6.04b Find centre of mass: using symmetry

\includegraphics{figure_1} A uniform lamina \(ABCD\) is made by taking a uniform sheet of metal in the form of a rectangle \(ABED\), with \(AB = 3a\) and \(AD = 2a\), and removing the triangle \(BCE\), where \(C\) lies on \(DE\) and \(CE = a\), as shown in Fig. 1.

1. Find the distance of the centre of mass of the lamina from \(AD\). [5]

The lamina has mass \(M\). A particle of mass \(m\) is attached to the lamina at \(B\). When the loaded lamina is freely suspended from the mid-point of \(AB\), it hangs in equilibrium with \(AB\) horizontal.

1. Find \(m\) in terms of \(M\). [4]

Figure 1 \includegraphics{figure_1} Figure 1 shows four uniform rods joined to form a rigid rectangular framework \(ABCD\), where \(AB = CD = 2a\), and \(BC = AD = 3a\). Each rod has mass \(m\). Particles, of mass \(6m\) and \(2m\), are attached to the framework at points \(C\) and \(D\) respectively.

1. Find the distance of the centre of mass of the loaded framework from
   1. \(AB\),
   2. \(AD\).
   [7]

The loaded framework is freely suspended from \(B\) and hangs in equilibrium.

1. Find the angle which \(BC\) makes with the vertical. [3]

\includegraphics{figure_1} A triangular frame is formed by cutting a uniform rod into 3 pieces which are then joined to form a triangle \(ABC\), where \(AB = AC = 10\) cm and \(BC = 12\) cm, as shown in Figure 1.

1. Find the distance of the centre of mass of the frame from \(BC\). (5)

The frame has total mass \(M\). A particle of mass \(M\) is attached to the frame at the mid-point of \(BC\). The frame is then freely suspended from \(B\) and hangs in equilibrium.

1. Find the size of the angle between \(BC\) and the vertical. (4)

Three particles of masses 2 kg, 3 kg and \(m\) kg are positioned at the points with coordinates \((a, 3)\), \((3, -1)\) and \((-2, 4)\) respectively. Given that the centre of mass of the particles is at the point with coordinates \((0, 2)\), find

1. the value of \(m\), [4]
2. the value of \(a\). [4]

\includegraphics{figure_2} The uniform L-shaped lamina \(OABCDE\), shown in Figure 2, is made from two identical rectangles. Each rectangle is 4 metres long and \(a\) metres wide. Giving each answer in terms of \(a\), find the distance of the centre of mass of the lamina from

1. \(OE\), [4]
2. \(OA\). [4]

The lamina is freely suspended from \(O\) and hangs in equilibrium with \(OE\) at an angle \(\theta\) to the downward vertical through \(O\), where \(\tan \theta = \frac{4}{3}\).

1. Find the value of \(a\). [4]

1. Use algebraic integration to show that the centre of mass of a uniform solid right circular cone of height \(h\) is at a distance \(\frac{3}{4}h\) from the vertex of the cone. [You may assume that the volume of a cone of height \(h\) and base radius \(r\) is \(\frac{1}{3}\pi r^2 h\)] [5]

\includegraphics{figure_2} A uniform solid \(S\) consists of a right circular cone, of radius \(r\) and height \(5r\), fixed to a hemisphere of radius \(r\). The centre of the plane face of the hemisphere is \(O\) and this plane face coincides with the base of the cone, as shown in Figure 2.

1. Find the distance of the centre of mass of \(S\) from \(O\). [5]

The point \(A\) lies on the circumference of the base of the cone. The solid is suspended by a string attached at \(A\) and hangs freely in equilibrium.

1. Find the size of the angle between \(OA\) and the vertical. [3]

The mass of the hemisphere is \(M\). A particle of mass \(kM\) is fixed to the surface of the hemisphere on the axis of symmetry of \(S\). The solid is again suspended by the string attached at \(A\) and hangs freely in equilibrium. The axis of symmetry of \(S\) is now horizontal.

1. Find the value of \(k\). [4]

The diagram shows a uniform lamina \(ABCDEFGHIJKL\). \includegraphics{figure_3}

1. Explain why the centre of mass of the lamina is \(35\) cm from \(AL\). [1 mark]
2. Find the distance of the centre of mass from \(AF\). [4 marks]
3. The lamina is freely suspended from \(A\). Find the angle between \(AB\) and the vertical when the lamina is in equilibrium. [3 marks]
4. Explain, briefly, how you have used the fact that the lamina is uniform. [1 mark]

\(PQR\) is a triangular lamina with \(PQ = 18\) cm, \(QR = 24\) cm and \(PR = 30\) cm.

1. Verify that angle \(PQR\) is a right angle and find the distances of the centre of mass of the lamina from
   1. \(PQ\),
   2. \(QR\).
   [5 marks]

\includegraphics{figure_6} The lamina is held in a vertical plane and placed on a line of greatest slope of a rough plane inclined at an angle \(\theta\) to the horizontal, as shown.

1. Find the largest value of \(\theta\) for which equilibrium will not be broken by toppling. [4 marks]

\includegraphics{figure_1} A uniform lamina \(ABDC\) is bounded by two semicircular arcs \(AB\) and \(CD\), each with centre \(O\) and of radii \(3a\) and \(a\) respectively, and two straight edges, \(AC\) and \(DB\), which lie on the line \(AOB\) (see Fig. 1).

1. Show that the distance of the centre of mass of the lamina from \(O\) is \(\frac{13a}{3\pi}\). [5]

\includegraphics{figure_2} The lamina has mass 3 kg and is freely pivoted to a fixed point at \(A\). The lamina is held in equilibrium with \(AB\) vertical by means of a light string attached to \(B\). The string lies in the same plane as the lamina and is at an angle of \(40°\) below the horizontal (see Fig. 2).

1. Calculate the tension in the string. [3]
2. Find the direction of the force acting on the lamina at \(A\). [4]

A box is to be assembled in the shape of the cuboid shown in Fig. 3.1. The lengths are in centimetres. All the faces are made of the same uniform, rigid and thin material. All coordinates refer to the axes shown in this figure. \includegraphics{figure_3.1}

1. The four vertical faces OAED, ABFE, FGCB and CODG are assembled first to make an open box without a base or a top. Write down the coordinates of the centre of mass of this open box. [1]

The base OABC is added to the vertical faces.

1. Write down the \(x\)- and \(y\)-coordinates of the centre of mass of the box now. Show that the \(z\)-coordinate is now 1.875. [5]

The top face FGDE is now added. This is a lid hinged to the rest of the box along the line FG. The lid is open so that it hangs in a vertical plane touching the face FGCB.

1. Show that the coordinates of the centre of mass of the box in this situation are \((10, 2.4, 2.1)\). [6]

[This question is continued on the facing page.] The box, with the lid still touching face FGCB, is now put on a sloping plane with the edge OA horizontal and the base inclined at \(30°\) to the horizontal, as shown in Fig. 3.2. \includegraphics{figure_3.2} The weight of the box is 40 N. A force \(P\) N acts parallel to the plane and is applied to the mid-point of FG at \(90°\) to FG. This force tends to push the box down the plane. The box does not slip and is on the point of toppling about the edge AO.

1. Show that the clockwise moment of the weight of the box about the edge AO is about 0.411 Nm. [4]
2. Calculate the value of \(P\). [2]

A lamina is made from uniform material in the shape shown in Fig. 3.1. BCJA, DZOJ, ZEIO and FGHI are all rectangles. The lengths of the sides are shown in centimetres. \includegraphics{figure_3}

1. Find the coordinates of the centre of mass of the lamina, referred to the axes shown in Fig. 3.1. [5]

The rectangles BCJA and FGHI are folded through 90° about the lines CJ and FI respectively to give the fire-screen shown in Fig. 3.2.

1. Show that the coordinates of the centre of mass of the fire-screen, referred to the axes shown in Fig. 3.2, are (2.5, 0, 57.5). [4]

The \(x\)- and \(y\)-axes are in a horizontal floor. The fire-screen has a weight of 72 N. A horizontal force \(P\) N is applied to the fire-screen at the point Z. This force is perpendicular to the line DE in the positive \(x\) direction. The fire-screen is on the point of tipping about the line AH.

1. Calculate the value of \(P\). [5]

The coefficient of friction between the fire-screen and the floor is \(\mu\).

1. For what values of \(\mu\) does the fire-screen slide before it tips? [4]

You are given that the centre of mass, G, of a uniform lamina in the shape of an isosceles triangle lies on its axis of symmetry in the position shown in Fig. 4.1. \includegraphics{figure_4_1} Fig. 4.2 shows the cross-section OABCD of a prism made from uniform material. OAB is an isosceles triangle, where OA = AB, and OBCD is a rectangle. The distance OD is \(h\) cm, where \(h\) can take various positive values. All coordinates refer to the axes Ox and Oy shown. The units of the axes are centimetres. \includegraphics{figure_4_2}

1. Write down the coordinates of the centre of mass of the triangle OAB. [1]
2. Show that the centre of mass of the region OABCD is \(\left(\frac{12-h^2}{2(h+3)}, 2.5\right)\). [6]

The \(x\)-axis is horizontal. The prism is placed on a horizontal plane in the position shown in Fig. 4.2.

1. Find the values of \(h\) for which the prism would topple. [3]

The following questions refer to the case where \(h = 3\) with the prism held in the position shown in Fig. 4.2. The cross-section OABCD contains the centre of mass of the prism. The weight of the prism is 15 N. You should assume that the prism does not slide.

1. Suppose that the prism is held in this position by a vertical force applied at A. Given that the prism is on the point of tipping clockwise, calculate the magnitude of this force. [3]
2. Suppose instead that the prism is held in this position by a force in the plane of the cross-section OABCD, applied at 30° below the horizontal at C, as shown in Fig. 4.3. Given that the prism is on the point of tipping anti-clockwise, calculate the magnitude of this force. [4]

\includegraphics{figure_4_3}

A uniform rod \(AB\) has mass \(2m\) and length \(4a\).

1. Show by integration that the moment of inertia of the rod about an axis perpendicular to the rod through \(A\) is \(\frac{32}{3}ma^2\). [4]

The rod is initially at rest with \(B\) vertically below \(A\) and it is free to rotate in a vertical plane about a smooth fixed horizontal axis through \(A\). A particle of mass \(m\) is moving horizontally in the plane in which the rod is free to rotate. The particle has speed \(v\), and strikes the rod at \(B\). In the subsequent motion the particle adheres to the rod and the combined rigid body \(Q\), consisting of the rod and the particle, starts to rotate.

1. Find, in terms of \(v\) and \(a\), the initial angular speed of \(Q\). [4]

At time \(t\) seconds the angle between \(Q\) and the downward vertical is \(\theta\) radians.

1. Show that \(\dot{\theta}^2 = k\frac{g}{a}(\cos \theta - 1) + \frac{9v^2}{400a^2}\), stating the value of the constant \(k\). [4]
2. Find, in terms of \(a\) and \(g\), the set of values of \(v^2\) for which \(Q\) makes complete revolutions. [2]

When \(Q\) is horizontal, the force exerted by the axis on \(Q\) has vertically upwards component \(R\).

1. Find \(R\) in terms of \(m\) and \(g\). [4]

\includegraphics{figure_6} A compound pendulum consists of a uniform rod \(AB\) of length 1 m and mass 3 kg, a particle of mass 1 kg attached to the rod at \(A\) and a circular disc of radius \(\frac{1}{5}\) m, mass 6 kg and centre \(C\). The end \(B\) of the rod is rigidly attached to a point on the circumference of the disc in such a way that \(ABC\) is a straight line. The pendulum is initially at rest with \(B\) vertically below \(A\) and it is free to rotate in a vertical plane about a smooth fixed horizontal axis passing through the point \(P\) on the rod where \(AP = x\) m and \(x < \frac{1}{3}\) (see diagram).

1. Show that the moment of inertia of the pendulum about the axis of rotation is \((10x^2 - 19x + 12)\) kg m\(^2\). [6]

The pendulum is making small oscillations about the equilibrium position, such that at time \(t\) seconds the angular displacement that the pendulum makes with the downward vertical is \(\theta\) radians.

1. Find the angular acceleration of the pendulum, in terms of \(x\), \(g\) and \(\theta\). [4]
2. Show that the motion is approximately simple harmonic, and show that the approximate period of oscillations, in seconds, is given by \(2\pi\sqrt{\frac{20x^2 - 38x + 24}{(19-20x)g}}\). [2]
3. Hence find the value of \(x\) for which the approximate period of oscillations is least. [3]

A pendulum consists of a uniform rod \(AB\), of length \(4a\) and mass \(2m\), whose end \(A\) is rigidly attached to the centre \(O\) of a uniform square lamina \(PQRS\), of mass \(4m\) and side \(a\). The rod \(AB\) is perpendicular to the plane of the lamina. The pendulum is free to rotate about a fixed smooth horizontal axis \(L\) which passes through \(B\). The axis \(L\) is perpendicular to \(AB\) and parallel to the edge \(PQ\) of the square.

1. Show that the moment of inertia of the pendulum about \(L\) is \(75ma^2\). [4]

The pendulum is released from rest when \(BA\) makes an angle \(\alpha\) with the downward vertical through \(B\), where \(\tan \alpha = \frac{3}{4}\). When \(BA\) makes an angle \(\theta\) with the downward vertical through \(B\), the magnitude of the component, in the direction \(AB\), of the force exerted by the axis \(L\) on the pendulum is \(X\).

1. Find an expression for \(X\) in terms of \(m\), \(g\) and \(\theta\). [9]

Using the approximation \(\theta \approx \sin \theta\),

1. find an estimate of the time for the pendulum to rotate through an angle \(\alpha\) from its initial rest position. [6]

\includegraphics{figure_2} **Figure 1** A uniform circular disc has mass \(4m\), centre \(O\) and radius \(4a\). The line \(POQ\) is a diameter of the disc. A circular hole of radius \(2a\) is made in the disc with the centre of the hole at the point \(R\) on \(PQ\) where \(QR = 5a\), as shown in Figure 1. The resulting lamina is free to rotate about a fixed smooth horizontal axis \(L\) which passes through \(Q\) and is perpendicular to the plane of the lamina.

1. Show that the moment of inertia of the lamina about \(L\) is \(69ma^2\). [7]

The lamina is hanging at rest with \(P\) vertically below \(Q\) when it is given an angular velocity \(\Omega\). Given that the lamina turns through an angle \(\frac{2\pi}{3}\) before it first comes to instantaneous rest,

1. find \(\Omega\) in terms of \(g\) and \(a\). [6]

A uniform circular disc has mass \(m\), centre \(O\) and radius \(2a\). It is free to rotate about a fixed smooth horizontal axis \(L\) which lies in the same plane as the disc and which is tangential to the disc at the point \(A\). The disc is hanging at rest in equilibrium with \(O\) vertically below \(A\) when it is struck at \(O\) by a particle of mass \(m\). Immediately before the impact the particle is moving perpendicular to the plane of the disc with speed \(3\sqrt{ag}\). The particle adheres to the disc at \(O\).

1. Find the angular speed of the disc immediately after the impact. [5]

1. Find the magnitude of the force exerted on the disc by the axis immediately after the impact. [6]

A uniform rod \(PQ\), of mass \(m\) and length \(2a\), is made to rotate in a vertical plane with constant angular speed \(\sqrt{\frac{g}{a}}\) about a fixed smooth horizontal axis through the end \(P\) of the rod. Show that, when the rod is inclined at an angle \(\theta\) to the downward vertical, the magnitude of the force exerted on the axis by the rod is \(2mg|\cos(\frac{1}{2}\theta)|\). [8]

A uniform rod \(AB\) of mass \(4m\) is free to rotate in a vertical plane about a fixed smooth horizontal axis, \(L\), through \(A\). The rod is hanging vertically at rest when it is struck at its end \(B\) by a particle of mass \(m\). The particle is moving with speed \(u\), in a direction which is horizontal and perpendicular to \(L\), and after striking the rod it rebounds in the opposite direction with speed \(v\). The coefficient of restitution between the particle and the rod is 1. Show that \(u = 7v\). [7]

A pendulum consists of a uniform rod \(PQ\), of mass \(3m\) and length \(2a\), which is rigidly fixed at its end \(Q\) to the centre of a uniform circular disc of mass \(m\) and radius \(a\). The rod is perpendicular to the plane of the disc. The pendulum is free to rotate about a fixed smooth horizontal axis \(L\) which passes through the end \(P\) of the rod and is perpendicular to the rod.

1. Show that the moment of inertia of the pendulum about \(L\) is \(\frac{33}{4}ma^2\). [5]

The pendulum is released from rest in the position where \(PQ\) makes an angle \(\alpha\) with the downward vertical. At time \(t\), \(PQ\) makes an angle \(\theta\) with the downward vertical.

1. Show that the angular speed, \(\dot{\theta}\), of the pendulum satisfies $$\dot{\theta}^2 = \frac{40g(\cos \theta - \cos \alpha)}{33a}$$ [4]

1. Hence, or otherwise, find the angular acceleration of the pendulum. [3]

Given that \(\alpha = \frac{\pi}{20}\) and that \(PQ\) has length \(\frac{8}{33}\) m,

1. find, to 3 significant figures, an approximate value for the angular speed of the pendulum 0.2 s after it has been released from rest. [5]

1. A uniform lamina of mass \(m\) is in the shape of a triangle \(ABC\). The perpendicular distance of \(C\) from the line \(AB\) is \(h\). Prove, using integration, that the moment of inertia of the lamina about \(AB\) is \(\frac{1}{6}mh^2\). [7]

1. Deduce the radius of gyration of a uniform square lamina of side \(2a\), about a diagonal. [3]

The points \(X\) and \(Y\) are the mid-points of the sides \(RQ\) and \(RS\) respectively of a square \(PQRS\) of side \(2a\). A uniform lamina of mass \(M\) is in the shape of \(PQXYS\).

1. Show that the moment of inertia of this lamina about \(XY\) is \(\frac{79}{84}Ma^2\). [6]

\(ABCD\) is a uniform lamina in the shape of a kite with \(BA = BC = 0.37\) m, \(DA = DC = 0.91\) m and \(AC = 0.7\) m (see diagram). The centre of mass of \(ABCD\) is \(G\). \includegraphics{figure_4}

1. Explain why \(G\) lies on \(BD\). [1]
2. Show that the distance of \(G\) from \(B\) is \(0.36\) m. [4]

The lamina \(ABCD\) is freely suspended from the point \(A\).

1. Determine the acute angle that \(CD\) makes with the horizontal, stating which of \(C\) or \(D\) is higher. [4]

[In this question, you may use the fact that the volume of a right circular cone of base radius \(r\) and height \(h\) is \(\frac{1}{3}\pi r^2 h\).]

1. By using integration, show that the centre of mass of a uniform solid right circular cone of height \(h\) and base radius \(r\) is at a distance \(\frac{3}{4}h\) from the vertex. [5]

\includegraphics{figure_8} Fig. 8 shows the side view of a toy formed by joining a uniform solid circular cylinder of radius \(r\) and height \(2r\) to a uniform solid right circular cone, made of the same material as the cylinder, of radius \(r\) and height \(r\). The toy is placed on a horizontal floor with the curved surface of the cone in contact with the floor.

1. Determine whether the toy will topple. [7]
2. Explain why it is not necessary to know whether the floor is rough or smooth in answering part (b). [1]

\includegraphics{figure_2} The uniform L-shaped lamina \(OABCDE\), shown in Figure 2, is made from two identical rectangles. Each rectangle is 4 metres long and \(a\) metres wide. Giving each answer in terms of \(a\), find the distance of the centre of mass of the lamina from

1. \(OE\). [4]
2. \(OA\). [4]

The lamina is freely suspended from \(O\) and hangs in equilibrium with \(OE\) at an angle \(\theta\) to the downward vertical through \(O\), where \(\tan \theta = \frac{4}{3}\).

1. Find the value of \(a\). [4]