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Normal reaction in a rolling object

  1. Aug 5, 2009 #1
    Why is the normal reaction to a rolling object shown to the side of the CM?
    Forces must be equal and opposite
    Is there a real problem if you assume it through the C?
     
  2. jcsd
  3. Aug 5, 2009 #2

    Doc Al

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    Staff: Mentor

    What exactly are you talking about? The normal force exerted by the surface of an incline on a sphere rolling down it? If so, why do you say that the normal force is shown "to the side of the CM"?
     
  4. Aug 5, 2009 #3
    I thought this should be clear
    Thishttp://webphysics.davidson.edu/faculty/dmb/py430/friction/rolling.html" [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. Aug 5, 2009 #4

    Doc Al

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    In the first two diagrams, the normal force arrow is shown to the side just for clarity. (Otherwise it would be right on top of the weight force arrow.) There's no implication that the normal force actually acts off to the side.
     
    Last edited by a moderator: May 4, 2017
  6. Aug 5, 2009 #5
    :smile:
     
  7. Aug 5, 2009 #6
    Maybe this will help. If a solid sphere of radius R and moment of inertia I = (2/5) m R2 is rolling (without slipping) down a slope of angle theta, then the accelerating force (parallel to the slope) on the sphere at its center of mass is mg sin(theta). But there is an uphill force at the point of contact with the slope (not at the sphere's center of mass) that is accelerating the rotation of the sphere. Also, for completeness, the normal (perpendicular) force of the sphere against the slope is mg cos(theta).
     
  8. Aug 5, 2009 #7

    rcgldr

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    Just under the diagram is this text:

    ( The normal and gravitational forces produce no effect because their line of action is through the center of rotation. )

    So although the diagram has the normal vector drawn a bit to side, it actually goes through the center of mass.
     
    Last edited by a moderator: May 4, 2017
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