Tender Embrace: A Mother’s Love Captured in Holding Her Precious Bundle of Joy

November 17, 2023November 18, 2023 / Michael Xue / Leave a comment

Generated from Rikitake’s dynamo model

$\begin{cases} \frac{dx}{dt}=-\mu x + yz \\ \frac{dy}{dt} = -\mu y + (z-\mu(k^2-\frac{1}{k^2}))x \\\frac{dz}{dt}=1-xy \end{cases}$

by Omega CAS Explorer with $\mu=1.05$ and $k = 1.5$ :

The after-math of “Tamed in a fleeting glance”

November 15, 2023November 16, 2023 / Michael Xue / Leave a comment

A high-schooler would have found the braking distance in “Tamed in a fleeting glance” differently:

$F = ma\quad \implies \quad -\mu mg = ma \quad \implies \quad a = -\mu g$

$v(t) = v_0 + at\quad \implies \quad 0 = v_0 + a t_b \quad \implies \quad t_b = \frac{-v_0}{a} = \frac{-v_0}{-\mu g} = \frac{v_0}{\mu g}$

$D_b = v_0 t_b + \frac{1}{2}at_b^2 = v_0\cdot\frac{v_0}{\mu g} + \frac{1}{2}\cdot(-\mu g)\cdot(\frac{v_0}{\mu g})^2 = \frac{v_0^2}{\mu g} - \frac{v_0^2}{2\mu g} = \frac{v_0^2}{2\mu g}$

Tamed in a fleeting glance

November 10, 2023December 16, 2023 / Michael Xue / Leave a comment

The effect of braking the vehicle can be modeled by the introduction of a resisting frictional force upon pressing the brakes.

Fig. 1

Suppose $m$ is the mass of the vehicle and $\mu$ is the coefficient of friction. Then by definition, the braking force on the vehicle is $\mu\cdot mg$ where $g$ is the gravitational acceleration (see Fig. 1). The distance traveled in stopping, called braking distance, can be found by solving the differential equation of motion in a straight line, subject to the braking force opposite to the direction of motion. Namely,

$m\frac{d^2x}{dt^2} = -\mu\cdot mg \quad\implies\quad \frac{d^2x}{dt^2}=-\mu g.\quad\quad\quad(1)$

The conditions to impose on function $x=x(t)$ are

$x(0)=0$ and $\frac{dx(0)}{dt} = v_0.\quad\quad\quad(2)$

The braking distance, then, is the distance traveled when $\frac{dx}{dt} = 0$ .

Integrate (1) gives

$\displaystyle\int\limits_{0}^{t}\frac{d}{dt}\left(\frac{dx}{dt}\right)\;dt= \int\limits_{0}^{t}-\mu g\;dt\;\implies\;\frac{dx}{dt}\bigg|_{0}^{t} = -\mu g t\bigg|_{0}^{t}\;\overset{(2)}{\implies} \;\frac{dx}{dt} - v_0 = -\mu g t$ .

That is,

$\frac{dx}{dt} = -\mu g t + v_0.\quad\quad\quad(3)$

Hence, the vehicle stops when

$\boxed{t = t_b = \frac{v_0}{\mu g}}$

Integrate (3) subject to $x(0)=0$ stated in (2), we have

$\displaystyle\int\limits_{0}^{t} \frac{dx}{dt}\;dt = \int\limits_{0}^{t}-\mu gt+v_0\;dt \quad\implies\quad x(t) = -\frac{\mu g t^2}{2} + v_0 t.$

It follows that $D_b$ , the braking distance is the value of $x$ at $t=t_b$ . i.e.,

$\boxed{D_b = x(t_b) = \frac{v_0^2}{2\mu g}}$

Exercise-1 A vehicle traveling on dry, leveled pavement at 40 mi/h has the brakes applied. The vehicle travels 270 ft before stopping. What is the coefficient of friction?

Hint: 1 mi = 5280 ft, 1 h = 3600 s and g = 32 ft/s^2

Exercise-2 Find the braking distance of a vehicle traveling up a grade of 1 ft/100 ft (see Fig. 2)

Fig. 2

Hint: The normal force is reduced, and an additional stopping force due to gravity is present.

The Kiss of Steel

November 6, 2023November 7, 2023 / Michael Xue / Leave a comment

A bullet hits a plate of thickness $h$ with speed $v_0$ , then penetrates it, and leaves the plate with speed $v_1 < v_0$ . Assume that the resistance of the plate to the bullet is proportional to the square of the bullet’s speed, find the time it takes for the bullet to pass through the plate.

Let $m, v$ and $\tau$ denotes respectively the mass of the bullet, its speed and the time it takes for the bullet to pass through the plate. By Newton’s second law,

$m\frac{dv}{dt} = -\alpha v^2 \implies \frac{dv}{dt} = -\frac{\alpha}{m} v^2\quad\quad\quad(1)$

Let $\beta = \frac{\alpha}{m}$ , (1) becomes

$\frac{dv}{dt}=-\beta v^2\quad\quad\quad(2)$

Integrate (2) from $t=0$ to $\tau$ , we have

$\displaystyle\int\limits_{0}^{\tau}\frac{1}{v^2}\frac{dv}{dt}\;dt = -\int\limits_{0}^{\tau}\beta\;dt \implies \int\limits_{v_0}^{v_1}\frac{1}{v^2}\;dv = -\beta \tau$ .

That is,

$-\frac{1}{v}\bigg|_{v_0}^{v1} = -\beta \tau \implies \frac{1}{v_1}-\frac{1}{v_0} = \beta \tau$

Therefore,

$\tau = \frac{1}{\beta}(\frac{1}{v_1}-\frac{1}{v_0}).\quad\quad\quad(3)$

To find $\beta$ , we write (2) as

$\frac{1}{v}\;\frac{dv}{dt} = -\beta v.\quad\quad\quad(4)$

Integrate (4) from 0 to $\tau$ , we have

$\displaystyle\int\limits_{0}^{\tau}\frac{1}{v}\frac{dv}{dt}\;dt = -\beta\underline{\int_{0}^{\tau}v\;dt}.$

Since

$\displaystyle\underline{\int_{0}^{\tau}v\;dt}= h$ ,

it gives

$\displaystyle\int\limits_{v_0}^{v_1}\frac{1}{v}dv = -\beta h$ .

i.e.,

$\log(v)\bigg|_{v_0}^{v_1} = -\beta h \;\implies\; \log(v_1) - \log(v_0) = -\beta h\;\implies\; \beta= \frac{\log(v_0)-\log(v_1)}{h}\quad\quad(5)$

Substitute (5) into (3), we obtain

$\tau = \frac{h}{\log(\frac{v_0}{v_1})}(\frac{1}{v_1}-\frac{1}{v_0})$

Exercise-1 Explain $\int\limits_{0}^{\tau}\frac{1}{v^2}\frac{dv}{dt}\;dt = \int\limits_{v_0}^{v_1}\frac{1}{v^2}\;dv$ .

Exercise-2 Explain $\int\limits_{0}^{\tau}v\;dt = h$ .

A Whispering Chamber

November 1, 2023February 7, 2024 / Michael Xue / Leave a comment

Question: A gallery has two fixed points A and B. It is a room so shaped such that a visitor standing at A can hear even the slightest whisper emitted by another visitor at B. What is the smooth curve that represents the wall of the gallery?

Answer:

In order for a visitor at A to hear even the slightest whisper emitted at B, the wall must be represented by a curve such that the sound waves reflected off the wall will focus at A (see Fig. 1).

Fig. 1 The unknown curve’s tangent line at (x, y) is in red

The law of sound reflection states that sound waves bounce off a surface in a manner such that the angle at which they approach the surface is equal to the angle at which they are reflected away from it. i.e.,

$\frac{\pi}{2}-\alpha = \frac{\pi}{2}-\beta.\quad\quad\quad(1)$

We deduce from (1) that $\alpha = \beta$ and hence,

$\cos(\alpha) = \cos(\beta).$

Since $\alpha$ and $\beta$ are the angles between $\overrightarrow{CA}, \overrightarrow{BC}$ and $\overrightarrow{CD}$ respectively, it follows that

$\underbrace {\frac{<\overrightarrow{CA},\overrightarrow{CD}>}{|\overrightarrow{CA}|\cdot|\overrightarrow{CD}|}}_{\cos(\alpha)} = \underbrace{\frac{<\overrightarrow{BC},\overrightarrow{CD}>}{|\overrightarrow{BC}|\cdot |\overrightarrow{CD}|}}_{\cos(\beta)}.\quad\quad\quad(2)$

We have

$\overrightarrow{OB} + \overrightarrow{BC} = \overrightarrow{OC} \implies \overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix}x \\y \end{pmatrix} - \begin{pmatrix}c \\ 0 \end{pmatrix}$ so

$\overrightarrow{BC}=\begin{pmatrix}x-c \\y \end{pmatrix}.\quad\quad\quad(3)$

Similarly,

$\overrightarrow{OC} + \overrightarrow{CA} = \overrightarrow{OA} \implies \overrightarrow{CA} = \overrightarrow{OA}-\overrightarrow{OC}=\begin{pmatrix}-c \\0 \end{pmatrix}-\begin{pmatrix}x \\y \end{pmatrix}$ or

$\overrightarrow{CA} =\begin{pmatrix}-c-x\\-y \end{pmatrix}. \quad\quad\quad(4)$

And,

$\overrightarrow{OC} + \overrightarrow{CD} = \overrightarrow{OD}\implies \overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC}=\begin{pmatrix} x_* \\y_* \end{pmatrix}-\begin{pmatrix}x \\y \end{pmatrix}$ yields

$\overrightarrow{CD} = \begin{pmatrix} x_*-x \\ y_*-y \end{pmatrix}.\quad\quad\quad(5)$

Notice that

$\frac{y_*-y}{x_*-x} = y'.$

It means

$y_*-y = (x_*-x)y'.\quad\quad\quad(6)$

Let $\delta = x_*-x$ , (5) becomes

$\overrightarrow{CD} = \begin{pmatrix} \delta \\ \delta y' \end{pmatrix}.\quad\quad\quad(7)$

Therefore, (2) gives

$\frac{<\overrightarrow{CA}, \overrightarrow{CD}>}{|\overrightarrow{CA}|} = \frac{<\overrightarrow{BC},\overrightarrow{CD}>}{|\overrightarrow{BC}|}$

$\implies \frac{(-c-x)\cdot \delta + (-y\cdot ky')}{\sqrt{(-c-x)^2+y^2}} = \frac{(x-c)\cdot \delta + (y\cdot \delta y')}{\sqrt{(x-c)^2+y^2}}$

$\implies \frac{(-c-x)-y\cdot y'}{\sqrt{(c+x)^2+y^2}} = \frac{(x-c)+y\cdot y'}{\sqrt{(x-c)^2+y^2}}.$

That is,

$\frac{(x+c)+y\cdot y'}{\sqrt{(c+x)^2+y^2}} = \frac{(c-x)-y\cdot y'}{\sqrt{(x-c)^2+y^2}}.\quad\quad\quad(8)$

Fig. 2

Solving (8) for $y$ (see Fig. 2), we obtain

$\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = k$

where $k=\%c$ is a constant. It is equivalent to

$\sqrt{(x-(-c))^2+(y-0)^2} + \sqrt{(x-c)^2+(y-0)^2} = k.\quad\quad\quad(*)$

Recall the definition of Ellipse:

An ellipse is the set of points in a plane such that the sum of their distances from two fixed points is a constant.

From (*) we recognize the plane curve represents the gallery’s wall must be an ellipse:

For the points on the curve, the sum of their distances from two fixed points $(-c, 0)$ and $(c, 0)$ is a constant $k$ .

From https://www.dam.brown.edu/people/mumford/beyond/papers/Atlantic14.pdf :