Recent content by eddiezhang

Thanks, I will investigate this. I have to write this for a math class. This was the scenario I chose to investigate, but if it doesn't resolve itself nicely or is too much effort, I can always just change topics.

AI is technically allowed, but the point of the report is to show a) understanding of the math used and b) solving things from the ground up (which is why I was discouraged from using energy concepts).

Ooops. Just ##\mu u ## at the end, right?

Ah right... ##\frac{du}{d\theta} = \frac{d}{d\theta}\omega^2 = 2\omega \frac{d\omega}{d\theta} = 2\ddot{\theta}## So the ODE turns into: ##\frac{1}{2} \times \frac{du}{d\theta} = \frac{g}{R} \left(\cos\theta - \mu \sin\theta \right) - \mu u^2## And if I'm not mistaken, the strategy is to...

So ##\dot{\theta} = \omega## and ##\ddot{\theta} = \omega \frac{d\omega}{d\theta}## (I was clearly having a brain fart before with this part) Substituting into ##\ddot{\theta} = \frac{g}{R} \left( \cos\theta - \mu \sin\theta \right) - \mu \dot{\theta}^2## Produces ##\omega...

Oh wait 🤦‍♂️I don't know why I worded it that way but it's literally just the chain rule. Oops. Thanks for the help so far... I might come back here if the algebra gets too rough.

It does... I see. OK I'll see if I can solve the ODE with the subs. and (maybe?) get the time of descent (too optimistic?)

OK... I'm not sure how exactly: ##\ddot{\theta} = \frac{d}{dt} (\omega \frac{d \omega}{d\theta})## Bear with me because I haven't done much calculus... multiplying infinitesimals still gives me the heebie-jeebies. Is this right? ##\ddot{\theta} = \frac{d\omega}{dt} \times \frac{d...

Oh yay! OK so looking at the DE again: ##\ddot{\theta} = \frac{g}{R} \left( \cos\theta - \mu \sin\theta \right) - \mu \dot{\theta}^2## ##\dot{\theta}## substitutes out directly, but how do I deal with ##\ddot{\theta}##? Plus don't the trig terms still make it non-linear? Edit: I don't have...

Attempt: ##\frac{d \omega}{dt} = \frac{d \omega}{d\theta} \times \frac{d \theta}{dt} = \omega \frac{d \omega}{d\theta}## Am I barking up the wrong tree? I'll probably kick myself hard, but I'm just not seeing it yet...

I think I follow the chain rule part - would I be correct in saying ## \frac{du}{d \theta} =2\ddot{\theta}##? You may need to walk me through the rest of that approach... how exactly do you get it first-order linear? Similarly, if you start me off on the ## \omega ( \theta) = \frac{d...

oops :oops: that's pretty embarrassing. I take it the correct DE is gcos(φ) - φ'' = μ(φ')^2 +μgsin(φ)?

This is for a math report that I'm supposed to write, which means I'm not supposed to use conservation of energy. This makes life much harder... so please bear with me. I am interested to see how you'd solve this purely kinematically though (if it can be solved that way). Please tell me if this...

Continued grinding away on it; I think I get it now.

Forgot to mention: the author is Russian and I don't know how I might contact him (if necessary... or even welcome). Would anyone know how to and should I try? Thanks.