1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rolling and sliding sphere

  1. May 9, 2009 #1
    If we have an inclined plane at angle [tex]\alpha[/tex] and a sphere on that plane it will have angular acceleration due to the friction force. If [tex]F_{lim}<mg\sin \alpha[/tex] , where [tex]F_{lim}[/tex] is the limiting friction will the sphere also slide along the plane?
  2. jcsd
  3. May 9, 2009 #2
    I think this is basically the same question as "will a ball roll down a frictionless slope", which I believe is asked here many many times. If you have a frictionless slope and you release a ball, it will not roll, since there is no net torque.
  4. May 9, 2009 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Yes. And the required static friction for rolling without slipping depends on the angle. And the maximum available static friction also depends on the angle and the coefficient of friction.
    That's not the exact relationship, but your overall idea is correct. If the maximum available static friction is insufficient to create the needed torque, the sphere will slide as well as roll. (I think that's what you are getting at.)
  5. May 10, 2009 #4


    User Avatar
    Homework Helper

    To simplify things you could assume that static and dynamic friction are the same and independent of speed. Assuming that friction force is less than mg sin(theta), and an experiment that operates in a vacuum, then the rate of both linear and angular acceleration would be linear. If the friction force equals mg sin(theta), with an initial linear velocity, the linear velocity would remain constant, but there would a be a constant rate of angular acceleration (until the sphere started rolling in which case both linear and angular acceleration would occur due to the reduced opposing friction force once the sphere starts rolling).

    Depending on angular inertia, initial velocity, and coefficient of dynamic friction, if the rate of angular acceleration times radius is greater than the rate of linear acceleration, eventually the sphere starts rolling without sliding. Take the simple case of a sphere sliding from a frictionless horizontal plane to a non-frictionless horizontal plane. Then angle the plane until it reaches the point where tan(theta) = coefficient of dynamic friction where the ball slides at constant speed but increases angular velocity until it starts rolling. Once tan(theta) > coefficient of dynamic friction, then both linear and angular velocity increase and if angular inertia and/or angle of plane is high enough (all the mass at the surface of the sphere, like a ping pong ball), the sphere may never transition into rolling.
    Last edited: May 10, 2009
  6. May 13, 2009 #5
    The only way I know to deal with the simultaneously rolling and sliding ball down an inclined plane is to use the Lagrangian formulation. It's non-trivial.
  7. May 13, 2009 #6
    Ok, let's give it a try.
    If the slope angle is theta, the force along the slope is
    Fs= mg sin (theta)
    The force perpendicular to the slope is
    Fperp= mg cos(theta)
    If the coefficient of friction is Cf, then the max uphill force due to torque before slipping is
    Ftorque = Cf mg cos(theta)
    If the mass were a point then the total kinetic energy is
    -mgh = 1/2 m v2, where h = starting elevation. But because the moment of inertia of the ball is
    I = (2/5) m R2, so then the total energy is
    - mgh = (7/10) m v2
    So there is an opposing frictional force slowing down the acceleration. By equating the two equations, the actual (effective) downhill acceleration force is F = 5/7 mg sin (theta). Thus there must be an uphill frictional (opposite to mg sin(theta)) force
    Fuphill= (2/7) mg sin(theta)
    So the ball will slide if (2/7) mg sin(theta) > Cf mg cos(theta)
    or if
    tan(theta) > (7/2) Cf.
    This is a WAG (wild *** guess).
  8. May 14, 2009 #7
    Ya. i got the same result. 7/2 Cf. So it's a SAG rather. :smile:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook