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Rolling friction force and velocity

  1. Jul 5, 2014 #1
    I have a concept about rolling friction which someone might be able to confirm for me. Let's say there is a steel ball rolling on a steel surface. It seems to me that the rolling friction force would be proportional to the rolling object's velocity.

    My understanding is rolling friction is caused by deformation of the rolling ball or deformation of the surface or both. As a ball rolls in one direction if there is a "ploughing" effect of the surface, for example, then it makes sense that a faster moving ball would compress or deform the surface deeper in front of the moving ball, thereby generating greater elastic restoring forces, than a slower moving ball which would compress or deform the surface less, and therefore, would generate a smaller elastic restoring force. Therefore, the rolling friction force would be less.

    I cannot imagine a ball rolling at a mere .5 cm/sec about ready to come to a stop having the same rolling friction force as a ball rolling at 1 m/sec. I visualize a graph with the horizontal line being velocity and the vertical line being the rolling friction force, and that the line would begin at the origin and slope up.

    Am I basically right about this concept?
  2. jcsd
  3. Jul 5, 2014 #2


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    Hi e2m2a. If you look at 'friction coefficients' for rolling element bearings, you'll find they are generally given as a constant regardless of what rotational speed the bearing operates at1. That isn't to say there are no other 'frictional' loads on a rotating shaft. There are2. Oil for example may produce a resisting load that's a function of the square of velocity (ie: square of rotational speed). But generally, rolling element bearing friction is modeled as if there were some coefficient of friction and additional loads are negligable.

    Consider your 'plough' analogy. Clearly there is going to be some deformation between 2 pieces of metal in contact that have some force pushing them together. Per Hooke's law, one can imagine that the amount of deformation between the parts in contact is a function of contact stress and material properties but nothing more. Basically, the ball and race are springs that get compressed and are released as the rolling element passes. As the ball or roller goes past some point on the race, you might imagine the material deforming as it accepts the compressive load and as it passes the point of contact it de-compresses or springs back to it's unstressed state. How fast it can 'spring back' after it rolls past the point of contact can be compared to the speed of sound in the metal. How much it deforms the surface is strictly a function of compressive stress which is independent of rotational speed. Metal deforms at some limiting speed (speed of sound in materail) which is generally much faster than the rate of a bearing that's rolling in a race, so you should find it springing back pretty much as fast as the contact stress decreases.

    There is some conversion of mechanical energy to thermal energy in the bearing which occurs because of internal 'friction' inside the balls and races just as the coefficient of restitution in material doesn't allow all energy to remain in a bouncing ball for example. But there isn't any 'ploughing' going on in a bearing. It's more like a spring being compressed as the ball rolls through the contact point and then decompresses as it rolls away from that contact point.

    1. http://www.roymech.co.uk/Useful_Tables/Tribology/Bearing Friction.html
    2. http://medias.ina.com/medias/en!hp.tg.cat/tg_hr*ST4_102160011
  4. Jul 5, 2014 #3
    I kind of grasp your explanation, but I don't think my questing was answered directly. Overall, is the rolling friction force velocity dependent? Would a rolling sphere whose center of mass is moving at .5 cm/sec have the same rolling friction force as a rolling sphere moving 1 meters/sec?
  5. Jul 6, 2014 #4


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    The coefficient of friction and therefore the rolling frictional force does not change with rotational velocity.
    Last edited: Jul 6, 2014
  6. Jul 7, 2014 #5
    I will reexamine my assumptions. Thank you for your explanation.
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