Recent content by rivendell

got it. Thank you TSny!

I thought f was just going against θ, so I had written that. Should I have written dv/dt = f/M - gsinθ instead? If I did that, then trying again subbing, dv/dt = u(Mgcos + Mv^2/R)/M - gsinθ = u(gcosθ + v^2/R) - gsinθ = 0. Does that work? and just if I wanted to solve for θ, would I use the sin...

thanks for noting that! I'll correct my writeup: -f - Wsinθ = MR(θ") = M(dv/dt) --> dv/dt = -f/M - gsinθ --> N = Wcosθ + M(v^2/R) Subst: dv/dt = (-ugcosθ + v^2/R)/M - gsinθ = 0 I guess I can't solve this eq, which is ok, as long as it's right. friction acts in direction of decreasing θ, which I...

Homework Statement An automobile of mass M drives onto a loop-the-loop, as shown. (click here for diagram) The minimum speed for going completely around the loop without falling off is v0. However, the automobile drives at constant speed v, where v < v0. The coefficient of friction between the...

hi, I did this problem too and got sqrt((2d)/(ug)) instead of what the first post said, 2sqrt(d/(ug)). The way I did it was just F=ma=friction, so umg=ma and a = ug. So using d = 1/2 (at^2), plugged a=ug into get d = 1/2 (ug)t^2. finally, solved for t to get sqrt((2d)/(ug)). The first post's way...

oh. thanks for catching that mistake! :)

lol, I do mean b. It's given in this problem's diagram, where the setup is just like another problem in the book shown here (though I don't know if I can assume that motion exerted by the rod is in line with it as was stated there): https://www.physicsforums.com/attachments/kk-2-7-jpg.72173/...

ok. just wondering, aren't we supposed to use b since it was given? and isn't this problem exactly the same as 2.7 in kleppner and kolenkow (the other 'leaning pole' one, I know this one came from 3.1) or am I just missing something here? Is it supposed to be done in a different way in this...

I'm working on this problem now in my book, and I'm not really sure if y^2+x^2=L^2 is really the way to go on this problem. I'd just like to know why it isn't (b-y)^2 + x^2 = L^2 since y is really the distance the top of the pole moves. If that's the case, then the answer the first guy who...