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Magnitude of force of a disk on a circular track

  1. Apr 18, 2017 #1
    1. The problem statement, all variables and given/known data
    A small disc, radius r and mass m = 7.9 g, rolls on its edge. The friction with the track is enough to prevent slipping. When released, it rolls down the track (sketch) and reaches a circular section with radius R = 5.1 cm, which is very much greater than r. The initial height of m above the lowest point of the track is h = 30.4 cm. The disc reaches the top of the circular part of the track. When it does so, what is the magnitude of the force the disc exerts on the track? (Hints: how fast is it going? and don't forget to draw a free body diagram)

    2. Relevant equations
    ½mvi2+mghi = ½mvf2+mghf
    Fc=mv2/r

    3. The attempt at a solution
    The initial height is 30.4cm from the bottom while the final height is 10.2cm from the bottom, therefore the change in height is 20.2cm = 0.202m

    At the top, the force of the disk on the track would be the centrifugal force pushing it against the track which is equal to the centripetal force

    Subbing into conservation of energy to get the velocity:
    Initial v is 0, therefore:
    mghi = ½mvf2 + mghf
    divide by m and rearrange
    v = √2*g*(hi-hf)
    sub in values
    v = √2*9.8*0.202
    = 1.98977

    subbing it into mv2/r:
    (0.0079*1.989772)/0.051 = 0.61N

    What did I do incorrectly?
     

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  3. Apr 18, 2017 #2

    haruspex

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    The centripetal force is the radial component of the net force. What other forces act on the disc?
     
  4. Apr 18, 2017 #3
    Gravity and the Normal Force, both pointing downwards and equal in magnitude
     
  5. Apr 18, 2017 #4

    haruspex

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    Why would they be equal in magnitude?
     
  6. Apr 18, 2017 #5
    Nevermind they aren't
     
  7. Apr 18, 2017 #6
    So at the top the normal, gravity and centripetal force are all acting downwards while the centrifugal force acts outwards? So the force of the ball on the track would be the centrifugal force - normal force since the centripetal force should be provided by gravity?
     
    Last edited: Apr 18, 2017
  8. Apr 18, 2017 #7

    haruspex

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    Don't mix centrifugal and centripetal in the same analysis. They are two different ways of viewing the same thing.

    In an inertial frame, there is centripetal, not centrifugal. But centripetal force is not an actual force acting on the system. It is the radial component of the resultant of the actual forces. The actual forces are the normal force and gravity:
    Centripetal force = radial component of ( normal + gravity).

    In the frame of reference of the accelerated body, there is no acceleration, by definition. Yet there are forces (such as normal and gravity). To explain this you have to invent a balancing force, which we call centrifugal. If you are being spun around in a drum, you feel the normal force from the wall of the drum, and to explain it you feel there must be an outward force pushing you against the wall.

    For more on the subject, see https://www.physicsforums.com/insights/frequently-made-errors-pseudo-resultant-forces/
     
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