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What angle does it seperate from the bowl?

  1. Jul 8, 2008 #1
    1. The problem statement, all variables and given/known data
    an ice cube is placed on top of an overturned spherical bowl of radious r. if the cube slides downward from rest at the top of the bowl, at what angle does it seperate from the bowl?


    2. Relevant equations



    3. The attempt at a solution
    i'm stuck with this problem. i have thought about it for a loong time and i am deperate for a huge hint. i know the answer but i need to know how they got it. the question came from the energy chapter and i simply dont know how to approach it. plz help.
     
  2. jcsd
  3. Jul 8, 2008 #2

    Doc Al

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    Staff: Mentor

    Hint: In additional to using conservation of energy, you'll need to analyze forces and apply Newton's 2nd law.
     
  4. Jul 8, 2008 #3
    yeah, ive thought about that. it will fall off when the normal force goes to zero. or when the net force is just the weight. but that still gets me no where. i am not noticing somehting very obvious apparently. plz say more.
     
  5. Jul 8, 2008 #4

    Doc Al

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    Take advantage of the fact that the bowl is spherical. What kind of motion does that imply for the ice cube?
     
  6. Jul 8, 2008 #5
    yeah, circular motion. i know the cube follows a circle. i thought about using a=(v^2)/r but that equation is only valid for constant speed. the cube's speed increases as it falls. i have thought of everything, i think. what am i missing?
     
  7. Jul 8, 2008 #6

    Doc Al

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    Why do you think that? :wink:
     
  8. Jul 8, 2008 #7
    hmm, my book says that "when an object moves in a circle of radius r with constant speed v, its cent accel is." i guess not then. let me play around with that then, i will return tomorrow if i am still stuck. ty for helping me so far. now i will attack some more.
     
  9. Jul 9, 2008 #8
    ok, ive thought about it and i cannot get any closer. i need the answer. this problem is driving me crazy. the net force acting on it before it falls off is -W+N=(mv^2)/r thats all i see. need help.
     
  10. Jul 10, 2008 #9

    Doc Al

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    Staff: Mentor

    Another hint: Consider forces in the radial direction (normal to the surface). What's the component of the weight in the radial direction?
     
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