Particle in hemispherical bowl

4 A particle \(P\) of mass \(m\) is moving in a horizontal circle with angular speed \(\omega\) on the smooth inner surface of a hemispherical shell of radius \(r\). The angle between the vertical and the normal reaction of the surface on \(P\) is \(\theta\).

1. Show that \(\cos \theta = \frac { \mathrm { g } } { \omega ^ { 2 } \mathrm { r } }\).
   The plane of the circular motion is at a height \(x\) above the lowest point of the shell. When the angular speed is doubled, the plane of the motion is at a height \(4 x\) above the lowest point of the shell.
2. Find \(x\) in terms of \(r\).

2 A hollow hemispherical bowl of radius \(a\) has a smooth inner surface and is fixed with its axis vertical. A particle \(P\) of mass \(m\) moves in horizontal circles on the inner surface of the bowl, at a height \(x\) above the lowest point of the bowl. The speed of \(P\) is \(\sqrt { \frac { 8 } { 3 } } \mathrm { ga }\). Find \(x\) in terms of \(a\).

2. \begin{figure}[h] \end{figure} A hemispherical bowl of internal radius \(6 r\) is fixed with its circular rim horizontal. The centre of the circular rim is \(O\) and the point \(A\) on the surface of the bowl is vertically below \(O\). A particle \(P\) moves in a horizontal circle, with centre \(C\), on the smooth inner surface of the bowl. The particle moves with constant angular speed \(\sqrt { \frac { g } { 4 r } }\). The point \(C\) lies on \(O A\), as shown in Figure 1. Find, in terms of \(r\), the distance \(O C\)

6. \begin{figure}[h] \end{figure} Figure 3 represents the path of a skier of mass 70 kg moving on a ski-slope \(A B C D\). The path lies in a vertical plane. From \(A\) to \(B\), the path is modelled as a straight line inclined at \(60 ^ { \circ }\) to the horizontal. From \(B\) to \(D\), the path is modelled as an arc of a vertical circle of radius 50 m . The lowest point of the \(\operatorname { arc } B D\) is \(C\). At \(B\), the skier is moving downwards with speed \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). At \(D\), the path is inclined at \(30 ^ { \circ }\) to the horizontal and the skier is moving upwards. By modelling the slope as smooth and the skier as a particle, find

1. the speed of the skier at \(C\),
2. the normal reaction of the slope on the skier at \(C\),
3. the speed of the skier at \(D\),
4. the change in the normal reaction of the slope on the skier as she passes \(B\). The model is refined to allow for the influence of friction on the motion of the skier.
5. State briefly, with a reason, how the answer to part (b) would be affected by using such a model. (No further calculations are expected.)

2
\includegraphics[max width=\textwidth, alt={}, center]{7fcedc6d-8dc1-4159-8a72-be0f6a3f659b-2_698_737_484_703} A particle \(P\) travels on a smooth surface whose vertical cross-section is in the form of two arcs of circles. The first arc \(A B\) is a quarter of a circle of radius \(\frac { 1 } { 8 } a\) and centre \(O\). The second arc \(B C\) is a quarter of a circle of radius \(a\) and centre \(Q\). The two arcs are smoothly joined at \(B\). The point \(Q\) is vertically below \(O\) and the two arcs are in the same vertical plane. The particle \(P\) is projected vertically downwards from \(A\) with speed \(u\). When \(P\) is on the \(\operatorname { arc } B C\), angle \(B Q P\) is \(\theta\) (see diagram). Given that \(P\) loses contact with the surface when \(\cos \theta = \frac { 5 } { 6 }\), find \(u\) in terms of \(a\) and \(g\).

2 A fixed hollow sphere with centre O has an inside radius of 2.7 m . A particle P of mass 0.4 kg moves on the smooth inside surface of the sphere. At first, P is moving in a horizontal circle with constant speed, and OP makes a constant angle of \(60 ^ { \circ }\) with the vertical (see Fig. 2.1). \begin{figure}[h] \end{figure}

1. Find the normal reaction acting on P .
2. Find the speed of P . The particle P is now placed at the lowest point of the sphere and is given an initial horizontal speed of \(9 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It then moves in part of a vertical circle. When OP makes an angle \(\theta\) with the upward vertical and P is still in contact with the sphere, the speed of P is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the normal reaction acting on P is \(R \mathrm {~N}\) (see Fig. 2.2). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{39e14918-5017-43c0-9b74-7c68717ad5f3-3_716_778_1653_696} \captionsetup{labelformat=empty} \caption{Fig. 2.2}
   \end{figure}
3. Find \(v ^ { 2 }\) in terms of \(\theta\).
4. Show that \(R = 4.16 - 11.76 \cos \theta\).
5. Find the speed of P at the instant when it leaves the surface of the sphere.

2 A hollow hemisphere has internal radius 2.5 m and is fixed with its rim horizontal and uppermost. The centre of the hemisphere is O . A small ball B of mass 0.4 kg moves in contact with the smooth inside surface of the hemisphere. At first, B is moving at constant speed in a horizontal circle with radius 1.5 m , as shown in Fig. 2.1. \begin{figure}[h] \end{figure}

1. Find the normal reaction of the hemisphere on \(B\).
2. Find the speed of \(\mathbf { B }\). The ball B is now released from rest on the inside surface at a point on the same horizontal level as O . It then moves in part of a vertical circle with centre O and radius 2.5 m , as shown in Fig. 2.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c93aed95-f655-45cb-805f-7114a15acccf-3_378_663_1427_740} \captionsetup{labelformat=empty} \caption{Fig. 2.2}
   \end{figure}
3. Show that, when \(B\) is at its lowest point, the normal reaction is three times the weight of \(B\). For an instant when the normal reaction is twice the weight of \(\mathbf { B }\), find
4. the speed of \(\mathbf { B }\),
5. the tangential component of the acceleration of \(\mathbf { B }\).

6. The diagram shows a slide, \(A B C\), at a water park. The shape of the slide may be modelled by two circular arcs, \(A B\) and \(B C\), in the same vertical plane. Arc \(A B\) has radius 7 m and subtends an angle \(\alpha\) at its centre \(D\), where \(\cos \alpha = \frac { 9 } { 10 }\). Arc \(B C\) has radius 5 m and subtends an angle of \(45 ^ { \circ }\) at its centre, \(O\). The straight line \(D B O\) is vertical.
\includegraphics[max width=\textwidth, alt={}, center]{b6b77d92-acb6-4cc3-8328-1899ed4a87cd-7_1171_940_630_552} Users of the slide are required to sit in a rubber ring and are released from rest at point \(A\). A girl decides to use the slide. The combined mass of the girl and the rubber ring is 50 kg .

1. When the rubber ring is at a point \(P\) on the circular arc \(B C\), its speed is \(v \mathrm {~ms} ^ { - 1 }\) and \(O P\) makes an angle \(\theta\) with the upward vertical.
   1. Show that \(v ^ { 2 } = 111.72 - 98 \cos \theta\).
   2. Find, in terms of \(\theta\), the reaction between the rubber ring and the slide at \(P\).
   3. Show that, according to this model, the rubber ring loses contact with the slide before reaching \(C\).
   4. In reality, there will be resistive forces opposing the motion of the rubber ring. Explain how this fact will affect your answer to (iii).
2. Show that the rubber ring will remain in contact with the slide along the arc \(A B\).

8 The diagram shows part of a water park slide, \(A B C\).
The slide is in the shape of two circular arcs, \(A B\) and \(B C\), each of radius \(r\).
The point \(A\) is at a height of \(\frac { r } { 4 }\) above \(B\).
The circular \(\operatorname { arc } B C\) has centre \(O\) and \(B\) is vertically above \(O\).
These points are joined as shown in the diagram.
\includegraphics[max width=\textwidth, alt={}, center]{4fdb2637-6368-422c-99da-85b80efe31c5-12_590_1173_756_443} A child starts from rest at \(A\), moves along the slide past the point \(B\) and then loses contact with the slide at a point \(D\). The angle between the vertical, \(O B\), and \(O D\) is \(\theta\)
Assume that the slide is smooth. 8

1. Show that the speed \(v\) of the child at \(D\) is given by \(v = \sqrt { \frac { g r } { 2 } ( 5 - 4 \cos \theta ) }\), where \(g\) is the acceleration due to gravity. 8
2. Find \(\theta\), giving your answer to the nearest degree.
   8
3. A refined model takes into account air resistance. Explain how taking air resistance into account would affect your answer to part (b).
   [0pt] [2 marks]
   8
4. In reality the slide is not smooth. It has a surface with the same coefficient of friction between the slide and the child for its entire length. Explain why the frictional force experienced by the child is not constant.
   [0pt] [1 mark]