6.04c Composite bodies: centre of mass

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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3. \begin{figure}[h] \end{figure} The uniform lamina \(A B C D\) is a square of side \(6 a\). The template \(T\), shown shaded in Figure 1, is formed by removing the right-angled triangle \(E F G\) and the circle, centre \(H\) and radius \(a\), from the square lamina. Triangle \(E F G\) has \(E F = E G = 4 a\), with \(E F\) parallel to \(A B\) and \(E G\) parallel to \(A D\). The distance between \(A B\) and \(E F\) is \(a\) and the distance between \(A D\) and \(E G\) is \(a\). The point \(H\) lies on \(A C\) and the distance of \(H\) from \(B C\) is \(2 a\).

1. Show that the centre of mass of \(T\) is a distance \(\frac { 4 ( 67 - 3 \pi ) } { 3 ( 28 - \pi ) } a\) from \(A D\). The template \(T\) is suspended from the ceiling by two light inextensible vertical strings. One string is attached to \(T\) at \(A\) and the other string is attached to \(T\) at \(B\) so that \(T\) hangs in equilibrium with \(A B\) horizontal. The weight of \(T\) is \(W\). The tension in the string attached to \(T\) at \(B\) is \(k W\), where \(k\) is a constant.
2. Find the value of \(k\), giving your answer to 2 decimal places.

7. \begin{figure}[h] \end{figure} The template shown in Figure 3 is formed by joining together three separate laminas. All three laminas lie in the same plane.

• PQUV is a uniform square lamina with sides of length \(3 a\)
• URST is a uniform square lamina with sides of length \(6 a\)
• \(Q R U\) is a uniform triangular lamina with \(U Q = 3 a , U R = 6 a\) and angle \(Q U R = 90 ^ { \circ }\)

The mass per unit area of \(P Q U V\) is \(k\), where \(k\) is a constant.
The mass per unit area of URST is \(k\).
The mass per unit area of \(Q R U\) is \(2 k\).
The distance of the centre of mass of the template from \(Q T\) is \(d\).

1. Show that \(d = \frac { 29 } { 14 } a\) The template is freely suspended from the point \(Q\) and hangs in equilibrium with \(Q R\) at \(\theta ^ { \circ }\) to the downward vertical.
2. Find the value of \(\theta\)

3. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} The uniform triangular lamina \(A B C\), shown in Figure 1, has height \(9 y\), base \(B C = 6 x\), and \(A B = A C\) The points \(P\) and \(Q\) are such that \(A P : P C = A Q : Q B = 2 : 1\) The lamina is folded along \(P Q\) to form the folded lamina \(F\) The distance of the centre of mass of \(F\) from \(P Q\) is \(d\)

1. Show that \(d = \frac { 16 } { 9 } y\) The folded lamina is suspended from \(P\) and hangs freely in equilibrium with \(P Q\) at an angle \(\alpha\) to the downward vertical.
   Given that \(\tan \alpha = \frac { 64 } { 81 }\)
2. find \(x\) in terms of \(y\)

3. \begin{figure}[h] \end{figure} A uniform circular disc \(C\) has centre \(X\) and radius \(R\).
A disc with centre \(Y\) and radius \(r\), where \(0 < r < R\) and \(X Y = R - r\), is removed from \(C\) to form the template shown shaded in Figure 1. The centre of mass of the template is a distance \(k r\) from \(X\).

1. Show that \(r = \frac { k } { 1 - k } R\)
2. Hence find the range of possible values of \(k\). The point \(P\) is on the outer edge of the template and \(P X\) is perpendicular to \(X Y\).
   The template is freely suspended from \(P\) and hangs in equilibrium.
   Given that \(k = \frac { 4 } { 9 }\)
3. find the angle that \(X Y\) makes with the vertical. The mass of the template is \(M\).
4. Find, in terms of \(M\), the mass of the lightest particle that could be attached to the template so that it would hang in equilibrium from \(P\) with \(X Y\) horizontal.

1. Three particles of masses \(m , 4 m\) and \(k m\) are placed at the points whose coordinates are \(( - 3,2 ) , ( 4,3 )\) and \(( 6 , - 4 )\) respectively. The centre of mass of the three particles is at the point with coordinates \(( c , 0 )\).

Find

1. the value of \(k\),
2. the value of \(c\).

6. \begin{figure}[h] \end{figure} The uniform lamina \(L\) shown shaded in Figure 2 is formed by removing two circular discs, \(C _ { 1 }\) and \(C _ { 2 }\), from a circular disc with centre \(O\) and radius \(8 a\). Disc \(C _ { 1 }\) has centre \(A\) and radius \(a\). Disc \(C _ { 2 }\) has centre \(B\) and radius \(2 a\). The diameters \(P R\) and \(Q S\) are perpendicular. The midpoint of \(P O\) is \(A\) and the midpoint of \(O R\) is \(B\).

1. Show that the centre of mass of \(L\) is \(\frac { 484 } { 59 } a\) from \(R\). The mass of \(L\) is \(M\). A particle of mass \(k M\) is attached to \(L\) at \(S\). The lamina with the attached particle is suspended from \(R\) and hangs freely in equilibrium with the diameter \(P R\) at an angle of arctan \(\left( \frac { 1 } { 4 } \right)\) to the downward vertical through \(R\).
2. Find the value of \(k\).

5. \begin{figure}[h] \end{figure} The uniform lamina \(A B C\) is in the shape of an equilateral triangle with sides of length \(4 a\). The midpoint of \(B C\) is \(D\). The point \(E\) lies on \(A D\) with \(D E = \frac { 3 a } { 2 }\). A circular hole, with centre \(E\) and radius \(a\), is made in the lamina \(A B C\) to form the lamina \(L\), shown shaded in Figure 2.

1. Find the distance of the centre of mass of \(L\) from \(D\). The lamina \(L\) is freely suspended from the point \(B\) and hangs in equilibrium.
2. Find, to the nearest degree, the size of the acute angle between \(A D\) and the downward vertical.

5. \begin{figure}[h] \end{figure} Figure 3 shows a uniform rectangular lamina \(A B C D\) with sides of length \(3 a\) and \(k a\), where \(k > 3\). The point \(E\) on side \(A D\) is such that \(D E = 3 a\). Rectangle \(A B C D\) is folded along the line \(C E\) to produce the folded lamina \(L\) shown in Figure 4. \begin{figure}[h] \end{figure} Find, in terms of \(a\) and \(k\),

1. the distance of the centre of mass of \(L\) from \(A B\),
2. the distance of the centre of mass of \(L\) from \(A E\). The folded lamina \(L\) is freely suspended from \(A\) and hangs in equilibrium with \(A B\) at \(45 ^ { \circ }\) to the downward vertical.
3. Find, to 3 significant figures, the value of \(k\).

7. In this question you may use, without proof, the formula for the centre of mass of a uniform sector of a circle, as given in the formulae book. \begin{figure}[h] \end{figure} The uniform lamina \(A B C D E\), shown shaded in Figure 3, is formed by joining a rectangle to a sector of a circle.

• The rectangle \(A B C E\) has \(A B = E C = a\) and \(A E = B C = d\)
• The sector \(C D E\) has centre \(C\) and radius \(a\)
• Angle \(E C D = \frac { \pi } { 3 }\) radians

The centre of mass of the lamina lies on EC.

1. Show that \(a = \sqrt { 3 } d\) The lamina is freely suspended from \(B\) and hangs in equilibrium with \(B C\) at an angle \(\beta\) radians to the downward vertical.
2. Find the value of \(\beta\)

6. \begin{figure}[h] \end{figure} The uniform lamina \(P Q R S T U V\) shown in Figure 2 is formed from two identical rectangles, \(P Q U V\) and \(Q R S T U\).
The rectangles have sides \(P Q = R S = 2 a\) and \(P V = Q R = k a\).

1. Show that the centre of mass of the lamina is \(\left( \frac { 6 + k } { 4 } \right) a\) from \(P V\) The lamina is freely suspended from \(P\) and hangs in equilibrium with \(P R\) at an angle of \(\alpha\) to the downward vertical. Given that \(\tan \alpha = \frac { 7 } { 15 }\)
2. find the value of \(k\).

2. \begin{figure}[h] \end{figure} Figure 1 shows a template where

• PQUY is a uniform square lamina with sides of length \(4 a\)
• RSTU is a uniform square lamina with sides of length \(2 a\)
• VWXY is a uniform square lamina with sides of length \(2 a\)
• the three squares all lie in the same plane
• the mass per unit area of \(V W X Y\) is double the mass per unit area of \(P Q U Y\)
• the mass per unit area of \(R S T U\) is double the mass per unit area of \(P Q U Y\)
• the distance of the centre of mass of the template from \(P X\) is \(d\)
  1. Show that \(d = \frac { 5 } { 2 } a\)

The template is freely pivoted about \(Q\) and hangs in equilibrium with \(P Q\) at an angle of \(\theta\) to the downward vertical.

Find the value of \(\tan \theta\) The mass of the template is \(M\) The template is still freely pivoted about \(Q\), but it is now held in equilibrium, with \(P Q\) vertical, by a horizontal force of magnitude \(F\) which acts on the template at \(X\). The line of action of the force lies in the same plane as the template.

Find \(F\) in terms of \(M\) and \(g\)

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}

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

1. Find the distance of the centre of mass of the frame from \(B C\). 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.
2. Find the size of the angle between BC and the vertical.

7. \begin{figure}[h] \end{figure} A loaded plate \(L\) is modelled as a uniform rectangular lamina \(A B C D\) and three particles. The sides \(C D\) and \(A D\) of the lamina have lengths \(5 a\) and \(2 a\) respectively and the mass of the lamina is \(3 m\). The three particles have mass \(4 m , m\) and \(2 m\) and are attached at the points \(A , B\) and \(C\) respectively, as shown in Fig. 3.

1. Show that the distance of the centre of mass of \(L\) from \(A D\) is \(2.25 a\).
2. Find the distance of the centre of mass of \(L\) from \(A B\). The point \(O\) is the mid-point of \(A B\). The loaded plate \(L\) is freely suspended from \(O\) and hangs at rest under gravity.
3. Find, to the nearest degree, the size of the angle that \(A B\) makes with the horizontal. A horizontal force of magnitude \(P\) is applied at \(C\) in the direction \(C D\). The loaded plate \(L\) remains suspended from \(O\) and rests in equilibrium with \(A B\) horizontal and \(C\) vertically below \(B\).
4. Show that \(P = \frac { 5 } { 4 } \mathrm { mg }\).
5. Find the magnitude of the force on \(L\) at \(O\).

5. \begin{figure}[h] \end{figure} A uniform lamina \(A B C D\) is made by joining a uniform triangular lamina \(A B D\) to a uniform semi-circular lamina \(D B C\), of the same material, along the edge \(B D\), as shown in Figure 2. Triangle \(A B D\) is right-angled at \(D\) and \(A D = 18 \mathrm {~cm}\). The semi-circle has diameter \(B D\) and \(B D = 12 \mathrm {~cm}\).

1. Show that, to 3 significant figures, the distance of the centre of mass of the lamina \(A B C D\) from \(A D\) is 4.69 cm . Given that the centre of mass of a uniform semicircular lamina, radius \(r\), is at a distance \(\frac { 4 r } { 3 \pi }\) from the centre of the bounding diameter,
2. find, in cm to 3 significant figures, the distance of the centre of mass of the lamina \(A B C D\) from \(B D\). The lamina is freely suspended from \(B\) and hangs in equilibrium.
3. Find, to the nearest degree, the angle which \(B D\) makes with the vertical.

5. \begin{figure}[h] \end{figure} The uniform L-shaped lamina \(A B C D E F\), shown in Figure 2, has sides \(A B\) and \(F E\) parallel, and sides \(B C\) and \(E D\) parallel. The pairs of parallel sides are 9 cm apart. The points \(A , F\), \(D\) and \(C\) lie on a straight line. \(A B = B C = 36 \mathrm {~cm} , F E = E D = 18 \mathrm {~cm} . \angle A B C = \angle F E D = 90 ^ { \circ }\), and \(\angle B C D = \angle E D F = \angle E F D = \angle B A C = 45 ^ { \circ }\).

1. Find the distance of the centre of mass of the lamina from
   1. side \(A B\),
   2. side \(B C\). The lamina is freely suspended from \(A\) and hangs in equilibrium.
2. Find, to the nearest degree, the size of the angle between \(A B\) and the vertical.

3. \begin{figure}[h] \end{figure} Figure 1 shows a decoration which is made by cutting the shape of a simple tree from a sheet of uniform card. The decoration consists of a triangle \(A B C\) and a rectangle \(P Q R S\). The points \(P\) and \(S\) lie on \(B C\) and \(M\) is the mid-point of both \(B C\) and \(P S\). The triangle \(A B C\) is isosceles with \(A B = A C , B C = 4 \mathrm {~cm} , A M = 6 \mathrm {~cm} , P S = 2 \mathrm {~cm}\) and \(P Q = 3 \mathrm {~cm}\).

1. Find the distance of the centre of mass of the decoration from \(B C\). The decoration is suspended from \(Q\) and hangs freely.
2. Find, in degrees to one decimal place, the angle between \(P Q\) and the vertical.

3. \begin{figure}[h] \end{figure} A uniform lamina \(A B C D E F\) is formed by taking a uniform sheet of card in the form of a square \(A X E F\), of side \(2 a\), and removing the square \(B X D C\) of side \(a\), where \(B\) and \(D\) are the mid-points of \(A X\) and \(X E\) respectively, as shown in Figure 1.

1. Find the distance of the centre of mass of the lamina from \(A F\). The lamina is freely suspended from \(A\) and hangs in equilibrium.
2. Find, in degrees to one decimal place, the angle which \(A F\) makes with the vertical.

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.

5. \begin{figure}[h] \end{figure} A shop sign \(A B C D E F G\) is modelled as a uniform lamina, as illustrated in Figure 2. \(A B C D\) is a rectangle with \(B C = 120 \mathrm {~cm}\) and \(D C = 90 \mathrm {~cm}\). The shape \(E F G\) is an isosceles triangle with \(E G = 60 \mathrm {~cm}\) and height 60 cm . The mid-point of \(A D\) and the mid-point of \(E G\) coincide.

1. Find the distance of the centre of mass of the sign from the side \(A D\). The sign is freely suspended from \(A\) and hangs at rest.
2. Find the size of the angle between \(A B\) and the vertical.

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 \(k\).

4. \begin{figure}[h] \end{figure} The uniform lamina \(A B C D E F\) is a regular hexagon with centre \(O\) and sides of length 2 m , as shown in Figure 1. \begin{figure}[h] \end{figure} The triangles \(O A F\) and \(O E F\) are removed to form the uniform lamina \(O A B C D E\), shown in Figure 2.

1. Find the distance of the centre of mass of \(O A B C D E\) from \(O\). The lamina \(O A B C D E\) is freely suspended from \(E\) and hangs in equilibrium.
2. Find the size of the angle between \(E O\) and the downward vertical.