September 28, 2022 – Physicspupil

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 $v=\sqrt{gh}$ .