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

2. \begin{figure}[h] \end{figure} A uniform solid right circular cone \(R\), with vertex \(V\), has base radius \(4 r\) and height \(4 h\). A right circular cone \(S\), also with vertex \(V\) and the same axis of symmetry as \(R\), has base radius \(3 r\) and height \(3 h\). The cone \(S\) is cut away from the cone \(R\) leaving a solid \(T\). The centre of the larger plane face of \(T\) is \(O\). Figure 1 shows the solid \(T\).

1. Find the distance from \(O\) to the centre of mass of \(T\). The point \(A\) lies on the circumference of the smaller plane face of \(T\). The solid is freely suspended from \(A\) and hangs in equilibrium. Given that \(h = r\)
2. find the size of the angle between \(O A\) and the downward vertical.

6. A uniform solid right circular cone has base radius \(r\) and height \(h\).

1. Use algebraic integration to show that the distance of the centre of mass of the cone from its vertex is \(\frac { 3 } { 4 } h\).
   [0pt] [You may assume that the volume of a cone of base radius \(r\) and height \(h\) is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\) ] \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{2273ca38-1e16-44ab-ae84-f4c576cbb8f9-20_394_716_632_621} \captionsetup{labelformat=empty} \caption{Figure 3}
   \end{figure} A solid \(S\) is formed by joining a uniform right circular solid cone of mass \(5 m\) to a uniform solid hemisphere, of radius \(r\) and mass \(k m\) where \(k < 20\). The cone has base radius \(r\) and height \(6 r\). The plane face of the cone coincides with the plane face of the hemisphere. The centre of the plane face of the cone is \(O\) and the point \(A\) is on the circular edge of this plane face, as shown in Figure 3.
2. Find the distance from \(O\) to the centre of mass of \(S\). The solid is suspended from \(A\) and hangs freely in equilibrium. The angle between the axis of the cone and the horizontal is \(30 ^ { \circ }\).
3. Find, to the nearest whole number, the value of \(k\).

4. (a) Use algebraic integration to show that the centre of mass of a uniform solid hemisphere of radius \(a\) is a distance \(\frac { 3 } { 8 } a\) from the centre of its plane face.
[0pt] [You may assume that the volume of a sphere of radius \(r\) is \(\frac { 4 } { 3 } \pi r ^ { 3 }\) ] \begin{figure}[h] \end{figure} A uniform solid hemisphere has mass \(m\) and radius \(a\). A particle of mass \(k m\) is attached to a point \(A\) on the circumference of the plane face of the hemisphere to form the loaded solid \(S\). The centre of the plane face of the hemisphere is the point \(O\), as shown in Figure 4. The loaded solid \(S\) is placed on a horizontal plane. The curved surface of \(S\) is in contact with the plane and \(S\) rests in equilibrium with \(O A\) making an angle \(\alpha\) with the horizontal, where \(\tan \alpha = \sqrt { 3 }\) (b) Find the exact value of \(k\). \includegraphics[max width=\textwidth, alt={}, center]{ace84823-db30-463e-b24b-f0cd7df73746-09_2255_50_314_34}

1. \begin{figure}[h] \end{figure} A hollow toy is formed by joining a uniform right circular conical shell \(C\), with radius \(4 a\) and height \(3 a\), to a uniform hemispherical shell \(H\), with radius \(4 a\). The circular edge of \(C\) coincides with the circular edge of \(H\), as shown in Figure 1. The mass per unit area of \(C\) is \(\lambda\) and the mass per unit area of \(H\) is \(k \lambda\) where \(k\) is a constant.
Given that the centre of mass of the toy is a distance \(4 a\) from the vertex of the cone, find the value of \(k\).

1. The finite region enclosed by the curve with equation \(y = 3 - \sqrt { x }\) and the lines \(x = 0\) and \(y = 0\) is rotated through \(2 \pi\) radians about the \(x\)-axis, to form a uniform solid \(S\).

Use algebraic integration to

1. show that the volume of \(S\) is \(\frac { 27 } { 2 } \pi\)
2. find the \(x\) coordinate of the centre of mass of \(S\).

1. (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 \(\frac { 4 } { 3 } \pi r ^ { 3 }\) ]

\begin{figure}[h] \end{figure} A uniform solid hemisphere of radius \(r\) is joined to a uniform solid right circular cone made of the same material to form a toy. The cone has base radius \(r\) and height \(k r\). The centre of the base of the cone is \(O\). The plane face of the cone coincides with the plane face of the hemisphere, as shown in Figure 3. The toy can rest in equilibrium on a horizontal plane with any point of the curved surface of the hemisphere in contact with the plane.
(b) Find the exact value of \(k\)

1. In this question you must show all stages in your working.

Solutions relying on calculator technology are not acceptable. \begin{figure}[h] \end{figure} The finite region \(R\), shown shaded in Figure 1, is bounded by the \(x\)-axis, the line with equation \(x = 3\), the curve with equation \(y = \sqrt { ( x + 1 ) }\) and the \(y\)-axis.
Find the \(\boldsymbol { y }\) coordinate of the centre of mass of a uniform lamina in the shape of \(R\).

1. A uniform solid right circular cone \(C\) has base radius \(r\), height \(H\) and vertex \(V\). A uniform solid \(S\), shown in Figure 3, is formed by removing from \(C\) a uniform solid right circular cone of height \(h ( h < H )\) that has the same base and axis of symmetry as \(C\).

\begin{figure}[h] \end{figure}

1. Show that the distance of the centre of mass of \(S\) from \(V\) is $$\frac { 1 } { 4 } ( 3 H - h )$$ The solid \(S\) is suspended by two vertical light strings. The first string is attached to \(S\) at \(V\) and the second string is attached to \(S\) at a point on the circumference of the circular base of \(S\).
   The solid \(S\) hangs freely in equilibrium with its axis of symmetry horizontal.
   The tension in the first string is \(T _ { 1 }\) and the tension in the second string is \(T _ { 2 }\)
2. Find \(\frac { T _ { 1 } } { T _ { 2 } }\), giving your answer in terms of \(H\) and \(h\), in its simplest form.

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]{bb73b211-7629-4ed7-9b71-91841c29bb85-20_483_469_1402_767} \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.

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   Q6

   VIIIV SIHI NI JAIIM ION OC

   VIIIV SIHI NI JIHMM ION OO

   VI4V SIHI NI JIIYM IONOO

6. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by the curve with equation \(y = \frac { 1 } { 2 x ^ { 2 } }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\), as shown in Figure 4. The unit of length on each axis is 1 m . A uniform solid \(S\) has the shape made by rotating \(R\) through \(360 ^ { \circ }\) about the \(x\)-axis.

1. Show that the centre of mass of \(S\) is \(\frac { 2 } { 7 } \mathrm {~m}\) from its larger plane face. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 5} \includegraphics[alt={},max width=\textwidth]{ab85ec29-b1fc-45a9-9343-09feb33ab6c5-010_616_431_1420_778}
   \end{figure} A sporting trophy \(T\) is a uniform solid hemisphere \(H\) joined to the solid \(S\). The hemisphere has radius \(\frac { 1 } { 2 } \mathrm {~m}\) and its plane face coincides with the larger plane face of \(S\), as shown in Figure 5. Both \(H\) and \(S\) are made of the same material.
2. Find the distance of the centre of mass of \(T\) from its plane face.

6. Figure 2 \includegraphics[max width=\textwidth, alt={}, center]{c4b453e7-8a32-458b-8041-58c9e4ef9533-5_691_1067_241_584} A uniform solid cylinder has radius \(2 a\) and height \(\frac { 3 } { 2 } a\). A hemisphere of radius \(a\) is removed from the cylinder. The plane face of the hemisphere coincides with the upper plane face of the cylinder, and the centre \(O\) of the hemisphere is also the centre of this plane face, as shown in Fig. 2. The remaining solid is \(S\).

1. Find the distance of the centre of mass of \(S\) from \(O\).
   (6) The lower plane face of \(S\) rests in equilibrium on a desk lid which is inclined at an angle \(\theta\) to the horizontal. Assuming that the lid is sufficiently rough to prevent \(S\) from slipping, and that \(S\) is on the point of toppling when \(\theta = \alpha\),
2. find the value of \(\alpha\).
   (3) Given instead that the coefficient of friction between \(S\) and the lid is 0.8 , and that \(S\) is on the point of sliding down the lid when \(\theta = \beta\),
3. find the value of \(\beta\).
   (3)

6. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by the curve with equation \(y = \frac { 1 } { 2 x ^ { 2 } }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\), as shown in Figure 4. The unit of length on each axis is 1 m . A uniform solid \(S\) has the shape made by rotating \(R\) through \(360 ^ { \circ }\) about the \(x\)-axis.

1. Show that the centre of mass of \(S\) is \(\frac { 2 } { 7 } \mathrm {~m}\) from its larger plane face. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 5} \includegraphics[alt={},max width=\textwidth]{25b3ece7-69ed-4ec4-a6c7-4cd83ec2cc5e-09_616_431_1420_778}
   \end{figure} A sporting trophy \(T\) is a uniform solid hemisphere \(H\) joined to the solid \(S\). The hemisphere has radius \(\frac { 1 } { 2 } \mathrm {~m}\) and its plane face coincides with the larger plane face of \(S\), as shown in Figure 5. Both \(H\) and \(S\) are made of the same material.
2. Find the distance of the centre of mass of \(T\) from its plane face.

6. \begin{figure}[h] \end{figure} The region \(R\) is bounded by part of the curve with equation \(y = 4 - x ^ { 2 }\), the positive \(x\)-axis and the positive \(y\)-axis, as shown in Figure 3. The unit of length on both axes is one metre. A uniform solid \(S\) is formed by rotating \(R\) through \(360 ^ { \circ }\) about the \(x\)-axis.

1. Show that the centre of mass of \(S\) is \(\frac { 5 } { 8 } \mathrm {~m}\) from \(O\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8374fa0f-cb28-497f-8696-877d7d0762f1-09_702_584_1138_676} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} Figure 4 shows a cross section of a uniform solid \(P\) consisting of two components, a solid cylinder \(C\) and the solid \(S\). The cylinder \(C\) has radius 4 m and length \(l\) metres. One end of \(C\) coincides with the plane circular face of \(S\). The point \(A\) is on the circumference of the circular face common to \(C\) and \(S\). When the solid \(P\) is freely suspended from \(A\), the solid \(P\) hangs with its axis of symmetry horizontal.
2. Find the value of \(l\).

3. \begin{figure}[h] \end{figure} A bowl \(B\) consists of a uniform solid hemisphere, of radius \(r\) and centre \(O\), from which is removed a solid hemisphere, of radius \(\frac { 2 } { 3 } r\) and centre \(O\), as shown in Figure 1.

1. Show that the distance of the centre of mass of \(B\) from \(O\) is \(\frac { 65 } { 152 } r\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{d831556d-fdf3-4639-9a89-6d3b372d3446-05_526_1014_1292_478} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} The bowl \(B\) has mass \(M\). A particle of mass \(k M\) is attached to a point \(P\) on the outer rim of \(B\). The system is placed with a point \(C\) on its outer curved surface in contact with a horizontal plane. The system is in equilibrium with \(P , O\) and \(C\) in the same vertical plane. The line \(O P\) makes an angle \(\theta\) with the horizontal as shown in Figure 2. Given that \(\tan \theta = \frac { 4 } { 5 }\),
2. find the exact value of \(k\). January 2010

7. Diagram NOT accurately drawn \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by the curve with equation \(y = \frac { 1 } { 2 } x ( 6 - x )\), the \(x\)-axis and the line \(x = 2\), as shown in Figure 1. The unit of length on both axes is 1 cm . A uniform solid \(P\) is formed by rotating \(R\) through \(360 ^ { \circ }\) about the \(x\)-axis.

1. Show that the centre of mass of \(P\) is, to 3 significant figures, 1.42 cm from its plane face. The uniform solid \(P\) is placed with its plane face on an inclined plane which makes an angle \(\theta\) with the horizontal. Given that the plane is sufficiently rough to prevent \(P\) from sliding and that \(P\) is on the point of toppling when \(\theta = \alpha\),
2. find the angle \(\alpha\). Given instead that \(P\) is on the point of sliding down the plane when \(\theta = \beta\) and that the coefficient of friction between \(P\) and the plane is 0.3 ,
3. find the angle \(\beta\).

2. \begin{figure}[h] \end{figure} A uniform solid consists of a right circular cone of radius \(r\) and height \(k r\), where \(k > \sqrt { } 3\), 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 1.

1. Show that the distance of the centre of mass of the solid from \(O\) is $$\frac { \left( k ^ { 2 } - 3 \right) r } { 4 ( k + 2 ) }$$ 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. The angle between \(A O\) and the vertical is \(\theta\), where \(\tan \theta = \frac { 11 } { 14 }\)
2. Find the value of \(k\).

5. \begin{figure}[h] \end{figure} A toy is formed by joining a uniform solid right circular cone, of base radius \(3 r\) and height \(4 r\), to a uniform solid cylinder, also of radius \(3 r\) and height \(4 r\). The cone and the cylinder are made from the same material, and the plane face of the cone coincides with a plane face of the cylinder, as shown in Fig. 2. The centre of this plane face is \(O\).

1. Find the distance of the centre of mass of the toy from \(O\). The point \(A\) lies on the edge of the plane face of the cylinder which forms the base of the toy. The toy is suspended from \(A\) and hangs in equilibrium.
2. Find, in degrees to one decimal place, the angle between the axis of symmetry of the toy and the vertical. The toy is placed with the curved surface of the cone on horizontal ground.
3. Determine whether the toy will topple.
   (4)

2. A closed container \(C\) consists of a thin uniform hollow hemispherical bowl of radius \(a\), together with a lid. The lid is a thin uniform circular disc, also of radius \(a\). The centre \(O\) of the disc coincides with the centre of the hemispherical bowl. The bowl and its lid are made of the same material.

1. Show that the centre of mass of \(C\) is at a distance \(\frac { 1 } { 3 } a\) from \(O\). The container \(C\) has mass \(M\). A particle of mass \(\frac { 1 } { 2 } M\) is attached to the container at a point \(P\) on the circumference of the lid. The container is then placed with a point of its curved surface in contact with a horizontal plane. The container rests in equilibrium with \(P , O\) and the point of contact in the same vertical plane.
2. Find, to the nearest degree, the angle made by the line \(P O\) with the horizontal.

4. \begin{figure}[h] \end{figure} A uniform solid hemisphere, of radius \(6 a\) and centre \(O\), has a solid hemisphere of radius \(2 a\), and centre \(O\), removed to form a bowl \(B\) as shown in Figure 3.

1. Show that the centre of mass of \(B\) is \(\frac { 30 } { 13 } a\) from \(O\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{f07b8a65-ccb5-423f-96cc-b303bd05ad1f-07_735_614_1126_662} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} The bowl \(B\) is fixed to a plane face of a uniform solid cylinder made from the same material as \(B\). The cylinder has radius \(2 a\) and height \(6 a\) and the combined solid \(S\) has an axis of symmetry which passes through \(O\), as shown in Figure 4.
2. Show that the centre of mass of \(S\) is \(\frac { 201 } { 61 } a\) from \(O\). The plane surface of the cylindrical base of \(S\) is placed on a rough plane inclined at \(12 ^ { \circ }\) to the horizontal. The plane is sufficiently rough to prevent slipping.
3. Determine whether or not \(S\) will topple. \section*{
   \includegraphics[max width=\textwidth, alt={}]{f07b8a65-ccb5-423f-96cc-b303bd05ad1f-08_56_366_251_178}
   }

2. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by the curve with equation \(y = 9 - x ^ { 2 }\), 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. Find the \(x\)-coordinate of the centre of mass of \(S\).

6. (a) A uniform lamina is in the shape of a quadrant of a circle of radius \(a\). Show, by integration, that the centre of mass of the lamina is at a distance of \(\frac { 4 a } { 3 \pi }\) from each of its straight edges. \begin{figure}[h] \end{figure} A second uniform lamina \(A B C D E F A\) is shown shaded in Figure 3. The straight sides \(A C\) and \(A E\) are perpendicular and \(A C = A E = 2 a\). In the figure, the midpoint of \(A C\) is \(B\), the midpoint of \(A E\) is \(F\), and \(A B D F\) and \(D G E F\) are squares of side \(a\). \(B C D\) is a quadrant of a circle with centre \(B\). \(D G E\) is a quadrant of a circle with centre \(G\).
(b) Find the distance of the centre of mass of the lamina from the side \(A E\). The lamina is smoothly hinged to a horizontal axis which passes through \(E\) and is perpendicular to the plane of the lamina. The lamina has weight \(W\) newtons. The lamina is held in equilibrium in a vertical plane, with \(A\) vertically above \(E\), by a horizontal force of magnitude \(X\) newtons applied at \(C\).
(c) Find \(X\) in terms of \(W\).

5. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by the curve with equation \(y = ( x + 1 ) ^ { 2 }\), the \(x\)-axis, the \(y\)-axis and the line with equation \(x = 2\), as shown in Figure 3. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid \(S\).

1. Use algebraic integration to find the \(x\) coordinate of the centre of mass of \(S\).
   (8) \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{f6ab162c-8473-4464-ad62-87a359d85ab3-08_558_492_1263_703} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} A uniform solid hemisphere is fixed to \(S\) to form a solid \(T\). The hemisphere has the same radius as the smaller plane face of \(S\) and its plane face coincides with the smaller plane face of \(S\), as shown in Figure 4. The mass per unit volume of the hemisphere is 10 times the mass per unit volume of \(S\). The centre of the circular plane face of \(T\) is \(A\). All lengths are measured in centimetres.
2. Find the distance of the centre of mass of \(T\) from \(A\).

5. \begin{figure}[h] \end{figure} Figure 4 shows the region \(R\) bounded by part of the curve with equation \(y = \cos x\), the \(x\)-axis and the \(y\)-axis. 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 \(\frac { \pi ^ { 2 } } { 4 }\)
2. Find, using algebraic integration, the \(x\) coordinate of the centre of mass of \(S\).