2018 Paper Ⅳ
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  1. Let $\mathbf{u}, \mathbf{v} \in \mathbb{R}^3$ be two vectors with Cartesian components $u_1,u_2,u_3$ and $v_1,v_2,v_3$ respectively. The scalar product of $\mathbf{u}$ and $\mathbf{v}$ is defined by$$\mathbf{u} \cdot \mathbf{v}=\sum_{i=1}^3u_iv_i$$(a) [1 mark] Define $\mathbf{u} \wedge \mathbf{v}$, that is the vector product of $\mathbf{u}$ and $\mathbf{v}$.
    (b) [5 marks] By considering $|\mathbf{u}+t \mathbf{v}|^2$ for $t \in \mathbb{R}$, or otherwise, show that$$
    |\mathbf{u} \cdot \mathbf{v}| \leqslant|\mathbf{u}||\mathbf{v}|
    $$(c) [4 marks] Given $\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}|| \mathbf{v}| \cos θ$, with $θ \in[0, \pi]$, show that$$
    |\mathbf{u} \wedge \mathbf{v}|=|\mathbf{u}||\mathbf{v}| \sin θ
    $$(d) [10 marks] Let $\mathbf{p}, \mathbf{q} \in \mathbb{R}^{3}$. The line $L_{1}$ is defined by$$
    \mathbf{r}=\mathbf{p}+\lambda \mathbf{q}, \quad \text { with } \quad \lambda \in \mathbb{R}
    $$Let $\mathbf{a}, \mathbf{b} \in \mathbb{R}^3$ be two vectors with $\mathbf{a} \cdot \mathbf{b}=0$, and $|\mathbf{a}|=1 .$ The line $L_2$ is defined by$$
    \mathbf{r} \wedge \mathbf{a}=\mathbf{b}$$You are also given that $\mathbf{a} \wedge \mathbf{q} \neq \mathbf{0}$.
    (i) Show that line $L_2$ can be written in the form$$\mathbf{r}=\mathbf{a} \wedge \mathbf{b}+\mu \mathbf{a}, \quad \mu \in \mathbb{R}$$and hence show that the lines $L_1$ and $L_2$ intersect at most once.
    (ii) Show that the lines $L_1$ and $L_2$ intersect if and only if $\mathbf{q} \cdot \mathbf{b}=\mathbf{p} \cdot(\mathbf{a} \wedge \mathbf{q})$
    Solution.
    (a)$\mathbf u\wedge\mathbf v=\left(\begin{array}{lll}\mathbf e_1&\mathbf e_2&\mathbf e_3\\u_{1} & u_{2} & u_{3} \\v_{1} & v_{2} & v_{3}\end{array}\right)=\left(\begin{array}{c}u_2 v_3-u_3v_2\\u_3v_1-u_1v_3\\u_1v_2-u_2v_1\end{array}\right)$
    (b)For any $t \in \mathbb{R}$, $0 \leqslant|\mathbf{u}+t\mathbf{v}|^2=\mathbf{u}^2+t^2\mathbf{v}^2+2t\mathbf{u}·\mathbf{v}$, quadratic in $t$ with either no real roots or a double root(with equality)
    $\therefore 4(\mathbf{u}·\mathbf{v})^2\leqslant 4|\mathbf{u}|^{2}|\mathbf{v}|^{2},\therefore|\mathbf{u} \cdot \mathbf{v}| \leqslant|\mathbf{u}||\mathbf{v}|.$
    (c)$\mathbf{u}·\mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos θ$, with $\mathbf u=(u_1,0,0),\mathbf v=(v_1,v_2,0)$, without loss by choice of axes.
    $|\mathbf u∧\mathbf v|^2=\left|\pmatrix{0\\0\\u_1v_2}\right|^2=u_1^2v_2^2=u_1^2(v_1^2+v_2^2)-u_1^2v_1^2=|\mathbf u|^2|\mathbf v|^2-(\mathbf u·\mathbf v)^2=|\mathbf u|^2|\mathbf v|^2(1-\cos^2θ)=|\mathbf u|^2|\mathbf v|^2\sin^2θ,∴|\mathbf u∧\mathbf v|=|\mathbf u||\mathbf v|\sinθ$, noting $\sinθ≥0$ as $θ∈[0,π]$.
    (d)$\mathbf r=\mathbf p+λ\mathbf q,\quad|\mathbf q|=1$ WLOG ($L_1$)
         $\mathbf r∧\mathbf a=\mathbf b,\quad|\mathbf a|=1$        ($L_2$)
    (i) Line $L_2∥\mathbf a,∴\mathbf r=\mathbf s-μ\mathbf a,μ∈\mathbb R$ for some $\mathbf s$
    $\mathbf s=\mathbf a∧\mathbf b$, as $(\mathbf a∧\mathbf b)∧\mathbf a=\mathbf a^2\mathbf b-(\mathbf a·\mathbf b)\mathbf a=\mathbf b$ since $\mathbf a^2=1,\mathbf a·\mathbf b=0$.
    $∴\mathbf r=(\mathbf a∧\mathbf b)-μ\mathbf a,μ∈\mathbb R,$ on $L_2$. ($μ→-μ$)
    The lines intersect at most once as $\bf a$ not parallel to $\bf q$ since $\mathbf a∧\mathbf q≠0$.
    The lines intersect⇔$\mathbf a∧\mathbf b-μ\mathbf a=\mathbf p+λ\mathbf q,\quadμ,λ∈\Bbb R$⇔$(\mathbf a∧\mathbf b)-\mathbf p=λ\mathbf q+μ\mathbf a$⇔$(\mathbf a∧\mathbf b)-\mathbf p$ is in the plane spanned by $\mathbf a,\mathbf q$
    ⇔$(\mathbf a∧\mathbf b)-\mathbf p$ has no component in the direction $\mathbf a∧\mathbf q$⇔$(\mathbf a∧\mathbf b-\mathbf p)·(\mathbf a∧\mathbf q)=0$⇔$(\mathbf a∧\mathbf b)·(\mathbf a∧\mathbf q)-\mathbf p·(\mathbf a∧\mathbf q)=0$
    Noting $({\bf a}∧{\bf b})·({\bf a}∧{\bf q})=\underbrace{{\bf a}^2}_1({\bf b}·{\bf q})-({\bf a}·{\bf q})\underbrace{({\bf a}·{\bf b})}_0={\bf b}·{\bf q}$, we have the lines intersect iff $\mathbf{q} \cdot \mathbf{b}=\mathbf{p} \cdot(\mathbf{a} \wedge \mathbf{q})$.
  2. A transformation $T:\mathbb{R}^3→\mathbb{R}^3$ is defined by $T(\mathbf{v})=A \mathbf{v}+\mathbf{b}$, for some orthogonal $3 \times 3$ matrix $A$ and vector $\mathbf{b} \in \mathbb{R}^3$.
    (a) [8 marks] Show that $T$ is an isometry.
    Show further that if two lines in $\mathbb{R}^{3}$ intersect with angle $\alpha$, their images under $T$ also intersect with angle $\alpha$.
    (b) [5 marks] Suppose $\mathbf{b}=\mathbf{0}$ and $T$ represents a rotation by angle $θ$ about a line through the origin. Explain why, with a suitable choice of orthonormal basis, the matrix representing $T$ may be taken to have the form$$\left(\begin{array}{ccc}1 & 0 & 0 \\0 & \cos θ & -\sin θ \\0 & \sin θ & \cos θ\end{array}\right)$$Hence give a condition on $A$ (i.e. independent of the basis) under which $T$ represents such a rotation about the origin, and give a formula relating $θ$ to an invariant of $A$.
    (c) [7 marks] Show that the transformation given by$$
    A=\left(\begin{array}{ccc}
    0 & 1 / \sqrt{2} & -1 / \sqrt{2} \\
    -1 / \sqrt{2} & 1 / 2 & 1 / 2 \\
    1 / \sqrt{2} & 1 / 2 & 1 / 2
    \end{array}\right), \quad \mathbf{b}=\left(\begin{array}{c}
    1 \\
    1 / \sqrt{2} \\
    -1 / \sqrt{2}
    \end{array}\right)
    $$represents a rotation. Find the invariant line and the angle of rotation. [You may assume that the given $A$ is orthogonal.]
    Solution.
    (a)$T$ is an isometry if $|T(\mathbf u)-T(\mathbf v)|=|\mathbf u-\mathbf v|$ for every $\mathbf u,\mathbf v$.
    $|T(\mathbf u)-T(\mathbf v)|^2=|A(\mathbf u-\mathbf v)|^2=A(\mathbf u-\mathbf v)^{\sf T}A(\mathbf u-\mathbf v)=(\mathbf u-\mathbf v)^{\sf T}\underbrace{A^{\sf T}A}_I(\mathbf u-\mathbf v)=|\mathbf u-\mathbf v|^2$, so $T$ is an isometry.
    Without loss of generality, suppose lines intersect at origin, then direction vectors $\mathbf u,\mathbf v$ have $\mathbf u·\mathbf v=\cosα$,
    The intersection point is mapped to $\bf b$, so angle between image lines has
    $\cosβ=(T(\mathbf u)-\mathbf b)·(T(\mathbf v)-\mathbf b)$
    $\hphantom{\cosβ}=(A\mathbf u)·(A\mathbf v)$
    $\hphantom{\cosβ}=\mathbf u·\mathbf v$
    $\hphantom{\cosβ}=\cosα⇒β=α$
    (b) Choose a basis with $\mathbf e_1$ pointing along the line, $\mathbf e_2$ and $\mathbf e_3$ perpendicular to this and orthogonal to each other. The $\mathbf e_2,\mathbf e_3$ plane is rotated by $θ$, while $\mathbf e_1$ is left alone. So the rotation with respect to this basis has matrix $\tilde A=\pmatrix{1&0&0\\0&\cosθ&-\sinθ\\0&\sinθ&\cosθ}$
    Hence $A=QAQ^{\sf T}$ for some orthogonal $Q$ so $\det A=1$ and $\operatorname{tr}(A)=1+2\cosθ$.
    (c) This transformation does not map the origin to itself but the invariant points have $\mathbf v=T(\mathbf v)=A\mathbf v+\mathbf b⇒(I-A)\mathbf v=\mathbf b$ i.e. $\pmatrix{1&-{1\over\sqrt2}&{1\over\sqrt2}\\{1\over\sqrt2}&\frac12&-\frac12\\-\frac1{\sqrt2}&-\frac12&\frac12}\mathbf v=\pmatrix{1\\\frac1{\sqrt2}\\-\frac1{\sqrt2}}$
    $⇒\mathbf v=\pmatrix{1\\0\\0}+λ\pmatrix{0\\1\\1}$ is the invariant line.
    Translate coordinates so that origin lies on line, i.e. let $\tilde{\mathbf v}=\mathbf v-\pmatrix{1\\0\\0}$. Then $T(\mathbf v)-\pmatrix{1\\0\\0}=A$.
    so $\tilde{\bf v}$ is mapped to $A\tilde{\bf v}$, i.e. a rotation or reflection around the origin in the translated coordinate
    $\det A=\frac1{\sqrt2}\left(\frac1{2\sqrt2}+\frac1{2\sqrt2}\right)+\frac1{\sqrt2}\left(\frac1{2\sqrt2}+\frac1{2\sqrt2}\right)=1⇒$ this is a rotation.
    $\operatorname{tr}(A)=1=1+2\cosθ⇒\cosθ=0⇒θ=\fracπ2$
  3. (a) [4 marks] A curve in $\mathbb{R}^2$ is given parametrically by $(x(t),y(t))$ where $a\leqslant t\leqslant b$ and $x,y$ are differentiable functions of $t$.
    (i) Write down a formula for the curve's arc length.
    (ii) Assuming $y(t)>0$, write down a formula for the area of the surface of revolution formed by rotating the curve around the $x$-axis.
    (b) [10 marks] Let $a>0$. A cardioid has equation $r=a(1+\cos θ)$ in polar co-ordinates.
    (i) Sketch the cardioid.
    (ii) Let $s$ denote arc length on the cardioid. Show that$$\frac{\mathrm{d} s}{\mathrm{d}θ}=2a\left|\cos\frac{θ}{2}\right|$$Hence find the total arc length of the cardioid.
    (c) [6 marks] An astroid can be parametrized as$$x(t)=b\cos^3t,\quad y(t)=b\sin^3t, \quad 0 \leqslant t<2 \pi$$where $b>0$. This astroid is rotated around the $x$-axis so as to form a surface of revolution. Find the total area of this surface of revolution.
    Proof.
    (a)(i)The curve’s arc length equals $\int_{a}^{b} \sqrt{x'(t)^{2}+y'(t)^{2}}\mathrm dt$.
    (ii)The area of the surface of revolution generated equals $2 \pi \int_{a}^{b} y(t) \sqrt{x'(t)^{2}+y'(t)^{2}} \mathrm{~d} t$.
    (b)
    (i)      
    (ii)We can parametrize the cardioid as $x(θ)=a(1+\cos θ) \cos θ, \quad y(θ)=a(1+\cos θ) \sin θ$, and we then have\begin{aligned} s'(θ)^{2} &=x'(θ)^{2}+y'(θ)^{2} \\ &=a^{2}(-\sin θ-2 \sin θ \cos θ)^{2}+a^{2}\left(\cos θ+\cos^2θ-\sin^{2}θ\right)^{2} \\ &=a^{2}(-\sin θ-\sin 2 θ)^{2}+a^{2}(\cos θ+\cos 2 θ)^{2} \\ &=a^{2}[2+2 \cos 2 θ \cos θ+2 \sin 2 θ \sin θ] \\ &=2 a^{2}[1+\cos (2 θ-θ)] \\ &=2 a^{2}\left[1+2 \cos ^{2}(θ / 2)-1\right] \\ &=4 a^{2} \cos ^{2}(θ / 2) \end{aligned}and the result follows.
    We then have that the cardioid’s length equals$$2 a \int_{0}^{2 \pi}\left|\cos \frac{θ}{2}\right| \mathrm{d} θ=4 a \int_{0}^{\pi} \cos \frac{θ}{2} \mathrm{~d} θ=4 a\left[2 \sin \frac{θ}{2}\right]_{0}^{\pi}=8 a$$(c)(i)Again we have that\begin{aligned} s'(t)^{2} &=x'(t)^{2}+y'(t)^{2} \\ &=b^{2}\left(-3 \cos ^{2} t \sin t\right)^{2}+b^{2}\left(3 \sin ^{2} t \cos t\right)^{2} \\ &=9 b^{2} \sin ^{2} t \cos ^{2} t\left(\cos^2t+\sin^2t\right) \\ &=9 b^{2} \sin ^{2} t \cos ^{2} t \end{aligned}We then have that the area of the surface generated equals\begin{aligned} 2 \pi \int_0^πy(t) s'(t) \mathrm{d} t &=2π×3b\int_0^π\left(b\sin^3t\right)\left|\sin t \cos t\right|\mathrm{d} t \\ &=12 b^2\pi \int_0^{\pi / 2} \sin^4t \cos t \mathrm{~d} t \\ &=12 b^2 \pi\left[\frac{\sin ^{5} t}{5}\right]_0^{\pi / 2} \\ &=\frac{12b^2\pi}5\end{aligned}
  4. (a) [5 marks] The position of a particle $x(t)$ evolves according to $\ddot{x}=-V'(x)$ for a potential function $V(x)$. Give the definition of an equilibrium point $x_{0}$, and use linear theory to show how its stability depends upon the sign of $V''\left(x_{0}\right)$.
    A bead of mass $m$ slides along a smooth straight horizontal wire that passes through the origin $O$. The bead is attached by a light elastic spring of natural length $l$ and spring constant $k$, to a fixed point $P$ a distance $l / 2$ directly above the origin.
    (b) [5 marks] Show that the equation of motion is$$\ddot{x}=-\frac{k}{m} x\left(1-\frac{2 l}{\sqrt{4 x^{2}+l^{2}}}\right),$$where $x(t)$ denotes the coordinate of the bead relative to $O$.
    (c) [5 marks] Find the potential function $V(x)$ satisfying $V(0)=0$, and sketch its graph. Hence determine the location and stability of all equilibrium points.
    (d) [5 marks] Suppose that the bead is released from rest at $x=d>0$. Find the minimum value of $d$ such that the bead enters into symmetric oscillations about the origin.
    Solution.
    (a)Equilibrium point $x_0$ is where $V'(x_0)=0$.
    $\ddot x=-V'(x)$ small pertubation $x=x_0+ξ$⇒$\ddotξ=-\cancel{V'(x_0)}-V''(x_0)ξ+⋯$⇒linearized equation: $\ddotξ+V''(x_0)ξ=0$
    ⇒stable if $V''(x_0)>0$(oscillatory solution); unstable if $V''(x_0)<0$(exponential solution)
    (b)$T=k\left(\left(x^2+\frac{l^2}4\right)^{1/2}-l\right)$
    By Newton's second law, $m\ddot x=-T\cosθ=-T\frac x{\left(x^2+\frac{l^2}4\right)^{1/2}}=-kx\left(1-\frac{l}{\left(x^2+\frac{l^2}4\right)^{1/2}}\right)⇒\ddot x=-\frac kmx\left(1-\frac{2l}{\left(4x^2+l^2\right)^{1/2}}\right)$
    (c)$V'(x)=\frac kmx\left(1-\frac{2l}{(4x^2+l^2)^{1/2}}\right)⇒V=\frac km\frac{x^2}2-\frac km\int_0^x\frac{2lx}{(4x^2+l^2)^{1/2}}\mathrm dx$$=\frac km\frac{x^2}2-\left[\frac{kl}{2m}(4x^2+l^2)^{1/2}\right]^x_0$$=\frac km\left[\frac{x^2}2-\frac l2(4x^2+l^2)^{1/2}+\frac{l^2}2\right]$
    $V'(x)=0$⇒$x=0$ and $4x^2+l^2=4l^2⇒x=±\frac{\sqrt3}2l$
    Equilibrium point $x=0$ unstable(since $V''<0$ there, from graph); $x=±\frac{\sqrt3}2l$ stable(since $V''>0$)
    (d)Note that $\frac12{\dot x}^2+V(x)=\text{constant}=V(d)$ if particle released from $x=d$ at rest, so $\frac12\dot x^2=V(d)-V(x)$
    From the graph, need $V(d)>V(0)=0$ in order for particle to reach(and pass through) $x=0$
    so minimum $d$ is when $V(d)=0⇒\left(\frac{d^2}2+\frac{l^2}2\right)^2=\frac{l^2}4(4d^2+l^2)$⇒$d^4+2d^2l^2+l^4=4d^2l^2+l^4$⇒$d^2(d^2-2l^2)=0$⇒$d=\sqrt2l$
  5. A sphere of radius $a$ has its centre at the origin. A particle of mass $m$ moves smoothly on its interior surface under the action of constant gravitational acceleration $g$ acting downwards. Let $(r, θ, z)$ be cylindrical polar co-ordinates, with the $z$-axis upwards, so that the equation of the sphere is given by $r^2+z^2=a^2$.
    (a) [10 marks] Show from Newton's second law that the quantities$$E=\frac12\left(\dot{r}^2+r^2\dot{θ}^2+\dot{z}^2\right)+g z, \quad \text{ and }\quad h=r^2\dot{θ}$$are constants of the motion, where $r(t),θ(t)$ and $z(t)$ are the co-ordinates of the particle at time $t$.
    Show that the magnitude of the normal reaction force equals $2 m E / a-3 m g z / a$. Hence show that the particle will lose contact with the surface if it attains a height $z$ greater than $2 E / 3 g$.
    (b) [10 marks] Suppose that the particle is initially at $r=a,z=0$, and is projected with speed $V$ at an angle $\alpha$ to the horizontal (and tangent to the sphere). Show that$$a^2\dot{z}^2=a^2V^2\sin^2α-V^2z^2-2 gz\left(a^2-z^2\right)$$Show graphically that motion in contact with the sphere is confined between two heights $z_-$ and $z_+$(which you are not expected to determine).
    What value of $\alpha$ is required for the particle to reach $z=a$? Find the least value of $V$ that will allow the particle to reach $z=a$ in this case.
    Solution.
    (a)$r^2+z^2=a^2⇒\dot r=-\frac zr\dot z$
    $\mathbf r=r\mathbf e_r+z\mathbf e_z$
    $\dot{\bf r}=\dot r\mathbf e_r+r\dotθ\mathbf e_θ+\dot z\mathbf e_z$
    $\ddot{\bf r}=(\ddot r-r\dotθ^2)\mathbf e_r+\frac1r\frac{\rm d}{\mathrm dt}(r^2\dotθ)\mathbf e_θ+\ddot z\mathbf e_z$
    From Newton's second law, $m\ddot{\bf r}=-mg\mathbf e_z+{\bf R}$⋯⋯①
    Reaction force $\bf R$ must be normal to $\dot{\bf r}$, so ${\bf R}·\dot{\bf r}=0⇒m\dot{\bf r}·\ddot{\bf r}=-mg\dot z⇒\frac{\rm d}{\mathrm dt}(\frac12|\dot r|^2+gz)=0$ i.e. $\frac12(\dot r^2+r^2\dotθ^2+\dot z^2)+gz=E$ is constant.
    (${\bf R}·{\bf e}_θ=0$)Dotting ① with $\mathbf e_θ$ gives $m\ddot{\bf r}·{\bf e}_θ=0+0⇒\frac{\rm d}{\mathrm dt}(r^2\dotθ)=0$ i.e. $r^2\dotθ=h$ is constant.
    $N=-\frac{\bf r}a·{\bf R}$        (from $\mathbf R=-N\frac{\bf r}a$, since $\bf R$ is parallel to $-\mathbf r$)
    $\hphantom{N}=\frac1a(-m\ddot{\bf r}·{\bf r}-mg\mathbf e_z·\mathbf r)$     (from ①)
    $\hphantom{N}=\frac1a(-m(\ddot rr-r^2\dotθ^2+z\ddot z)-mgz)$
    $\hphantom{N}=\frac1a(m(\dot r^2+r^2\dotθ^2+\dot z^2)-mgz)$     ($r\dot r+z\dot z=0⇒r\ddot r+z\ddot z=-\dot r^2-\dot z^2$)
    $\hphantom{N}=\frac ma(2E-3gz)$
    loses contact if $N$ becomes negative, which happens if $z>\frac{2E}{3g}$
    (b) Initially $r=a,z=0,\dot r=0,\dot z=V\sinα,a\dotθ=V\cosα⇒h=aV\cosα,E=\frac12V^2$
    so $\displaystyle\frac12V^2=\frac12\underbrace{\left(\frac{z^2}{r^2}+1\right)}_{a^2\over a^2-z^2}\dot z^2+\frac12\underbrace{\frac{a^2V^2\cos^2α}{r^2}}_{a^2V^2\cos^2α\over a^2-z^2}+gz$
    $⇒a^2\dot z^2=(V^2-2gz)(a^2-z^2)-a^2V^2\cos^2α$
    Since $\dot z^2\ge0$ throughout the motion, it is clear from the graph that the motion be confined between $z_-$ and $z_+$. The heights are the same regardless of the sign of $α$(upwards/downwards).
    To reach $z=a$, need $α=\fracπ2$, and $a^2\dot z^2=(V^2-2gz)(a^2-z^2)$ must not reach zero for any $z<a$. This happens if $V^2\ge2ga$, but note that particle leaves surface at $z=\frac{V^2}{3g}$, so we need $V^2\ge3ga$.
  6. A circular hoop of radius $a$ is forced to rotate with constant angular velocity $\omega$ about a vertical diameter. The centre of the hoop is labelled $O$. A bead slides smoothly on the hoop, and its position vector from $O$ makes an angle $\varphi(t)$ with the downward vertical. Gravitational acceleration $g$ acts vertically downwards.
    (a) [10 marks] Show that the equation of motion is$$\ddot{\varphi}+\left(\frac{g}{a}-\omega^2\cos \varphi\right) \sin \varphi=0 .$$[You may make use of Newton's second law in a rotating frame,$$m \ddot{\mathbf{r}}+2 m \boldsymbol{\omega} \wedge \dot{\mathbf{r}}+m \boldsymbol{\omega} \wedge(\boldsymbol{\omega} \wedge \mathbf{r})=\mathbf{F}$$where $\mathbf{r}=a\left(\sin \varphi \mathbf{e}_{1}-\cos \varphi \mathbf{e}_{3}\right), \dot{\mathbf{r}}$ and $\ddot{\mathbf{r}}$ are the position, velocity and acceleration respectively, measured in the rotating frame with fixed orthonormal basis $\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$, and where $\boldsymbol{\omega}=\omega \mathbf{e}_{3}$ is the angular velocity of the rotating frame.]
    (b) [6 marks] Find the equilibrium positions and determine their stability, distinguishing between the cases $\omega^{2}<\frac{g}{a}$ and $\omega^{2}>\frac{g}{a}$.
    (c) [4 marks] If the bead is initially perturbed from rest at the lowest point of the hoop, find how large $\omega$ must be for the bead to exceed the height of $O$ at some point during its motion.
    Solution.
    (a)${\bf r}=a(\sinφ{\bf e}_r-\cosφ{\bf e}_z)$
    ${\bfω}=ω{\bf e}_z$
    Note ${\bf e}_r$ is fixed in rotating frame.
    $\dot{\bf r}=a\dotφ(\cosφ{\bf e}_r+\sinφ{\bf e}_z)$
    $\ddot{\bf r}=a\dotφ^2(-\sinφ{\bf e}_r+\cosφ{\bf e}_z)+a\ddotφ(\cosφ{\bf e}_r+\sinφ{\bf e}_z)$
    From Newton's second law, $m\ddot{\bf r}+2m{\bfω}∧\dot{\bf r}+m{\bfω}∧({\bfω}∧{\bf r})=-mg{\bf e}_z+{\bf N}$⋯⋯②
    Note ${\bf N}·\dot{\bf r}=0$, since $\bf N$ is orthogonal to hoop and $\dot{\bf r}$ is tangent to it.
    ${\bfω}∧\dot{\bf r}=a\dotφω\cosφ{\bf e}_z∧{\bf e}_r=a\dotφω\cosφ{\bf e}_θ$
    ${\bfω}∧({\bfω}∧{\bf r})=aω^2\sinφ{\bf e}_z∧({\bf e}_z∧{\bf e}_r)=-aω^2\sinφ{\bf e}_r$
    so dotting ② with $\dot{\bf r}$ gives (also note $\dot{\bf r}·{\bf e}_θ=0$)
    $m\dot{\bf r}·\ddot{\bf r}+0-amω^2\sinφ{\bf e}_r·\dot{\bf r}=-mg{\bf e}_z·\dot{\bf r}$
    $⇒a^2\dotφ[(\cos^2φ+\sin^2φ)\ddotφ]-a^2\dotφω^2\sinφ\cosφ=-ga\dotφ\sinφ$
    i.e. $\ddotφ+\left(\frac ga-ω^2\cosφ\right)\sinφ=0$
    (b)Equilibria have $\sinφ(\frac ga-ω^2\cosφ)=0⇒\sinφ=0$(⇒$φ=0$ or $φ=π$)or $\cosφ=\frac g{aω^2}$(⇒if $ω^2>\frac ga$,$φ=\cos^{-1}\left(g\over aω^2\right)$)
    linear stability, $φ=φ_0+ξ\;(ξ\ll1)$
    $φ_0=0$: $\ddotξ+\left(\frac ga-ω^2\right)ξ=0⇒$stable if $ω^2<\frac ga$; unstable if $ω^2>\frac ga$
    $φ_0=π$: $\ddotξ-\left(\frac ga+ω^2\right)ξ=0⇒$unstable
    $φ_0=\cos^{-1}\left(g\over aω^2\right)$: $\ddotξ+ω^2\sin^2φ_0ξ=0$⇒stable(if it exists)
    (c)If particle is initially at rest at $φ=0$, multiply equation of motion by $\dotφ$ and integrate, $\frac12\dotφ^2-\frac ga\cosφ-\frac{ω^2}2\sin^2φ=\text{constant}=-\frac ga$
    so $\frac12\dotφ^2=\frac ga(\cosφ-1)+\frac{ω^2}2\sin^2φ=\frac{ω^2}2-\frac ga+\frac ga\cosφ-\frac{ω^2}2\cos^2φ=(1-\cosφ)(\frac{ω^2}2-\frac ga+\frac{ω^2}2\cosφ)$
    Maximum height where $\dotφ=0$⇒$\cosφ=\frac{2g}{ω^2a}-1$
    so max $φ$ is larger than $\fracπ2(\cosφ<0)$ if $ω^2>\frac{2g}a$.
  7. (a) [8 marks] Let $g:[a,b]→[a,b]$ be continuous.
    (i) Define what is meant by a fixed point of $g$, and the associated fixed-point iteration.
    (ii) State the Contraction Mapping Theorem for uniqueness of a fixed point and convergence of the fixed-point iteration.
    (iii) If $g$ is differentiable, state and prove a sufficient condition for convergence in terms of the derivative $g'$.
    (b) [6 marks] Let $f(x)=x^3+4 x-1$. Find a suitable function $g$ and interval $[a,b]$ for which a fixed-point iteration is guaranteed to converge to the root $\alpha$ of $f$.
    (c) [6 marks] State what it means for a fixed-point iteration to have convergence of order $p$. Find a function $g$, which you should define explicitly, such that the fixed-point iteration to determine $\alpha$ in part (b) has quadratic convergence (i.e. of order 2).
    Solution.
    (a)(i)Fixed point is point $α$ such that $α=g(α)$.
    Fixed point iteration is defined by $x_{k+1}=g(x_k),k=0,1,2,⋯$ starting with some $x_0∈[a,b]$.
    (ii)Suppose $g:[a,b]→[a,b]$ satisfies $|g(x)-g(y)|≤γ|x-y|$ for some $0≤γ<1$ for all $x,y∈[a,b]$($g$ is a contraction). Then there is a unique fixed point and the fixed point iteration converges to it.
    (iii)If $g$ is differentiable, $|g'(α)|≤γ$ with $0≤γ<1$ is sufficient for convergence.
    Since, by mean value theorem, $g(x)-g(y)=(x-y)g'(ξ)$ for some $ξ∈[y,x]$, so $|g(x)-g(y)|=|g'(ξ)||x-y|≤γ|x-y|$ and $g$ is a contraction in this case.
    (b)$f=x^3+4x-1$ is monotonic so has one root.
    Since $f(0)=-1,f(1)=4,$ the root is between 0 and 1.
    Let $g(x)=\frac14(1-x^3)$. For $x∈[0,1]$, $g(x)∈[0,1]$, and $g'(x)=-\frac34x^2$ has $|g'(x)|≤\frac34<1$. By (a), the fixed point iteration will converge, i.e. $g(x)=\frac14(1-x^3),[a,b]=[0,1]$
    ($g(x)=(1-4x)^{1/3}$ doesn't converge as has too large a gradient)
    (c)$x_{k+1}=g(x_k)$ has order of convergence $p$ if ∃ constant $k$ with $|x_{k+1}-α|≤k|x_k-α|^p$
    Use a Newton iteration for quadratic convergence.
    i.e. $x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}=g(x_k)$
    $g(x)=x-\frac{x^3+4x-1}{3x^2+4}=\frac{2x^3+1}{3x^2+4}$.