Recent content by stephenklein

Wow, I feel like a dummy for not just looking up acceleration components in polar coordinates. That was immensely helpful, thank you. So to flip the argument around, can we say that because ##m\dot{\vec{v}}## has only a radial component, that ##a_{\phi} = 0##, meaning ##2\dot{r} \dot{\phi} =...

I actually have worked through the solution just fine by taking the derivative of \vec{L}: \frac{d \vec{L}}{dt} = \dot{\vec{v}} \times \vec{M} - \alpha \left(\frac{\vec{v}}{r} - \frac{\left(\vec{v} \cdot \vec{r}\right)\vec{r}}{r^{3}}\right) I permuted the double cross product: \dot{\vec{v}}...

Mark, thank you very much. PAllen -- In my attempt, I was trying to perform the dot product in the basis of the inertial frame. But ##\bar{e_{t'}}## isn't given to me in that frame. However, I do know that ##\bar{e_{t'}}## in the rotating frame is just ##\bar{e_{0'}}##, so in order to obtain...

I know my LaTeX is not executing how it's supposed to. The code works fine in Overleaf, so I'm not sure where the issue is arising from. Hopefully it's clear what I'm asking.

**I realize some of my inline math delimiters '\(' and '\)' are not acting on the text for some reason, and it looks clunky. I spend 20-30 minutes trying to understand why this is, but I can't. My limited LaTeX experience is in Overleaf, and these delimiters work fine in that compiler. My...

I was able to work through parts (a) and (b). For part (c), I got $$\frac {\partial f} {\partial y} = \sqrt {1-y'^2}$$ and $$\frac {\partial f} {\partial y'} = \frac {-y y'} {\sqrt {1-y'^2}}$$ Taking ##\frac {d} {ds}## of the latter, I used the product rule for all three terms ##y, y'...

I'm an undergraduate physics major in my junior year (US). I'm hoping to improve on what I learn in class, as well as learn more about topics in physics I haven't seen in class