SPS SPS FM Mechanics (SPS FM Mechanics) 2021 January

1. A disc, of mass \(m\) and radius \(r\), rotates about an axis through its centre, perpendicular to the plane face of the disc. The angular speed of the disc is \(\omega\).
A possible model for the kinetic energy \(E\) of the disc is $$E = k m ^ { a } r ^ { b } \omega ^ { c }$$ where \(a\), \(b\) and \(c\) are constants and \(k\) is a dimensionless constant.
Find the values of \(a , b\) and \(c\).

2. The triangular region shown below is rotated through \(360 ^ { \circ }\) around the \(x\)-axis, to form a solid cone.
\includegraphics[max width=\textwidth, alt={}, center]{bab18666-3571-4906-ab2f-b85e87c6e8f4-3_316_716_479_671} The coordinates of the vertices of the triangle are \(( 0,0 ) , ( 8,0 )\) and \(( 0,4 )\).
All units are in centimetres.

1. State an assumption that you should make about the cone in order to find the position of its centre of mass.
   [0pt] [1 mark]
2. Using integration, prove that the centre of mass of the cone is 2 cm from its plane face.
   [0pt] [5 marks]
3. The cone is placed with its plane face on a rough board. One end of the board is lifted so that the angle between the board and the horizontal is gradually increased. Eventually the cone topples without sliding.
4. 1. Find the angle between the board and the horizontal when the cone topples, giving your answer to the nearest degree.
      [0pt] [2 marks]
5. (ii) Find the range of possible values for the coefficient of friction between the cone and the board.

3. \begin{figure}[h] \end{figure} Figure 1 represents the plan of part of a smooth horizontal floor, where \(W _ { 1 }\) and \(W _ { 2 }\) are two fixed parallel vertical walls. The walls are 3 metres apart. A particle lies at rest at a point \(O\) on the floor between the two walls, where the point \(O\) is \(d\) metres, \(0 < d \leqslant 3\), from \(W _ { 1 }\) At time \(t = 0\), the particle is projected from \(O\) towards \(W _ { 1 }\) with speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a direction perpendicular to the walls. The coefficient of restitution between the particle and each wall is \(\frac { 2 } { 3 }\)
The particle returns to \(O\) at time \(t = T\) seconds, having bounced off each wall once.

1. Show that \(T = \frac { 45 - 5 d } { 4 u }\) The value of \(u\) is fixed, the particle still hits each wall once but the value of \(d\) can now vary.
2. Find the least possible value of \(T\), giving your answer in terms of \(u\). You must give a reason for your answer. \section*{4.} A car of mass 600 kg pulls a trailer of mass 150 kg along a straight horizontal road. The trailer is connected to the car by a light inextensible towbar, which is parallel to the direction of motion of the car. The resistance to the motion of the trailer is modelled as a constant force of magnitude 200 N . At the instant when the speed of the car is \(v \mathrm {~ms} ^ { - 1 }\), the resistance to the motion of the car is modelled as a force of magnitude \(( 200 + \lambda v ) \mathrm { N }\), where \(\lambda\) is a constant. When the engine of the car is working at a constant rate of 15 kW , the car is moving at a constant speed of \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)
3. Show that \(\lambda = 8\) Later on, the car is pulling the trailer up a straight road inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac { 1 } { 15 }\)
   The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude 200 N at all times. At the instant when the speed of the car is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the resistance to the motion of the car from non-gravitational forces is modelled as a force of magnitude \(( 200 + 8 v ) \mathrm { N }\). The engine of the car is again working at a constant rate of 15 kW .
   When \(v = 10\), the towbar breaks. The trailer comes to instantaneous rest after moving a distance \(d\) metres up the road from the point where the towbar broke.
4. Find the acceleration of the car immediately after the towbar breaks.
5. Use the work-energy principle to find the value of \(d\).

5. \begin{figure}[h] \end{figure} 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. \section*{Numerical (calculator) integration is not acceptable in this question.} \begin{figure}[h] \end{figure} The shaded region \(O A B\) in Figure 2 is bounded by the \(x\)-axis, the line with equation \(x = 4\) and the curve with equation \(y = \frac { 1 } { 4 } ( x - 2 ) ^ { 3 } + 2\). The point \(A\) has coordinates \(( 4,4 )\) and the point \(B\) has coordinates \(( 4,0 )\). A uniform lamina \(L\) has the shape of \(O A B\). The unit of length on both axes is one centimetre. The centre of mass of \(L\) is at the point with coordinates \(( \bar { x } , \bar { y } )\). Given that the area of \(L\) is \(8 \mathrm {~cm} ^ { 2 }\),

1. show that \(\bar { y } = \frac { 8 } { 7 }\) The lamina is freely suspended from \(A\) and hangs in equilibrium with \(A B\) at an angle \(\theta ^ { \circ }\) to the downward vertical.
2. Find the value of \(\theta\).