Solid of revolution MI

2 A uniform solid of revolution is formed by rotating the region bounded by the \(x\)-axis, the line \(x = 1\) and the curve \(y = x ^ { 2 }\) for \(0 \leqslant x \leqslant 1\), about the \(x\)-axis. The units are metres, and the density of the solid is \(5400 \mathrm {~kg} \mathrm {~m} ^ { - 3 }\). Find the moment of inertia of this solid about the \(x\)-axis.

5 The region bounded by the \(x\)-axis, the line \(x = 8\) and the curve \(y = x ^ { \frac { 1 } { 3 } }\) for \(0 \leqslant x \leqslant 8\), is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid of revolution. The unit of length is the metre, and the density of the solid is \(350 \mathrm {~kg} \mathrm {~m} ^ { - 3 }\).

1. Show that the mass of the solid is \(6720 \pi \mathrm {~kg}\).
2. Find the \(x\)-coordinate of the centre of mass of the solid.
3. Find the moment of inertia of the solid about the \(x\)-axis.

5 In this question, \(a\) and \(k\) are positive constants.
The region enclosed by the curve \(y = a \mathrm { e } ^ { - \frac { x } { a } }\) for \(0 \leqslant x \leqslant k a\), the \(x\)-axis, the \(y\)-axis and the line \(x = k a\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid of mass \(m\). Show that the moment of inertia of this solid about the \(x\)-axis is \(\frac { 1 } { 4 } m a ^ { 2 } \left( 1 + \mathrm { e } ^ { - 2 k } \right)\).

5 The region bounded by the curve \(y = \sqrt { a x }\) for \(a \leqslant x \leqslant 4 a\) (where \(a\) is a positive constant), the \(x\)-axis, and the lines \(x = a\) and \(x = 4 a\), is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid of revolution of mass \(m\).

1. Show that the moment of inertia of this solid about the \(x\)-axis is \(\frac { 7 } { 5 } m a ^ { 2 }\). The solid is free to rotate about a fixed horizontal axis along the line \(y = a\), and makes small oscillations as a compound pendulum.
2. Find, in terms of \(a\) and \(g\), the approximate period of these small oscillations.
   \includegraphics[max width=\textwidth, alt={}, center]{a9e010ce-c3a8-4f95-a154-fd16ef3e5e5b-3_734_862_813_644} A uniform rectangular lamina \(A B C D\) has mass \(m\) and sides \(A B = 2 a\) and \(B C = 3 a\). The mid-point of \(A B\) is \(P\) and the mid-point of \(C D\) is \(Q\). The lamina is rotating freely in a vertical plane about a fixed horizontal axis which is perpendicular to the lamina and passes through the point \(X\) on \(P Q\) where \(P X = a\). Air resistance may be neglected. When \(Q\) is vertically above \(X\), the angular speed is \(\sqrt { \frac { 9 g } { 10 a } }\). When \(X Q\) makes an angle \(\theta\) with the upward vertical, the angular speed is \(\omega\), and the force acting on the lamina at \(X\) has components \(R\) parallel to \(P Q\) and \(S\) parallel to \(B A\) (see diagram).

3 The region \(R\) is bounded by the \(x\)-axis, the \(y\)-axis, the curve \(y = a \mathrm { e } ^ { \frac { x } { a } }\) and the line \(x = a \ln 2\) (where \(a\) is a positive constant). A uniform solid of revolution, of mass \(M\), is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis. Find, in terms of \(M\) and \(a\), the moment of inertia about the \(x\)-axis of this solid of revolution.
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4. \begin{figure}[h] \end{figure} A region \(R\) is bounded by the curve \(y ^ { 2 } = 4 a x ( y > 0 )\), the \(x\)-axis and the line \(x = a ( a > 0 )\), as shown in Figure 1. A uniform solid \(S\) of mass \(M\) is formed by rotating \(R\) about the \(x\)-axis through \(360 ^ { \circ }\). Using integration, prove that the moment of inertia of \(S\) about the \(x\)-axis is \(\frac { 4 } { 3 } M a ^ { 2 }\).
(You may assume without proof that the moment of inertia of a uniform disc, of mass \(m\) and radius \(r\), about an axis through its centre perpendicular to its plane is \(\frac { 1 } { 2 } m r ^ { 2 }\).)