6.04e Rigid body equilibrium: coplanar forces

\includegraphics{figure_1} A smooth wire \(PMQ\) is in the shape of a semicircle with centre \(O\) and radius \(a\). The wire is fixed in a vertical plane with \(PQ\) horizontal and the mid-point \(M\) of the wire vertically below \(O\). A smooth bead \(B\) of mass \(m\) is threaded on the wire and is attached to one end of a light elastic string. The string has modulus of elasticity \(4mg\) and natural length \(\frac{3}{4}a\). The other end of the string is attached to a fixed point \(P\) which is a distance \(a\) vertically above \(O\), as shown in Fig. 1.

1. Show that, when \(\angle BFO = \theta\), the potential energy of the system is $$\frac{1}{16}mga(8 \cos \theta - 5)^2 - 2mga \cos^2\theta + \text{constant}.$$ [6]
2. Hence find the values of \(\theta\) for which the system is in equilibrium. [6]
3. Determine the nature of the equilibrium at each of these positions. [5]

\includegraphics{figure_1} A smooth wire with ends \(A\) and \(B\) is in the shape of a semi-circle of radius \(a\). The mid-point of \(AB\) is \(O\) and is fixed in a vertical plane and hangs below \(AB\) which is horizontal. A small ring \(R\), of mass \(m\sqrt{2}\), is threaded on the wire and is attached to two light inextensible strings. The other end of each string is attached to a particle of mass \(\frac{3m}{2}\). The particles hang vertically under gravity, as shown in Figure 1.

1. Show that, when the radius \(OR\) makes an angle \(2\theta\) with the vertical, the potential energy, \(V\), of the system is given by $$V = \sqrt{2}mga(3 \cos \theta - \cos 2\theta) + \text{constant}.$$ [7]
2. Find the values of \(\theta\) for which the system is in equilibrium. [6]
3. Determine the stability of the position of equilibrium for which \(\theta > 0\). [4]

A non-uniform rod \(BC\) has mass \(m\) and length \(3l\). The centre of mass of the rod is at distance \(l\) from \(B\). The rod can turn freely about a fixed smooth horizontal axis through \(B\). One end of a light elastic string, of natural length \(l\) and modulus of elasticity \(\frac{mg}{6}\), is attached to \(C\). The other end of the string is attached to a point \(P\) which is at a height \(3l\) vertically above \(B\).

1. Show that, while the string is stretched, the potential energy of the system is $$mgl(\cos^2 \theta - \cos \theta) + \text{constant},$$ where \(\theta\) is the angle between the string and the downward vertical and \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\). [6]
2. Find the values of \(\theta\) for which the system is in equilibrium with the string stretched. [6]

\includegraphics{figure_1} A uniform rod \(PQ\) has mass \(m\) and length \(2l\). A small smooth light ring is fixed to the end \(P\) of the rod. This ring is threaded on to a fixed horizontal smooth straight wire. A second small smooth light ring \(R\) is threaded on to the wire and is attached by a light elastic string, of natural length \(l\) and modulus of elasticity \(kmg\), to the end \(Q\) of the rod, where \(k\) is a constant.

1. Show that, when the rod \(PQ\) makes an angle \(\theta\) with the vertical, where \(0 < \theta \leq \frac{\pi}{3}\), and \(Q\) is vertically below \(R\), as shown in Figure 1, the potential energy of the system is $$mgl[2k\cos^2\theta - (2k + 1)\cos\theta] + \text{constant}.$$ [7]

Given that there is a position of equilibrium with \(\theta > 0\),

1. show that \(k > \frac{1}{2}\). [5]

\includegraphics{figure_1} A framework consists of two uniform rods \(AB\) and \(BC\), each of mass \(m\) and length \(2a\), joined at \(B\). The mid-points of the rods are joined by a light rod of length \(a\sqrt{2}\), so that angle \(ABC\) is a right angle. The framework is free to rotate in a vertical plane about a fixed smooth horizontal axis. This axis passes through the point \(A\) and is perpendicular to the plane of the framework. The angle between the rod \(AB\) and the downward vertical is denoted by \(\theta\), as shown in Fig. 1.

1. Show that the potential energy of the framework is $$-mga(3 \cos \theta + \sin \theta) + \text{constant}.$$ [4]
2. Find the value of \(\theta\) when the framework is in equilibrium, with \(B\) below the level of \(A\). [4]
3. Determine the stability of this position of equilibrium. [4]

\includegraphics{figure_3} A small smooth peg \(P\) is fixed at a distance \(d\) from a fixed smooth vertical wire. A particle of mass \(3m\) is attached to one end of a light inextensible string which passes over \(P\). The particle hangs vertically below \(P\). The other end of the string is attached to a small ring \(R\) of mass \(m\), which is threaded on the wire, as shown in Figure 3.

1. Show that when \(R\) is at a distance \(x\) below the level of \(P\) the potential energy of the system is $$3mg \sqrt{(x^2 + d^2)} - mgx + \text{constant}$$ [4]
2. Hence find \(x\), in terms of \(d\), when the system is in equilibrium. [3]
3. Determine the stability of the position of equilibrium. [3]

\includegraphics{figure_1} A uniform rod \(AB\), of length \(2l\) and mass \(12m\), has its end \(A\) smoothly hinged to a fixed point. One end of a light inextensible string is attached to the other end \(B\) of the rod. The string passes over a small smooth pulley which is fixed at the point \(C\), where \(AC\) is horizontal and \(AC = 2l\). A particle of mass \(m\) is attached to the other end of the string and the particle hangs vertically below \(C\). The angle \(BAC\) is \(\theta\), where \(0 < \theta < \frac{\pi}{2}\), as shown in Figure 1.

1. Show that the potential energy of the system is $$4mgl\left(\sin\frac{\theta}{2} - 3\sin\theta\right) + \text{constant}$$ (4)

1. Find the value of \(\theta\) when the system is in equilibrium and determine the stability of this equilibrium position. (10)

\includegraphics{figure_2} A bead \(B\) of mass \(m\) is threaded on a smooth circular wire of radius \(r\), which is fixed in a vertical plane. The centre of the circle is \(O\), and the highest point of the circle is \(A\). A light elastic string of natural length \(r\) and modulus of elasticity \(kmg\) has one end attached to the bead and the other end attached to \(A\). The angle between the string and the downward vertical is \(\theta\), and the extension in the string is \(x\), as shown in Figure 2. Given that the string is taut,

1. show that the potential energy of the system is $$2mgr[(k-1)\cos^2 \theta - k\cos \theta] + \text{constant}$$ [6]

Given also that \(k = 3\),

1. find the positions of equilibrium and determine their stability. [9]

\includegraphics{figure_3} A uniform rod \(AB\) has mass \(m\) and length \(2a\). The end \(A\) is smoothly hinged at a fixed point on a fixed straight horizontal wire. A smooth light ring \(R\) is threaded on the wire. The ring \(R\) is attached by a light elastic string, of natural length \(a\) and modulus of elasticity \(mg\), to the end \(B\) of the rod. The end \(B\) is always vertically below \(R\) and angle \(\angle RAB = \theta\), as shown in Fig. 3.

1. Show that the potential energy of the system is $$mga(2\sin^2\theta - 3\sin\theta) + \text{constant}.$$ [6]
2. Hence determine the value of \(\theta\), \(0 < \frac{\pi}{2}\), for which the system is in equilibrium. [5]
3. Determine whether this position of equilibrium is stable or unstable. [5]

\includegraphics{figure_4} A uniform rod \(AB\), of mass \(m\) and length \(2a\), is freely hinged to a fixed point at \(A\). A particle of mass \(2m\) is attached to the rod at \(B\). A light elastic string, with natural length \(a\) and modulus of elasticity \(5mg\), passes through a fixed smooth ring \(R\). One end of the string is fixed to \(A\) and the other end is fixed to the mid-point \(C\) of \(AB\). The ring \(R\) is at the same horizontal level as \(A\), and is at a distance \(a\) from \(A\). The rod \(AB\) and the ring \(R\) are in a vertical plane, and \(RC\) is at an angle \(\theta\) above the horizontal, where \(0 < \theta < \frac{1}{2}\pi\), so that the acute angle between \(AB\) and the horizontal is \(2\theta\) (see diagram).

1. By considering the energy of the system, find the value of \(\theta\) for which the system is in equilibrium. [7]
2. Determine whether this position of equilibrium is stable or unstable. [3]

\includegraphics{figure_7} A uniform rod \(AB\) has mass \(m\) and length \(6a\). It is free to rotate in a vertical plane about a smooth fixed horizontal axis passing through the point \(C\) on the rod, where \(AC = a\). The angle between \(AB\) and the upward vertical is \(\theta\), and the force acting on the rod at \(C\) has components \(R\) parallel to \(AB\) and \(S\) perpendicular to \(AB\) (see diagram). The rod is released from rest in the position where \(\theta = \frac{1}{4}\pi\). Air resistance may be neglected.

1. Find the angular acceleration of the rod in terms of \(a\), \(g\) and \(\theta\). [4]
2. Show that the angular speed of the rod is \(\sqrt{\frac{2g(1 - 2\cos\theta)}{7a}}\). [3]
3. Find \(R\) and \(S\) in terms of \(m\), \(g\) and \(\theta\). [6]
4. When \(\cos\theta = \frac{1}{3}\), show that the force acting on the rod at \(C\) is vertical, and find its magnitude. [4]

\includegraphics{figure_3} Two uniform rods \(AB\) and \(BC\), each of length \(a\) and mass \(m\), are rigidly joined together so that \(AB\) is perpendicular to \(BC\). The rod \(AB\) is freely hinged to a fixed point at \(A\). The rods can rotate in a vertical plane about a smooth fixed horizontal axis through \(A\). One end of a light elastic string of natural length \(a\) and modulus of elasticity \(\lambda mg\) is attached to \(B\). The other end of the string is attached to a fixed point \(D\) vertically above \(A\), where \(AD = a\). The string \(BD\) makes an angle \(\theta\) radians with the downward vertical (see diagram).

1. Taking \(D\) as the reference level for gravitational potential energy, show that the total potential energy \(V\) of the system is given by $$V = \frac{1}{2}mga(\sin 2\theta - 3\cos 2\theta) + \frac{1}{2}\lambda mga(2\cos \theta - 1)^2 - 2mga.$$ [5]
2. Given that \(\theta = \frac{1}{3}\pi\) is a position of equilibrium, find the exact value of \(\lambda\). [4]
3. Find \(\frac{d^2V}{d\theta^2}\) and hence determine whether the position of equilibrium at \(\theta = \frac{1}{3}\pi\) is stable or unstable. [4]

\includegraphics{figure_6} A compound pendulum consists of a uniform rod \(AB\) of length 1 m and mass 3 kg, a particle of mass 1 kg attached to the rod at \(A\) and a circular disc of radius \(\frac{1}{5}\) m, mass 6 kg and centre \(C\). The end \(B\) of the rod is rigidly attached to a point on the circumference of the disc in such a way that \(ABC\) is a straight line. The pendulum is initially at rest with \(B\) vertically below \(A\) and it is free to rotate in a vertical plane about a smooth fixed horizontal axis passing through the point \(P\) on the rod where \(AP = x\) m and \(x < \frac{1}{3}\) (see diagram).

1. Show that the moment of inertia of the pendulum about the axis of rotation is \((10x^2 - 19x + 12)\) kg m\(^2\). [6]

The pendulum is making small oscillations about the equilibrium position, such that at time \(t\) seconds the angular displacement that the pendulum makes with the downward vertical is \(\theta\) radians.

1. Find the angular acceleration of the pendulum, in terms of \(x\), \(g\) and \(\theta\). [4]
2. Show that the motion is approximately simple harmonic, and show that the approximate period of oscillations, in seconds, is given by \(2\pi\sqrt{\frac{20x^2 - 38x + 24}{(19-20x)g}}\). [2]
3. Hence find the value of \(x\) for which the approximate period of oscillations is least. [3]

A uniform lamina \(ABC\) of mass \(m\) is in the shape of an isosceles triangle with \(AB = AC = 5a\) and \(BC = 8a\).

1. Show, using integration, that the moment of inertia of the lamina about an axis through \(A\), parallel to \(BC\), is \(\frac{9}{2}ma^2\). [6]

The foot of the perpendicular from \(A\) to \(BC\) is \(D\). The lamina is free to rotate in a vertical plane about a fixed smooth horizontal axis which passes through \(D\) and is perpendicular to the plane of the lamina. The lamina is released from rest when \(DA\) makes an angle \(\alpha\) with the downward vertical. It is given that the moment of inertia of the lamina about an axis through \(D\), perpendicular to \(BC\) and in the plane of the lamina, is \(\frac{8}{3}ma^2\).

1. Find the angular acceleration of the lamina when \(DA\) makes an angle \(\theta\) with the downward vertical. [8]

Given that \(\alpha\) is small,

1. find an approximate value for the period of oscillation of the lamina about the vertical. [2]

A body consists of a uniform plane circular disc, of radius \(r\) and mass \(2m\), with a particle of mass \(3m\) attached to the circumference of the disc at the point \(P\). The line \(PQ\) is a diameter of the disc. The body is free to rotate in a vertical plane about a fixed smooth horizontal axis, \(L\), which is perpendicular to the plane of the disc and passes through \(Q\). The body is held with \(QP\) making an angle \(\beta\) with the downward vertical through \(Q\), where \(\sin \beta = 0.25\), and released from rest. Find the magnitude of the component, perpendicular to \(PQ\), of the force acting on the body at \(Q\) at the instant when it is released. [You may assume that the moment of inertia of the body about \(L\) is \(15mr^2\).] [6]

A uniform circular disc, of mass \(m\), radius \(a\) and centre \(O\), is free to rotate in a vertical plane about a fixed smooth horizontal axis. The axis passes through the mid-point \(A\) of a radius of the disc.

1. Find an equation of motion for the disc when the line \(AO\) makes an angle \(\theta\) with the downward vertical through \(A\). [5]
2. Hence find the period of small oscillations of the disc about its position of stable equilibrium. [2]

When the line \(AO\) makes an angle \(\theta\) with the downward vertical through \(A\), the force acting on the disc at \(A\) is \(\mathbf{F}\).

1. Find the magnitude of the component of \(\mathbf{F}\) perpendicular to \(AO\). [5]

A pendulum consists of a uniform rod \(PQ\), of mass \(3m\) and length \(2a\), which is rigidly fixed at its end \(Q\) to the centre of a uniform circular disc of mass \(m\) and radius \(a\). The rod is perpendicular to the plane of the disc. The pendulum is free to rotate about a fixed smooth horizontal axis \(L\) which passes through the end \(P\) of the rod and is perpendicular to the rod.

1. Show that the moment of inertia of the pendulum about \(L\) is \(\frac{33}{4}ma^2\). [5]

The pendulum is released from rest in the position where \(PQ\) makes an angle \(\alpha\) with the downward vertical. At time \(t\), \(PQ\) makes an angle \(\theta\) with the downward vertical.

1. Show that the angular speed, \(\dot{\theta}\), of the pendulum satisfies $$\dot{\theta}^2 = \frac{40g(\cos\theta - \cos\alpha)}{33a}$$ [4]
2. Hence, or otherwise, find the angular acceleration of the pendulum. [3]

Given that \(\alpha = \frac{\pi}{20}\) and that \(PQ\) has length \(\frac{8}{33}\) m,

1. find, to 3 significant figures, an approximate value for the angular speed of the pendulum \(0.2\) s after it has been released from rest. [5]

A uniform rod \(PQ\) of mass \(m\) and length \(3a\), is free to rotate about a fixed smooth horizontal axis \(L\), which passes through the end \(P\) of the rod and is perpendicular to the rod. The rod hangs at rest in equilibrium with \(Q\) vertically below \(P\). One end of a light inextensible string of length \(2a\) is attached to the rod at \(P\) and the other end is attached to a particle of mass \(3m\). The particle is held with the string taut, and horizontal and perpendicular to \(L\), and is then released. After colliding, the particle sticks to the rod forming a body \(B\).

1. Show that the moment of inertia of \(B\) about \(L\) is \(15ma^2\). [2]
2. Show that \(B\) first comes to instantaneous rest after it has turned through an angle \(\arccos\left(\frac{9}{25}\right)\). [10]

A body consists of a uniform plane circular disc, of radius \(r\) and mass \(2m\), with a particle of mass \(3m\) attached to the circumference of the disc at the point \(P\). The line \(PQ\) is a diameter of the disc. The body is free to rotate in a vertical plane about a fixed smooth horizontal axis, \(L\), which is perpendicular to the plane of the disc and passes through \(Q\). The body is held with \(QP\) making an angle \(\beta\) with the downward vertical through \(Q\), where \(\sin \beta = 0.25\), and released from rest. Find the magnitude of the component, perpendicular to \(PQ\), of the force acting on the body at \(Q\) at the instant when it is released. [You may assume that the moment of inertia of the body about \(L\) is \(15mr^2\).] [6]

1. A uniform lamina of mass \(m\) is in the shape of a triangle \(ABC\). The perpendicular distance of \(C\) from the line \(AB\) is \(h\). Prove, using integration, that the moment of inertia of the lamina about \(AB\) is \(\frac{1}{6}mh^2\). [7]
2. Deduce the radius of gyration of a uniform square lamina of side \(2a\), about a diagonal. [3]

The points \(X\) and \(Y\) are the mid-points of the sides \(RQ\) and \(RS\) respectively of a square \(PQRS\) of side \(2a\). A uniform lamina of mass \(M\) is in the shape of \(PQXYS\).

1. Show that the moment of inertia of this lamina about \(XY\) is \(\frac{79}{84}Ma^2\). [6]

A uniform rod is resting on two fixed supports at points \(A\) and \(B\). \(A\) lies at a distance \(x\) metres from one end of the rod. \(B\) lies at a distance \((x + 0.1)\) metres from the other end of the rod. The rod has length \(2L\) metres and mass \(m\) kilograms. The rod lies horizontally in equilibrium as shown in the diagram below. \includegraphics{figure_17} The reaction force of the support on the rod at \(B\) is twice the reaction force of the support on the rod at \(A\). Show that $$L - x = k$$ where \(k\) is a constant to be found. [4 marks]

A uniform lamina has the shape of the region enclosed by the curve \(y = x^2 + 1\) and the lines \(x = 0\), \(x = 4\) and \(y = 0\) The diagram below shows the lamina. \includegraphics{figure_5}

1. Find the coordinates of the centre of mass of the lamina, giving your answer in exact form. [4 marks]
2. The lamina is suspended from the point where the curve intersects the line \(x = 4\) and hangs in equilibrium. Find the angle between the vertical and the longest straight edge of the lamina, giving your answer correct to the nearest degree. [3 marks]

The finite region enclosed by the line \(y = kx\), the \(x\)-axis and the line \(x = 5\) is rotated through 360° around the \(x\) axis to form a solid cone.

1. 1. Use integration to show that the position of the centre of mass of the cone is independent of \(k\) [4 marks]
   2. State the distance between the base of the cone and its centre of mass. [1 mark]
2. State one assumption that you have made about the cone. [1 mark]
3. The plane face of the cone is placed on a rough inclined plane. The coefficient of friction between the cone and the plane is 0.8 The angle between the plane and the horizontal is gradually increased from 0° Find the range of values of \(k\) for which the cone slides before it topples. [4 marks]

\(ABCD\) is a uniform lamina in the shape of a kite with \(BA = BC = 0.37\) m, \(DA = DC = 0.91\) m and \(AC = 0.7\) m (see diagram). The centre of mass of \(ABCD\) is \(G\). \includegraphics{figure_4}

1. Explain why \(G\) lies on \(BD\). [1]
2. Show that the distance of \(G\) from \(B\) is \(0.36\) m. [4]

The lamina \(ABCD\) is freely suspended from the point \(A\).

1. Determine the acute angle that \(CD\) makes with the horizontal, stating which of \(C\) or \(D\) is higher. [4]

\includegraphics{figure_4} Fig. 4 shows a uniform lamina ABCDE such that ABDE is a rectangle and BCD is an isosceles triangle. AB = 5a, AE = 4a and BC = CD. The point F is the midpoint of BD and FC = a.

1. Find, in terms of \(a\), the exact distance of the centre of mass of the lamina from AE. [4]

The lamina is freely suspended from B and hangs in equilibrium.

1. Find the angle between AB and the downward vertical. [2]