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Homework Help: Help please more energy questions

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data
    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.




    2. Relevant equations




    3. 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
     
  2. jcsd
  3. Jan 7, 2010 #2

    rock.freak667

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    Start by drawing a free body diagram at an angle θ
     
  4. Jan 7, 2010 #3
    I have a free body diagram, I don't know what equations to use :S

    thanks
     
  5. Jan 7, 2010 #4

    ideasrule

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    Now, write out Newton's second law for the radial direction.
     
  6. Jan 7, 2010 #5
    okay, so m v^2/ r = mgsintheta - N? When it falls off, N will be 0

    thanks
     
  7. Jan 7, 2010 #6

    rock.freak667

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    Yes so now find what mv2/r works out to be using conservation of energy.
     
  8. Jan 7, 2010 #7
    hmm.. so should I express v in terms of kinetic energy and stuff? like from 1/2mv^2.. et c

    thanks
     
  9. Jan 7, 2010 #8

    rock.freak667

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    yes that will help.
     
  10. Jan 7, 2010 #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
     
  11. Jan 7, 2010 #10

    rock.freak667

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

    So you can find what mv2 is equal to.
     
  12. Jan 7, 2010 #11
    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
     
  13. Jan 7, 2010 #12

    rock.freak667

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    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
     
  14. Jan 7, 2010 #13
    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
     
  15. Jan 8, 2010 #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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