Ex 1.29 A particle of mass m slides off a horizontal cylinder, 2.1

Posted on June 28, 2023 by rpflee

Structure and Interpretation of Classical Mechanics

A particle of mass $m$ slides off a horizontal cylinder of radius $R$ in a uniform gravitational field with acceleration $g$ . If the particle starts close to the top of the cylinder with zero initial speed, with what angular velocity does it leave the cylinder?

~~~

Kinetic energy:

$\displaystyle{\begin{aligned} T(r(t), \theta(t)) &= \frac{1}{2} m \left( v_x^2 + v_y^2 \right) \\ &= \frac{1}{2} m \left[ \left( \frac{d}{dt} r \sin \theta \right)^2 + \left( \frac{d}{dt} r \cos \theta \right)^2 \right] \\ &= \frac{1}{2} m \left[ \left( \dot r \sin \theta + r \dot \theta \cos \theta \right)^2 + \left( \dot r \cos \theta - r \dot \theta \sin \theta \right)^2 \right] \\ &= \frac{1}{2} m \left[ \dot r^2 + r^2 \dot \theta^2 \right] \\ \end{aligned}}$

Potential energy:

$\displaystyle{\begin{aligned} V(r(t), \theta(t)) &= mgr \cos \theta \\ \end{aligned}}$

Constraint:

$\displaystyle{\begin{aligned} \phi(t) &= r(t) - R \\ &= 0 \\ \end{aligned}}$

The Lagrangian:

$\displaystyle{\begin{aligned} L(r(t), \theta(t)) &= T - V + \lambda \phi \\ &= \frac{1}{2} m ( \dot r^2 + r^2 \dot \theta^2 ) - mg r \cos \theta + \lambda (r - R) \\ \end{aligned}}$

The Lagrange equations:

$\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot r} \right) - \frac{\partial L}{\partial r} &= 0 \\ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot \theta} \right) - \frac{\partial L}{\partial \theta} &= 0 \\ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot \lambda} \right) - \frac{\partial L}{\partial \lambda} &= 0 \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot r} \right) - \frac{\partial L}{\partial r} &= 0 \\ \frac{d}{dt} m \dot r - \left( m r \dot \theta^2 - mg \cos \theta + \lambda \right) &= 0 \\ m \ddot r &= m r \dot \theta^2 - mg \cos \theta + \lambda \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot \theta} \right) - \frac{\partial L}{\partial \theta} &= 0 \\ \frac{d}{dt} \left( m r^2 \dot \theta \right) - mgr \sin \theta &= 0 \\ m \left( 2 r \dot r \dot \theta + r^2 \ddot \theta \right) - mgr \sin \theta &= 0 \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot \lambda} \right) - \frac{\partial L}{\partial \lambda} &= 0 \\ - ( r - R ) &= 0 \\ r(t) &= R \\ \end{aligned}}$

By the constraint $\displaystyle{r(t) = R}$ ,

$\displaystyle{ \begin{aligned} \dot r(t) &= 0 \\ \ddot r(t) &= 0 \\ \end{aligned}}$

— Me@2023-06-25 07:47:21 PM

Reality 4.2

Posted on June 25, 2023 by rpflee

One’s own pain cannot be fake, because by definition, one’s own pain is personal.

Pain itself is a feeling, not a description of one’s external reality, although the source can be.

— Me@2015-12-30 11:46:28 PM

— Me@2023-06-25 12:02:45 PM

Civilization

Posted on June 24, 2023 by rpflee

Privacy 3

Technology/civilization is the increase of the ability to ignore other people, so that you can focus all your time on yourself and the ones you like.

— Me@2016-02-11 02:39:18 PM

未來騙局

Posted on June 23, 2023 by rpflee

這段改編自 2023 年 06 月 20 日的對話。

當年，有位女同事說，她當年曾到過，某間名校面試，做英文老師。

面試後，校方想聘請她，尤其是科主任。但是，校長卻說，按該校的慣例，初時不會簽永久合約，只會簽二年約。在那裡有相當的年資，即是幾年後，才可能會有長約。我那女同事，堅持要長約。結果，沒有接受該校那職位。

年輕時的我，就不懂那樣，守衞自己的權利，保證自己的生存。

那時，我可能會想，難得可以入了名校任教，長遠對我的履歷，很有幫助。另外，市面上的講法是，先爭取累積工作經驗，長遠自然會賺得多。

那大錯特錯。

微觀而言：

1. 假設你月薪少了 5000 元，你並不只是「少了 5000 元」，而是「每月少了 5000 元」。

2. 你薪金低於公道之價，僱主和同事就會覺得，你的時間不值錢。他們自然會對你再差。

何謂「公道之價」呢？

客觀的公平，就是雙方也覺得，這次合作對雙方有利。

3. 薪金公道的工作，難道就不可以，累積到工作經驗嗎？

宏觀來說：

承諾 = 騙局

1. 首先，剛才只是說，幾年後，才「可能」有長約，並沒有承諾。

2. 即使有承諾，例如承諾三年後對我好，為何不現在就對我好？

你現在已經對我不好，為何我還要相信，你三年後會對我好？

3. 即使會，為何三年後之前，我要委屈自己？

可能僱主想現在對你好，但暫時資金不夠給你，公平的待遇。

4. 那暫到幾時？

現在既然資金不夠，為何你會知道，三年後會夠？

三年後會夠的話，那我三年後之前的所有損失，會賠償給我嗎？

只有本來，沒有未來。

本來無未來。

承諾皆廢話，未來皆騙局。

除非，那潛在的未來事件，化成了現在有法律效力的合約，即是未來如果任何一方不遵守，另一方會獲得賠償。亦即是話，所謂的「未來事件」，透過合約，「翻譯」成了現在事件。

— Me@2023-06-20 12:47:31 PM

Emacs theme

Posted on June 18, 2023 by rpflee

(custom-set-variables
 '(cua-mode t nil (cua-base))
 '(custom-enabled-themes '(leuven))
 '(global-display-line-numbers-mode t))

(custom-set-faces
 '(default ((t (:family "DejaVu Sans Mono"
                        :foundry "PfEd"
                        :slant normal
                        :weight normal
                        :height 120
                        :width normal)))))

(prefer-coding-system 'utf-8)

(global-visual-line-mode t)

— Me@2023-06-18 01:27:37 PM

Ex 1.29 A particle of mass m slides off a horizontal cylinder, 1.3

Posted on June 15, 2023 by rpflee

Structure and Interpretation of Classical Mechanics

…

$\displaystyle{\begin{aligned} \dot \theta^2 &= \pm \frac{2 g (\beta \sin \theta - \cos \theta) }{R (\beta^2 + 1)} + \pm \frac{2 g }{R (\beta^2 + 1)} e^{-\beta \theta} \\ &= \pm \frac{2 g }{R (\beta^2 + 1)} \left[ \beta \sin \theta - \cos \theta + e^{-\beta \theta} \right] \\ \end{aligned}}$

Let us reconsider the equation of motion along the normal direction:

$\displaystyle{\begin{aligned} m g \cos \theta - F_R &= \frac{m v^2}{R} \\ \end{aligned}}$ ,

where $\displaystyle{F_R}$ is the normal reaction force.

When the mass $m$ leaves the cylinder with $\theta=\theta_1$ , $F_R = 0$ .

$\displaystyle{\begin{aligned} m g \cos \theta_1 &= \frac{m R^2 \dot \theta^2}{R} \\ \dot \theta^2 &= \frac{g}{R} \cos \theta_1 \\ \\ \frac{g}{R} \cos \theta_1 &= \pm \frac{2 g }{R (\beta^2 + 1)} \left[ (\beta \sin \theta - \cos \theta) + e^{-\beta \theta} \right] \\ (1+ \beta^2) \cos \theta_1 &= 2 \left| ( \beta \sin \theta_1 - \cos \theta_1 + e^{-\beta \theta_1} ) \right| \\ \end{aligned}}$

If there is no air resistance, $\beta = 0$ ,

$\displaystyle{\begin{aligned} (1+ \beta^2) \cos \theta_1 &= 2 \left| ( \beta \sin \theta_1 - \cos \theta_1 + e^{-\beta \theta_1} ) \right| \\ \cos \theta_1 &= 2 ( 1 - \cos \theta_1) \\ \cos \theta_1 &= \frac{2}{3} \\ \end{aligned}}$

— Me@2023-05-23 11:02:25 AM

Visualize, 3

Posted on June 13, 2023 by rpflee

Algebra requires sequential understanding.

Geometry requires parallel understanding.

space ~ parallel

time ~ series

By the definition of “at once”, you cannot see objects or the same object at different times at once.

— Me@2016-01-04 08:44:21 PM

FirstNotes, 1

Posted on June 10, 2023 by rpflee

16[.]12[.]2002

Cary Grant
I began by acting like the person I want to be, and 
eventually I became that person. 

10122002

Be strong in tough moment

04122002

Nothing Less: 
M. C. Chan: Do not have an escape character: 
Learn and get it now! 

Get the point now! 
 
03122002 

When you are good, don't think that you are very good; 
when you are bad, don't think that you are very bad. 

"You must first jump, so that while you are falling, 
you can create your wings."

Rather than hoping things become easier, 
why not make yourself stronger and tougher?  

Emerson 
What we persist doing becomes easier, 
not because the thing itself is easier, 
but because our capacity has increased. 

80 20 principle 

One way to build up your faith is to spend time with 
somebody who truly believe in you. And the important 
thing is that be that person to others; also, be that 
person to yourself.

離婚 3.2

Posted on June 9, 2023 by rpflee

離婚協議，在感情還好時，即是結婚之前，才可以達成。

正如人壽保險，在一個人還健康時，才可以買到。

— Me@2023-03-10 11:17:40 AM

Euler problem 13.2

Posted on June 6, 2023 by rpflee

e13 :: IO ()
e13 = do
  str <- readFile "n.txt"
  let sList = lines str
      nList = map (\x -> read x :: Integer) sList
   in print $ take 10 . show . sum $ nList

— Me@2023-06-06 11:37:32 AM

Ex 1.29 A particle of mass m slides off a horizontal cylinder, 1.2

Posted on June 3, 2023 by rpflee

Structure and Interpretation of Classical Mechanics

…

$\displaystyle{\begin{aligned} R \frac{d^2 \theta}{dt^2} &= g \sin \theta - \frac{\beta R \dot \theta^2}{2} \\ \end{aligned}}$

DSolve[
    R theta''[t]
        == g Sin[theta[t]]
            - (beta*R (theta'[t])^2)/2,

    theta[t],
    t
] // Simplify // TeXForm

Let

$\displaystyle{\begin{aligned} u &= \dot \theta^2 \\ \end{aligned}}$

Then

$\displaystyle{\begin{aligned} \frac{du}{dt} &= 2 \left( \pm \sqrt{u} \right) \ddot \theta \\ \ddot \theta &= \frac{\pm 1}{2} \frac{du}{d \theta} \\ \frac{du}{d \theta} + \beta u &= \pm \frac{2 g}{R} \sin \theta \\ \end{aligned}}$

This is a first-order linear differential equation. The integrating factor is

$\displaystyle{\begin{aligned} e^{\int \beta d \theta} &= e^{\beta \theta} \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \frac{du}{d \theta} + \beta u &= \pm \frac{2 g}{R} \sin \theta \\ \frac{d}{d \theta} \left( u e^{\beta \theta} \right) &= \pm \frac{2 g e^{\beta \theta} }{R} \sin \theta \\ \end{aligned}}$

$\displaystyle{\begin{aligned} u e^{\beta \theta} &= \pm \frac{2 g }{R} \int e^{\beta \theta} \sin \theta d \theta \\ u &= \pm \frac{2 g (\beta \sin \theta - \cos \theta) }{R (\beta^2 + 1)} + C e^{-\beta \theta} \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \dot \theta^2 &= \pm \frac{2 g (\beta \sin \theta - \cos \theta) }{R (\beta^2 + 1)} + C e^{-\beta \theta} \\ \end{aligned}}$

When $t=0$ , $\theta=0$ and $\dot \theta = 0$ :

$\displaystyle{\begin{aligned} 0 &= \pm \frac{- 2 g }{R (\beta^2 + 1)} + C \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \dot \theta^2 &= \pm \frac{2 g (\beta \sin \theta - \cos \theta) }{R (\beta^2 + 1)} + \pm \frac{2 g }{R (\beta^2 + 1)} e^{-\beta \theta} \\ &= \pm \frac{2 g }{R (\beta^2 + 1)} \left( \beta \sin \theta - \cos \theta + e^{-\beta \theta} \right) \\ \end{aligned}}$

…

— Me@2023-05-23 11:02:25 AM

They don’t exist under normal condition

Posted on June 2, 2023 by rpflee

rickreynoldssf 2 days ago | next [–]

I’m an airchair physics person so I’m curious how close my layman understanding of this matches reality… So Higgs particles don’t exist under normal conditions, they’re just proof that the Higgs field exists and explains how mass exists. When the energy that’s perturbing the Higgs field dissipates it does so through other fields perturbing them to create one of their particles and so on.

Sci-Fi or Fact or somewhere in the middle?

RickyS 2 days ago | parent | next [–]

Experimental particle physicist here. What you say about Higgs particles “they don’t exist under normal condition” is loosely true of all particles in nature in the sense that a particle is nothing but a “quantum” of a “field”. Fields pervade all physical space and can vary in time. Particles (or quanta) simply represent a local state of observable things. A field can only do certain things to certain physical states and at a probabilistic level. Notice that fields do things even with the vacuum which is just another state from which particles can be “extracted”.

The peculiar experimental challenge about the Higgs field is that it can extract its quanta from certain physical states (certain initial conditions in a particle physics reaction) only at very high energy and with low probability, but that is true also for other particles. Its truly peculiar thing is that the presence of the Higgs field, in addition to the fields of all other particles that we know of, explains why quanta in general have a mass (although this is not clear for neutrinos) through a mechanism where the Higgs field interacts with the quanta of other particles.

— LHC experiments see first evidence of a rare Higgs boson decay

— Hacker News

2023.06.02 Friday ACHK

If, 2

Posted on June 1, 2023 by rpflee

To keep her

you must be willing to lose her.

Most wouldn’t understand

— Dee Sage | King Maker

2023.05.31 Wednesday ACHK

Physics Town

continues

Monthly Archives: June 2023

Ex 1.29 A particle of mass m slides off a horizontal cylinder, 2.1

Reality 4.2

Civilization

未來騙局

Emacs theme

Ex 1.29 A particle of mass m slides off a horizontal cylinder, 1.3

Visualize, 3

FirstNotes, 1

離婚 3.2

Euler problem 13.2

Ex 1.29 A particle of mass m slides off a horizontal cylinder, 1.2

They don’t exist under normal condition

If, 2