Solid with removed cone from cone or cylinder

7 \includegraphics[max width=\textwidth, alt={}, center]{8dda6c21-7cb5-43b6-9a34-485bdf4042c4-12_732_581_260_774} A uniform solid cone has height 1.2 m and base radius 0.5 m . A uniform object is made by drilling a cylindrical hole of radius 0.2 m through the cone along the axis of symmetry (see diagram).

1. Show that the height of the object is 0.72 m and that the volume of the cone removed by the drilling is \(0.0352 \pi \mathrm {~m} ^ { 3 }\).
   [0pt] [The volume of a cone is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\).]
2. Find the distance of the centre of mass of the object from its base.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

3 One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(4 m g\), is attached to a fixed point \(O\). The other end of the string is attached to a particle of mass \(m\). The particle moves in a horizontal circle with a constant angular speed \(\sqrt { \frac { \mathrm { g } } { \mathrm { a } } }\) with the string inclined at an angle \(\theta\) to the downward vertical through \(O\). The length of the string during this motion is \(( \mathrm { k } + 1 ) \mathrm { a }\).

1. Find the value of \(k\).
2. Find the value of \(\cos \theta\). \includegraphics[max width=\textwidth, alt={}, center]{1c53c407-25ea-43fc-a571-74ba1fffea8f-06_584_695_264_667} The diagram shows the cross-section \(A B C D\) of a uniform solid object which is formed by removing a cone with cross-section \(D C E\) from the top of a larger cone with cross-section \(A B E\). The perpendicular distance between \(A B\) and \(D C\) is \(h\), the diameter \(A B\) is \(6 r\) and the diameter \(D C\) is \(2 r\).
   1. Find an expression, in terms of \(h\), for the distance of the centre of mass of the solid object from \(A B\).
      The object is freely suspended from the point \(B\) and hangs in equilibrium. The angle between \(A B\) and the downward vertical through \(B\) is \(\theta\).
   2. Given that \(h = \frac { 13 } { 4 } r\), find the value of \(\tan \theta\).

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.

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.

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.

3. \begin{figure}[h] \end{figure} A uniform solid \(S\) is formed by taking a uniform solid right circular cone, of base radius \(2 r\) and height \(2 h\), and removing the cone, with base radius \(r\) and height \(h\), which has the same vertex as the original cone, as shown in Figure 1.

1. Show that the distance of the centre of mass of \(S\) from its larger plane face is \(\frac { 11 } { 28 } h\). The solid \(S\) lies with its larger plane face on a rough table which is inclined at an angle \(\theta ^ { \circ }\) to the horizontal. The table is sufficiently rough to prevent \(S\) from slipping. Given that \(h = 2 r\),
2. find the greatest value of \(\theta\) for which \(S\) does not topple.

4. \begin{figure}[h] \end{figure} A stand used to reach high shelves in a storeroom is in the shape of a frustum of a cone. It is modelled as a uniform solid formed by removing a right circular cone of height \(2 h\) from a similar cone of height \(3 h\) and base radius \(3 r\) as shown in Figure 1.

1. Show that the centre of mass of the stand is a distance of \(\frac { 33 } { 76 } h\) from its larger plane face.
   (7 marks)
   The stand is stored hanging in equilibrium from a point on the circumference of the larger plane face. Given that \(h = 2 r\),
2. find, correct to the nearest degree, the acute angle which the plane faces of the stand make with the vertical.
   (4 marks)

2. You are given that the centre of mass of a uniform solid cone of height \(h\) and base radius \(r\) is at a height of \(\frac { 1 } { 4 } h\) above its base. The diagram shows a solid conical frustum. It is formed by taking a uniform right circular cone, of base radius \(3 x\) and height \(6 y\), and removing a smaller cone, of base radius \(x\), with the same vertex. \includegraphics[max width=\textwidth, alt={}, center]{d7f600c5-af4a-4708-bfd9-92b37a95c634-3_490_903_1937_575} Show that the distance of the centre of mass of the frustum from its base along the axis of symmetry is \(\frac { 18 } { 13 } y\).

2. (a) Prove that the centre of mass of a uniform solid cone of height \(h\) and base radius \(b\) is at a height of \(\frac { 1 } { 4 } h\) above its base.
(b) A uniform solid cone \(C _ { 1 }\) has height 3 m and base radius 2 m . A smaller cone \(C _ { 2 }\) of height 2 m and base radius 1 m is contained symmetrically inside \(C _ { 1 }\). The bases of \(C _ { 1 }\) and \(C _ { 2 }\) have a common centre and the axis of \(C _ { 2 }\) is part of the axis of \(C _ { 1 }\). If \(C _ { 2 }\) is removed from \(C _ { 1 }\), show that the centre of mass of the remaining solid is at a distance of \(\frac { 11 } { 5 } \mathrm {~m}\) from the vertex of \(C _ { 1 }\).
(c) The remaining solid is suspended from a string which is attached to a point on the outer curved surface at a distance of \(\frac { 1 } { 3 } \sqrt { 13 } \mathrm {~m}\) from the vertex of \(C _ { 1 }\). Given that the axis of symmetry is inclined at an angle of \(\alpha\) to the vertical, find \(\tan \alpha\).

\includegraphics{figure_7} A uniform solid cone has height \(1.2 \text{ m}\) and base radius \(0.5 \text{ m}\). A uniform object is made by drilling a cylindrical hole of radius \(0.2 \text{ m}\) through the cone along the axis of symmetry (see diagram).

1. Show that the height of the object is \(0.72 \text{ m}\) and that the volume of the cone removed by the drilling is \(0.0352\pi \text{ m}^3\). [4]

[The volume of a cone is \(\frac{1}{3}\pi r^2 h\).]

1. Find the distance of the centre of mass of the object from its base. [6]

\includegraphics{figure_4} The diagram shows the cross-section \(ABCD\) of a uniform solid object which is formed by removing a cone with cross-section \(DCE\) from the top of a larger cone with cross-section \(ABE\). The perpendicular distance between \(AB\) and \(DC\) is \(h\), the diameter \(AB\) is \(6r\) and the diameter \(DC\) is \(2r\).

1. Find an expression, in terms of \(h\), for the distance of the centre of mass of the solid object from \(AB\). [4]

The object is freely suspended from the point \(B\) and hangs in equilibrium. The angle between \(AB\) and the downward vertical through \(B\) is \(\theta\).

1. Given that \(h = \frac{13}{4}r\), find the value of \(\tan\theta\). [2]

\includegraphics{figure_3} A container is formed by removing a right circular solid cone of height \(4l\) from a uniform solid right circular cylinder of height \(6l\). The centre \(O\) of the plane face of the cone coincides with the centre of a plane face of the cylinder and the axis of the cone coincides with the axis of the cylinder, as shown in Figure 3. The cylinder has radius \(2l\) and the base of the cone has radius \(l\).

1. Find the distance of the centre of mass of the container from \(O\). [6]

\includegraphics{figure_4} The container is placed on a plane which is inclined at an angle \(\theta°\) to the horizontal. The open face is uppermost, as shown in Figure 4. The plane is sufficiently rough to prevent the container from sliding. The container is on the point of toppling.

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

\includegraphics{figure_3} An ornament \(S\) is formed by removing a solid right circular cone, of radius \(r\) and height \(\frac{1}{2}h\), from a solid uniform cylinder, of radius \(r\) and height \(h\), as shown in Fig. 3.

1. Show that the distance of the centre of mass \(S\) from its plane face is \(\frac{19}{30}h\). [7]

The ornament is suspended from a point on the circular rim of its open end. It hangs in equilibrium with its axis of symmetry inclined at an angle \(\alpha\) to the horizontal. Given that \(h = 4r\),

1. find, in degrees to one decimal place, the value of \(\alpha\). [4]

\includegraphics{figure_2} Figure 2 shows the cross-section \(AVBC\) of the solid \(S\) formed when a uniform right circular cone of base radius \(a\) and height \(a\), is removed from a uniform right circular cone of base radius \(a\) and height \(2a\). Both cones have the same axis \(VCO\), where \(O\) is the centre of the base of each cone.

1. Show that the distance of the centre of mass of \(S\) from the vertex \(V\) is \(\frac{5}{4}a\). [5]

The mass of \(S\) is \(M\). A particle of mass \(kM\) is attached to \(S\) at \(B\). The system is suspended by a string attached to the vertex \(V\), and hangs freely in equilibrium. Given that \(VA\) is at an angle \(45°\) to the vertical through \(V\),

1. find the value of \(k\). [5]

A uniform solid right circular cone has height \(h\) and base radius \(r\). The top part of the cone is removed by cutting through the cone parallel to the base at a height \(\frac{h}{2}\). \includegraphics{figure_5}

1. Show that the centre of mass of the remaining solid is at a height \(\frac{11h}{56}\) above the base, along its axis of symmetry. [7 marks]

The remaining part of the solid is suspended from the point \(D\) on the circumference of its smaller circular face, and the axis of symmetry then makes an angle \(\alpha\) with the vertical, where \(\tan \alpha = \frac{1}{2}\).

1. Find the value of the ratio \(h : r\). [5 marks]