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Pulling a plank across cylinders - Rotational Dynamics

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data

    A plank with a mass M = 6.10 kg rides on top of two identical, solid, cylindrical rollers that have R = 4.10 cm and m = 2.00 kg. The plank is pulled by a constant horizontal force F of magnitude 4.80 N applied to the end of the plank and perpendicular to the axes of the cylinders (which are parallel). The cylinders roll without slipping on a flat surface. There is also no slipping between the cylinders and the plank.

    (a) Find the initial acceleration of the plank at the moment the rollers are equidistant from the ends of the plank.

    (b) Find the acceleration of the rollers at this moment.

    (c) What friction forces are acting at this moment? (Let fp be the frictional force exerted by each roller on the plank, and let fg be the rolling friction exerted by the ground on each roller.)

    fp
    fg

    http://www.webassign.net/serpse8/10-p-088.gif

    2. Relevant equations

    F = ma
    T=I∂
    I = (1/2)mR2 (Moment of inertia for a cylinder)


    3. The attempt at a solution

    From f = ma, I understand that the net force pulling the system is equal to the net inertial mass multiplied by the acceleration of the system.

    Net force would be equal to F pulling minus force of friction giving me the equation:
    F - Ff = ma

    In order to find Ff I used the T=I∂ equation
    T=I∂
    (fp - fg)R = (1/2 * mR2) * (a/R)

    Simplifying this results in
    (fp - fg) = (1/2)ma
    which I can plug back into F = ma netting me

    F - (1/2)ma = mneta

    So I plug in all my values:
    4.8 - a = (6.1 + (2)(1/2)(2)(.041^2))a
    and I get a = .676 for part (a)
    and a = .338 for part (b)

    Both of these answers are close, but wrong. The correct answers for a and b are .632 and .316, respectively. What am I doing wrong/missing?

    Then, I'm unsure where to begin for part (c). If I plug in acceleration back into my net torque equations, I'm left with 2 unknowns and 1 equation.
     
  2. jcsd
  3. Nov 10, 2012 #2

    haruspex

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    A couple of problems there. In which rotational directions do the forces fp and fg act on the rollers? You haven't defined a. What is it the linear acceleration of?
    What are you equating to Ff in order to do that? Is m in the above equation the given m (mass of one roller) or mass of the system?
     
  4. Nov 10, 2012 #3
    Woops, I have fp and fg mixed up, so it should be fg - fp because friction with the ground acts in the direction of the applied force. Then a is the linear acceleration of one of the rollers. Since there is no slipping between the cylinders and the plank, linear acceleration should be equal to system acceleration, correct?

    ff = fg - fp = (1/2)ma where m is the mass of a roller.

    So I realize that I should be doing F - 2Ff = ma because I have 2 rollers, and thus, 2 sources of friction. But that still yields me an answer of .59 for acceleration, which is still incorrect.
     
  5. Nov 10, 2012 #4

    haruspex

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    I'll ask again: in which rotational directions do these two forces act on the rollers?
    So what is the linear acceleration of the plank? Hint: think twice before answering.
     
  6. Nov 11, 2012 #5
    Both of them act in the clockwise direction, I think.

    The linear acceleration of the plank should be two times that of an individual roller.
     
  7. Nov 11, 2012 #6

    haruspex

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    Right. So instead of fg - fp or fp - fg it should be?
    Yes.
     
  8. Nov 12, 2012 #7

    haruspex

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    One small correction/clarification to my earlier response.
    It is not clear at the outset which way fg acts. But that doesn't matter as long as it is treated consistently. We can take it as acting in the same direction as F, in which case it will oppose fp in torque but work with it in linear acceleration of the rollers, or take it as being in the opposite direction. If we make the wrong guess we'll simply get a negative value. The important thing is that we'll have fp+fg in one equation and fp-fg in the other.
    You may by now be aware of two other current threads on essentially the same problem, but perhaps with different numbers.
     
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