Lamina with particle attached

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 \(k\).

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}

5. A uniform solid hemisphere has radius \(r\). The centre of the plane face of the hemisphere is \(O\).

1. Use algebraic integration to show that the distance from \(O\) to the centre of mass of the hemisphere is \(\frac { 3 } { 8 } r\).
   [0pt] [You may assume that the volume of a sphere of radius \(r\) is \(\frac { 4 } { 3 } \pi r ^ { 3 }\) ] \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{2cf74ba3-857a-4ce9-ab5b-e6203b279161-14_378_602_740_671} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} A solid \(S\) is formed by joining a uniform solid hemisphere of radius \(a\) to a uniform solid hemisphere of radius \(\frac { 1 } { 2 } a\). The plane faces of the hemispheres are joined together so that their centres coincide at \(O\), as shown in Figure 1. The mass per unit volume of the smaller hemisphere is \(k\) times the mass per unit volume of the larger hemisphere.
2. Find the distance from \(O\) to the centre of mass of \(S\). When \(S\) is placed on a horizontal plane with any point on the curved surface of the larger hemisphere in contact with the plane, \(S\) remains in equilibrium.
3. Find the value of \(k\).

2. A metal sign is formed by removing triangle \(B C D\) from a rectangular lamina \(A C E F\) made of uniform material, and adding a quarter circle XYZ, made of the same uniform material, with a particle attached to its vertex at \(Y\). The sign is supported by two light chains fixed at \(E\) and \(F\). The quarter circle has radius 24 cm and the particle at \(Y\) has a mass equal to half of that of the removed triangle. \(X D\) is parallel to \(A C\) and \(B Z\) is parallel to \(A F\). The dimensions, in cm , are as shown in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{3578a810-46da-4d9e-a98f-248be72a517a-3_885_636_712_715}

1. Calculate the distance of the centre of mass of the sign from
   1. \(A F\),
   2. \(A C\).
2. The support at \(F\) comes loose so that the sign is freely suspended at \(E\) by one chain alone. Given that it then hangs in equilibrium, calculate the angle that \(E F\) makes with the vertical.