Recent content by randy

Oh you're totally right I meant error, thanks for correcting

Hi yungman, return NegativeWidth(w) will in fact create an object of type NegativeWidth and then return it to be thrown. So you do throw an object in both cases (line 26 and 31). Note: this is equivalent to writing NegativeWidth error = NegativeWidth(w); throw error; If you call the...

Thanks a lot! This was very helpful, I think I understood the concepts of that task now ways better :)

Looking at ##\dot x## it seems to be monotonically increasing for any choice of k. So the particle would in fact never have to slow down. Does that then mean it never leaves the curve?

Okay, so from conversation of energy I have ## E = T + U = \frac{1}{2}m(\dot x^2 + \dot y^2) -mgkx^2= 0## (due to kinetic and potential energy being 0 at the start) and from ##y=-kx^2## I get ##\dot y= -2kx\dot x##. Now solving for ##\dot x##: \begin{align*} &\frac{1}{2}m(\dot x^2 + \dot y^2)...

Thanks a lot! This approach is so much easier! Ended up with ##v(x) = \sqrt{2gk}x## for part 1, which seems appropriate. However, now I am kinda lost for part 2... If I'm not mistaken, the normal force on the particle must vanish in order for it to leave the curve. So the normal vector at a...

I tried 1. using the Lagrangian method: From ##y=-kx^2## I got ##\dot y = -2kx \dot x## and ##\ddot y = -2k \dot x^2 - 2 kx \dot x##. (Can I use ##\dot y = g## here due to gravity?) This gives for kinetic energy: $$T = \frac{1}{2} mv^2 = \frac{1}{2} m (\dot x^2 + \dot y^2) = \frac{1}{2} m (\dot...