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Pushing a cylinder along a surface with friction.

  1. Apr 11, 2012 #1
    If I push a cylinder along a surface with friction then am I right in saying that it would almost instantly start a pure rolling motion with a combination of translational and rotational kinetic energy?
    Or would it start initially with pure translational motion and gradually accelerate rotationally to a pure rolling motion over time?

    I am stuggling to prove with equations either way could someone help with this?
     
  2. jcsd
  3. Apr 11, 2012 #2

    Filip Larsen

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    Welcome to PF!

    The cylinder cannot instantly acquire a roll velocity, if that is what you are asking.

    Depending on how much horizontal force you apply to the cylinder it will either begin to roll without slipping (if the force is low enough, see below) or it will begin to roll with slipping and then only later (assuming a constant push force) go back to roll without slipping when the cylinder rolls fast enough for the surface of it to "keep up" with the ground.

    Usually one models the difference between the two types of roll (non-slipping and slipping) using static and dynamic coefficient of friction. Only if the magnitude of your initial push exceeds the magnitude of the normal force times the static coefficient of friction you would have rolling with slipping and the reaction force (which determines how much angular acceleration the cylinder gets) then equals the magnitude of the normal force times the coefficient of dynamic friction. If the cylinder rolls without slipping the magnitude of the reaction force is equal to the magnitude of the normal force times the coefficient of static friction. Since the coefficient of static friction is larger than the coefficient of dynamic friction you would get the highest angular acceleration when the cylinder rolls without slipping.

    Also note, that a perfectly round cylinder on a perfectly flat surface is not physical possible since it distributes its normal force from the surface over an area of size zero. A physical cylinder will of course have non-zero contact area with the surface, but depending on how smooth the surfaces are the size of the contact area can vary as the cylinder rolls giving rise to variations in a practical experiment that you would not see if you slide a flat block over a surface.
     
  4. Apr 11, 2012 #3
    Thanks for the welcome.

    I was trying to understand it in terms of kinetic energy and work done by the friction as it rolls.
    Using the forces on the cylinder instead makes more sense.

    Also, a slightly related question.

    Would the vcm of a rolling cylinder with slipping be the same as the vcm of a rolling cylinder without slipping?
    Or would the vcm be higher when rolling without slipping?
     
  5. Apr 11, 2012 #4
    This problem is very similar to a pool cue hitting a billiard ball. If the cue hits the billiard ball at its center, the ball will initially slide. If the cue hits the ball at a sweet spot above the center of gravity, it will transfer both a linear momentum impulse and a rotational momentum impulse, hence no slipping.

    See discussion of sweet spot in http://www.real-world-physics-problems.com/physics-of billiards.html [Broken]. Scroll down to Physics Of Billiards — The Sweet Spot
     
    Last edited by a moderator: May 5, 2017
  6. Apr 12, 2012 #5

    Filip Larsen

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    Well, it is difficult to give a direct answer on that because you need to have some difference in the two situations so that the cylinders end up slipping and non-slipping respectively, and since such difference will influence the acceleration it may not make sense to compare the velocities directly in those two situations.

    By the way, I can see that I've made a blunder in my previous post that may have you confused. I said that in the case of rolling without slipping the reaction force is equal to the normal force times the coefficient of static friction. This is not correct. What I should have said was that the reaction force have a maximum at that point. As long as the push force is less that this maximum the reaction force will be equal to the push force. If you look at a block instead of a cylinder, this means that the sum of the push force and the reaction force is zero and the block doesn't accelerate as will be expected. For the cylinder to accelerate when it is not slipping then fully comes from the rolling of the cylinder, that is, the torque on the cylinder from the reaction force will induce an angular acceleration on the cylinder which then lead to the cylinder start rolling.
     
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