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A Rolling Disc on a Slab

  1. Apr 9, 2012 #1
    A uniform solid cylinder of mass M and radius R is at rest on a slab of mass m, which in turn rests on a horizontal, frictionless table. If a horizontal force F is applied to the slab, it accelerates and the cylinder rolls without slipping. Find the acceleration of the slab in terms of M, R, and F.

    Okay, I'm not usually one to be stumped but this one really has me frustrated. The solution according to the text that I got this problem from says do two free body diagrams (see image), then for the cylinder do sum of forces in the x and the sum of torques about the center of mass and do ƩF on the slab. They then suggest that I relate the acceleration of the cylinder from sum of forces (let's call it ac) to the acceleration of the slab (as) by subtracting the acceleration of the center of mass of the cylinder due to it's rolling, acs (this is aquired from the sum of torques equation done for the cylinder and then using the no slip condition relating angular and tangential acceleration).

    I guess I'm okay with the last part -- it is just vector addition (even if I am sceptical about treating a non-inertial reference frame with a that type of tranformation). But there's a big red flag in this explanation (thus this post): the sum of forces in the x on the cylinder is f = m*ac. This is the linear acceleration an outside observer would see the cylinder moving at. I would expect the exact same equation for a block of equal mass resting on the slab. However, my intuition is firmly convinced that the block would not be accelerating at the same rate as the cylinder because the cylinder is rolling back with respect to the slab. In other words, if the slab was invisible, we would see the block pull out ahead of the cylinder.

    So, how are we allowed to use the same equation for two obviously different circumstances? Is this the right solution? I brought up the non-inertial reference frame thing earlier because it is the only thing in the solution that rubs me wrong mathematically.

    Any help would be appreciated. Thanks!
     

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  2. jcsd
  3. Apr 9, 2012 #2

    rcgldr

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    In the diagram, the mass of the cylinder is M and the force is f, so f = M * acylinder, where acylinder is the acceleration relative to the frictionless table, not relative to the slab.
     
    Last edited: Apr 9, 2012
  4. Apr 9, 2012 #3
    Right, sorry. "m" instead of "M" was a typo in my sum of forces. However, I did say that "This is the linear acceleration an outside observer would see the cylinder moving at." The problem is still unresolved. How is this different than a block resting on the slab?
     
  5. Apr 9, 2012 #4

    tiny-tim

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    welcome to pf!

    hi el professor! welcome to pf! :smile:
    for a block, you have an equation for ac, an equation for as, and ac = as

    for a cylinder, you have the same equation for ac, the same equation for as, and a different equation relating ac and as …

    where's the paradox? :confused:
     
  6. Apr 9, 2012 #5
    Re: welcome to pf!

    The problem is that they do not have the same acceleration (by observation).

    However, if we take the scenario where their masses are the same, then their accelerations should also be the same...

    UNLESS their forces of friction are different. This doesn't make sense though. Let's assume that the slab is accelerating at such a rate so that we are about to slip. For this condition, they should have the same force of friction and the same acceleration.

    I feel like the friction on the wheel is where I am tripping, here. I have a feeling that it is not actually equal to the friction on the block but can't give a good reason as to why.
     
  7. Apr 9, 2012 #6

    rcgldr

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    Re: welcome to pf!

    I think you mean the acceleration of the cylinder versus the acceleration of the block. The linear acceleration of the cylinder will be less because the cylinder's surface will move in the same direction as the slab because of the angular velocity and acceleration of the cylinder.

    Note the linear acceleration of the cylinder is equal to "f" / M, even though the cylinder also experiences angular acceleration.
     
    Last edited: Apr 9, 2012
  8. Apr 10, 2012 #7
    Ok, I figured it out.

    The problem was how friction accelerated a block (no slip) vs a rolling cylinder (no slip).

    I was using f = ma for both (correct) and using it in the same reference plane for both (incorrect).

    If I am standing on the slab, f = ma accelerates the cylinder and causes it to roll. An outside observer must take the slab's acceleration into account to determine the net acceleration of the cylinder.

    For a block, there is no acceleration observed if you are watching it from the slab. From outside of the slab, its acceleration is caused by the friction between it and the slab.


    Phew. A student asked me this question and had me running confused in no time. Thanks for letting me talk it out!
     
  9. Apr 10, 2012 #8

    rcgldr

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    You want to use the outside observer's frame of reference for both slab and cylinder.

    For the cylinder, you probably want to calculate the force "f" as a function of acceleration of the slab wrt (with respect to) the outside observer. The linear acceleration of the cylinder = "f" / M, and the angular acceleration = "f" x R / I, where I is angular inertia of cylinder. Then the linear acceleration of the surface of the cylinder = linear acceleration of cylinder + angular acceleration x R = acceleration of slab. This ends up allowing you to calculate the linear acceleration of the cylinder as some fraction (< 1) of the acceleration of the slab.
     
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