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Homework Help: Mechanical problem -- a mass sliding down a curved ramp

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  1. Dec 24, 2017 #1
    1. The problem statement, all variables and given/known data
    IMG_20171220_195152~2.jpg
    An object which has mass = m is released from point A which has height of h. The object moves along the rail with no friction. The rail OB is a semicircular rail which C is the center and R is the radius. If the object falls from the rail at the point B, find the following.
    1) h/R
    2) Distance of the object from point O if the object lands on the surface without hitting the rail

    2. Relevant equations


    3. The attempt at a solution
    Using the law of energy conservation

    mgh + 1/2mv2 = mgh + 1/2mv2

    The object should have velocity of sqrt(2gh) at point O

    What should I do next at the semicircular rail?
    How to calculate the reaction from the rail since the object will fall if the reaction is 0?
     
    Last edited by a moderator: Dec 24, 2017
  2. jcsd
  3. Dec 24, 2017 #2

    Doc Al

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    What speed must the object have at B to just start leaving the track at that point?
     
  4. Dec 24, 2017 #3
    The vertical speed is 0?

    I haven't studied the centripetal force yet. Is it used in the calculation?
     
  5. Dec 24, 2017 #4

    Doc Al

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    What's the horizontal speed?

    Yes. To figure out the minimum speed it must have to just barely make it to point B, you must consider centripetal force & acceleration.
     
  6. Dec 24, 2017 #5

    Doc Al

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    FYI: If the object isn't moving fast enough, it will leave the track before getting to point B.
     
  7. Dec 24, 2017 #6
    I just read about the centripetal force on wikipedia. The formula is F = mv2/R which v2/R is the centripetal acceleration.
    How can I use this to solve this problem?
     
  8. Dec 24, 2017 #7

    Doc Al

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    Good. Your textbook should have plenty of examples to learn from as well.

    Apply Newton's 2nd law at point B. What forces act?
     
  9. Dec 24, 2017 #8
    There are the gravitational force and centripetal force which are mg and mv2/R. Reaction force = 0 means the centripetal force (from object to center) is equal to gravitational force. Is that correct?
     
  10. Dec 24, 2017 #9

    Doc Al

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    Good! Now set up an equation to solve for the speed/energy needed at that point.
     
  11. Dec 24, 2017 #10
    I have another question. v at point B is equal to v and point O?
     
  12. Dec 24, 2017 #11

    Doc Al

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    Why would you think that? Energy is conserved! (Those points are at different heights.)
     
  13. Dec 24, 2017 #12
    Oh! I forgot about that.

    1/2m(2gh) = 1/2mv2 + mg(2R)
    v2 = 2g(h - 2R)

    Then mg = mv2/R

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

    h/R = 5/2

    Then use v at point B to calculate distance from O after landing using projectile. Is that correct?
     
  14. Dec 24, 2017 #13

    Doc Al

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    Good!

    Exactly.
     
  15. Dec 24, 2017 #14
    Thank you very much.
     
  16. Dec 26, 2017 #15

    CWatters

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    Yes but that's not quite the way I would think about it. The two forces aren't gravity and the centripetal force. The two forces are gravity and the normal force from the track.

    In order to move in a circle something must provide a net centripetal force = mv^2/, no more and no less. In this case the two forces that combine to produce the centripetal force are gravity and the normal force produced by the track. The normal force originates from the inertia of the object which tries to make it move in a straight line.

    If gravity provides less than mv^2/r then the normal force from the track will provide the rest.

    If gravity provides more than mv^2/r the radius will reduce and it will fall down.
     
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