Potential energy with elastic strings/springs

6 \includegraphics[max width=\textwidth, alt={}, center]{fb79f949-567c-4dbb-8533-7b7278cad21c-3_200_639_1754_753} A particle \(P\) of mass 1.6 kg is attached to one end of each of two light elastic strings. The other ends of the strings are attached to fixed points \(A\) and \(B\) which are 2 m apart on a smooth horizontal table. The string attached to \(A\) has natural length 0.25 m and modulus of elasticity 4 N , and the string attached to \(B\) has natural length 0.25 m and modulus of elasticity 8 N . The particle is held at the mid-point \(M\) of \(A B\) (see diagram).

1. Find the tensions in the strings.
2. Show that the total elastic potential energy in the two strings is 13.5 J . \(P\) is released from rest and in the subsequent motion both strings remain taut. The displacement of \(P\) from \(M\) is denoted by \(x \mathrm {~m}\). Find
3. the initial acceleration of \(P\),
4. the non-zero value of \(x\) at which the speed of \(P\) is zero. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{fb79f949-567c-4dbb-8533-7b7278cad21c-4_529_542_269_804} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure} A uniform solid body has a cross-section as shown in Fig. 1.
5. Show that the centre of mass of the body is 2.5 cm from the plane face containing \(O B\) and 3.5 cm from the plane face containing \(O A\).
6. The solid is placed on a rough plane which is initially horizontal. The coefficient of friction between the solid and the plane is \(\mu\).
   1. \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{fb79f949-567c-4dbb-8533-7b7278cad21c-4_332_469_1320_918} \captionsetup{labelformat=empty} \caption{Fig. 2}
      \end{figure} The solid is placed with \(O A\) in contact with the plane, and then the plane is tilted so that \(O A\) lies along a line of greatest slope with \(A\) higher than \(O\) (see Fig. 2). When the angle of inclination is sufficiently great the solid starts to topple (without sliding). Show that \(\mu > \frac { 5 } { 7 }\).
      [0pt] [5]
   2. \includegraphics[max width=\textwidth, alt={}, center]{fb79f949-567c-4dbb-8533-7b7278cad21c-4_291_465_1987_918} Instead, the solid is placed with \(O B\) in contact with the plane, and then the plane is tilted so that \(O B\) lies along a line of greatest slope with \(B\) higher than \(O\) (see Fig. 3). When the angle of inclination is sufficiently great the solid starts to slide (without toppling). Find another inequality for \(\mu\).

4. \begin{figure}[h] \end{figure} Four identical uniform rods, each of mass \(m\) and length \(2 a\), are freely jointed to form a rhombus \(A B C D\). The rhombus is suspended from \(A\) and is prevented from collapsing by an elastic string which joins \(A\) to \(C\), with \(\angle B A D = 2 \theta , 0 \leq \theta \leq \frac { 1 } { 3 } \pi\), as shown in Fig. 2. The natural length of the elastic string is \(2 a\) and its modulus of elasticity is \(4 m g\).

1. Show that the potential energy, \(V\), of the system is given by $$V = 4 m g a \left[ ( 2 \cos \theta - 1 ) ^ { 2 } - 2 \cos \theta \right] + \text { constant } .$$
2. Hence find the non-zero value of \(\theta\) for which the system is in equilibrium.
3. Determine whether this position of equilibrium is stable or unstable.

7. \begin{figure}[h] \end{figure} Figure 3 shows a framework \(A B C\), consisting of two uniform rods rigidly joined together at \(B\) so that \(\angle A B C = 90 ^ { \circ }\). The rod \(A B\) has length \(2 a\) and mass \(4 m\), and the rod \(B C\) has length \(a\) and mass \(2 m\). The framework is smoothly hinged at \(A\) to a fixed point, so that the framework can rotate in a fixed vertical plane. One end of a light elastic string, of natural length \(2 a\) and modulus of elasticity \(3 m g\), is attached to \(A\). The string passes through a small smooth ring \(R\) fixed at a distance \(2 a\) from \(A\), on the same horizontal level as \(A\) and in the same vertical plane as the framework. The other end of the string is attached to \(B\). The angle \(A R B\) is \(\theta\), where \(0 < \theta < \frac { \pi } { 2 }\).

1. Show that the potential energy \(V\) of the system is given by $$V = 8 a m g \sin 2 \theta + 5 a m g \cos 2 \theta + \text { constant }$$
2. Find the value of \(\theta\) for which the system is in equilibrium.
3. Determine the stability of this position of equilibrium.

6. \begin{figure}[h] \end{figure} Figure 3 shows a uniform rod \(A B\), of length \(2 l\) and mass \(4 m\). A particle of mass \(2 m\) is attached to the rod at \(B\). The rod can turn freely in a vertical plane about a fixed smooth horizontal axis through \(A\). One end of a light elastic spring, of natural length \(2 l\) and modulus of elasticity \(k m g\), where \(k > 4\), is attached to the rod at \(B\). The other end of the spring is attached to a fixed point \(C\) which is vertically above \(A\), where \(A C = 2 l\). The angle \(B A C\) is \(2 \theta\), where \(\frac { \pi } { 6 } < \theta \leqslant \frac { \pi } { 2 }\)

1. Show that the potential energy of the system is $$4 m g l \left\{ ( k - 4 ) \sin ^ { 2 } \theta - k \sin \theta \right\} + \text { constant }$$ Given that there is a position of equilibrium with \(\theta \neq \frac { \pi } { 2 }\)
2. show that \(k > 8\) Given that \(k = 10\)
3. determine the stability of this position of equilibrium.

7. \begin{figure}[h] \end{figure} Figure 2 shows four uniform rods, each of mass \(m\) and length \(2 a\). The rods are freely hinged at their ends to form a rhombus \(A B C D\). Point \(A\) is attached to a fixed point on a ceiling and the rhombus hangs freely with \(C\) vertically below \(A\). A light elastic spring of natural length \(2 a\) and modulus of elasticity \(7 m g\) connects the points \(A\) and \(C\). A particle of mass \(3 m\) is attached to point \(C\).

1. Show that, when \(A D\) is at an angle \(\theta\) to the downward vertical, the potential energy \(V\) of the system is given by $$V = 28 m g a \cos ^ { 2 } \theta - 48 m g a \cos \theta + \text { constant }$$ Given that \(\theta > 0\)
2. find the value of \(\theta\) for which the system is in equilibrium,
3. determine the stability of this position of equilibrium.

1. \begin{figure}[h] \end{figure} A uniform rod \(A B\) has mass \(m\) and length 4a. The end \(A\) of the rod is freely hinged to a fixed point. One end of a light elastic string, of natural length \(a\) and modulus \(\frac { 1 } { 4 } m g\), is attached to the end \(B\) of the rod. The other end of the string is attached to a small light smooth ring \(R\). The ring can move freely on a smooth horizontal wire which is fixed at a height \(a\) above \(A\), and in a vertical plane through \(A\). The angle between the rod and the horizontal is \(\theta\), where \(0 < \theta < \frac { \pi } { 2 }\), as shown in Figure 1. Given that the elastic string is vertical,

1. show that the potential energy of the system is $$2 m g a \left( \sin ^ { 2 } \theta - \sin \theta \right) + \text { constant }$$
2. Show that when \(\theta = \frac { \pi } { 6 }\) the rod is in stable equilibrium.

6 \includegraphics[max width=\textwidth, alt={}, center]{fb9e4e4a-953b-4e52-858e-438b4009e79c-3_428_595_221_806} A uniform rod \(A B\), of mass \(m\) and length \(2 a\), is free to rotate in a vertical plane about a fixed horizontal axis through \(A\). A light elastic string has natural length \(a\) and modulus of elasticity \(m g\); one end is attached to \(B\) and the other end is attached to a light ring \(R\) which can slide along a smooth vertical wire. The wire is in the same vertical plane as \(A B\), and is at a distance \(a\) from \(A\). The rod \(A B\) makes an angle \(\theta\) with the upward vertical, where \(0 < \theta < \frac { 1 } { 2 } \pi\) (see diagram).

1. Give a reason why the string \(R B\) is always horizontal.
2. By considering potential energy, find the value of \(\theta\) for which the system is in equilibrium.
3. Determine whether this position of equilibrium is stable or unstable.

2 A uniform rigid rod AB of mass \(m\) and length \(4 a\) is freely hinged at the end A to a horizontal rail. The end B is attached to a light elastic string BC of modulus \(\frac { 1 } { 2 } m g\) and natural length \(a\). The end C of the string is attached to a ring which is small, light and smooth. The ring can slide along the rail and is always vertically above B . The angle that AB makes below the rail is \(\theta\). The system is shown in Fig. 2. \begin{figure}[h] \end{figure}

1. Find the potential energy, \(V\), of the system when the string is stretched and show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 4 m g a \cos \theta ( 2 \sin \theta - 1 )$$
2. Hence find any positions of equilibrium of the system and investigate their stability.

3 A uniform rigid rod AB of length \(2 a\) and mass \(m\) is smoothly hinged to a fixed point at A so that it can rotate freely in a vertical plane. A light elastic string of modulus \(\lambda\) and natural length \(a\) connects the midpoint of AB to a fixed point C which is vertically above A with \(\mathrm { AC } = a\). The rod makes an angle \(2 \theta\) with the upward vertical, where \(\frac { 1 } { 3 } \pi \leqslant 2 \theta \leqslant \pi\). This is shown in Fig. 3. \begin{figure}[h] \end{figure}

1. Find the potential energy, \(V\), of the system relative to A in terms of \(m , \lambda , a\) and \(\theta\). Show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 2 a \cos \theta ( 2 \lambda \sin \theta - 2 m g \sin \theta - \lambda ) .$$ Assume now that the system is set up so that the result (*) continues to hold when \(\pi < 2 \theta \leqslant \frac { 5 } { 3 } \pi\).
2. In the case \(\lambda < 2 m g\), show that there is a stable position of equilibrium at \(\theta = \frac { 1 } { 2 } \pi\). Show that there are no other positions of equilibrium in this case.
3. In the case \(\lambda > 2 m g\), find the positions of equilibrium for \(\frac { 1 } { 3 } \pi \leqslant 2 \theta \leqslant \frac { 5 } { 3 } \pi\) and determine for each whether the equilibrium is stable or unstable, justifying your conclusions.

2 A uniform rod AB of length 0.5 m and mass 0.5 kg is freely hinged at A so that it can rotate in a vertical plane. Attached at B are two identical light elastic strings BC and BD each of natural length 0.5 m and stiffness \(2 \mathrm {~N} \mathrm {~m} ^ { - 1 }\). The ends C and D are fixed at the same horizontal level as A and with \(\mathrm { AC } = \mathrm { CD } = 0.5 \mathrm {~m}\). The system is shown in Fig. 2.1 with the angle \(\mathrm { BAC } = \theta\). You may assume that \(\frac { 1 } { 3 } \pi \leqslant \theta \leqslant \frac { 5 } { 3 } \pi\) so that both strings are taut. \begin{figure}[h] \end{figure}

1. Show that the length of BC in metres is \(\sin \frac { 1 } { 2 } \theta\).
2. Find the potential energy, \(V \mathrm {~J}\), of the system relative to AD in terms of \(\theta\). Hence show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 1.5 \sin \theta - 1.225 \cos \theta - \frac { 0.5 \sin \theta } { \sqrt { 1.25 - \cos \theta } } - 0.5 \cos \frac { 1 } { 2 } \theta .$$
3. Fig. 2.2 shows a graph of the function \(\mathrm { f } ( \theta ) = 1.5 \sin \theta - 1.225 \cos \theta - \frac { 0.5 \sin \theta } { \sqrt { 1.25 - \cos \theta } } - 0.5 \cos \frac { 1 } { 2 } \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-2_453_1264_2021_397} \captionsetup{labelformat=empty} \caption{Fig. 2.2}
   \end{figure} Use the graph both to estimate, correct to 1 decimal place, the values of \(\theta\) for which the system is in equilibrium and also to determine their stability.

3 A uniform rigid rod AB of mass \(m\) and length \(2 a\) is freely hinged to a horizontal floor at A . The end B is attached to a light elastic string of modulus \(\lambda\) and natural length \(5 a\). The other end of the string is attached to a small, light, smooth ring C which can slide along a horizontal rail. The rail is a distance \(7 a\) above the floor and C is always vertically above B . The angle that AB makes with the floor is \(\theta\). The system is shown in Fig. 3. \begin{figure}[h] \end{figure}

1. Find the potential energy, \(V\), of the system and hence show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = a \cos \theta \left( m g - \frac { 4 \lambda } { 5 } ( 1 - \sin \theta ) \right) .$$
2. Show that there is a position of equilibrium when \(\theta = \frac { 1 } { 2 } \pi\) and determine whether or not it is stable. There are two further positions of equilibrium when \(0 < \theta < \pi\).
3. Find the magnitude of the tension in the string and the vertical force of the hinge on the rod in these positions.
4. Show that \(\lambda > \frac { 5 m g } { 4 }\).
5. Show that these positions of equilibrium are stable.

2 Fig. 2 shows a system in a vertical plane. A uniform rod AB of length \(2 a\) and mass \(m\) is freely hinged at A . The angle that AB makes with the horizontal is \(\theta\), where \(- \frac { 2 } { 3 } \pi < \theta < \frac { 2 } { 3 } \pi\). Attached at B is a light spring BC of natural length \(a\) and stiffness \(\frac { m g } { a }\). The other end of the spring is attached to a small light smooth ring C which can slide freely along a vertical rail. The rail is at a distance of \(a\) from A and the spring is always horizontal. \begin{figure}[h] \end{figure}

1. Find the potential energy, \(V\), of the system and hence show that \(\frac { \mathrm { d } V } { \mathrm {~d} \theta } = m g a \cos \theta ( 1 - 4 \sin \theta )\).
2. Hence find the positions of equilibrium of the system and investigate their stability.

\includegraphics{figure_3} A uniform rod \(AB\), of mass \(m\) and length \(2a\), can rotate freely in a vertical plane about a fixed smooth horizontal axis through \(A\). The fixed point \(C\) is vertically above \(A\) and \(AC = 4a\). A light elastic string, of natural length \(2a\) and modulus of elasticity \(\frac{1}{4}mg\), joins \(B\) to \(C\). The rod \(AB\) makes an angle \(\theta\) with the upward vertical at \(A\), as shown in Fig. 3.

1. Show that the potential energy of the system is $$-mga[\cos \theta + \sqrt{(5 - 4 \cos \theta)}] + \text{constant}.$$ [9]
2. Hence determine the values of \(\theta\) for which the system is in equilibrium. [6]

\includegraphics{figure_1} Figure 1 shows a uniform rod \(AB\), of mass \(m\) and length \(4a\), resting on a smooth fixed sphere of radius \(a\). A light elastic string, of natural length \(a\) and modulus of elasticity \(\frac{1}{4}mg\), has one end attached to the lowest point \(C\) of the sphere and the other end attached to \(A\). The points \(A\), \(B\) and \(C\) lie in a vertical plane with \(\angle BAC = 2\theta\), where \(\theta < \frac{\pi}{4}\). Given that \(AC\) is always horizontal,

1. show that the potential energy of the system is $$\frac{mga}{8}(16\sin 2\theta + 3\cot^2 \theta - 6\cot \theta) + \text{constant}.$$ [7]
2. show that there is a value of \(\theta\) for which the system is in equilibrium such that \(0.535 < \theta < 0.545\). [6]
3. Determine whether this position of equilibrium is stable or unstable. [3]

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_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_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]