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Friction and Rotation

  1. Apr 10, 2008 #1
    I have a question regarding rotation and kinetic friction that I cant find anywhere else on the web.

    Say you have a ball of radius r and mass m rotating on a flat surface without any forces acting on it but the force of friction. How does the deceleration of the ball relate to the frictional coefficient of the surface?

    Is there a simple equation that describes the force of kinetic friction acting on the ball?
     
  2. jcsd
  3. Apr 12, 2008 #2
    Nobody have any ideas? I just need to know what knowledge is required to solve this problem.
     
  4. Apr 12, 2008 #3

    Hootenanny

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    Yes,

    [tex]F=\mu R[/tex]
     
  5. Apr 12, 2008 #4
    What is R? And how to calculate the deceleration of the ball?
     
  6. Apr 12, 2008 #5

    Hootenanny

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    R is the normal reaction force. One can calculate the deceleration of the ball by applying Newton's Second law (and it's rotational analogy).
     
  7. Apr 12, 2008 #6
    So what is the difference between the ball sliding and the ball rotating? How is the deceleration diffenent?
     
  8. Apr 12, 2008 #7

    Hootenanny

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    If the motion is rotational with no sliding, then there will be no acceleration, assuming that both the ball and surface a rigid. If the ball is sliding there will be both linear and angular acceleration unit the no-slip condition is met (i.e. the velocity of the point of the ball in contact with the surface is zero relative to the surface), at which point it will cease sliding and the ball will roll with a constant angular and linear velocity indefinitely.
     
    Last edited: Apr 12, 2008
  9. Apr 12, 2008 #8
    say the frictional coefficent, the radius, the mass of the ball, and an initial force was given. How can teh distance that the ball travel be calculated?
     
  10. Apr 12, 2008 #9

    Hootenanny

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    Quite easily using Newton's second law (and it's rotational analogue).
     
  11. Apr 12, 2008 #10
    I dont know what the rotational analogue is. What is it?
     
  12. Apr 12, 2008 #11

    Hootenanny

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    For a rigid body,

    [tex]\sum_i\underline{OP}_i\times\underline{F}_i = \frac{d}{dt}\underline{L}_0[/tex]

    In words: The sum of the external torques about the point O is equal to the time derivative of the angular momentum about that same point.
     
  13. Apr 12, 2008 #12
    I can find the angular and linear deceleration of the ball, but what do they represent? Which one do I use to find the time it takes for the ball to stop?

    Also, the force of friction for both the angular and linear deceleration are the same?
     
  14. Apr 13, 2008 #13

    Hootenanny

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    The angular acceleration represents the change in angular velocity and the linear acceleration represents the change in linear velocity, as one would expect. However, at some point the ball will stop accelerating and have a constant angular and linear velocity. If the ball and surface are rigid, then it will never stop. See my post #7 above.
    Yes.

    If your solving a specific question, why don't you post it so we can help you out?
     
  15. Apr 13, 2008 #14
    I don't know what you mean by this.

    I am not trying to solve a specific problem, I am just trying to figure out the relation ship between the friction coefficient and the time it takes for the ball to stop rolling.
     
  16. Apr 13, 2008 #15

    Hootenanny

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    Simply put, if the surface is horizontal and both the ball and surface do not deform, then the ball will never stop.
     
  17. Apr 13, 2008 #16
    You mean the ball will roll on forever even though a friction force is acted on it?
     
  18. Apr 13, 2008 #17

    Hootenanny

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    Yes, once the ball has finished sliding the frictional force is a static frictional force and does no work on the ball since the point of contact on the ball doesn't move relative to the surface.

    This is only the case for a rigid ball and surface.
     
  19. Apr 13, 2008 #18
    What happens in real life? Eg. When a basketball or pool ball is rolled on a smooth surface.
    It obviously stops rolling at some point. How is that time calculated?
     
  20. Apr 13, 2008 #19

    Shooting Star

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    Sorry to intrude, Hoot, but right in the first post the OP says there's no force acting on the ball. I presume he meant an earthly scenario where the weight of the ball was acting as the reaction R. If there is actually no force pressing the ball to the surface, there will be no friction.

    In reality, rolling friction will slow the ball down, as Hootenanny will explain.
     
  21. Apr 13, 2008 #20

    Hootenanny

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    Again using Newton's second law (in rotational form), however to take into account the deformation of the ball and surface one must include the action of an additional force, the so-called rolling resistance, which is usually of the form,

    [tex]F = K_rR[/tex]

    and acts in the same direction as the frictional force. More information can be found here.
     
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