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Penny sits on top of a frictionless sphere, please 
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#1
Mar2309, 05:38 PM

P: 20

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? 


#2
Mar2309, 05:47 PM

HW Helper
P: 5,343

Welcome to PF.
What considerations do you think need to be made? Maybe start with what condition determines when it will lose contact? 


#3
Mar2309, 05:47 PM

P: 133

Normal reaction = 0 when the block has left the surface. Did you try that out?



#4
Mar2309, 05:51 PM

P: 20

Penny sits on top of a frictionless sphere, please
I have no clue how to start this problem, I look at a similar problem here (http://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?



#5
Mar2309, 05:58 PM

HW Helper
P: 5,343

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. 


#6
Mar2309, 06:00 PM

P: 20

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?



#7
Mar2309, 06:02 PM

P: 133

You have to work on it. If you don't, then what is the point in doing homework? You should at least try.



#8
Mar2309, 06:10 PM

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P: 5,343




#9
Mar2309, 06:11 PM

P: 20

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 mgcosxma=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 


#10
Mar2309, 06:22 PM

P: 20

Am I getting closer to figuring this out? haha



#11
Mar2309, 06:31 PM

HW Helper
P: 5,343

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 ? 


#12
Mar2309, 06:40 PM

P: 20

Ok so this is independent of the mass of the penny and of g. So would i have
mgcosx= mgh + mv^2/R?? 


#13
Mar2309, 06:42 PM

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P: 5,343




#14
Mar2309, 06:46 PM

P: 20

Gotcha so it would only be
mgcosx= mv^2/R? But what would i solve for? 


#15
Mar2309, 07:00 PM

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#16
Mar2309, 07:14 PM

P: 20

Alright, so since v^2 = 2gh you substitute and get Cosx= 2h/r. But how is theta a function of h??



#17
Mar2309, 07:28 PM

HW Helper
P: 5,343

Looks like you can do a lot of substituting. 


#18
Mar2309, 07:31 PM

P: 20

Yeah but with what? what can i substitute cosx with?



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