Vertical circle: string becomes slack

6
\includegraphics[max width=\textwidth, alt={}, center]{5bb3bd29-a2eb-4124-802c-fb17b68c50e4-3_598_839_1480_706} One end of a light inextensible string of length 0.5 m is attached to a fixed point \(O\). A particle \(P\) of mass 0.3 kg is attached to the other end of the string. With the string taut and at an angle of \(60 ^ { \circ }\) to the upward vertical, \(P\) is projected with speed \(2 \mathrm {~ms} ^ { - 1 }\) (see diagram). \(P\) begins to move without air resistance in a vertical circle with centre \(O\). When the string makes an angle \(\theta\) with the upward vertical, the speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that \(v ^ { 2 } = 8.9 - 9.8 \cos \theta\).
2. Find the tension in the string in terms of \(\theta\).
3. \(P\) does not move in a complete circle. Calculate the angle through which \(O P\) turns before \(P\) leaves the circular path.

6
\includegraphics[max width=\textwidth, alt={}, center]{7e0f600a-18f1-458b-8549-27fca592b19c-4_497_524_276_804} A particle \(P\) of mass 0.4 kg is attached to one end of a light inextensible string of length 2 m . The other end of the string is attached to a fixed point \(O\). With the string taut the particle is travelling in a circular path in a vertical plane. The angle between the string and the downward vertical is \(\theta ^ { \circ }\) (see diagram). When \(\theta = 0\) the speed of \(P\) is \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. At the instant when the string is horizontal, find the speed of \(P\) and the tension in the string.
2. At the instant when the string becomes slack, find the value of \(\theta\).

7 One end of a light inextensible string of length 0.8 m is attached to a fixed point \(O\). A particle \(P\) of mass 0.3 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 \(5.6 \mathrm {~m} \mathrm {~s} ^ { - 1 } . P\) starts to move in a vertical circle with centre \(O\). The speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) 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 \(\theta\).
2. For the instant when the string becomes slack, find the value of \(\theta\) and the value of \(v\).
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. OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download from our public website (\href{http://www.ocr.org.uk}{www.ocr.org.uk}) after the live examination series.
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3 A particle P of mass 0.6 kg is connected to a fixed point O by a light inextensible string of length 1.25 m . When it is 1.25 m vertically below \(\mathrm { O } , \mathrm { P }\) is set in motion with horizontal velocity \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and then moves in part of a vertical circle with centre O and radius 1.25 m . When OP makes an angle \(\theta\) with the downward vertical, the speed of P is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), as shown in Fig. 3.1. \begin{figure}[h] \end{figure}

1. Show that \(v ^ { 2 } = 11.5 + 24.5 \cos \theta\).
2. Find the tension in the string in terms of \(\theta\).
3. Find the speed of P at the instant when the string becomes slack. A second light inextensible string, of length 0.35 m , is attached to P , and the other end of this string is attached to a point C which is 1.2 m vertically below O . The particle P now moves in a horizontal circle with centre C and radius 0.35 m , as shown in Fig. 3.2. The speed of P is \(1.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{023afdfb-21b6-40fe-9a09-e6769667ee7b-3_518_488_1701_826} \captionsetup{labelformat=empty} \caption{Fig. 3.2}
   \end{figure}
4. Find the tension in the string OP and the tension in the string CP.

3. \begin{figure}[h] \end{figure} A particle of mass \(m\) is suspended at a point \(A\) vertically below a fixed point \(O\) by a light inextensible string of length \(a\) as shown in Figure 2. The particle is given a horizontal velocity \(u\) and subsequently moves along a circular arc until it reaches the point \(B\) where the string becomes slack. Given that the point \(B\) is at a height \(\frac { 1 } { 2 } a\) above the level of \(O\),

1. show that \(\angle B O A = 120 ^ { \circ }\),
2. show that \(u ^ { 2 } = \frac { 7 } { 2 } g a\).

6. \begin{figure}[h] \end{figure} A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(l\). The other end of the string is attached to a fixed point \(O\). The particle is held with the string taut and \(O P\) horizontal. The particle is then projected vertically downwards with speed \(u\), where \(u ^ { 2 } = \frac { 9 } { 5 } \mathrm { gl }\). When \(O P\) has turned through an angle \(\alpha\) and the string is still taut, the speed of \(P\) is \(v\), as shown in Figure 5. At this instant the tension in the string is \(T\).

1. Show that \(T = 3 m g \sin \alpha + \frac { 9 } { 5 } m g\)
2. Find, in terms of \(g\) and \(l\), the speed of \(P\) at the instant when the string goes slack.
3. Find, in terms of \(l\), the greatest vertical height reached by \(P\) above the level of \(O\).