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I’m still working through the exercises to the Feynman’s Lectures. There are 36 of them to the Chapter 4! Doing exercises turned out to be the best way to understand something. I want to move forward faster with lectures and I have to stop myself constantly. It’s easy to fool yourself and think you understand the material after reading a lecture, only I open a new exercise and realize that there are gaps.

Mochi helps a lot. It’s surprising that this isn’t taught in schools and universities. Spaced repetition might be the only reliable and proven way to enhance memory. As Michael Nilsen says, “Anki makes memory a choice, rather than a haphazard event, to be left to chance.” It feels like superpower.

Movies watched

• First Man. Damien Chazelle brought depth to the story and focused on personal drama, stepping away from all the stories surrounding a well-known person.
• Oppenheimer. I rewatched it again with subtitles this time. It was better. Still, this movie lacks complexity and depth.
• Flow. An indie animation without any dialogues and with interesting style (probably shaped by its budget).
• The Wild Robot. As someone on Letterboxd put it, “the best Pixar cartoon was made by Dreamworks”. It’s an overstatement, but it’s a good cartoon.

Reading

Finished reading Bullshit Jobs. It resonated with my thoughts that many modern jobs are mind-numbing, and many more make the world a worse place. Still, it was quite repetitive and could’ve been half its size.

Dropped The Invention of Science after a few chapters. This turned out not to be a history of inventions, but more a philosophical work on how that was happening — how people talked and thought about it. It might be good, but it wasn’t what I expected.

Currently reading The Life of Isaac Newton, which has been good so far.

The easiest way to solve 2.19 (Plank Weight Trough)

A plank of weight W and length $R$ lies in a smooth circular trough of radius $R$. At one end of the plank is a weight $W/2$. Calculate the angle $\theta$ at which the plank lies when it is in equilibrium.

There are already a few solutions at the Feynman’s Lectures website, but my solution is simpler.

Since the plank is in equilibrium, it must be at its lowest possible position. This means the center of mass of the plank lies directly below the center of the trough.

We can calculate the center of mass $c$ given the length of the plank $L = R$:

${\displaystyle \begin{array}{rl}{\displaystyle c}& {\displaystyle =\frac{1}{1.5W}(W\frac{L}{2}+\frac{W}{2}L)}\\ {\displaystyle }& {\displaystyle =\frac{L}{1.5}(\frac{L}{2}+\frac{L}{2})}\\ {\displaystyle }& {\displaystyle =\frac{L}{1.5}=\frac{\sqrt{3}R}{1.5}=\frac{2R}{\sqrt{3}}}\end{array}}$

Since $c$ forms the hypotenuse from the left point of contact to the vertical line beneath the trough center, we have:

${\displaystyle \cos \theta =\frac{R}{c}=\frac{R\sqrt{3}}{2R}=\frac{\sqrt{3}}{2}}$

This gives $= 30^{}$

Solving 2.17 and 2.18 Using the Principle of Virtual Work

It is possible to solve 2.17 and 2.18 using torques, but since the chapter was about using the virtual work principle, let’s use it.

Let’s consider the ladder from the exercise 2.18 (2.17 uses similar approach) rotating clockwise due to the reactive force of the wall.

Let’s imagine a ladder rotating clockwise under the influence of the reaction force of the wall. As the ladder rotates by a small angle $\delta \theta$ radians, it displaces a distance $s = r$ (by the definition of a radian).

For a small angle $\theta$, it is possible to approximate that any point on ladder moves in a straight line, not in a circular path (see figure below).

This linear movement allows us to compute displacement of each point on the ladder as the following:

${\displaystyle \begin{array}{rl}{\displaystyle \mathrm{\Delta }x}& {\displaystyle =s\sin \theta =\delta \theta \cdot r\cdot \sin \theta }\\ {\displaystyle \mathrm{\Delta }y}& {\displaystyle =s\cos \theta =\delta \theta \cdot r\cdot \cos \theta }\end{array}}$

Now we can calculate the work done by a reactive force of the wall $T$:

${\displaystyle W_T=T(\delta \theta \cdot L\cdot \sin \theta )}$

Changes in potential energies of the weight $W$ and the ladder $\omega$ are:

${\displaystyle \begin{array}{rl}{\displaystyle E_W}& {\displaystyle =(\delta \theta \cdot 0.75L\cdot \cos \theta )\cdot W}\\ {\displaystyle E_{\omega }}& {\displaystyle =(\delta \theta \cdot 0.5L\cdot \cos \theta )\cdot \omega }\end{array}}$

Equating the work done by $T$ to the total change in potential energy yields:

${\displaystyle \begin{array}{rl}{\displaystyle T(\delta \theta \cdot L\cdot \sin \theta )}& {\displaystyle =(\delta \theta \cdot 0.5L\cdot \cos \theta )\cdot \omega }\\ {\displaystyle }& {\displaystyle +(\delta \theta \cdot 0.75L\cdot \cos \theta )\cdot W}\end{array}}$

${\displaystyle T=\mathrm{ctg}\theta (0.5\omega +0.75W)}$

Feynman’s Lectures Exercises 2.16

I spent more time on this exercise than needed because I didn’t notice that the masses are equal. Otherwise, the application of the virtual work principle is straightforward.

The work is done by the gravitation force and the force accelerates the entire system, that is $M = m_1 + m_2 = 2m$.

Let’s use the positive sign the direction of gravity acting on $m_2$.

The work can be calculate as:

${\displaystyle \begin{array}{rl}{\displaystyle W}& {\displaystyle =F\mathrm{\Delta }s}\\ {\displaystyle }& {\displaystyle =Ma\cdot \mathrm{\Delta }s}\\ {\displaystyle }& {\displaystyle =Ma\cdot \mathrm{\Delta }{y}_2}\\ {\displaystyle }& {\displaystyle =2m\cdot \mathrm{\Delta }{y}_2}\end{array}}$

This work is equal to the change of the potential energy in the system as follows:

${\displaystyle \begin{array}{rl}{\displaystyle \mathrm{\Delta }{E}_k}& {\displaystyle =m_2\cdot g\cdot \mathrm{\Delta }{y}_2-m_1\cdot g\cdot \mathrm{\Delta }{y}_1}\\ {\displaystyle }& {\displaystyle =m\cdot g\cdot \mathrm{\Delta }{y}_2-m\cdot g\cdot \mathrm{\Delta }{y}_1}\\ {\displaystyle }& {\displaystyle =m\cdot g\cdot (\mathrm{\Delta }{y}_2-\mathrm{\Delta }{y}_1)}\\ {\displaystyle }& {\displaystyle =m\cdot g\cdot (\mathrm{\Delta }{y}_2-\mathrm{\Delta }{y}_2\cdot \sin \frac{\pi }{4})}\end{array}}$

Thus, we get:

${\displaystyle \begin{array}{rl}{\displaystyle a}& {\displaystyle =g\frac{m-m\cdot \sin \frac{\pi }{4}}{2m}}\\ {\displaystyle }& {\displaystyle =g\frac{(1-\sin \frac{\pi }{4})\cdot m}{2m}}\\ {\displaystyle }& {\displaystyle =g\frac{1-\frac{\sqrt{2}}{2}}{2}}\\ {\displaystyle }& {\displaystyle =\frac{1}{2}(1-\frac{1}{\sqrt{2}})g}\end{array}}$

Feynman’s Lectures Exercises 2.14 and 2.15

If the weights $W_1$ and $W_2$ move the distance $s$ along the plane to the left, their corresponding difference in height are:

${\displaystyle \mathrm{\Delta }{h}_1=s\cdot \sin \theta ,\mathrm{\Delta }{h}_2=-s\cdot \sin \theta }$

Then, the difference in potential energy of the system is:

${\displaystyle \mathrm{\Delta }{E}_p=\mathrm{\Delta }{h}_1{W}_1-\mathrm{\Delta }{h}_2{W}_2}$

By the definition of work:

${\displaystyle F\cdot s=\mathrm{\Delta }{E}_p}$

Then:

${\displaystyle \begin{array}{rl}{\displaystyle F}& {\displaystyle =\frac{\mathrm{\Delta }{E}_p}{s}}\\ {\displaystyle }& {\displaystyle =\sin \theta \cdot m_1\cdot g-\sin \theta \cdot m_2\cdot g}\\ {\displaystyle }& {\displaystyle =\sin \theta \cdot g\cdot (m_1-m_2)}\end{array}}$

For the acceleration along the plane, using $a =$ and the total mass $m = m_1 + m_2$, the acceleration is:

${\displaystyle a=\sin \theta \cdot g\cdot \frac{m_1-m_2}{m_1+m_2}}$

For the motion over distance $D$, starting from rest:

${\displaystyle D=\frac{a{t}^2}{2}}$

Thus:

${\displaystyle \begin{array}{rl}{\displaystyle t^2}& {\displaystyle =\frac{2D}{a}}\\ {\displaystyle t}& {\displaystyle =\sqrt{\frac{2D}{a}}}\end{array}}$

The speed at time $t$ is given by:

${\displaystyle \begin{array}{rl}{\displaystyle v}& {\displaystyle =at}\\ {\displaystyle }& {\displaystyle =a\cdot \sqrt{\frac{2D}{a}}}\\ {\displaystyle }& {\displaystyle =\sqrt{2Da}}\\ {\displaystyle }& {\displaystyle =\sqrt{2D\cdot \sin \theta \cdot g\cdot \frac{m_1-m_2}{m_1+m_2}}}\end{array}}$

For Exercise 2.15, the acceleration is given by:

${\displaystyle a=\frac{1}{2}\cdot g\cdot (\sin \theta -\sin \varphi )}$

Substituting this into the expression for speed, we get:

${\displaystyle v=\sqrt{g\cdot D\cdot (\sin \theta -\sin \varphi )}}$

I finally bought the Feynman Lectures on Physics, something I wanted to do for a very long time. Ever since reading You’re Surely Joking Mr. Feynman. After watching some of his interviews on YouTube and going through a few chapters online, I realized there’s something special about how Feynman explains things. He doesn’t just teach formulas — he teaches understanding. Once you grasp that, you realize what true understanding is. Here are a few quotes from the introduction to the New Millennium edition:

It was like going to church. The lectures were a transformational experience, the experience of a lifetime, probably the most important thing I got from Caltech. I was a biology major but Feynman’s lectures stand out as a high point in my undergraduate experience … though I must admit I couldn’t do the homework at the time and I hardly turned any of it in. I was among the least promising of students in this course, and I never missed a lecture. … I remember and can still feel Feynman’s joy of discovery. … His lectures had an … emotional impact that was probably lost in the printed Lectures.

The book is beautiful. I don’t choose books solely on looks, but when there are different options, it’s better to go for the one you like. I’ve noticed that this works for anything — if I like something that’s part of an activity, I’m more likely to engage with it. So, in that sense, it is rational to choose a book by its cover. One sad thing is the lettering from the original edition has been lost in the New Millennium version:

Almost all other popular physics books look very average. Compare another popular textbook with the austere design of Feynman’s lectures:

My plan is to work through all three volumes over the winter and solve all the exercises from the companion workbook. I will buy a refresher for math once I hit a wall. There are a few books I have in mind: Spivak’s Calculus and The Princeton Companion to Applied Mathematics. Both are as beautifully designed as Feynman’s lectures.

Made a few changes to the site, and now I can have posts without titles. It gives a sense of freedom, similar to what you might have unconsciously experienced on Twitter. Having a title adds a layer of seriousness to a post.

The Archive layout has changed to accommodate missing titles — now the dates link to posts, and the dates are in ISO format set in tabular figures for proper alignment.

With each iteration, it’s becoming more brutalist, and I like it.

Finally watched Carl Sagan’s Cosmos, a TV series produced in the 1980s. I don’t think there are many people like Carl Sagan. The show he created is so delicate, with many complex emotions. It’s captivating, fun, sad, and melancholic. The original music was composed by Vangelis, and it’s hard to imagine how it could be better.

We’re now rewatching the sequel by one of his students — Neil deGrasse Tyson. While it covers similar topics and has better graphics, it feels a little shallower.

THE GAP by Ira Glass

Feynman on Playing

A few hours after I wrote the last post, I remembered one chapter from Surely you’re joking, Mr. Feynman:

Then I had another thought: Physics disgusts me a little bit now, but I used to enjoy doing physics. Why did I enjoy it? I used to play with it. I used to do whatever I felt like doing - it didn’t have to do with whether it was important for the development of nuclear physics, but whether it was interesting and amusing for me to play with. When I was in high school, I’d see water running out of a faucet growing narrower, and wonder if I could figure out what determines that curve. I found it was rather easy to do. I didn’t have to do it; it wasn’t important for the future of science; somebody else had already done it. That didn’t make any difference. I’d invent things and play with things for my own entertainment.

So I got this new attitude. Now that I am burned out and I’ll never accomplish anything, I’ve got this nice position at the university teaching classes which I rather enjoy, and just like I read the Arabian Nights for pleasure, I’m going to play with physics, whenever I want to, without worrying about any importance whatsoever.

Within a week I was in the cafeteria and some guy, fooling around, throws a plate in the air. As the plate went up in the air I saw it wobble, and I noticed the red medallion of Cornell on the plate going around. It was pretty obvious to me that the medallion went around faster than the wobbling. I had nothing to do, so I start to figure out the motion of the rotating plate. I discover that when the angle is very slight, the medallion rotates twice as fast as the wobble rate — two to one. It came out of a complicated equation! Then I thought, “Is there some way I can see in a more fundamental way, by looking at the forces or the dynamics, why it’s two to one?” I don’t remember how I did it, but I ultimately worked out what the motion of the mass particles is, and how all the accelerations balance to make it come out two to one.

I still remember going to Hans Bethe and saying, “Hey, Hans! I noticed something interesting. Here the plate goes around so, and the reason it’s two to one is …” and I showed him the accelerations.

He says, “Feynman, that’s pretty interesting, but what’s the importance of it? Why are you doing it?”

“Hah!” I say. “There’s no importance whatsoever. I’m just doing it for the fun of it.” His reaction didn’t discourage me; I had made up my mind I was going to enjoy physics and do whatever I liked.

I went on to work out equations of wobbles. Then I thought about how electron orbits start to move in relativity. Then there’s the Dirac Equation in electrodynamics. And then quantum electrodynamics. And before I knew it (it was a very short time) I was “playing” — working, really — with the same old problem that I loved so much, that I had stopped working on when I went to Los Alamos: my thesis-type problems; all those old-fashioned, wonderful things.

It was effortless. It was easy to play with these things. It was like uncorking a bottle: Everything flowed out effortlessly. I almost tried to resist it! There was no importance to what I was doing, but ultimately there was. The diagrams and the whole business that I got the Nobel Prize for came from that piddling around with the wobbling plate.