1. A particle moves on the inside surface of a smooth cone with its axis vertical, defined by the equation $z=r$ in cylindrical polar coordinates $(r, \theta, z)$. Initially the particle is at height $z=a$, and its velocity is horizontal, speed $v$, in the $\mathbf{e}_{\theta}$ direction. Starting from Newton's second law show that $r^2\dot{\theta}$ is constant. Explain why the total energy is conserved, and deduce that $\dot{z}^{2}+\frac{1}{2} \frac{a^{2} v^{2}}{z^{2}}+g z=\frac{1}{2} v^{2}+g a$. Hence show that the particle remains at all times between two heights, which should be determined.
   Solution.
   By Newton's second law, $m\left[\left(\ddot{r}-r \dot{\theta}^{2}\right) \mathbf{e}_{r}+\frac{1}{r} \frac{\mathrm{d}}{\mathrm{d} t}\left(r^{2} \dot{\theta}\right) \mathbf{e}_{\theta}+\ddot{z} \mathbf{e}_{z}\right]=m \ddot{\mathbf{r}}=\mathbf{F}=-m g \mathbf{e}_{z}+\mathbf{N}$.
   Since there is no θ-component of force we have $0=\mathbf{e}_{\theta} \cdot m \ddot{\mathbf{r}}=\frac{m}{r} \frac{\mathrm{d}}{\mathrm{d} t}\left(r^{2} \dot{\theta}\right)$, so that $r^2\dot{\theta}$ is constant, so $r^2\dot{\theta}=av$  ①
   The work done by the normal reaction is 0, so the total energy is conserved, so $\frac{1}{2} m \dot{\mathbf{r}}^{2}+m g z=$constant, so $\frac{1}{2}\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}+\dot{z}^{2}\right)+g z=\frac{1}{2} v^{2}+g a$.
   By the equation of the cone, $z=r$, and using ①, we get $\dot{z}^{2}+\frac{1}{2} \frac{a^{2} v^{2}}{z^{2}}+g z=\frac{1}{2} v^{2}+g a$.
   $\dot{z}^{2}≥0$⇒$\frac{1}{2} \frac{a^{2} v^{2}}{z^{2}}+g z≤\frac{1}{2} v^{2}+g a$⇒$(a-z) \left(a v^2-2 g z^2+v^2 z\right)≤0$ but $a-z≥0$, so $a v^2-2 g z^2+v^2 z≤0$, which has a negative root $z_-=\frac{-\sqrt{8 a g v^2+v^4}+v^2}{4 g}$ and a positive root $z_+=\frac{\sqrt{8 a g v^2+v^4}+v^2}{4 g}$.
   So either $z≥a,z≤z_+$⇒$a≤z≤z_+$ or $z≤a,z≥z_+$⇒$z_+≤z≤a$(which of these happens is determined by whether $z_+$ is less than or greater than $a$)
2. A particle of mass $m$, moving under gravity, is disturbed from rest at the highest point on the outside of a smooth sphere of radius $a$.
   (a) Explain why the particle subsequently moves on a great circle.
   (b) By introducing plane polar coordinates in the vertical plane of this circle (or otherwise), show that
   $\ddot{\theta}=\frac{g}{a} \sin \theta, \quad N=m g \cos \theta-m a \dot{\theta}^2$.
   Here $\theta(t)$ denotes the angle between the upward vertical axis of the sphere and the straight line from the particle to the centre of the sphere (the usual polar angle for a sphere), and $N$ is the magnitude of the normal reaction.
   (c) Show that the normal reaction is given by $N=m g\left(\frac{3 z}{a}-2\right)$, where $z$ is the height of the particle above the centre of the sphere. At what height does the particle lose contact with the sphere?
   Solution.
   (a)By Newton's second law, $m\ddot{\bf r}=-mg\mathbf k+\mathbf N$. Dot multiply with $\mathbf e_φ,$
   (b)$\dot{\mathbf{r}}=a \dot{\theta} \mathbf{e}_{\theta}, \quad \ddot{\mathbf{r}}=-a \dot{\theta}^2\mathbf{e}_{r}+a\ddot{\theta}\mathbf{e}_{\theta}$
   By Newton's second law, $m\ddot{\bf r}=-mg\mathbf e_z+N\mathbf e_r$,
   Resolving in the ${\mathbf e}_θ$ direction gives $ma\ddot{\theta}=m\ddot{\mathbf{r}}\cdot{\mathbf e}_{\theta}=m g \sin \theta$, thus $\ddot{\theta}=\frac{g}{a} \sin \theta$.
   Resolving in the $\mathbf e_r$ direction gives $-m a \dot{\theta}^{2}=m \ddot{\mathbf{r}} \cdot \mathbf{e}_{r}=-m g \cos \theta+N$, thus $N=m g \cos \theta-m a \dot{\theta}^2$.
   (c)$\ddot{\theta}=\frac{g}{a} \sin \theta⇒2\dotθ\ddot{\theta}=\frac{2g}{a}\dotθ\sin\theta$, integrating from 0 to θ, ${\dotθ}^2=\frac{2g}a(1-\cosθ)$, therefore $N=m g \cos \theta-m a \dot{\theta}^2=3mg\cosθ-2mg=m g\left(\frac{3 z}{a}-2\right)$
   The normal force gets smaller and smaller as the particle speeds up, and at the point where the particle leaves the sphere, $N=0$, so that $z=\frac{2a}3$.
3. A bead of mass $m$ is free to slide on a smooth wire that is made to rotate at constant angular velocity $\omega$ about the vertical axis through a fixed point $O$ on the wire. The wire is bent into the shape of a parabola, $z=r^2/2a$, where $z$ is measured vertically upwards from $O$, and $r$ is the horizontal distance from $O$.
   (a) Show that if $z(t)$ and $r(t)$ are the vertical and horizontal distances of the bead from $O$, then$$m\left[\left(\ddot{r}-r \omega^{2}\right) \mathbf{e}_r+2 \omega \dot{r} \mathbf{e}_{\theta}+\ddot{z} \mathbf{e}_{z}\right]=-m g \mathbf{e}_{z}+\mathbf{N},$$where $\mathbf{N}$ is the normal reaction.
   (b) Hence deduce that$$\left(a^{2}+r^{2}\right) \ddot{r}+r \dot{r}^{2}=\left(a^{2} \omega^{2}-g a\right) r\tag{*}$$(c) Show that $r=0$ is an equilibrium point. The linearized equation of motion about $r=0$ is $a^2\ddot{\xi}=\left(a^{2} \omega^{2}-g a\right) \xi$, where we have written $r=\xi$, and kept only the linear terms in $\xi, \dot{\xi}$ in equation (*), in a Taylor expansion around $\xi=0$. Discuss the stability of the equilibrium point.
   Solution.
   (a)By Newton's second law, $m\left[\left(\ddot{r}-r \omega^{2}\right) \mathbf{e}_r+2 \omega \dot{r} \mathbf{e}_{\theta}+\ddot{z} \mathbf{e}_{z}\right]=m \ddot{\mathbf{r}}=\mathbf{F}=-m g \mathbf{e}_{z}+\mathbf{N}$
   (b)Resolving in the direction of the tangent of the wire, since $\bf N$ and $\mathbf e_θ$ are perpendicular to tangent,$$(a\mathbf e_r+r\mathbf e_z)·\left[\left(\ddot{r}-r \omega^{2}\right) \mathbf{e}_r+\ddot{z} \mathbf{e}_{z}\right]=-g \mathbf{e}_z·(a\mathbf e_r+r\mathbf e_z)⇒a\left(\ddot{r}-r \omega^{2}\right)+r\ddot z=-rg$$Using $\ddot z=\frac{r\ddot r+\dot r^2}a$,$$a^2\left(\ddot{r}-r \omega^{2}\right)+r^2\ddot r+r\dot r^2=-rga$$$$⇒\left(a^{2}+r^{2}\right) \ddot{r}+r \dot{r}^{2}=\left(a^{2} \omega^{2}-g a\right) r$$(c)When $r=\dot r=\ddot r=0$, both sides of the equation (*) is 0, so $r=0$ is an equilibrium point.
   When $aω^2>g$, unstable; when $aω^2<g$, stable.
4. A particle of mass $m$ is released from rest at a very large height $z=z_{0}$ above the Earth. The Newtonian gravitational potential energy of the particle is $V(z)=-\frac{G_{N} M m}{z}$,
   where $M$ is the mass of the Earth, and $G_{N}$ is Newton's gravitational constant.
   (a) Using conservation of energy show that the trajectory $z(t)$ satisfies $\sqrt{2 G_{N} M} t=-\int_{z_{0}}^{z}\left(\frac{1}{s}-\frac{1}{z_{0}}\right)^{-1 / 2} \mathrm{~d} s$.
   (b) Using the substitution $s=z_{0} \sin ^{2} \theta$, show $z(t)$ satisfies the unlikely looking equation$$\frac{\pi}{2}-\sin ^{-1}\left(\sqrt{\frac{z(t)}{z_{0}}}\right)+\frac{1}{2} \sin \left[2 \sin ^{-1}\left(\sqrt{\frac{z(t)}{z_{0}}}\right)\right]=\sqrt{\frac{2 G_{N} M}{z_{0}^{3}}} t.$$This is a radial Kepler trajectory (c.f. section 6.2 of the lecture notes).
   Proof.
   (a) Using conservation of energy, $\frac12m{\dot z}^2+V(z)=V(z_0)$⇒$\frac12m{\dot z}^2=G_{N} M m\left(\frac1{z}-\frac1{z_0}\right)$⇒$\dot z=-\sqrt{2G_{N} M\left(\frac1{z}-\frac1{z_0}\right)}$(taking negative square root because $z$ is decreasing)⇒$\sqrt{2G_{N} M}\mathrm dt=-\left(\frac1{z}-\frac1{z_0}\right)^{-1/2}\mathrm dz$, integrating, $\sqrt{2 G_{N} M} t=-\int_{z_{0}}^{z}\left(\frac{1}{s}-\frac{1}{z_{0}}\right)^{-1 / 2} \mathrm{~d} s$.
   (b) Using the substitution $s=z_{0} \sin ^2\theta$, $\frac{\mathrm ds}{\mathrm dθ}=2z_0\sinθ\cosθ$⇒$\sqrt{2 G_{N} M\over z_0^3} t=-2\int_{π/2}^θ\sin^2\vartheta~\mathrm d\vartheta=\left[-\vartheta+\frac12\sin2\vartheta\right]_{π/2}^θ$$=\frac{\pi}{2}-\sin ^{-1}\left(\sqrt{\frac{z}{z_{0}}}\right)+\frac{1}{2} \sin \left[2 \sin ^{-1}\left(\sqrt{\frac{z}{z_{0}}}\right)\right].$