6.04d Integration: for centre of mass of laminas/solids

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.

3. \begin{figure}[h] \end{figure} The uniform lamina \(A B C D E F\), shown shaded in Figure 1, is symmetrical about the line through \(B\) and \(E\). It is formed by removing the isosceles triangle \(F E D\), of height \(6 a\) and base \(8 a\), from the isosceles triangle \(A B C\) of height \(9 a\) and base \(12 a\).

1. Find, in terms of \(a\), the distance of the centre of mass of the lamina from \(A C\). 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 downward vertical.

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 \(k\).

3. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} The uniform rectangular lamina \(A B D E\), shown in Figure 1, has side \(A B\) of length \(2 a\) and side \(B D\) of length \(6 a\). The point \(C\) divides \(B D\) in the ratio 1:2 and the point \(F\) divides \(E A\) in the ratio \(1 : 2\). The rectangular lamina is folded along \(F C\) to produce the folded lamina \(L\), shown in Figure 2.

1. Show that the centre of mass of \(L\) is \(\frac { 16 } { 9 } a\) from \(E F\). The folded lamina, \(L\), is freely suspended from \(C\) and hangs in equilibrium.
2. Find the size of the angle between \(C F\) and the downward vertical.

3. [The centre of mass of a semicircular lamina of radius \(r\) is \(\frac { 4 r } { 3 \pi }\) from the centre.] \begin{figure}[h] \end{figure} Figure 2 shows the uniform lamina \(A B C D E\), such that \(A B D E\) is a square with sides of length \(2 a\) and \(B C D\) is a semicircle with diameter \(B D\).

1. Show that the distance of the centre of mass of the lamina from \(B D\) is \(\frac { 20 a } { 3 ( 8 + \pi ) }\). The lamina is freely suspended from \(D\) and hangs in equilibrium.
2. Find, to the nearest degree, the angle that \(D E\) makes with the downward vertical.

3. \begin{figure}[h] \end{figure} A uniform plane lamina is in the shape of an isosceles triangle \(A B C\), where \(A B = A C\). The mid-point of \(B C\) is \(M , A M = 30 \mathrm {~cm}\) and \(B M = 40 \mathrm {~cm}\). The mid-points of \(A C\) and \(A B\) are \(D\) and \(E\) respectively. The triangular portion \(A D E\) is removed leaving a uniform plane lamina \(B C D E\) as shown in Fig. 2.

1. Show that the centre of mass of the lamina \(B C D E\) is \(6 \frac { 2 } { 3 } \mathrm {~cm}\) from \(B C\).
   (6 marks)
   The lamina \(B C D E\) is freely suspended from \(D\) and hangs in equilibrium.
2. Find, in degrees to one decimal place, the angle which \(D E\) makes with the vertical.
   (3 marks)

7. \includegraphics[max width=\textwidth, alt={}, center]{0d3d35b1-e3c5-47ac-b05e-78cdf1eb3083-4_360_472_1105_815} A uniform plane lamina \(A B C D E\) is formed by joining a uniform square \(A B D E\) with a uniform triangular lamina \(B C D\), of the same material, along the side \(B D\), as shown in Fig. 2. The lengths \(A B , B C\) and \(C D\) are \(18 \mathrm {~cm} , 15 \mathrm {~cm}\) and 15 cm respectively.

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

6. (a) Use algebraic integration to show that the centre of mass of a uniform solid hemisphere of radius \(r\) is at a distance \(\frac { 3 } { 8 } r\) from the centre of its plane face.
[0pt] [You may assume that the volume of a sphere of radius \(r\) is \(\frac { 4 } { 3 } \pi r ^ { 3 }\) ]
(5) \begin{figure}[h] \end{figure} A uniform solid hemisphere of mass \(m\) and radius \(r\) is joined to a uniform solid right circular cone to form a solid \(S\). The cone has mass \(M\), base radius \(r\) and height \(4 r\). The vertex of the cone is \(O\). The plane face of the cone coincides with the plane face of the hemisphere, as shown in Figure 3.
(b) Find the distance of the centre of mass of \(S\) from \(O\). The point \(A\) lies on the circumference of the base of the cone. The solid is placed on a horizontal table with \(O A\) in contact with the table. The solid remains in equilibrium in this position.
(c) Show that \(M \geqslant \frac { 1 } { 10 } m\)

1. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by the curve with equation \(y ^ { 2 } = 9 ( 4 - x )\), the positive \(x\)-axis and the positive \(y\)-axis, as shown in Figure 1. A uniform solid \(S\) is formed by rotating \(R\) through \(360 ^ { \circ }\) about the \(x\)-axis.
Use algebraic integration to find the \(x\) coordinate of the centre of mass of \(S\).

3. \begin{figure}[h] \end{figure} A uniform right circular solid cylinder has radius \(4 a\) and height \(6 a\). A solid hemisphere of radius \(3 a\) is removed from the cylinder forming a solid \(S\). The upper plane face of the cylinder coincides with the plane face of the hemisphere. The centre of the upper plane face of the cylinder is \(O\) and this is also the centre of the plane face of the hemisphere, as shown in Figure 2. Find the distance from \(O\) to the centre of mass of \(S\).
(6)

5. \begin{figure}[h] \end{figure} Figure 3 shows the finite region \(R\) which is bounded by part of the curve with equation \(y = \sin x\), the \(x\)-axis and the line with equation \(x = \frac { \pi } { 2 }\). A uniform solid \(S\) is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis. Using algebraic integration,

1. show that the volume of \(S\) is \(\frac { \pi ^ { 2 } } { 4 }\)
2. find, in terms of \(\pi\), the \(x\) coordinate of the centre of mass of \(S\).

5. \begin{figure}[h] \end{figure} The region \(R\), shown shaded in Figure 3, is bounded by the circle with centre \(O\) and radius \(r\), the line with equation \(x = \frac { 3 } { 5 } r\) and the \(x\)-axis. The region is rotated through one complete revolution about the \(x\)-axis to form a uniform solid \(S\).

1. Use algebraic integration to show that the \(x\) coordinate of the centre of mass of \(S\) is \(\frac { 48 } { 65 } r\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{ae189c40-0071-4a6b-91eb-8ffebe082a04-16_394_643_1311_653} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} A bowl is made from a uniform solid hemisphere of radius 6 cm by removing a hemisphere of radius 5 cm . Both hemispheres have the same centre \(A\) and the same axis of symmetry. The bowl is fixed with its open plane face uppermost and horizontal. Liquid is poured into the bowl. The depth of the liquid is 2 cm , as shown in Figure 4. The mass of the empty bowl is \(5 M \mathrm {~kg}\) and the mass of the liquid is \(2 M \mathrm {~kg}\).
2. Find, to 3 significant figures, the distance from \(A\) to the centre of mass of the bowl with its liquid.

1. \begin{figure}[h] \end{figure} The region \(R\), shown shaded in Figure 1, is bounded by the curve with equation \(y = \frac { 1 } { x }\), the line with equation \(x = 1\), the positive \(x\)-axis and the line with equation \(x = a\) where \(a > 1\) A uniform solid \(S\) is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis.

1. Show that the volume of \(S\) is $$\pi \left( 1 - \frac { 1 } { a } \right)$$
2. Find the \(x\) coordinate of the centre of mass of \(S\).

4. \begin{figure}[h] \end{figure} A uniform right solid cone \(C\) has diameter \(6 a\) and height \(8 a\), as shown in Figure 3.
The solid \(S\) is formed by removing a cone of height \(4 a\) from the top of \(C\) and then removing an identical, inverted cone. The vertex of the removed cone is at the point \(O\) in the centre of the base of \(C\), as shown in Figure 4. \begin{figure}[h] \end{figure}

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 \(S\) and the point \(B\) lies on the circumference of the top of \(S\). The points \(O\), \(A\) and \(B\) all lie in the same vertical plane, as shown in Figure 5. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8a687d17-ec7e-463f-84dd-605f5c230db1-12_248_449_1845_749} \captionsetup{labelformat=empty} \caption{Figure 5}
   \end{figure} The solid \(S\) is freely suspended from the point \(B\) and hangs in equilibrium.
2. Find the size of the angle that \(A B\) makes with the downward vertical.

1. \begin{figure}[h] \end{figure} A uniform lamina is in the shape of the region \(R\).
Region \(R\) is bounded by the curve with equation \(y = x ( x + a )\) where \(a\) is a positive constant, the positive \(x\)-axis and the line with equation \(x = a\), as shown shaded in Figure 1. Find the \(\boldsymbol { y }\) coordinate of the centre of mass of the lamina.

4. \begin{figure}[h] \end{figure} A thin uniform right hollow cylinder, of radius \(2 a\) and height \(k a\), has a base but no top. A thin uniform hemispherical shell, also of radius \(2 a\), is made of the same material as the cylinder. The hemispherical shell is attached to the end of the cylinder forming a container \(C\). The open circular rim of the cylinder coincides with the rim of the hemispherical shell. The centre of the base of \(C\) is \(O\), as shown in Figure 3.

1. Show that the distance from \(O\) to the centre of mass of \(C\) is $$\frac { \left( k ^ { 2 } + 4 k + 4 \right) } { 2 ( k + 3 ) } a$$ The container is placed with its circular base on a plane which is inclined at \(30 ^ { \circ }\) to the horizontal. The plane is sufficiently rough to prevent \(C\) from sliding. The container is on the point of toppling.
2. Find the value of \(k\).

1. \begin{figure}[h] \end{figure} The shaded region R is bounded by the x -axis, the line with equation \(\mathrm { x } = 1\), the curve with equation \(y = 1 + \sqrt { x }\) and the y-axis, as shown in Figure 1. The unit of length on both of the axes is 1 m . The region R is rotated through \(2 \pi\) radians about the x-axis to form a solid of revolution which is used to model a uniform solid \(S\). Show, using the model and algebraic integration, that

1. the volume of \(S\) is \(\frac { 17 \pi } { 6 } \mathrm {~m} ^ { 3 }\)
2. the centre of mass of \(S\) is \(\frac { 49 } { 85 } \mathrm {~m}\) from 0 . \includegraphics[max width=\textwidth, alt={}, center]{631b78c4-2763-4a1e-9d30-2f301fe3af2e-02_2264_41_314_1987}

3. A square ABCD of side 4a is made from thin uniform cardboard. The centre of the square is 0 . A circle with centre 0 and radius \(\frac { 7 a } { 4 }\) is then removed from the square to form a template T, shown shaded in Figure 3.
A right conical shell, with no base, has radius \(\frac { 7 a } { 4 }\) and perpendicular height \(6 a\).
The shell is made of the same thin uniform cardboard as T.
The shell is attached to T so that the circumference of the end of the shell coincides with the circumference of the circle centre 0 , to form the hat H , shown in Figure 4.
[0pt] [The surface area of a right conical shell of radius r and slant height I is \(\pi r l\).]

1. Show that the exact distance of the centre of mass of H from O is $$\frac { 175 \pi a } { ( 63 \pi + 128 ) }$$ A fixed rough plane is inclined to the horizontal at an angle \(\alpha\). The hat H is placed on the plane, with ABCD in contact with the plane, and AB parallel to a line of greatest slope of the plane. The plane is sufficiently rough to prevent the hat from sliding down the plane. Given that the hat is on the point of toppling,
2. find the exact value of \(\tan \alpha\), giving your answer in simplest form.

3. \begin{figure}[h] \end{figure} The shaded region in Figure 2 is bounded by the \(x\)-axis, the line with equation \(x = 2\) and the curve with equation \(y = \frac { 1 } { 4 } x ( 3 - x )\).
This region is rotated through \(2 \pi\) radians about the \(x\)-axis, to form a solid of revolution which is used to model a uniform solid \(S\). The volume of \(S\) is \(\frac { 2 } { 5 } \pi\)

1. Use the model and algebraic integration to show that the \(x\) coordinate of the centre of mass of \(S\) is \(\frac { 31 } { 24 }\) The solid \(S\) is placed with its circular face on a rough plane which is inclined at \(\alpha ^ { \circ }\) to the horizontal. The plane is sufficiently rough to prevent \(S\) from sliding. The solid \(S\) is on the point of toppling.
2. Find the value of \(\alpha\)

1. (a) Use algebraic integration to show that the centre of mass of a uniform semicircular disc of radius \(r\) and centre \(O\) is at a distance \(\frac { 4 r } { 3 \pi }\) from the diameter through \(O\) [You may assume, without proof, that the area of a circle of radius \(r\) is \(\pi r ^ { 2 }\) ]

A uniform lamina L is in the shape of a semicircle with centre \(B\) and diameter \(A C = 8 a\). The semicircle with diameter \(A B\) is removed from \(L\) and attached to the straight edge \(B C\) to form the template \(T\), shown shaded in Figure 4. \begin{figure}[h] \end{figure} The distance of the centre of mass of \(T\) from \(A C\) is \(d\).
(b) Show that \(d = \frac { 4 a } { \pi }\) The template \(T\) is freely suspended from \(A\) and hangs in equilibrium with \(A C\) at an angle \(\theta\) to the downward vertical.
(c) Find the exact value of \(\tan \theta\)

2. \begin{figure}[h] \end{figure} A uniform lamina \(L\) is in the shape of an equilateral triangle of side \(2 a\). The lamina is placed in the \(x y\)-plane with one vertex at the origin \(O\) and an axis of symmetry along the \(x\)-axis, as shown in Figure 1. Use algebraic integration to find the \(x\) coordinate of the centre of mass of \(L\).

7. \begin{figure}[h] \end{figure} Diagram not drawn to scale A uniform right circular solid cylinder has radius \(3 a\) and height \(2 a\). A right circular cone of height \(\frac { 3 a } { 2 }\) and base radius \(2 a\) is removed from the cylinder to form a solid \(S\), as shown in Figure 4. The plane face of the cone coincides with the upper plane face of the cylinder and the centre \(O\) of the plane face of the cone is also the centre of the upper plane face of the cylinder.

1. Show that the distance of the centre of mass of \(S\) from \(O\) is \(\frac { 69 a } { 64 }\). The point \(A\) is on the open face of \(S\) such that \(O A = 3 a\), as shown in Figure 4. The solid is now suspended from \(A\) and hangs freely in equilibrium.
2. Find the angle between \(O A\) and the horizontal.
   (3) \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{e5b08946-7311-4cf7-9c5f-5f309a1feed7-13_543_826_1653_557} \captionsetup{labelformat=empty} \caption{Figure 5}
   \end{figure} The solid is now placed on a rough inclined plane with the face through \(A\) in contact with the inclined plane, as shown in Figure 5. The solid rests in equilibrium on this plane. The coefficient of friction between the plane and \(S\) is 0.6 and the plane is inclined at an angle \(\phi ^ { \circ }\) to the horizontal. Given that \(S\) is on the point of sliding down the plane,
3. show that \(\phi = 31\) to 2 significant figures.

6. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by part of the curve with equation \(y = x ^ { 2 } + 3\), the \(x\)-axis, the \(y\)-axis and the line with equation \(x = 2\), as shown in Figure 4. The unit of length on each axis is one centimetre. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid \(S\).
Using algebraic integration,

1. show that the volume of \(S\) is \(\frac { 202 } { 5 } \pi \mathrm {~cm} ^ { 3 }\),
2. show that, to 2 decimal places, the centre of mass of \(S\) is 1.30 cm from \(O\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{b7cfcf0a-8f54-4350-8e07-a3b51d94d0f2-11_478_472_1407_762} \captionsetup{labelformat=empty} \caption{Figure 5}
   \end{figure} A uniform right circular solid cone, of base radius 7 cm and height 6 cm , is joined to \(S\) to form a solid \(T\). The base of the cone coincides with the larger plane face of \(S\), as shown in Figure 5. The vertex of the cone is \(V\).
   The mass per unit volume of \(S\) is twice the mass per unit volume of the cone.
3. Find the distance from \(V\) to the centre of mass of \(T\). The point \(A\) lies on the circumference of the base of the cone. The solid \(T\) is suspended from \(A\) and hangs freely in equilibrium.
4. Find the size of the angle between \(V A\) and the vertical.

1. The region enclosed by the curve with equation \(y = \frac { 1 } { 2 } \sqrt { x }\), the \(x\)-axis and the lines \(x = 2\) and \(x = 4\), is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid \(S\). Use algebraic integration to find the exact value of the \(x\) coordinate of the centre of mass of \(S\).
   (6)

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