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Solid sphere rolling on rough surface

  1. Aug 3, 2009 #1
    Hello !
    Is there an expression for the friction force exerted on a solid sphere rolling (or rolling and slipping) on a rough solid inclined plane ?
    I have searched a lot but couldnt find anything, most studies are related to caoutchouc spheres or plane that will deform anbd cause viscoplastic friction which is not my case since both the sphere and the plane are non deformable solids.
    thanks !
     
  2. jcsd
  3. Aug 3, 2009 #2
    I was confused about this earlier.
    If none of the surfaces deform there are two possibilities either there is kinetic fiction or no friction.
    In a rigid solid sphere with slipping, kinetic friction comes into play and with no slipping, there is no friction
     
  4. Aug 3, 2009 #3

    Doc Al

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    If the sphere rolls down the incline without slipping, then static friction acts.
     
  5. Aug 3, 2009 #4

    rcgldr

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    If the sphere is sliding, then the dynamic friction force is just the normal force times the coefficient of dynamic (sliding) friction. The normal force is m g cos(θ).

    If the sphere is rolling, then the static friction force opposes the component force from gravity (m g sin(θ)) and is equal to m g sin(θ) Ia / (Ia + Ih), where Ia = angular inertia of the object, and Ih = angular interia of a hoop, the equivalent to the linear inertia (mass) of the object. In this case the object is a sphere, so you have m g sin(θ) ((2/5) m r2) / ((2/5) m r2 + m r2) = 2/7 m g sin(θ) .

    For a hoop or hollow cylinder it's 1/2 m g sin(θ).
    For a hollow sphere it's 2/5 m g sin(θ).
    For a solid disk or cylinder it's 1/3 m g sin(θ).
    For a solid sphere it's 2/7 m g sin(θ) .
     
    Last edited: Aug 3, 2009
  6. Aug 5, 2009 #5
    Thanks everybody!

    @ vin300 : Is the kinetic friction written with it's usual form (i.e. sliding kinetic friction) ?
    @ Jeff Reid : I couldn't find out how you managed to develop that expression for static friction, could you explain it more please

    but I imagine that if the surface is rough the sphere cannot roll without sliding, I cannot explain that physically but thats how i see it
    In that case the roughness of the surface is taken into by using an adequat coefficient of friction ?
     
  7. Aug 5, 2009 #6
    No problem
     
    Last edited: Aug 5, 2009
  8. Aug 5, 2009 #7

    Doc Al

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    Just the opposite: If there is no friction between sphere and surface, then the sphere cannot roll. It will just slide. It's static friction providing the torque about the center of mass that causes the sphere to roll.

    To derive the static friction (as Jeff did), just apply Newton's 2nd law to both translation and rotation. You'll get two equations, which will allow you to solve for the acceleration down the incline and the force of friction.
     
  9. Aug 5, 2009 #8

    rcgldr

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    I did the math for the cylinder case in this post:

    https://www.physicsforums.com/showpost.php?p=2238684&postcount=6


    Ih = m r2

    static friction = f = (Ia/Ih) m a

    eventually you get to this equation

    linear force = gravitational component - friction force

    m a = m g sin(θ) - (Ia/Ih) m a

    m a + (Ia/Ih) m a = m g sin(θ)
    (m a) (1 + (Ia/Ih) = m g sin(θ)
    m a = m g sin(θ) / (1 + (Ia/Ih)
    m a = m g sin(θ) / ((Ih + Ia)/Ih)
    m a = m g sin(θ) Ih / (Ia + Ih)


    f = (Ia/Ih) m a
    f = (Ia/Ih) m g sin(θ) Ih / (Ia + Ih)
    f = m g sin(θ) (Ia/Ih) Ih / (Ia + Ih)
    f = m g sin(θ) Ia / (Ia + Ih)
     
    Last edited: Aug 5, 2009
  10. Aug 5, 2009 #9
    Last edited by a moderator: Apr 24, 2017
  11. Aug 9, 2009 #10
    Jeff concerning the link you provided. I follow the derivation and it is fine. However I have trouble reconciling something.

    When a object is rolling I have always thought of the frictional force to be the rolling resistance = mg*b/r on a flat surface. b is the coefficient of rolling resistance had has units of m, r is the radius of the wheel. An odd equation but their it is.

    In your derivation you come with the frictional force for a solid cylinder = (1/3)*g*sin*(theta).

    This is fixed depending of the angle. The materials the slope and the cylinder are made from would have a effect of the friction experienced but your result does not allow for this. Can you reconcile this?

    The reason why I ask is two similar questions have been asked and I posted a derivation in response. My method was different and the result is an different form as the derivation gave an answer for the final velocity.

    https://www.physicsforums.com/showpost.php?p=2302293&postcount=4

    If I where to use your answer for the acceleration produced by friction I would get a different result. How can our approaches be reconciled or has one or both of us made a mistake? I am not assuming you have made a mistake Jeff.
     
  12. Aug 9, 2009 #11

    Doc Al

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    For an object rolling without slipping down an incline, static friction will be whatever it needs to be to prevent slipping (up to its maximum value, which depends on the materials). The amount of friction required depends on the angle. (Rolling friction is ignored.)
     
  13. Aug 9, 2009 #12

    Doc Al

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    In that link you do a calculation for a solid ball, not a cylinder, so it's unclear which of Jeff's calculations you are questioning.
     
  14. Aug 9, 2009 #13
    While his case does not directly apply to mine. If I was to rework his sums for a sphere then I would come up with a similar equation for the acceleration up the slope caused by friction which would still be material independent.

    If the condition is no slipping would not rolling friction be important? I can see the roll of static friction. In fact the torque generated by it must be related somehow to rolling friction.
    Jeff's derivation is in italics. I have not quite worked out how to quote individual lines yet.

    a = α / r

    for a solid cylinder, i = 1/2 m r2

    for the downhill rolling case:

    m a = m g sin(θ) - f

    Here the force that a particle on the surface of the cylinder experiences (I am aware this is not the centripetal acceleration as that would be radial) is equated to the component of weight acting down the slope minus frictional forces. In a sliding situation this cannot be argued with. I have never thought about it for a rolling case before so I can argue it both ways. That is not satisfactory one must be faulty. I assume the above equation is correct as it is newtons second law but because a is affected by the torque created by the friction acting at a distance r from the centre of the cylinder some confusion in my head results.
    t = f r = i α = 1/2 m r2 (a / r)

    f = t / r = 1/2 m a

    m a = m g sin(θ) - 1/2 m a
    a = 2/3 g sin(θ)

    f = 1/3 m g sin(θ)


    This is the bit that puzzles me, the frictional force is fixed with the only variable being the angle of the slope. Now while the angle of the slope will determine whether an objects slips or starts rolling the material that the two contacting surfaces are made from has part to play. For a rough surface at a particular angle there will be more friction than if the same material has a smooth surface (I am sure their is an exception to this but that is a special case), yet f it seems to be fixed for a particular angle. I find it difficult to reconcile my thinking with the maths. This does not mean the maths is wrong.

    ag = g sin(θ)
    af = 1/3 g sin(θ)
    a = ag - af = 2/3 g sin(θ)

    Also then would my derivation or approach to the problem I answered (I the above is flawless) be in error, if so I need to know how. I hope my confusion is clear.
     
  15. Aug 9, 2009 #14

    rcgldr

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    Rollling resistance is being ignored (b == 0 in mg*b/r) in these simplified cases for objects rolling down inclinded planes. So is aerodynamic drag. Only the angular acceleration aspect of rolling is being considered.

    Also depending on the material, rolling resistance isn't constant, but usually has a linear relationship to speed. See figure 4:

    http://www.tut.fi/plastics/tyreschool/moduulit/moduuli_8/hypertext_1/3/3_3.html
     
  16. Aug 9, 2009 #15
    OK so my derivation ignored static resistance hence the difference. So by incorporating aspects of your derivation into mine a more accurate solution can be found (I deliberately left out air resistance as for low speeds and heavy objects it not so significant),

    Thanks for the reply Jeff.
     
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