Vertical circle: string becomes slack

A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\) and \(P\) is held with the string taut and horizontal. The particle \(P\) is projected vertically downwards with speed \(\sqrt{(2ag)}\) so that it begins to move along a circular path. The string becomes slack when \(OP\) makes an angle \(\theta\) with the upward vertical through \(O\).

1. Show that \(\cos \theta = \frac{2}{3}\). [5]
2. Find the greatest height, above the horizontal through \(O\), reached by \(P\) in its subsequent motion. [4]

One end of a light inextensible string of length \(a\) is attached to a fixed point \(O\). A particle of mass \(m\) is attached to the other end of the string. The particle is held at the point \(A\) with the string taut. The angle between \(OA\) and the downward vertical is equal to \(\alpha\), where \(\cos \alpha = \frac{4}{5}\). The particle is projected from \(A\), perpendicular to the string in an upwards direction, with a speed \(\sqrt{3ga}\). It then moves along a circular path in a vertical plane. The string first goes slack when it makes an angle \(\theta\) with the upward vertical through \(O\). Find the value of \(\cos \theta\). [5]

A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\). The particle \(P\) is held at the point \(A\), where \(OA\) makes an angle \(\theta\) with the downward vertical through \(O\), and with the string taut. The particle \(P\) is projected perpendicular to \(OA\) in an upwards direction with speed \(u\). It then starts to move along a circular path in a vertical plane. The string goes slack when \(P\) is at \(B\), where angle \(AOB\) is \(90°\) and the speed of \(P\) is \(\sqrt{\frac{1}{3}ag}\).

1. Find the value of \(\sin\theta\). [2]
2. Find, in terms of \(m\) and \(g\), the tension in the string when \(P\) is at \(A\). [5]

A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\). The particle \(P\) is held at the point \(A\), where \(OA\) makes an angle \(\alpha\) with the downward vertical through \(O\), and with the string taut. The particle \(P\) is projected perpendicular to \(OA\) in an upwards direction with speed \(\sqrt{3ag}\). It then starts to move along a circular path in a vertical plane. The string goes slack when \(P\) is at \(B\), where \(OB\) makes an angle \(\theta\) with the upward vertical. Given that \(\cos \alpha = \frac{3}{5}\), find the value of \(\cos \theta\). [4]

A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\). When the particle is hanging vertically below \(O\), it is projected horizontally with speed \(u\) so that it begins to move along a circular path. When \(P\) is at the lowest point of its motion, the tension in the string is \(T\). When \(OP\) makes an angle \(\theta\) with the upward vertical, the tension in the string is \(S\).

1. Show that \(S = T - 3mg(1 + \cos\theta)\). [5]
2. Given that \(u = \sqrt{4ag}\), find the value of \(\cos\theta\) when the string goes slack. [2]

One end of a light inextensible string of length \(l\) is attached to a fixed point \(A\). The other end is attached to a particle \(P\) of mass \(m\), which is held at a point \(B\) with the string taut and \(AP\) making an angle arccos \(\frac{1}{4}\) with the downward vertical. The particle is released from rest. When \(AP\) makes an angle \(\theta\) with the downward vertical, the string is taut and the tension in the string is \(T\).

1. Show that $$T = 3mg \cos \theta - \frac{mg}{2}.$$ [6]

\includegraphics{figure_3} At an instant when \(AP\) makes an angle of \(60°\) to the downward vertical, \(P\) is moving upwards, as shown in Figure 3. At this instant the string breaks. At the highest point reached in the subsequent motion, \(P\) is at a distance \(d\) below the horizontal through \(A\).

1. Find \(d\) in terms of \(l\). [5]

A particle \(P\) is attached to one end of a light inextensible string of length \(l\) m. The other end of the string is attached to a fixed point \(O\). When \(P\) is hanging at rest vertically below \(O\), it is given a horizontal speed \(u\) ms\(^{-1}\) and starts to move in a vertical circle. Given that the string becomes slack when it makes an angle of 120° with the downward vertical through \(O\),

1. show that \(u^2 = \frac{7gl}{2}\). [10 marks]
2. Find, in terms of \(l\), the greatest height above \(O\) reached by \(P\) in the subsequent motion. [7 marks]

A particle \(P\) of mass 0.4 kg hangs by a light, inextensible string of length 20 cm whose other end is attached to a fixed point \(O\). It is given a horizontal velocity of 1.4 ms\(^{-1}\) so that it begins to move in a vertical circle. If in the ensuing motion the string makes an angle of \(\theta\) with the downward vertical through \(O\), show that

1. \(\theta\) cannot exceed 60°, [6 marks]
2. the tension, \(T\) N, in the string is given by \(T = 3.92(3 \cos \theta - 1)\). [4 marks]

If the string breaks when \(\cos \theta = \frac{3}{5}\) and \(P\) is ascending,

1. find the greatest height reached by \(P\) above the initial point of projection. [5 marks]

One end of a light inextensible string of length \(0.8\) m is attached to a fixed point \(O\). A particle \(P\) of mass \(0.5\) kg is attached to the other end of the string. \(P\) is projected horizontally from the point \(0.8\) m vertically below \(O\) with speed \(5.6\) m s\(^{-1}\). \(P\) starts to move in a vertical circle with centre \(O\). The speed of \(P\) is \(v\) m s\(^{-1}\) when the string makes an angle \(\theta\) with the downward vertical.

1. While the string remains taut, show that \(v^2 = 15.68(1 + \cos \theta)\), and find the tension in the string in terms of \(\theta\). [7]
2. For the instant when the string becomes slack, find the value of \(\theta\) and the value of \(v\). [3]
3. Find, in either order, the speed of \(P\) when it is at its greatest height after the string becomes slack, and the greatest height reached by \(P\) above its point of projection. [4]

\includegraphics{figure_5} One end of a light inextensible string of length \(a\) is attached to a fixed point \(O\). A particle \(P\) of mass \(m\) is attached to the other end of the string and hangs at rest. \(P\) is then projected horizontally from this position with speed \(2\sqrt{ag}\). When the string makes an angle \(\theta\) with the upward vertical \(P\) has speed \(v\) (see diagram). The tension in the string is \(T\).

1. Find an expression for \(T\) in terms of \(m\), \(g\) and \(\theta\), and hence find the height of \(P\) above its initial level when the string becomes slack. [6]

\(P\) is now projected horizontally from the same initial position with speed \(U\).

1. Find the set of values of \(U\) for which the string does not remain taut in the subsequent motion. [5]

A small sphere, of mass \(m\), is attached to one end of a light inextensible string of length \(a\) The other end of the string is attached to a fixed point \(O\) The sphere is at rest in equilibrium directly below \(O\) when it is struck, giving it a horizontal impulse of magnitude \(mU\) After the impulse, the sphere follows a circular path in a vertical plane containing the point \(O\) until the string becomes slack at the point \(C\) At \(C\) the string makes an angle of 30° with the upward vertical through \(O\), as shown in the diagram below. \includegraphics{figure_9}

1. Show that $$U^2 = \frac{ag}{2}\left(4 + 3\sqrt{3}\right)$$ where \(g\) is the acceleration due to gravity. [6 marks]
2. With reference to any modelling assumptions that you have made, explain why giving your answer as an inequality would be more appropriate, and state this inequality. [2 marks]

A particle \(P\), of mass \(m\) kg, is attached to one end of a light inextensible string of length \(l\) m. The other end of the string is attached to a fixed point \(O\). Initially, \(P\) is held at rest with the string just taut and making an angle of 60° with the downward vertical. It is then given a velocity \(u\text{ ms}^{-1}\) perpendicular to the string in a downward direction.

1. 1. When the string makes an angle \(\theta\) with the downward vertical, the velocity of the particle is \(v\) and the tension in the string is \(T\). Find an expression for \(T\) in terms of \(m\), \(l\), \(v^2\) and \(\theta\).
   2. Given that \(P\) describes complete circles in the subsequent motion, show that \(u^2 > 4lg\). [10]
2. Given that now \(u^2 = 3lg\), find the position of the string when circular motion ceases. Briefly describe the motion of \(P\) after circular motion has ceased. [3]
3. The string is replaced by a light rigid rod. Given that \(P\) describes complete circles in the subsequent motion, show that \(u^2 > klg\), where \(k\) is to be determined. [2]

One end of a light inextensible string of length \(l\) is attached to a fixed point \(O\). The other end is attached to a particle \(P\) of mass \(m\). The particle hangs at rest vertically below \(O\). The particle is then given a horizontal speed \(u\).

1. 1. Show that when \(OP\) has turned through an angle \(\theta\) the tension in the string is given by $$T = mg(3\cos \theta - 2) + \frac{mu^2}{l}$$ as long as the string remains taut. [5]
   2. Deduce that \(u^2 \geq 5gl\) in order for the particle to perform complete circles. [1]
2. 1. In the case \(u^2 = 3gl\), find the angle that \(OP\) makes with the downward vertical at \(O\) at the instant when the string becomes slack. [2]
   2. Describe the nature of the motion while the string is slack. [1]