Particle in hemispherical bowl

A question is this type if and only if a particle moves on the inner surface of a fixed hemispherical bowl or shell, either in horizontal circles or vertical motion.

10 questions · Standard +0.8

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CAIE Further Paper 3 2020 November Q4
7 marks Standard +0.8
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 \(r\).
Edexcel M3 2022 June Q2
9 marks Standard +0.3
2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2e837bb9-4ada-4f0f-8b21-2730611335f2-04_390_515_246_772} \captionsetup{labelformat=empty} \caption{Figure 1}
\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 \(O C\)
Edexcel M3 2004 June Q6
15 marks Standard +0.8
6. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 3} \includegraphics[alt={},max width=\textwidth]{b9e9b91c-7e6d-4b84-9f0e-180b626887c2-4_460_799_301_657}
\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.)
CAIE FP2 2013 June Q2
8 marks Challenging +1.8
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 \(g\).
OCR MEI M3 2007 June Q2
18 marks Standard +0.3
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]
\includegraphics[alt={},max width=\textwidth]{39e14918-5017-43c0-9b74-7c68717ad5f3-3_655_666_488_696} \captionsetup{labelformat=empty} \caption{Fig. 2.1}
\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.
OCR MEI M3 2010 June Q2
18 marks Standard +0.3
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]
\includegraphics[alt={},max width=\textwidth]{c93aed95-f655-45cb-805f-7114a15acccf-3_392_661_529_742} \captionsetup{labelformat=empty} \caption{Fig. 2.1}
\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 \(\mathbf { B }\).
AQA Further Paper 3 Mechanics Specimen Q8
8 marks Challenging +1.8
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]
OCR MEI Further Mechanics Major 2023 June Q10
16 marks Challenging +1.8
\includegraphics{figure_10} A hollow sphere has centre O and internal radius \(r\). A bowl is formed by removing part of the sphere. The bowl is fixed to a horizontal floor, with its circular rim horizontal and the centre of the rim vertically above O. The point A lies on the rim of the bowl such that AO makes an angle of \(30°\) with the horizontal (see diagram). A particle P of mass \(m\) is projected from A, with speed \(u\), where \(u > \sqrt{\frac{gr}{2}}\), in a direction perpendicular to AO and moves on the smooth inner surface of the bowl. The motion of P takes place in the vertical plane containing O and A. The particle P passes through a point B on the inner surface, where OB makes an acute angle \(\theta\) with the vertical.
  1. Determine, in terms of \(m\), \(g\), \(u\), \(r\) and \(\theta\), the magnitude of the force exerted on P by the bowl when P is at B. [7]
The difference between the magnitudes of the force exerted on P by the bowl when P is at points A and B is \(4mg\).
  1. Determine, in terms of \(r\), the vertical distance of B above the floor. [4]
It is given that when P leaves the inner surface of the bowl it does not fall back into the bowl.
  1. Show that \(u^2 > 2gr\). [5]
SPS SPS FM Mechanics 2021 January Q5
6 marks Standard +0.3
\includegraphics{figure_3} A hemispherical shell of radius \(a\) is fixed with its rim uppermost and horizontal. A small bead, \(B\), is moving with constant angular speed, \(\omega\), in a horizontal circle on the smooth inner surface of the shell. The centre of the path of \(B\) is at a distance \(\frac{1}{4}a\) vertically below the level of the rim of the hemisphere, as shown in Figure 1. Find the magnitude of \(\omega\), giving your answer in terms of \(a\) and \(g\). [6]
SPS SPS FM Mechanics 2026 January Q4
6 marks Standard +0.3
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 \(\sqrt{\frac{g}{2}a}\). Find \(x\) in terms of \(a\). [6]