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Angular acceleration at the bottom

  1. Oct 9, 2015 #1
    1. The problem statement, all variables and given/known data
    You hold a small ice cube near the top edge of a hemispherical bowl of radius 100 mm. You release the cube from rest. What is the magnitude of its acceleration at the instant it reaches the bottom of the bowl? Ignore friction.

    2. Relevant equations
    a=v^2/r delta U= delta Krot +delta Ktrans

    3. The attempt at a solution
    mgR=1/2Iw^2+1/2mv^2
    a=v^2/r


    So I'm not sure if it is so far so good. How do I find the rotational inertia here? How should I approach this question.
     
  2. jcsd
  3. Oct 9, 2015 #2

    andrewkirk

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    You can infer from the question saying 'small ice cube' that they want you to ignore any rotational energy of the cube. Such energy will be trivial if it is small, and cannot be calculated without knowing the size of the cube.
     
  4. Oct 9, 2015 #3
    It's not usual rotation problem, because of the source of force is only gravity. As you know, it always acts toward ground, so you don't have to consider the rotational things.
     
  5. Oct 10, 2015 #4
    does a=2g sound good?
     
    Last edited: Oct 10, 2015
  6. Oct 10, 2015 #5
    There is no horizontal forces acting the ice cube and only exist the vertical forces. The source of vertical force is only one. That is gravity.
     
  7. Oct 10, 2015 #6

    ehild

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    Yes, it is correct.
     
  8. Oct 10, 2015 #7
    Daeho Ro says it's only g?
     
  9. Oct 10, 2015 #8

    ehild

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    You do not need to believe anybody. How did you get your result?
     
  10. Oct 10, 2015 #9

    ehild

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    There is not only gravity acting on the ice cube in the bowl. What do you think it is, and in what direction, up or down?
     
  11. Oct 10, 2015 #10
    I think there's also a centripetal force that's upward. And that give the Fnet=mg=F-mg F=2mg=ma a=2g?
     
  12. Oct 10, 2015 #11

    ehild

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    The centripetal force does not "act": it is the resultant force needed to maintain circular motion with given radius and velocity. The wall of the bowl and the ice cube are in contact. Do they not exert force to each other? How this force is called?
     
  13. Oct 10, 2015 #12
    Yes. I think's it's the contact force?
     
  14. Oct 10, 2015 #13

    ehild

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    Yes, or normal force. What do you think its magnitude is now?
     
  15. Oct 10, 2015 #14
    I think my explanation is somewhat odd. Ignore this thing.

    The forces acting on the ice cube are gravity and normal force. Of course, the normal force is larger than the gravity and the difference of these forces is acting as the centripetal force.
     
    Last edited: Oct 10, 2015
  16. Oct 10, 2015 #15
    a=2g=19.6m/s^2?
     
  17. Oct 10, 2015 #16

    ehild

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    You gave the acceleration. I asked the normal force.
     
  18. Oct 10, 2015 #17
    3mg
     
  19. Oct 10, 2015 #18

    ehild

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    Yes. Upward or downward?
     
  20. Oct 10, 2015 #19
    upward
     
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