Particle attached to lamina - find mass/position

4. \begin{figure}[h] \end{figure} The uniform plane lamina \(A B C D E F\) shown in Figure 1 is made from two identical rhombuses. Each rhombus has sides of length \(a\) and angle \(B A D =\) angle \(D A F = \theta\). The centre of mass of the lamina is \(0.9 a\) from \(A\).

1. Show that \(\cos \theta = 0.8\) The weight of the lamina is \(W\). A particle of weight \(k W\) is fixed to the lamina at the point \(A\). The lamina is freely suspended from \(B\) and hangs in equilibrium with \(D A\) horizontal.
2. Find the value of \(k\).

2. \begin{figure}[h] \end{figure} A uniform lamina is in the shape of a trapezium \(A B C D\) with \(A B = a , D A = D C = 2 a\) and angle \(B A D =\) angle \(A D C = 90 ^ { \circ }\), as shown in Figure 1. The centre of mass of the lamina is at the point \(G\).

1. 1. Show that the distance of \(G\) from \(A B\) is \(\frac { 10 a } { 9 }\).
   2. Find the distance of \(G\) from \(A D\). The mass of the lamina is \(3 M\). A particle of mass \(k M\) is now attached to the lamina at \(B\). The lamina is freely suspended from the midpoint of \(A D\) and hangs in equilibrium with \(A D\) horizontal.
2. Find the value of \(k\).

4. \begin{figure}[h] \end{figure} The uniform lamina \(L\), shown shaded in Figure 1, is formed by removing a circular disc of radius \(2 a\) from a uniform circular disc of radius \(4 a\). The larger disc has centre \(O\) and diameter \(A B\). The radius \(O D\) is perpendicular to \(A B\). The smaller disc has centre \(C\), where \(C\) is on \(A B\) and \(B C = 3 a\)

1. Show that the centre of mass of \(L\) is \(\frac { 13 } { 3 } a\) from \(B\). The mass of \(L\) is \(M\) and a particle of mass \(k M\) is attached to \(L\) at \(B\). When \(L\), with the particle attached, is freely suspended from point \(D\), it hangs in equilibrium with \(A\) higher than \(B\) and \(A B\) at an angle \(\theta\) to the horizontal, where \(\tan \theta = \frac { 3 } { 4 }\)
2. Find the value of \(k\).

6. \begin{figure}[h] \end{figure} The uniform lamina \(A B C D\) is a square with sides of length \(6 a\). The point \(E\) is the midpoint of side \(A D\). The triangle \(C D E\) is removed from the square to form the uniform lamina \(L\), shown in Figure 3. The centre of mass of \(L\) is at the point \(G\).

1. Show that the distance of \(G\) from the side \(A B\) is \(\frac { 7 } { 3 } a\).
2. Find the distance of \(G\) from the side \(A E\). The mass of \(L\) is \(M\). A particle of mass \(k M\) is attached to \(L\) at the point \(E\). The lamina, with the particle attached, is freely suspended from \(A\) and hangs in equilibrium with the diagonal \(A C\) vertical.
3. Find the value of \(k\).

4. \begin{figure}[h] \end{figure} The uniform lamina \(L\), shown shaded in Figure 2, is formed by removing the square \(P Q R V\), of side \(2 a\), and the square \(R S T U\), of side \(4 a\), from a uniform square lamina \(A B C D\), of side \(8 a\). The lines \(Q R U\) and \(V R S\) are straight. The side \(A D\) is parallel to \(P V\) and the side \(A B\) is parallel to \(P Q\). The distance between \(A D\) and \(P V\) is \(a\) and the distance between \(A B\) and \(P Q\) is \(a\). The centre of mass of \(L\) is at the point \(G\).

1. Show that the distance of \(G\) from the side \(A D\) is \(\frac { 42 } { 11 } a\) The mass of \(L\) is \(M\). A particle of mass \(k M\) is attached to \(L\) at \(C\). The lamina, with the attached particle, is freely suspended from \(B\) and hangs in equilibrium with \(B C\) making an angle of \(45 ^ { \circ }\) with the horizontal.
2. Find the value of \(k\).

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5. Figure 2 The uniform lamina \(A B C D E F\), shown in Figure 2, consists of two identical rectangles with sides of length \(a\) and \(3 a\). The mass of the lamina is \(M\). A particle of mass \(k M\) is attached to the lamina at \(E\). The lamina, with the attached particle, is freely suspended from \(A\) and hangs in equilibrium with \(A F\) at an angle \(\theta\) to the downward vertical. Given that \(\tan \theta = \frac { 4 } { 7 }\), find the value of \(k\). \includegraphics[max width=\textwidth, alt={}, center]{f30ed5b8-880e-42de-860e-d1538fa68f11-16_677_677_244_580}

6. \begin{figure}[h] \end{figure} Figure 3 shows a rectangular lamina \(O A B C\). The coordinates of \(O , A , B\) and \(C\) are ( 0,0 ), \(( 8,0 ) , ( 8,5 )\) and \(( 0,5 )\) respectively. Particles of mass \(k m , 5 m\) and \(3 m\) are attached to the lamina at \(A , B\) and \(C\) respectively. The \(x\)-coordinate of the centre of mass of the three particles without the lamina is 6.4.

1. Show that \(k = 7\). The lamina \(O A B C\) is uniform and has mass \(12 m\).
2. Find the coordinates of the centre of mass of the combined system consisting of the three particles and the lamina. The combined system is freely suspended from \(O\) and hangs at rest.
3. Find the angle between \(O C\) and the horizontal.

4. \begin{figure}[h] \end{figure} A uniform circular disc has centre \(O\) and radius 4a. The lines \(P Q\) and \(S T\) are perpendicular diameters of the disc. A circular hole of radius \(2 a\) is made in the disc, with the centre of the hole at the point \(R\) on \(O P\) where \(O R = 2 a\), to form the lamina \(L\), shown shaded in Figure 2.

1. Show that the distance of the centre of mass of \(L\) from \(P\) is \(\frac { 14 a } { 3 }\). The mass of \(L\) is \(m\) and a particle of mass \(k m\) is now fixed to \(L\) at the point \(P\). The system is now suspended from the point \(S\) and hangs freely in equilibrium. The diameter \(S T\) makes an angle \(\alpha\) with the downward vertical through \(S\), where \(\tan \alpha = \frac { 5 } { 6 }\).
2. Find the value of \(k\).

2. \begin{figure}[h] \end{figure} The uniform lamina \(O A B C D\), shown in Figure 1, is formed by removing the triangle \(O A D\) from the square \(A B C D\) with centre \(O\). The square has sides of length \(2 a\).

1. Show that the centre of mass of \(O A B C D\) is \(\frac { 2 } { 9 } a\) from \(O\). The mass of the lamina is \(M\). A particle of mass \(k M\) is attached to the lamina at \(D\) to form the system \(S\). The system \(S\) is freely suspended from \(A\) and hangs in equilibrium with \(A O\) vertical.
2. Find the value of \(k\).

2. \begin{figure}[h] \end{figure} A thin uniform wire, of total length 20 cm , is bent to form a frame. The frame is in the shape of a trapezium \(A B C D\), where \(A B = A D = 4 \mathrm {~cm} , C D = 5 \mathrm {~cm}\), and \(A B\) is perpendicular to \(B C\) and \(A D\), as shown in Figure 1.

1. Find the distance of the centre of mass of the frame from \(A B\). The frame has mass \(M\). A particle of mass \(k M\) is attached to the frame at \(C\). When the frame is freely suspended from the mid-point of \(B C\), the frame hangs in equilibrium with \(B C\) horizontal.
2. Find the value of \(k\).

5. \begin{figure}[h] \end{figure} The uniform lamina \(A B C D E F\), shown in Figure 2, consists of two identical rectangles with sides of length \(a\) and \(3 a\). The mass of the lamina is \(M\). A particle of mass \(k M\) is attached to the lamina at \(E\). The lamina, with the attached particle, is freely suspended from \(A\) and hangs in equilibrium with \(A F\) at an angle \(\theta\) to the downward vertical. Given that \(\tan \theta = \frac { 4 } { 7 }\), find the value of \(k\).
(10)

2. \begin{figure}[h] \end{figure} A thin uniform wire, of total length 20 cm , is bent to form a frame. The frame is in the shape of a trapezium \(A B C D\), where \(A B = A D = 4 \mathrm {~cm} , C D = 5 \mathrm {~cm}\), and \(A B\) is perpendicular to \(B C\) and \(A D\), as shown in Figure 1.

1. Find the distance of the centre of mass of the frame from \(A B\). The frame has mass \(M\). A particle of mass \(k M\) is attached to the frame at \(C\). When the frame is freely suspended from the mid-point of \(B C\), the frame hangs in equilibrium with \(B C\) horizontal.
2. Find the value of \(k\).

4 A uniform rectangular lamina \(A B C D\) has a mass of 5 kg . The side \(A B\) has length 60 cm and the side \(B C\) has length 30 cm . The points \(P , Q , R\) and \(S\) are the mid-points of the sides, as shown in the diagram below. A uniform triangular lamina \(S R D\), of mass 4 kg , is fixed to the rectangular lamina to form a shop sign. The centre of mass of the triangular lamina \(S R D\) is 10 cm from the side \(A D\) and 5 cm from the side \(D C\). \includegraphics[max width=\textwidth, alt={}, center]{9d039ec3-fd0a-40ae-9afe-7627439081df-08_613_1086_660_518}

1. Find the distance of the centre of mass of the shop sign from \(A D\).
2. Find the distance of the centre of mass of the shop sign from \(A B\).
3. The shop sign is freely suspended from \(P\). Find the angle between \(A B\) and the horizontal when the shop sign is in equilibrium.
4. To ensure that the side \(A B\) is horizontal when the shop sign is freely suspended from point \(P\), a particle of mass \(m \mathrm {~kg}\) is attached to the shop sign at point \(B\). Calculate \(m\).
5. Explain how you have used the fact that the rectangular lamina \(A B C D\) is uniform in your solution to this question.
   (1 mark)
   \includegraphics[max width=\textwidth, alt={}]{9d039ec3-fd0a-40ae-9afe-7627439081df-10_2486_1714_221_153}
   \includegraphics[max width=\textwidth, alt={}]{9d039ec3-fd0a-40ae-9afe-7627439081df-11_2486_1714_221_153}

3

1. A uniform lamina made from rectangular parts is shown in Fig. 3.1. All the dimensions are centimetres. All coordinates are referred to the axes shown in Fig. 3.1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c1785fde-a6ce-4f8b-9948-4b4dd973ce84-4_691_529_427_762} \captionsetup{labelformat=empty} \caption{Fig. 3.1}
   \end{figure}
   1. Show that the \(x\)-coordinate of the centre of mass of the lamina is 6.5 and find the \(y\)-coordinate. A square of side 2 cm is to be cut from the lamina. The sides of the square are to be parallel to the coordinate axes and the centre of the square is to be chosen so that the \(x\)-coordinate of the centre of mass of the new shape is 6.4
   2. Calculate the \(x\)-coordinate of the centre of the square to be removed. The \(y\)-coordinate of the centre of the square to be removed is now chosen so that the \(y\)-coordinate of the centre of mass of the final shape is as large as possible.
   3. Calculate the \(y\)-coordinate of the centre of mass of the lamina with the square removed, giving your answer correct to three significant figures.
2. Fig. 3.2 shows a framework made from light rods of length 2 m freely pin-jointed at \(\mathrm { A } , \mathrm { B } , \mathrm { C }\), D and E. The framework is in a vertical plane and is supported at A and C. There are loads of 120 N at B and at E . The force on the framework due to the support at A is \(R \mathrm {~N}\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c1785fde-a6ce-4f8b-9948-4b4dd973ce84-5_448_741_459_662} \captionsetup{labelformat=empty} \caption{Fig. 3.2}
   \end{figure} each rod is 2 m long
   1. Show that \(R = 150\).
   2. Draw a diagram showing all the forces acting at the points \(\mathrm { A } , \mathrm { B } , \mathrm { D }\) and E , including the forces internal to the rods. Calculate the internal forces in rods AE and EB , and determine whether each is a tension or a thrust. [You may leave your answers in surd form.]
   3. Without any further calculation of the forces in the rods, explain briefly how you can tell that rod ED is in thrust.

3 The object shown shaded in Fig. 3.1 is cut from a flat sheet of thin rigid uniform material; LMJK, OAIJ, AEFH and CDEB are rectangles. The grid-lines in Fig. 3.1 are 1 cm apart. \begin{figure}[h] \end{figure}

1. Calculate the coordinates of the centre of mass of the object referred to the axes shown in Fig. 3.1. [5] The object is freely suspended from the point K and hangs in equilibrium.
2. Calculate the angle that KI makes with the vertical. The mass of the object is 0.3 kg .
   A particle of mass \(m \mathrm {~kg}\) is attached to the object at a point on the line OJ so that the new centre of mass is at the centre of the square OAIJ.
3. Calculate the value of \(m\) and the position of the particle referred to the axes shown in Fig. 3.1. The extra particle is now removed and the object shown in Fig. 3.1 is folded: LMJK is folded along JM so that it is perpendicular to OAIJ; ABCDEFH is folded along AH so that it is perpendicular to OAIJ and on the same side of OAIJ as LMJK. The folded object is placed on a horizontal table with the edges KL and FED in contact with the table. A plan view and a 3D representation are shown in Fig. 3.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{42b6ee17-f0ae-4687-8392-281ba724a607-4_609_648_1836_246} \captionsetup{labelformat=empty} \caption{Fig. 3.2}
   \end{figure} \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{42b6ee17-f0ae-4687-8392-281ba724a607-4_332_695_2001_1144} \captionsetup{labelformat=empty} \caption{Fig. 3.2}
   \end{figure}
4. On the plan, indicate the region corresponding to positions of the centre of mass for which the folded object is stable. You are given that the \(x\)-coordinate of the centre of mass of the folded object is 1.7 . Determine whether the object is stable.

3 A uniform heavy lamina occupies the region shaded in Fig. 3. This region is formed by removing a square of side 1 unit from a square of side \(a\) units (where \(a > 1\) ). \begin{figure}[h] \end{figure} Relative to the axes shown in Fig. 3, the centre of mass of the lamina is at \(( \bar { x } , \bar { y } )\).

1. Show that \(\bar { x } = \bar { y } = \frac { a ^ { 2 } + a + 1 } { 2 ( a + 1 ) }\).
   [0pt] [You may need to use the result \(\frac { a ^ { 3 } - 1 } { 2 \left( a ^ { 2 } - 1 \right) } = \frac { a ^ { 2 } + a + 1 } { 2 ( a + 1 ) }\).]
2. Show that the centre of mass of the lamina lies on its perimeter if \(a = \frac { 1 } { 2 } ( 1 + \sqrt { 5 } )\). In another situation, \(a = 4\).
   A particle of mass one third that of the lamina is attached to the lamina at vertex B ; the lamina with the particle is freely suspended from vertex A and hangs in equilibrium. The positions of A and B are shown in Fig. 3.
3. Calculate the angle that AB makes with the vertical.

6. \begin{figure}[h] \end{figure} Figure 3 shows a uniform rectangular lamina \(A B C D\) of mass \(8 m\) in which the sides \(A B\) and \(B C\) are of length \(a\) and \(2 a\) respectively. Particles of mass \(2 m , 6 m\) and \(4 m\) are fixed to the lamina at the points \(A , B\) and \(D\) respectively.

1. Write down the distance of the centre of mass from \(A D\).
2. Show that the distance of the centre of mass from \(A B\) is \(\frac { 4 } { 5 } a\). Another particle of mass \(k m\) is attached to the lamina at the point \(B\).
3. Show that the distance of the centre of mass from \(A D\) is now given by \(\frac { ( 10 + k ) a } { 20 + k }\).
   (4 marks)
   Given that when the lamina is suspended freely from the point \(A\) the side \(A B\) makes an angle of \(45 ^ { \circ }\) with the vertical,
4. find the value of \(k\).

3. The diagram below shows a lamina \(A B C D E\) which is made of a uniform material. It consists of a rectangle \(A B D E\) with \(A B = 6 a\) and \(A E = 8 a\), together with an isosceles triangle \(B C D\) with \(B C = D C = 5 a\). A semicircle, with its centre at the midpoint of \(A E\) and radius \(3 a\), is removed from \(A B D E\). \includegraphics[max width=\textwidth, alt={}, center]{b9c63cb4-d446-4548-be42-e30b10cb4b99-3_606_703_603_680}

1. Write down the distance of the centre of mass of the lamina \(A B C D E\) from \(A B\).
2. Show that the distance of the centre of mass of the lamina \(A B C D E\) from \(A E\) is \(\frac { 140 } { 40 - 3 \pi } a\).
3. The lamina \(A B C D E\) is freely suspended from the point \(D\) and hangs in equilibrium.
   1. Calculate the angle that \(B D\) makes with the vertical.
   2. The mass of the lamina is \(M\). When a particle of mass \(k M\) is attached at the point \(C\), the lamina hangs in equilibrium with \(A B\) horizontal. Determine the value of \(k\).

1. \begin{figure}[h] \end{figure} Figure 1 shows a uniform rectangular lamina \(A B C D\) with \(A B = 2 a\) and \(A D = a\) The mass of the lamina is \(6 m\). A particle of mass \(2 m\) is attached to the lamina at \(A\), a particle of mass \(m\) is attached to the lamina at \(B\) and a particle of mass \(3 m\) is attached to the lamina at \(D\), to form a loaded lamina \(L\) of total mass \(12 m\).

1. Write down the distance of the centre of mass of \(L\) from \(A B\). You must give a reason for your answer.
2. Show that the distance of the centre of mass of \(L\) from \(A D\) is \(\frac { 2 a } { 3 }\) A particle of mass \(k m\) is now also attached to \(L\) at \(D\) to form a new loaded lamina \(N\).
3. Show that the distance of the centre of mass of \(N\) from \(A B\) is \(\frac { ( k + 6 ) a } { ( k + 12 ) }\) When \(N\) is freely suspended from \(A\) and is hanging in equilibrium, the side \(A B\) makes an angle \(\alpha\) with the vertical, where \(\tan \alpha = \frac { 3 } { 2 }\)
4. Find the value of \(k\).

2. \begin{figure}[h] \end{figure} Uniform wire is used to form the framework shown in Figure 2.
In the framework

• \(A B C D\) is a rectangle with \(A D = 2 a\) and \(D C = a\)
• \(B E C\) is a semicircular arc of radius \(a\) and centre \(O\), where \(O\) lies on \(B C\)

The diameter of the semicircle is \(B C\) and the point \(E\) is such that \(O E\) is perpendicular to \(B C\). The points \(A , B , C , D\) and \(E\) all lie in the same plane.

1. Show that the distance of the centre of mass of the framework from \(B C\) is $$\frac { a } { 6 + \pi }$$ The framework is freely suspended from \(A\) and hangs in equilibrium with \(A E\) at an angle \(\theta ^ { \circ }\) to the downward vertical.
2. Find the value of \(\theta\). The mass of the framework is \(M\).
   A particle of mass \(k M\) is attached to the framework at \(B\).
   The centre of mass of the loaded framework lies on \(O A\).
3. Find the value of \(k\).

3. \begin{figure}[h] \end{figure} Nine uniform rods are joined together to form the rigid framework \(A B C D E F A\), with \(A B = B C = D F = 3 a , B F = C D = D E = 4 a\) and \(A F = F E = C F = 5 a\), as shown in Figure 1. All nine rods lie in the same plane. The mass per unit length of each of the rods \(B F , C F\) and \(D F\) is twice the mass per unit length of each of the other six rods.

1. Find the distance of the centre of mass of the framework from \(A C\) The mass of the framework is \(M\). A particle of mass \(k M\) is attached to the framework at \(E\) to form a loaded framework. When the loaded framework is freely suspended from \(F\), it hangs in equilibrium with \(C E\) horizontal.
2. Find the exact value of \(k\)