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Ball- Sliding or Rolling?

  1. Sep 26, 2010 #1
    Is it possible to make the coefficient of friction between a ball and an inclined plane low enough so that a ball will silde down the plane and not roll down the plane?
     
  2. jcsd
  3. Sep 26, 2010 #2

    Drakkith

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    It is possible to make it very very close. Ever go bowling? Notice how if you put a spin on the ball, it spins one way, but takes a long time for that spin to make the ball move on the lane? Same thing.
     
  4. Sep 27, 2010 #3
    Unless it is zero there will be a force between the ball and surface that will cause the ball to spin. That dosen't mean it won't slide, however any friction will make it spin.
     
  5. Sep 27, 2010 #4
    If you roll a ball down an inclined plane, the accelerating force (parallel to the inclined plane) is proportional to mg sin(θ), while the force pushing the ball against the inclined plane is mg cos(θ). If the ball has a moment of inertia (e.g., I = 2mR2/5 for a solid ball), a decelerating force parallel to the inclined plane (and opposite to mg sin(θ)) is required just to spin the ball. If the angle of the inclined plane exceeds a critical angle, the force pushing the ball against the inclined plane is insufficient to make the ball spin, even if the coefficient of friction is 1.

    Calculate the critical angle.

    Bob S
     
    Last edited: Sep 27, 2010
  6. Sep 28, 2010 #5
    Any unbalanced moment about the ball will cause rotation. Any friction will create a moment about the ball equal to Rmg cos(θ) times the coefficent of friction. Unless the inclined plane is a wall or the coefficient of friction is zero the ball will spin.
     
  7. Sep 28, 2010 #6
    Isn't this just a static friction equation? Round objects spin on a surface because they are not applying enough force to break the static friction. If the force required to break this static friction is so low that the ball can break it easily, it should just slide as opposed to rolling.
     
  8. Sep 28, 2010 #7

    rcgldr

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    The ball will slide and spin, but it's angular inertia combined with the rate of acceleration down a plane will prevent it from truly rolling (no sliding), if the plane is angled steep enough and/or if the coefficient of friction is low enough. The result is that the torque from friction is not enough to cause sufficient angular acceleration for the ball to transition into or maintain a true roll.
     
    Last edited: Sep 28, 2010
  9. Sep 28, 2010 #8
    So if I understand this correctly, since the coefficient of friction between the ball and the inclined plane is never exactly zero, the ball will never slide without rotating down an inclined plane no matter what the angle of inclination is.
     
  10. Sep 29, 2010 #9

    Chronos

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    An interesting experiment is to roll a rotating ball down an inclined plane. The loss of rotational momentum [velocity] from top to bottom is attributable to friction.
     
  11. Dec 2, 2010 #10
    I have a related question to this general area, but involving something I have not seen discussed elsewhere, so here goes (excuse me if this is the wrong thread, I am a newbie on this site):

    How would you diagram/describe all of the forces involved in a system where a solid sphere is rolling down a sloped belt (like a conveyor belt, say) that is moving up (i.e., the top surface of the belt where the sphere contacts it is moving toward the top of the slope), and given this construction, could there ever be an equilibrium reached where the forces imparted by the friction between the two objects (belt and sphere) and all the other forces involved (gravity, rotational inertia, etc.) cause the ball to remain stationary (but still rotating, of course) relative to a stationary observer?
     
  12. Dec 2, 2010 #11

    rcgldr

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    The sloped belt would have to be continously accelerating "uphill" in order to keep the ball in a stationary position. The belt speed and the ball's angular velocity would be constantly increasing.
     
  13. Dec 3, 2010 #12
    I would agree this would happen up to a point, i.e., where the sphere reaches it's terminal velocity. Terminal velocity in a system as described could vary largely based on the frictional forces between the sphere and the belt, the mass of the sphere, the angle of the belt, etc.

    The general formula I want to derive or find is one where all of the usual factors for mass, angle, coefficient of friction (CoF), etc., involved with the sphere, belt, and acceleration due to gravity are included and can be solved for various values given those known or derived experimentally (like CoF).

    Another factor that will need to be somehow factored in is the resistance to rotation of the belt due to the required rollers at either end, which would probably just be a constant in the formula - derived experimentally like CoF tends to be.
     
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