle).

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\begin{align*} \text{N2}(\uparrow): && 0 &= R_g - Mg - \underbrace{mg}_{R_p \text{ and } F_p} + \frac{1}{\sqrt{2}}R_r \\ \text{N2}(\rightarrow): && 0 &= \frac{1}{\sqrt{2}}R_r - F_g \\ \overset{\curvearrowleft}{O}: && 0 &= aF_p - aF_g \\ \Rightarrow && F_g &= mg \sin \theta \\ \Rightarrow && R_r &= \sqrt{2} mg \sin \theta \\ \Rightarrow && R_g &=(M+m)g + mg \sin \theta \\ \\ && F_g &\leq \mu R_g \\ \Rightarrow && mg \sin \theta &\leq \mu (M+m(1+\sin \theta))g \\ \Rightarrow && \mu &\geq \frac{m \sin \theta}{M+m(1+\sin \theta)} \end{align*} If \(m = 7M\) and \(\mu = \frac34\) we have: \begin{align*} && \tan \theta &\leq \frac34 \\ && 3(M+7M(1 + \sin \theta)) &\geq 4 \cdot 7 M \sin \theta \\ \Rightarrow && 10 + 7 \sin \theta & \geq 28 \sin \theta \\ \Rightarrow && 10 &\geq 21 \sin \theta \\ \Rightarrow && \sin \theta &\leq \frac{10}{21} \end{align*} If \(\tan \theta = \frac{3}{4}, \sin \theta = \frac35 > \frac{10}{21}\), so the critical bound is \(\sin \theta \leq \frac{10}{21}\), ie \( \theta \leq \sin^{-1} \frac{10}{21} \approx 30^{\circ}\)

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*}

The edges \(OA,OB,OC\) of a rigid cube are taken as coordinate axes and \(O',A',B',C'\) are the vertices diagonally opposite \(O,A,B,C,\) respectively. The four forces acting on the cube are \[ \begin{pmatrix}\alpha\\ \beta\\ \gamma \end{pmatrix}\mbox{ at }O\ (0,0,0),\ \begin{pmatrix}\lambda\\ 0\\ 1 \end{pmatrix}\mbox{ at }O'\ (a,a,a),\ \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix}\mbox{ at }B\ (0,a,0),\ \mbox{ and }\begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix}\mbox{ at }B'\ (a,0,a). \] The moment of the system about \(O\) is zero: find \(\lambda,\mu\) and \(\nu\).

1. Given that \(\alpha=\beta=\gamma=0,\) find the system consisting of a single force at \(B\) together with a couple which is equivalent to the given system.
2. Given that \(\alpha=2,\beta=3\) and \(\gamma=2,\) find the equation of the locus about each point of which the moment of the system is zero. Find the number of units of work done on the cube when it moves (without rotation) a distance in the direction of this line under the action of the given forces only.

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?

A uniform rectangular lamina of sides \(2a\) and \(2b\) rests in a vertical plane. It is supported in equilibrium by two smooth pegs fixed in the same horizontal plane, a distance \(d\) apart, so that one corner of the lamina is below the level of the pegs. Show that if the distance between this (lowest) corner and the peg upon which the side of length \(2a\) rests is less than \(a\), then the distance between this corner and the other peg is less than \(b\). Show also that \[ b\cos\theta-a\sin\theta=d\cos2\theta, \] where \(\theta\) is the acute angle which the sides of length \(2b\) make with the horizontal.

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Solution:

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We must have \(G\) between the two pegs (vertically), otherwise we will induce a moment. Considering moments about the peg, if the second peg is outside the centre then we must induce a moment and therefore we cannot be in equilibrium. \begin{align*} \text{N2}(\nearrow):&& 0 &= R_1-mg\sin\theta \\ \text{N2}(\nwarrow):&&0&= R_2-mg \cos \theta \\ \Rightarrow && R_1 &= mg \sin\theta \\ && R_2 &= mg \cos\theta \\ \\ \overset{\curvearrowleft}{G}: && 0 &= R_1(a-d\sin\theta) -R_2(b-d \cos\theta) \\ \Rightarrow && 0&= a \sin\theta -d \sin^2\theta - b\cos \theta+d \cos^2 \theta \\ \Rightarrow && b \cos \theta - a \sin \theta &= d \cos 2 \theta \end{align*}

A librarian wishes to pick up a row of identical books from a shelf, by pressing her hands on the outer covers of the two outermost books and lifting the whole row together. The covers of the books are all in parallel vertical planes, and the weight of each book is \(W\). With each arm, the librarian can exert a maximum force of \(P\) in the vertical direction, and, independently, a maximum force of \(Q\) in the horizontal direction. The coefficient of friction between each pair of books and also between each hand and a book is \(\mu.\) Derive an expression for the maximum number of books that can be picked up without slipping, using this method. {[}You may assume that the books are thin enough for the rotational effect of the couple on each book to be ignored.{]}

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Solution:

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The force acting vertically on each of the outer books must be (by symmetry) \(\frac{nW}{2}\). The force acting horizontally on the outer books (and between each book in the horizontal direction) will be the same (we might as well say \(Q\) since increasing this force doesn't make any task less achievable. Looking at an end book, it will have force \(\frac{nW}{2}\) acting on one side, but it this force needs to not slip, ie \(\frac{nW}{2} \leq \mu Q\) \begin{align*} && \frac{nW}{2} &\leq \mu Q \\ \Rightarrow && n &\leq \frac{2\mu Q}{W} \\ && \frac{nW}{2} & \leq P \\ && n & \leq \frac{2P}{W} \\ \Rightarrow && n &\leq \frac2{W}\min \left (P, \mu Q \right) \end{align*}

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.

A uniform ladder of length \(l\) and mass \(m\) rests with one end in contact with a smooth ramp inclined at an angle of \(\pi/6\) to the vertical. The foot of the ladder rests, on horizontal ground, at a distance \(l/\sqrt{3}\) from the foot of the ramp, and the coefficient of friction between the ladder and the ground is \(\mu.\) The ladder is inclined at an angle \(\pi/6\) to the horizontal, in the vertical plane containing a line of greatest slope of the ramp. A labourer of mass \(m\) intends to climb slowly to the top of the ladder.

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1. Find the value of \(\mu\) if the ladder slips as soon as the labourer reaches the midpoint.
2. Find the minimum value of \(\mu\) which will ensure that the labourer can reach the top of the ladder.

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Solution:

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1. \begin{align*} \text{N2}(\uparrow): && R_1 + R_2\sin(\frac{\pi}{6})-2mg &= 0 \\ \text{N2}(\rightarrow): && R_2 \cos (\frac{\pi}{6})-F_r &= 0 \\ \overset{\curvearrowleft}{X}: && lmg \cos \tfrac{\pi}{6} - l R_2 \cos \tfrac{\pi}{6} &= 0 \\ \\ \Rightarrow && R_2 &= mg \\ \Rightarrow && R_1 &= 2mg - \frac12mg \\ &&&=\frac32mg \\ \Rightarrow && \frac{\sqrt{3}}2mg - \mu\frac32mg &= 0 \\ \Rightarrow && \mu &= \frac{1}{\sqrt{3}} \end{align*}
2. \begin{align*} \text{N2}(\uparrow): && R_1 + R_2\sin(\frac{\pi}{6})-2mg &= 0 \\ \overset{\curvearrowleft}{X}: && \frac12 lmg \cos \tfrac{\pi}{6}+xmg \cos \tfrac{\pi}{6} - l R_2 \cos \tfrac{\pi}{6} &= 0 \\ \\ \Rightarrow && R_2 &= mg(\frac{1}2+\frac{x}{l}) \\ \Rightarrow && R_1 &= 2mg - \frac12mg(\frac{1}2+\frac{x}{l}) \\ &&&=(\frac74 - \frac{x}{2l})mg \\ &&&\geq \frac{5}{4}mg\\ \text{N2}(\rightarrow): && R_2 \cos (\frac{\pi}{6})-\mu R_1& \leq 0 \\ \Rightarrow && \frac{\sqrt{3}}2mg - \mu\frac54mg &\leq 0 \\ \Rightarrow && \mu &\geq \frac{2\sqrt{3}}{5} \end{align*}

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*}

A firework consists of a uniform rod of mass \(M\) and length \(2a\), pivoted smoothly at one end so that it can rotate in a fixed horizontal plane, and a rocket attached to the other end. The rocket is a uniform rod of mass \(m(t)\) and length \(2l(t)\), with \(m(t)=2\alpha l(t)\) and \(\alpha\) constant. It is attached to the rod by its front end and it lies at right angles to the rod in the rod's plane of rotation. The rocket burns fuel in such a way that \(\mathrm{d}m/\mathrm{d}t=-\alpha\beta,\) with \(\beta\) constant. The burnt fuel is ejected from the back of the rocket, with speed \(u\) and directly backwards relative to the rocket. Show that, until the fuel is exhausted, the firework's angular velocity \(\omega\) at time \(t\) satisfies \[ \frac{\mathrm{d}\omega}{\mathrm{d}t}=\frac{3\alpha\beta au}{2[Ma^{2}+2\alpha l(3a^{2}+l^{2})]}. \]

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Solution:

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The rocket principle states that the thrust generated by the rocket is \(-\frac{\d m}{\d t}u = \alpha \beta u\) This force is acting at a distance \(2a\) from \(O\) and therefore is generating a torque of \(2a \alpha \beta u\) on the system. Let's also consider the moments of inertia about \(O\). The fixed rod will have moment of inertia \(\frac13 M (2a)^2 = \frac43 M a^2\). The rocket will have moment of inertia \(I_{G} + md^2 = \frac1{12}m(t)(2l(t))^2 + m(t) ((2a)^2 + l(t)^2)= \frac43 ml^2+ 4ma^2\). Since our final equation doesn't involve \(m\), lets replace all the \(m\) with \(2al\) to obtain a total \(\displaystyle I = \frac43 Ma^2 + \frac83 \alpha l^3 + 8\alpha la^2\). Since \(\tau\) is constant, we can note that \(I\omega = 2a \alpha \beta u t\) (by integrating) and so \begin{align*} && \dot{\omega} &= \frac{\d }{\d t} \left ( \frac{2a \alpha \beta u t}{ \frac43 Ma^2 + \frac83 \alpha l^3 + 8\alpha la^2} \right) \\ &&&= \frac{\d }{\d t} \left ( \frac{3a \alpha \beta u t}{ 2Ma^2 +4\alpha l^3 + 4 \cdot 3 \cdot \alpha la^2} \right) \\ &&&= \frac{\d }{\d t} \left ( \frac{3a \alpha \beta u t}{ 2[Ma^2 +2\alpha l(l^2 + 3 a^2)]} \right) \\ \end{align*} This is, close, but not quite what they are after since the denominator also has a dependency on \(t\) we wont get exactly what they've asked for

A uniform rod, of mass \(3m\) and length \(2a,\) is freely hinged at one end and held by the other end in a horizontal position. A rough particle, of mass \(m\), is placed on the rod at its mid-point. If the free end is then released, prove that, until the particle begins to slide on the rod, the inclination \(\theta\) of the rod to the horizontal satisfies the equation \[ 5a\dot{\theta}^{2}=8g\sin\theta. \] The coefficient of friction between the particle and the rod is \(\frac{1}{2}.\) Show that, when the particle begins to slide, \(\tan\theta=\frac{1}{26}.\)

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Solution:

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While the particle is not sliding, we can consider the whole system. Considering the moment of inertia about the end, we have: \begin{align*} I &= \frac13 \cdot 3m \cdot (2a)^2 + m a^2 \\ &= 5ma^2 \end{align*} Taking the level of the pivot as the \(0\) GPE level, the initial energy is \(0\). The energy once it has rotated through an angle \(\theta\) is: \begin{align*} && 0 &= \text{rotational ke} + \text{gpe} \\ &&&= \frac12 I \dot{\theta}^2 - 4mg \sin \theta \\ &&&= \frac12 5am \dot{\theta}^2 -4mg \sin \theta \\ \Rightarrow && 5a\dot{\theta}^2 &= 8g \sin \theta \end{align*} as required. We also have \(5a \ddot{\theta} = 4g \cos \theta\) The acceleration towards the pivot required to maintain circular motion is \(m \frac{v^2}{r} = m a \dot{\theta}^2\). When we are on the point of sliding:

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\begin{align*} \text{N2}(\nearrow): && R - mg\cos \theta &= -ma \ddot{\theta} \\ \Rightarrow && R &= mg \cos \theta - ma \frac{4mg \cos \theta}{5a} \\ &&&= \frac15mg \cos \theta \end{align*} Therefore we must have: \begin{align*} \text{N2}(\nwarrow):&&\mu R - mg \sin \theta &= ma \dot{\theta}^2 \\ && \frac12 \cdot \frac 15 mg \cos \theta &= m \frac{13}5 g \sin \theta \\ \Rightarrow && \tan \theta &= \frac{1}{26} \end{align*}