September 2022 – Physicspupil

September 30, 2022October 24, 2022

202209300929 Theorem 4.1.3

Prove Theorem 4.1.3.

Let $(X,d)$ and $(Y,\rho )$ be metric spaces and $f:X\to Y$ be a function. Then, $f$ is continuous at a point $x_0$ if and only if $f(x_n)\to f(x_0)$ , for every sequence $\{ x_n\} \subset X$ with $x_n\to x_0$ .

_{Extracted from K. J. Pawan & A. Khalil. (2004). Metric Spaces.}

Background.

Define, by open spheres, continuity of a function $f$ at a point $x_0$ :

$\forall\, x,\exists\, x_0\in X, \forall\, \epsilon >0, \exists\,\delta >0,$ $d(x,x_0)<\delta\Rightarrow \rho (f(x),f(x_0))<\epsilon$

in other words,

$x\in S_\delta (x_0)\Rightarrow f(x)\in S_\epsilon (f(x_0))$ ,

or the same,

$f(S_\delta (x_0))\subset S_\epsilon (f(x_0))$ .

Proof.

$\begin{aligned} & f\textrm{ is continuous at }x_0 \\ \Leftrightarrow\enspace & f(S_\delta (x_0))\subset S_\epsilon (f(x_0))\textrm{ for some } \delta>0\textrm{ and any }\epsilon >0 \\ & \\ & \textrm{for every sequence }\{ x_n\}\subset X\textrm{ with }x_n\textrm{ tends to }x_0 \\ \Leftrightarrow\enspace & \textrm{there exists an }N>0\textrm{ such that }x_n\in S_{\delta}(x_0)\textrm{ for all }n>N \\ \end{aligned}$

$\begin{aligned} & x_n\in S_\delta (x_0) \Longrightarrow f(x_n)\in f(S_\delta (x_0))\subset S_\epsilon (f(x_0)) \\ \Leftrightarrow\enspace & f(x_n) \to f(x_0) \\ \end{aligned}$

September 29, 2022September 29, 2022

202209291831 Problem 1.1

Prove the parallelogram law

$|z_1+z_2|^2+|z_1-z_2|^2=2[|z_1|^2+|z_2|^2]$

and give a geometric interpretation.

_{Extracted from R. B. Ash & W. P. Novinger. (2004). Complex Variables.}

Draw a picture.

Apply the Law of Cosines,

$\begin{aligned} |z_1+z_2|^2 & =|z_1|^2+|z_2|^2-2(-z_1)\cdot z_2 \\ |z_1-z_2|^2 & =|z_1|^2+|z_2|^2-2z_1\cdot z_2 \\ \end{aligned}$

and that is all I could interpret.

September 28, 2022October 25, 2022

202209281521 Problem 5.37

A bead can slide without friction on a circular hoop of radius $R$ in a vertical plane. The hoop rotates at a constant rate of $\omega$ about a vertical diameter, as shown in the figure below.

(a) Find the angle $\theta$ at which the bead is in vertical equilibrium. (Of course it has a radial acceleration toward the axis.)
(b) Is it possible for the bead to “ride” at the same elevation as the centre of the hoop?
(c) What will happen if the hoop rotates at a slower rate $\omega' = \omega /2$ ?

_{Modified from H. D. Young. (1989). University Physics.}

Roughwork.

(a)

Kinetic energy $T$ :

$\begin{aligned} T & = \frac{1}{2}mv^2 \\ \dots\enspace v & = r\omega \enspace\dots \\ T & = \frac{1}{2}mr^2\omega^2 \\ \dots\enspace r & = R\sin\theta \enspace\dots \\ T & = \frac{1}{2}mR^2\omega^2\sin^2\theta \\ \end{aligned}$

Potential energy $V$ :

$\begin{aligned} V & = mg(R-R\cos\theta ) \\ & = mgR(1-\cos\theta ) \\ \end{aligned}$

Lagrangian $\mathcal{L}=T-V$ :

$\displaystyle{\mathcal{L}=\frac{1}{2}mR^2\omega^2\sin^2\theta-mgR(1-\cos\theta )}$

Euler–Lagrange equation:

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial \dot{\theta}}\bigg)-\frac{\partial\mathcal{L}}{\partial \theta}=0}$

Thereby

$\begin{aligned} \frac{\partial\mathcal{L}}{\partial\dot{\theta}} & = 0 \quad \textrm{and}\quad \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial\dot{\theta}}\bigg) = 0 \\ \frac{\partial\mathcal{L}}{\partial\theta} & = mR^2\omega^2\sin\theta\cos\theta -mgR\sin\theta \\ \end{aligned}$

The bead is in vertical equilibrium when:

$mR^2\omega^2\sin\theta\cos\theta -mgR\sin\theta =0$

solving for $\theta$ then :

$\begin{aligned} 0 & = (mR\sin\theta )(R\omega^2\cos\theta -g) \\ \theta & = 0\enspace\textrm{ (rej.)}\quad\textrm{\scriptsize{OR}}\quad \cos^{-1}\bigg(\frac{g}{R\omega^2}\bigg) \\ \end{aligned}$

(b)

$\theta\to 90^\circ$ iff $R\omega^2\gg g$ .

(c) This is left as an exercise to the reader.

September 28, 2022October 24, 2022

202209281050 Exercise 22.2 (Q23)

An electric field is given by $\overrightarrow{E}=E_0\,\hat{\jmath}$ , where $E_0$ is a constant. Find the potential as a function of position, taking $V=0$ at $y=0$ .

_{Extracted from R. Wolfson. (2016). Essential University Physics.}

Background.

Electric field vector $\overrightarrow{E}$ is related to electric potential $V$ by:

$\displaystyle{\overrightarrow{E}= -\bigg( \frac{\partial V}{\partial x}\,\hat{\mathbf{i}} + \frac{\partial V}{\partial y}\,\hat{\mathbf{j}} + \frac{\partial V}{\partial z}\,\hat{\mathbf{k}} \bigg)}$

or reversely,

$\displaystyle{V(r)=-\int\overrightarrow{E}\cdot\mathrm{d}\vec{r}}$ .

Write

$\displaystyle{(0,E_0,0) = \bigg( -\frac{\partial V}{\partial x},-\frac{\partial V}{\partial y},-\frac{\partial V}{\partial z}\bigg) }$

that implies $V=V(y)$ is $x$ – and $z$ -independent.

Hence

$\begin{aligned} V(y) & = -\int E_0\,\mathrm{d}y \\ & = -E_0y+C\qquad \textrm{for some constant }C \\ \because\enspace 0 & = V(y=0) = -E_0(0)+C \\ \Rightarrow C& = 0 \\ \therefore\enspace V&=V(y)=-E_0y\\ \end{aligned}$

September 28, 2022September 29, 2022

202209281002 Exercise 14.2 (Q24)

Analysis of waves in shallow water (depth much less than wavelength) yields the following wave equation:

$\displaystyle{\frac{\partial^2y}{\partial x^2}=\frac{1}{gh}\frac{\partial^2y}{\partial t^2}}$

where $h$ is the water depth and $g$ the gravitational acceleration. Give an expression for the wave speed.

_{Extracted from R. Wolfson. (2016). Essential University Physics.}

(top-down approach)

Write a traveling sinusoidal wave in the form

$y(x,t)=A\cos (kx\pm \omega t)$

the wave speed $v$ being such as

$\displaystyle{v=\frac{\lambda}{T}=\frac{2\pi /k}{2\pi /\omega}=\frac{\omega}{k}}$ .

$\begin{aligned} \textrm{LHS} & = \frac{\partial^2y}{\partial x^2} \\ & = \frac{\partial}{\partial x}\bigg(\frac{\partial}{\partial x}\big( A\cos (kx\pm\omega t)\big) \bigg) \\ & = \frac{\partial}{\partial x}\big( kA\sin (kx\pm\omega t)\big) \\ & = -k^2A\cos (kx\pm\omega t) \\ \textrm{RHS} & = \frac{1}{gh}\frac{\partial^2y}{\partial t^2} \\ & = \frac{1}{gh}\frac{\partial}{\partial t}\bigg(\frac{\partial}{\partial t}\big( A\cos (kx\pm\omega t)\big)\bigg) \\ & = \frac{1}{gh}\frac{\partial}{\partial t}\big(\omega A\sin (kx\pm\omega t)\big) \\ & = \frac{1}{gh}\big( -\omega^2A\cos (kx\pm\omega t)\big) \\ \textrm{LHS} & = \textrm{RHS} \\ \Rightarrow v = \frac{\omega}{k} & = \sqrt{gh} \\ \end{aligned}$

$\therefore$ The wave speed $v$ is given by $v=\sqrt{gh}$ .

September 27, 2022October 24, 2022

202209271040 Exercise 8.4 (Q53)

Neglecting the Earth’s rotation, show that the energy needed to launch a satellite of mass $m$ into circular orbit at altitude $h$ is

$\displaystyle{\bigg( \frac{GM_\textrm{E}m}{R_{\textrm{E}}}\bigg)\bigg(\frac{R_{\textrm{E}}+2h}{2(R_{\textrm{E}}+h)}\bigg)}$ .

_{Extracted from R. Wolfson. (2016). Essential University Physics.}

Abortive attempt.

(energy/work-done approach)

By conservation of mechanical energy,

$\begin{aligned} \Delta (\textrm{KE}+\textrm{PE}) & = 0 \\ (\textrm{KE}_f-\textrm{KE}_i) + (\textrm{PE}_f-\textrm{PE}_i) & = 0 \\ \bigg(\frac{1}{2}mv^2 - \frac{1}{2}mu^2\bigg) + \big(mg_f(R_\textrm{E}+h)-mg_iR_\textrm{E}\big) & = 0 \\ \end{aligned}$

$\begin{aligned} g_f & = G\frac{M_\textrm{E}}{(R_\textrm{E}+h)^2} \\ g_i & = G\frac{M_\textrm{E}}{(R_\textrm{E})^2} \\ \end{aligned}$

On the one hand, the gravitational pull provides the centripetal force for revolving at the orbital speed $v$ :

$\begin{aligned} \text{}_MF_m & = \text{}_mF_M \\ \frac{mv^2}{R_{\textrm{E}}+h} & = G\frac{mM_\textrm{E}}{(R_{\textrm{E}}+h)^2} \\ v & = \sqrt{\frac{GM_\textrm{E}}{R_{\textrm{E}}+h}} \\ \end{aligned}$

On the other hand, the escape speed $u$ of the satellite is

$\begin{aligned} \frac{1}{2}mu^2 & = G\frac{mM_\textrm{E}}{R_\textrm{E}} \\ u & = \sqrt{\frac{2GM_\textrm{E}}{R_\textrm{E}}} \\ \end{aligned}$

but what is this question asking for?

$\Delta\textrm{KE}=\displaystyle{\frac{m(v^2-u^2)}{2}}$

(to be continued)

September 27, 2022October 24, 2022

202209271004 Exercise 4.5 (Q31)

An elevator accelerates downward at $2.4\,\mathrm{m\, s^{-2}}$ . What force does the elevator’s floor exert on a $52\,\mathrm{kg}$ passenger?

_{Extracted from R. Wolfson. (2016). Essential University Physics.}

Roughwork.

Take downward positive, $\downarrow\textrm{+ve}$ . By Newton’s 2^nd Law,

$\mathbf{F}_{\textrm{net}}=m\mathbf{a}$ .

Given $m=52\,\mathrm{kg}$ . Let $F$ be the unknown magnitude of force:

$\begin{aligned} \mathbf{a} & = 2.4\,\hat{\mathbf{k}} \\ \mathbf{F}_\textrm{net} & = -F\,\hat{\mathbf{k}}+mg\,\hat{\mathbf{k}} \\ & = (-F+mg)\,\hat{\mathbf{k}} \\ \end{aligned}$

Then,

$\begin{aligned} -F+mg & = 2.4m \\ F & = (g-2.4)m \\ & = (9.81-2.4)(52) \\ & = 385\,\mathrm{N}\qquad \textrm{(3 s.f.)} \end{aligned}$

$\therefore$ The force the elevator’s floor exerts on the passenger is $385\,\mathrm{N}$ upward.

September 27, 2022October 24, 2022

202209270930 Exercise 5.3.B (Q1-Q6)

1. What is the potential energy of a $10\,\mathrm{kg}$ mass $25$ metres above the ground?
2. A $3\,\mathrm{kg}$ mass is $20\,\mathrm{m}$ above the ground. How much potential energy does it have?
3. How high must you raise a $25\,\mathrm{kg}$ mass before it has a potential energy of $8500\,\mathrm{J}$ ?
4. How high must you raise a $2\,\mathrm{kg}$ mass before it has a potential energy of $10\,\mathrm{J}$ ?
5. A mass has a potential energy of $1000\,\mathrm{J}$ when it is $40\,\mathrm{m}$ above ground level. What is the mass?
6. A mass has a potential energy of $20\,\mathrm{J}$ when it is $50\,\mathrm{cm}$ above ground level. What is the mass?

_{Extracted from B. Kennedy. (1999). Progressive Problems for ‘S’ Grade Physics.}

Background.

$\textrm{PE}=mgh$

Roughwork.

1.

$\begin{aligned} \textrm{PE} & = mgh \\ & = (10)(9.81)(25) \\ & = 2450\,\mathrm{J}\qquad \textrm{(3 s.f.)} \\ \end{aligned}$

2.

$\begin{aligned} \textrm{PE} & = mgh \\ & = (3)(9.81)(20) \\ & = 589\,\mathrm{J}\qquad \textrm{(3 s.f.)} \end{aligned}$

3.

$\begin{aligned} \textrm{PE} & = mgh \\ 8500 & = (25)(9.81)(h) \\ h & = 34.7\,\mathrm{m}\qquad \textrm{(3 s.f.)} \end{aligned}$

4.

$\begin{aligned} \textrm{PE} & = mgh \\ 10 & = (2)(9.81)(h) \\ h & = 51.0\,\mathrm{cm}\qquad \textrm{(3 s.f.)} \end{aligned}$

5.

$\begin{aligned} \textrm{PE} & = mgh \\ 1000 & = (m)(9.81)(40) \\ m & = 2.55\,\mathrm{kg}\qquad \textrm{(3 s.f.)} \end{aligned}$

6.

$\begin{aligned} \textrm{PE} & = mgh \\ 20 & = (m)(9.81)(0.5) \\ m & = 4.08\,\mathrm{kg}\qquad \textrm{(3 s.f.)} \end{aligned}$