Penny sits on top of a frictionless sphere, urgent please 1. The problem statement, all variables and given/known data At the top of a frictionless sphere of radius R a penny is given a push to speed x. At what angle, measured from the vertical does the penny leave the surface?
Re: Penny sits on top of a frictionless sphere, urgent please Welcome to PF. What considerations do you think need to be made? Maybe start with what condition determines when it will lose contact?
Re: Penny sits on top of a frictionless sphere, urgent please Normal reaction = 0 when the block has left the surface. Did you try that out?
Re: Penny sits on top of a frictionless sphere, urgent please I have no clue how to start this problem, I look at a similar problem here (https://www.physicsforums.com/showthread.php?t=260338) but i could not figure out what the last post tried to say. This is going to be a problem on my test tomorrow and I need to figure it out. Do any of you guys have aim or msn?
Re: Penny sits on top of a frictionless sphere, urgent please Well admittedly that was a brilliantly constructed suggested direction to go in solving the problem. Maybe you should consider using some of that to figure it out? Doing homework through other venues is not something encouraged here. And if that is too subtle, it's just not permitted, even through PM.
Re: Penny sits on top of a frictionless sphere, urgent please The thing is that i dont care much for the answer, I want help on starting the problem so I can work it on my own. Would you mind pointing me in the right direction please?
Re: Penny sits on top of a frictionless sphere, urgent please You have to work on it. If you don't, then what is the point in doing homework? You should at least try.
Re: Penny sits on top of a frictionless sphere, urgent please Well ... I'd start with that even if I did suggest it myself.
Re: Penny sits on top of a frictionless sphere, urgent please I have been working on it, I've been stuck on the damn problem for 3 days now, I know that energy and momentum are conserved, but I cant seem to translate that into actual equations : / Edit: ok i think im getting somewhere. I know that the point where the particle leaves is = 0, so mgcosx-ma=0 right? So Cosx = a/g Also Etot = Ekin + Epot= .5mv^2 - mgh= 0 =====> v^2= 2gh But im stuck in a loop now haha
Re: Penny sits on top of a frictionless sphere, urgent please Am I getting closer to figuring this out? haha
Re: Penny sits on top of a frictionless sphere, urgent please OK it's good knowing that momentum is conserved and perhaps you will need that elsewhere on your exam. But ... not on this problem. Now the mgCosθ term is the weight component of gravity. But isn't what you are interested in balancing the outward centripetal acceleration? mv^{2}/R ?
Re: Penny sits on top of a frictionless sphere, urgent please Ok so this is independent of the mass of the penny and of g. So would i have mgcosx= mgh + mv^2/R??
Re: Penny sits on top of a frictionless sphere, urgent please Part of the problem with that equation is that 2 of those terms are Force, the other energy.
Re: Penny sits on top of a frictionless sphere, urgent please Gotcha so it would only be mgcosx= mv^2/R? But what would i solve for?
Re: Penny sits on top of a frictionless sphere, urgent please Since V is, as you found, a function of h, and so is θ ...
Re: Penny sits on top of a frictionless sphere, urgent please Alright, so since v^2 = 2gh you substitute and get Cosx= 2h/r. But how is theta a function of h??
Re: Penny sits on top of a frictionless sphere, urgent please Even better then eh? Looks like you can do a lot of substituting.
Re: Penny sits on top of a frictionless sphere, urgent please Yeah but with what? what can i substitute cosx with?
Re: Penny sits on top of a frictionless sphere, urgent please For one thing it's not 2h/r. But if you make a drawing you can figure out what Cosθ is in terms of R and h.
Re: Penny sits on top of a frictionless sphere, urgent please Ok lets see now, so Cosx= a/g a= v^2/R So Cosx= a/g = v^2/ Rg = 2h/R Ok and Cosx=R-h/R So R-h/R = 2h/R thats what I could figure out, but i need to figure out the angle when it drops off the sphere