Recent content by PhDeezNutz

The distance depends on the distance……who would have thought? That article is exceptionally bad.

Man I’m so sorry I didn’t see this thread until now. Even more sorry that I didn’t know until now. R.I.P.

On a side note (or maybe not a side note?). I think I see the motivation for defining the Poisson Bracket. If we add a total time derivative ##\frac{dF(q,p,t)}{dt}## to the Lagrangian and ASSUME that Hamilton’s equations hold, that is ##\dot q = \frac{\partial H}{\partial p}## ##\dot p = -...

Based off of these MIT Notes: https://ocw.mit.edu/courses/8-09-classical-mechanics-iii-fall-2014/f00f7f68ac7ba346a0868efb7430582c_MIT8_09F14_Chapter_4.pdf 1) This set of notes starts with the premise that ##L’ = L + \frac{dF(q,t)}{dt} = L + \frac{\partial F}{\partial q} \dot q + \frac{\partial...

OP are you familiar with solving second order linear non-homogenous differential equations? I was able to get the correct answer (Which is A btw as others have pointed out) Use the ansatz ##x(t) = Ae^{\sqrt{5} t} + Be^{-\sqrt{5} t} + C## Apply the initial conditions ##x(0) = 0## and ##...

I think the answer for the charge across the second capacitor that was initially charged to ##C\epsilon## should be ##\frac{1}{3} C \epsilon \left(e^{-\frac{3t}{2CR}} + 2 \right)## Which i was able to get through Laplace Transforms. This answer at the very least supports the initial condition...

I think a partial answer to OP’s question “without integration or differentiation” is using Laplace transforms. I will give it some more thought as to how I would incorporate initial conditions.

What is the meaning of CE? (Capacitance times electric field?) I’m not sure that has units of charge as the problem statement indicates.

I believe ##x## is used for desktop diaries and ##y## is used for pocket diaries. They key word is “at least” “They will need at least twice as many pocket diaries as desktop diaries”……translation: “the number of pocket diaries must be greater than or equal to 2 times the number of desktop...

Did you just get your username changed to “Death Metal”? Nice.

nasum 😍

I have the third edition of Goldstein but I have heard that the 2nd edition is much better. Users have stated on this forum that there are problems with the treatment of non-holonomic constraints in the 3rd edition.

I believe your rationale is correct (it's the one I used) ##I = \int_{x=0}^{\pi} \int_{y=0}^{a \sin x} y^2 \, dy dx## ##I = \int_{x=0}^{\pi} \left.\frac{y^3}{3} \right|_{y=0}^{a\sin x} \, dx## etc....That ##\frac{1}{3}## turns to ##\frac{1}{9}## later when you integrate another "##u^2##" as...

Typo in last post fixed

I'm getting the same answer as the book. I disagree with your ##I## I agree with the integral you set up but I think you are going the wrong way about evaluating it. That is to say making it too complex/long and prone to errors. Try this ##I = \int_{0}^{\pi} \frac{a^3}{3} \sin^3 x \, dx =...