Help please more energy questions

  • Thread starter holezch
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In summary, the mass slides down the sphere and its velocity decreases as it falls. At a certain point, the normal force equals zero and the mass flies off the sphere.
  • #1
holezch
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Homework Statement


A point mass m starts from rest (at the top of a sphere) and slides down the surface of the frictionless solid sphere of radius r.

Find the angle at which the mass flies off the sphere.




Homework Equations






The Attempt at a Solution



Well, all I know is that at that point, the Normal force will be zero. Also, that it is moving in a circular way.. so that we will have mv^2/r for its radial components and some other tangential components..

please help.. thanks
 
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  • #2
Start by drawing a free body diagram at an angle θ
 
  • #3
I have a free body diagram, I don't know what equations to use :S

thanks
 
  • #4
Now, write out Newton's second law for the radial direction.
 
  • #5
okay, so m v^2/ r = mgsintheta - N? When it falls off, N will be 0

thanks
 
  • #6
holezch said:
okay, so m v^2/ r = mgsintheta - N? When it falls off, N will be 0

thanks

Yes so now find what mv2/r works out to be using conservation of energy.
 
  • #7
hmm.. so should I express v in terms of kinetic energy and stuff? like from 1/2mv^2.. et c

thanks
 
  • #8
holezch said:
hmm.. so should I express v in terms of kinetic energy and stuff? like from 1/2mv^2.. et c

thanks

yes that will help.
 
  • #9
okay, so I got -mgr(1-cos theta) as the change in PE in this case.. I'm having trouble expressing the KE though.. any hints? I need to express the velocity as the function of the position.. the angle..

thanks
 
  • #10
holezch said:
okay, so I got -mgr(1-cos theta) as the change in PE in this case.. I'm having trouble expressing the KE though.. any hints? I need to express the velocity as the function of the position.. the angle..

thanks

mgr(1-cosθ) = kinetic energy at the angle θ

So you can find what mv2 is equal to.
 
  • #11
rock.freak667 said:
mgr(1-cosθ) = kinetic energy at the angle θ

So you can find what mv2 is equal to.

ah right, of course.. do you mean by the conversation of energy, K = mgr(1-cos theta) at some point?

Thanks! (rats, I should've known)

By the way, thanks for replying back constantly :) there's nothing better than real-time help
 
  • #12
holezch said:
ah right, of course.. do you mean by the conversation of energy, K = mgr(1-cos theta) at some point?

Thanks! (rats, I should've known)

By the way, thanks for replying back constantly :) there's nothing better than real-time help

Well the change in potential energy= change in kinetic energy. Initially there is no kinetic energy so just the change in potential energy is 1/2mv2
 
  • #13
rock.freak667 said:
Well the change in potential energy= change in kinetic energy. Initially there is no kinetic energy so just the change in potential energy is 1/2mv2

ah, I see so:

change in KE = - change in PE..

there is no initial KE, so 1/2 mv^2 = - change in PE

but - change in PE = mgr(1-cos theta)

so 1/2mv^2 = mgr(1-cos theta)

:) woohoo
 
  • #14
awesome, I solved it.. just used mv^2/r and plugged in v from here 1/2mv^2 = mgr(1-cos theta) ...et c :)

Oh.. I had a question, if the tangential part of a force overcomes the radial force, will the object fling out of the orbit?

Thanks
 

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