January 2024 – Physicspupil

January 31, 2024January 31, 2024

202401311548 Pastime Exercise 008

Let there be a cone $R\,\mathrm{(unit)}$ in radius of its base and $h\,\mathrm{(unit)}$ in height.

Find the volume $V\,\mathrm{(cubic\, unit)}$ of the cone, using what are so-called i. the disc (/washer) method and ii. the (cylindrical) shell method, if the names my memory serves me right.

Answer. $\displaystyle{V=\frac{1}{3}\pi R^2h}$ .

Roughwork.

Visualize the cone.

i. By disc method,

$\begin{aligned} \frac{R}{h} & = \frac{r}{h-z} = \tan\theta \\ r & = R\bigg( 1-\frac{z}{h}\bigg) \\ V & = \pi R^2\int_{0}^{h}\bigg( 1-\frac{z}{h}\bigg)^2\,\mathrm{d}z \\ \cdots & \cdots\cdots\cdots\cdots\cdots\cdots\cdots \\ \textrm{let }& z'=1-\frac{z}{h} \\ \frac{\mathrm{d}z'}{\mathrm{d}z} & = -\frac{1}{h} \\ \mathrm{d}z & = -h\,\mathrm{d}z' \\ z=0 & \Leftrightarrow z'=1 \\ z=h & \Leftrightarrow z'= 0 \\ \cdots & \cdots\cdots\cdots\cdots\cdots\cdots\cdots \\ V& = \pi R^2\int_{1}^{0} z'^2(-h\,\mathrm{d}z') \\ & = -\pi R^2h \bigg[ \frac{z'^3}{3}\bigg]\bigg|_{1}^{0} \\ & = -\pi R^2h \bigg(\frac{(0)^3}{3} - \frac{(1)^3}{3}\bigg) \\ & = \frac{\pi R^2h}{3}\\ \end{aligned}$

ii. By shell method,

we are obtaining the volume of a solid of revolution about the $z$ -axis by integrating the slices (/cross sections) ranging between $\theta\in [0,2\pi ]$ .

It is left the reader as an exercise.

This problem is not to be attempted.

January 30, 2024January 30, 2024

202401301506 Solution to 2000-CE-PHY-II-28

Two insulated uncharged metal spheres $X$ and $Y$ are placed in contact. A positively-charged rod is brought near $X$ as shown below. $X$ is then earthed momentarily. The charged rod is removed and the two spheres are then separated. Describe the charges on $X$ and $Y$ . _(modified)

Roughwork.

Have some mental pictures (*make drawings if you don’t have ample RAM).

First, when a positively-charged rod is brought near $X$ , charges are induced:

Notice that on surface areas $X$ and $Y$ making contact, as boxed below:

the net charge is zero, $Q=0$ . Hence, the plus and the minus signs (representing positive and negative charges) are erased from the drawing:

Then, $X$ is earthed momentarily:

it is convenient to treat the two spheres $X$ and $Y$ as one body $X+Y$ :

The process of earthing enables a transfer of negative free charges (electrons being the carrier) unidirectionally between two bodies:

here from the earth to the body, but no reversed (why?); the plus and minus signs, as boxed below:

should cancel off each other, and are thus erased from the drawing:

once the charged rod is removed:

the negative charges will be redistributed over $X+Y$ such that the electrostatic repulsive forces between them are kept to a minimum:

after the two spheres are separated,

both spheres $X$ and $Y$ will be negative charged. The descriptions go complete.

This problem is not to be attempted.

January 30, 2024January 31, 2024

202401300926 Exercise 81.b.1

A particle is projected at an angle of $45^\circ$ with the horizon from a point on a horizontal plane, with a velocity $\textrm{1,000}$ feet per second. Find its range, and find its distance from the point of projection at the end of $\textrm{5}$ seconds.

_{Gage, Alfred Payson. (1887). The High School Physics.}

Roughwork.

Draw a picture for the scenario.

We set up the notation system below:

$\begin{aligned} \textrm{Displacement} &:\mathbf{s}(x(t),y(t))=s_x\,\hat{\mathbf{i}}+s_y\,\hat{\mathbf{j}} \\ \textrm{Velocity} &: \mathbf{v}(t)=v_x\,\hat{\mathbf{i}}+v_y\,\hat{\mathbf{j}} \\ \textrm{Acceleration} &: \mathbf{a}=-10\,\mathrm{m\,s^{-1}}\,\hat{\mathbf{j}} \\ \end{aligned}$

and resolve them into their components:

$\begin{aligned} v^2(t) & = v_x^2(t)+v_y^2(t) \\ v_x(t_0) & = v_y(t_0)=\frac{\sqrt{2}}{2}v(t_0) \\ v_y(t) & = (v_y(t_0))+(- a)(t)\\ s(t) & = s_x^2(t) + s_y^2(t) \\ s_x(t) & = (v_x(t_0))(t) \\ s_y(t) & = h+(v_y(t_0))(t)+\frac{1}{2}(-a)(t)^2 \end{aligned}$

its range being $s_x(t=5)$ and its distance from the point of projection $s(t=5)=|\mathbf{s}(t=5)|$ at the end of $\textrm{5}$ seconds.

This problem is not to be attempted.

January 4, 2024January 4, 2024

202401041119 Exercise 8.2.541

Evaluate the integral

$\displaystyle{\underset{\varGamma}{\int}(x\,\mathrm{d}y+y\,\mathrm{d}x)}$ ,

where (a) $\varGamma$ is an arc of the parabola $y=\sqrt{x}$ , (b) $\varGamma$ is a segment of a straight line, (c) $\varGamma$ is the arc of the parabola $y=x^2$ connecting the points $(0,0)$ and $(1,1)$ in the direction indicated by the arrows (see Figure below).

_{Extracted from Yakov Stepanovič Bugrov. (1984). A collection of problems.}

Answer. (a) 1; (b) 1; (c) 1.

Roughwork.

(a)

When a point moves along the curve $\varGamma :y=\sqrt{x}$ in the indicated direction,

$\begin{aligned} &\quad \underset{\varGamma}{\int}(x\,\mathrm{d}y+y\,\mathrm{d}x) \\ & = \int_{(0,0)}^{(1,1)}(y^2\,\mathrm{d}y+\sqrt{x}\,\mathrm{d}x) \\ & = \bigg[\frac{y^3}{3} + \frac{2x^{3/2}}{3}\bigg]\bigg|_{(0,0)}^{(1,1)}\\ & = \bigg(\frac{(1)^3}{3} + \frac{2(1)^{3/2}}{3}\bigg) - \bigg(\frac{(0)^3}{3} + \frac{2(0)^{3/2}}{3}\bigg) \\ & = \bigg(\frac{1}{3}+\frac{2}{3}\bigg) - (0+0) \\ & = 1 \\ \end{aligned}$

(b)

When a point moves along the curve $\varGamma :y=x$ in the indicated direction,

$\begin{aligned} &\quad \underset{\varGamma}{\int}(x\,\mathrm{d}y+y\,\mathrm{d}x) \\ & = \int_{(0,0)}^{(1,1)}(y\,\mathrm{d}y+x\,\mathrm{d}x) \\ & = \bigg[\frac{y^2}{2} + \frac{x^2}{2}\bigg]\bigg|_{(0,0)}^{(1,1)} \\ & = \bigg(\frac{(1)^2}{2} + \frac{(1)^2}{2}\bigg) - \bigg(\frac{(0)^2}{2} + \frac{(0)^2}{2}\bigg)\\ & = \bigg( \frac{1}{2} + \frac{1}{2}\bigg) - (0+0) \\ & = 1 \\ \end{aligned}$

(c)

When a point moves along the curve $\varGamma :y=x^2$ in the indicated direction,

$\begin{aligned} &\quad \underset{\varGamma}{\int}(x\,\mathrm{d}y+y\,\mathrm{d}x) \\ & = \int_{(0,0)}^{(1,1)}(\sqrt{y}\,\mathrm{d}y+x^2\,\mathrm{d}x) \\ & = \bigg[\frac{2y^{3/2}}{3} + \frac{x^3}{3}\bigg]\bigg|_{(0,0)}^{(1,1)} \\ & = \bigg(\frac{2(1)^{3/2}}{3} + \frac{(1)^3}{3}\bigg) - \bigg(\frac{2(0)^{3/2}}{3} + \frac{(0)^3}{3}\bigg) \\ & = \bigg(\frac{2}{3}+\frac{1}{3}\bigg) - (0+0) \\ & = 1 \\ \end{aligned}$

Afterword.

What am I doing?

$\displaystyle{\int (x\,\mathrm{d}y+y\,\mathrm{d}x) = \int x\,\mathrm{d}y + \int y\,\mathrm{d}x}$

Note the first term on RHS:

and the second term on RHS:

Many simple formulae in physics, such as the definition of work as $W=\mathbf{F}\cdot\mathbf{s}$ , have natural continuous analogues in terms of line integrals, in this case $\displaystyle{W=\underset{L}{\int}\mathbf{F}(\mathbf{s})\,\mathrm{d}\mathbf{s}}$ , which computes the work done on an object moving through an electric or gravitational field $\mathbf{F}$ along a path $L$ .

_{Cited from Wikipedia on Line integral}

$\begin{aligned} &\quad \int (x\,\mathrm{d}y+y\,\mathrm{d}x) \\ & = \int \begin{bmatrix} y & x \end{bmatrix} \begin{bmatrix} \mathrm{d}x \\ \mathrm{d}y \end{bmatrix} \\ & = \int \mathbf{F}(x,y)\,\mathrm{d}\mathbf{s} \\ \cdots & \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ & \mathbf{F}(x,y) = y\,\hat{\mathbf{i}} + x\,\hat{\mathbf{j}} \\ \cdots & \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \end{aligned}$

As an aside, work done $W$ is the scalar (/dot) product of force $\mathbf{F}$ and displacement $\mathbf{s}$ . A conservative vector field in physics is analogous to a path-independent vector field in mathematics. That is, over any paths, the line integral depends only upon the starting point and the end point.

This problem is not to be attempted.

January 3, 2024January 3, 2024

202401031807 Pastime Exercise 007

The blogger claims no originality of his problem below.

A ball of mass $m$ is being passed through a smooth extensible light string, the two ends of which being attached to walls a distance $d$ apart.

When time $t=t_0$ :

when $t=t_1$ :

when $t=t_2$ :

when $t=t_3$ :

when $t=t_4$ :

when $t=t_5$ :

when $t=t_6$ :

such that stroboscopically for $t\in [t_0,t_6]$ , we see the ball’s trajectory fits into a parabola as can be described by some quadratic equation:

The tension $T(t)$ of the string and the speed $v(t)$ of the ball are time-varying variables dependent on mass $m$ and distance $d$ as well. By considering the free-body diagrams of the ball and of the string in discrete time frames $t_i$ ‘s (where $i\in\{ 0,1,2,3,4,5,6\}$ ), find the equation of motion at continuous time intervals of $\Delta t=t_6-t_0$ .

This problem is not to be attempted.