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Direction of friction in rolling.

  1. Mar 26, 2013 #1

    bgq

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    Hi,

    Consider a ball rolling upward without slipping on an inclined plane. What is the direction of the force of static friction?

    Let me explain what confuses me. I know that the friction opposes the tendency of the motion. If we consider the whole motion of the ball as upward, then the friction opposes it, so it is downward. However, if we consider the rotational motion of the ball, the point of contact between the ball and the surface tends to move downward due to gravity, so in this case the friction tends to oppose this tendency; therefore the friction is upward!

    Thanks to any help.
     
  2. jcsd
  3. Mar 26, 2013 #2

    WannabeNewton

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    If the ball is rolling upward without slipping on the incline then there is no relative velocity between the contact point and the incline in which case friction is directed opposite the relative motion that ##would## occur in its absence. In this case, in its absence we would just have sliding motion up the incline so the relative velocity would be directed upwards hence friction would be directed downwards, opposite the relative velocity.
     
  4. Mar 26, 2013 #3
    Well, the point is that you talk about 'static' friction here. It doesn't quite work the way the 'sliding'(?) friction does.
    Meaning that since the body goes up it loses kinetic (rotational and translational) energy and gains potential energy. Translational energy is lost by the gravitational pull backwards. Rotational energy has to be lost somehow. ( I will work some better explanation than this. I mean -WHY rotational energy has to be lost -)
    That's how the friction vector goes upwards.
     
  5. Mar 26, 2013 #4

    rcgldr

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    Assuming that the only external force is gravity, then it doesn't matter if the ball is rolling upwards or downwards, the direction of the friction force exerted by the plane onto the ball is upwards (while the other half of the newton third law pair of forces, the direction of friction force exerted by the ball onto the plane is downwards). The magnitude of the force is equal to (angular acceleration) × (angular momentum) / (radius).
     
  6. Mar 30, 2013 #5

    bgq

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    Thank you all very much
     
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