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Rolling cylinder analysis

  1. Jul 19, 2006 #1

    quasar987

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    The setting is the same as in this thread:
    https://www.physicsforums.com/showthread.php?t=126623

    I understand the sliding mechanism, but not the rolling one. Suppose the force F is lesser than the maximum of the static friction.

    The force is applied, the atoms to the right of the cylinder reply by applying a force of static friction -F. Now the sum of the forces is 0 but the torque is not. It is -2RF and tends to make the cylinder rotate clockwise.

    Now what?

    I was tempted to say that the atoms of the table to the left of the cylinder try to oppose this torque by applying a force that creates a torque as close to +2RF as possible given the maximum static friction. That idea works out ok is the maximum static friction is less than 2F: there is still a net torque and there is a net translational force, so everything is explained. But in the case where the maximum static friction is more than 2F, then as a result of the -2RF torque, the atoms of the table to the left of the cylinder apply a torque of +2RF. Now the net torque is 0 but the net force is +F, so the cylinder slides but does not rotate! :(
     
  2. jcsd
  3. Jul 20, 2006 #2

    andrevdh

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    Rolling occurs over the contact point because the object actually "sticks" to the surface via rolling friction. That is any two surfaces tend to stick together a bit (differences in electron negativity of the two materials?). The rolling friction coefficient are usually quite small (~ factor 1/10 of static).
     
  4. Jul 20, 2006 #3

    Office_Shredder

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    quasar, the atoms to the left and right of the cylinder don't do anything, because the cylinder is not in contact with them.

    The cylinder is theoretically in contact with a thin line that stretches down the length of it (the floor is tangent to it). The surface the cylinder is in contact with applies a force that directly opposes the direction the cylinder is moving in... so if you push the cylinder to the right, the friction pushes to the left. So we have two cases:

    static friction maximum greater than Fapplied:
    In this case, the kinetic friction is completely ignored. What happens, is you have a force of F being applied to the top in the right direction, and a force of F being applied to the bottom in the left direction. Try taking a paper towel roll in between your hands, and hold it up with your palms. Then move your right hand away from your body, and your left hand towards your body. You see how the opposite forces on opposite sides make it rotate?

    Static friction max is less that Fapplied:
    Here, we have both kinetic and static friction. So how this works out, is that both forces again are applied to the object. But the kinetic friction actually works against the rolling motion. One common question is to find out, if a cylinder is initially accelerated with excessive force, what speed it will eventually reach and stop accelerating at (when the static and kinetic frictions balance out).
     
  5. Jul 20, 2006 #4

    Doc Al

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    Not so. Static friction acts just enough to oppose slipping between the surfaces; the friction force required to prevent slipping is less than F.

    Also realize that the direction of the static friction is the same direction as that of the applied force. Friction helps accelerate the cylinder!

    The sum of the forces is most certainly not zero!
     
  6. Jul 20, 2006 #5

    Andrew Mason

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    The problem here is choosing the proper reference frame in which analyse the motion. If you look at it from the frame of the rolling (accelerating) wheel or cylinder, it appears that the forces are balanced. But, as Doc Al points out, this is certainly not the case because the centre of mass of the cylinder accelerates.

    It has to be analysed in the inertial reference frame. If you look at the moment of inertia about the instantaneous point of contact, there is an unbalanced torque on the centre of mass which results in the acceleration of the centre of mass.

    AM
     
  7. Jul 20, 2006 #6

    quasar987

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    Two things are unclear to me:

    @Office_Shredder: I was told that the static friction and kinetic friction forces had the same origin: the atoms of the surface. The only difference between them being that static friction occurs when the maximum force of static friction is less than the applied force and kinetic friction occurs when the maximum force of static friction is greater than the applied force. So them being the same force under different circumstances, it is impossible that they act simultaneously, as you suggest they do in the case " Static friction max is less that F_applied".

    @Doc: You say that "static friction acts just enough to oppose slipping between the surfaces". How do you know how big "just enough" is? And why is "just enough" not -F? I have calculated that the acceleration of atoms at the "tip" of the cylinder is given by the translationnal acceleration of the body cause by the applied force minus the acceleration caused by the rotation of the body. And that amounts to a_net = -F/M. So to oppose that, a force of +F is necessary.
     
    Last edited: Jul 20, 2006
  8. Jul 20, 2006 #7

    Doc Al

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    That's correct.
    You figure it out by applying Newton's 2nd law and the constraint for rolling without slipping.
    "-F" is not even in the right direction! And -F would cancel out the applied F, leaving you with no net force and thus no acceleration. Do you really think that makes sense?
    I don't know what you mean by the "tip" of the cylinder. Do you mean the bottom? If so, that point has an instantaneous acceleration of zero: It doesn't move. Or do you mean the top? If so, that point has an instantaneous acceleration that is twice that of the center of mass. Note that the rotational and translational accelerations add at the top, but cancel at the bottom.
    I don't know what you mean. Show your calculations for both the translational and rotational accelerations. For the translational acceleration of the center of mass: a_net = F_net/M.

    Again, I don't know what you mean.

    If you really want to understand this problem, do this:
    - Identify the forces acting on the cylinder
    - Write Newton's 2nd law for translation and rotation
    - Solve for the needed friction force for rolling without slipping
     
  9. Jul 20, 2006 #8

    quasar987

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    I like how you respond to each point separately instead of in one large brick of text.
     
  10. Jul 20, 2006 #9

    quasar987

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    I will do that. By the "constraint of rolling without slipping", do you mean the constraint that the translational speed of the CM be at all time the equal to [itex]R\frac{d\theta}{dt}[/itex]?
     
  11. Jul 20, 2006 #10

    Doc Al

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    That's correct:
    [tex]V_{cm} = R \omega[/tex]
     
  12. Jul 23, 2006 #11

    quasar987

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    Torce of friction only has to be 1/3 of applied force to prevent slipping. Blows my mind. :p
     
  13. Jul 23, 2006 #12

    Andrew Mason

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    Not quite. Force of friction provides the linear acceleration, so it has to be equal to the mass of the cylinder x the acceleration of the centre of mass.

    The applied force has to provide the rotational acceleration to the cylinder.

    The angular acceleration is a/R where a is the linear acceleration of the center of mass. The torque on the centre of mass from the road is is:

    [tex]maR = mR^2\alpha = \mu_smgR[/tex]

    Since the moment of inertia is [itex]\frac{1}{2}mR^2[/itex], the torque (applied by the road) that is required to provide the linear acceleration is twice the torque required to provide the angular accleration. So the friction is 2/3 of the total force applied.

    AM
     
  14. Jul 23, 2006 #13

    Doc Al

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    That's right. That's why this is a fun (and instructive) problem to analyze, since the result can be quite unexpected.
    The linear acceleration is result of the total force on the cylinder, not just the friction: Applied force + friction = ma.

    The rotational acceleration is a result of the total torque on the cylinder: Friction also contributes (negatively) to the rotational acceleration.
     
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