Penny sits on top of a frictionless sphere, please

AI Thread Summary
A penny on top of a frictionless sphere is pushed with speed x, and the problem involves determining the angle at which it loses contact with the surface. The key condition for losing contact is when the normal force equals zero, leading to the equation mgcosx = mv²/R. The discussion emphasizes the conservation of energy and momentum, with the penny's speed related to its height on the sphere. Through substitution and manipulation of equations, the relationship between height and angle is established, ultimately leading to the conclusion that Cosθ = 2/3 when the penny leaves the sphere. This problem-solving process highlights the importance of understanding the dynamics involved in circular motion and energy conservation.
malta
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Penny sits on top of a frictionless sphere, urgent please

Homework Statement



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?
 
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Welcome to PF.

What considerations do you think need to be made?

Maybe start with what condition determines when it will lose contact?
 


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


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?
 


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.
 


The thing is that i don't 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?
 


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


malta said:
The thing is that i don't 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?

Well ... I'd start with that even if I did suggest it myself.
 


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 can't seem to translate that into actual equations : /

Edit: ok i think I am 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 I am stuck in a loop now haha
 
Last edited:
  • #10


Am I getting closer to figuring this out? haha
 
  • #11


malta said:
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 can't seem to translate that into actual equations : /

Edit: ok i think I am 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 I am stuck in a loop now haha

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? mv2/R ?
 
  • #12


Ok so this is independent of the mass of the penny and of g. So would i have

mgcosx= mgh + mv^2/R??
 
  • #13


malta said:
Ok so this is independent of the mass of the penny and of g. So would i have

mgcosx= mgh + mv^2/R??

Part of the problem with that equation is that 2 of those terms are Force, the other energy.
 
  • #14


Gotcha so it would only be

mgcosx= mv^2/R? But what would i solve for?
 
  • #15


malta said:
Gotcha so it would only be

mgcosx= mv^2/R? But what would i solve for?

Since V is, as you found, a function of h, and so is θ ...
 
  • #16


Alright, so since v^2 = 2gh you substitute and get Cosx= 2h/r. But how is theta a function of h??
 
  • #17


malta said:
... Cosx= 2h/r. But how is theta a function of h??

Even better then eh?

Looks like you can do a lot of substituting.
 
  • #18


Yeah but with what? what can i substitute cosx with?
 
  • #19


malta said:
Yeah but with what? what can i substitute cosx with?

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.
 
  • #20


Ok let's 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
 
  • #21


anyone?
 
  • #22


malta said:
Ok let's 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

That's almost right.

Cosθ is what you're solving for right? So don't you need to be eliminating h and not Cosθ?

And isn't Cosθ really = (R-h)/R ?
 
  • #23


Oh ok yeah I am solving for Cosx=(R-h)/R
So wouldn't the angle be arcCos of (R-h)/R?
 
  • #24


malta said:
Oh ok yeah I am solving for Cosx=(R-h)/R
So wouldn't the angle be arcCos of (R-h)/R?

Yes. But you need to eliminate h. Or eliminating R works too.

Cosθ is just a dimensionless ratio.
 
  • #25


How would I go by eliminating either h or R? Also would you mind looking at my other post about static friction, its driving me nuts trying to figure it out

Thanks for your help man
 
  • #26


You have it sitting right in front of you.

2h/R = (R-h)/R

2h = R - h

R = 3h

Now

Cosθ = (R - h)/R = (3h - h)/3h = 2h/3h = 2/3
 
  • #27


oh haha yeah that makes sense, thanks for all your help man, I am trying to pull an all nighter here for physics. Thanks again
 

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