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Falling chain on a cylinder

  1. Jan 5, 2016 #1
    Hi! I am struggling with this problem the last two days and I cannot decide which solution is correct.

    The problem statement, all variables and given/known data


    A cylinder of radius R is fixed horizontally on the floor. A uniform chain of mass M and length L (L<πR/2) is placed on the cylinder in such a way that one end of the chain is on a highest point of the cylinder. We release the chain. All drag and friction shall be negleted. Gravitational acceleration is g.
    (This is a 2D problem, the cylinder could also be a circular disk.)

    I am asked to find:

    1)The accelaration of the chain once released.

    2) An expression for the angular velocity and acceleration as a function of the total angle the chain has fallen.




    The attempt at a solution
    1)
    First Approach :
    We find the CM of the chain using the formula r=[Rsin(L/(2R))]/(L/((2R))
    We then decompose the weight to the tangential component. ( Mgsin(L/2R) )

    Acceleration of cm is gsin(L/2R). Since there is no velocity at this moment centripital acceleration is 0, so the magnitude of the accelerarion of each point of the chain is gsin(L/2R) * R/r .

    I dont know if this approach is correct. I believe that if we sum all the normal forces from the cylinder the net normal force will have a tangential component. ( for every point on the chain the normal force has different magnitude as the radial component of the weight gets smaller and smaller as we move down).

    Approach B:

    For each dm we write the N2L equation:

    F1 + dm1gsin(θο + θ1) - F2 = dm1a

    Where F1 and F2 are the contact forces from the "next" and "previous" dm.

    If we add all those equations for all dm's the F1,F2 forces cancell up. (By newton's 3rd law)

    We get:

    g * Integral(from 0 to M) of (sin[θο + θ(m)]dm =Ma

    Assuming a linear density λ = M/L
    We get θ(m) =m/(λR)

    Substituting back to the integral (with θο = 0 )we have

    a = gR/L (1-cos(L/R)

    I cant find a logical error in the second approach so i think it might be correct. But the first aproach seems simple and nice to turn down immediately :/


    For the question 2) for the acceleration we use exacly the same apporaches (but with θο != 0 )

    To find the velocity can we take the difference of the potential energies? (With respect to the postition of the cm). Do the normal forces lroduce any work? (If the is a tangential component of the total N force, shouldn't they?)

    This problem has made me very confused. I would be grateful if someone could help me understand the situation better.. :D
     
  2. jcsd
  3. Jan 5, 2016 #2

    Orodruin

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    Unless you have already learned Lagrangian mechanics, I would suggest that you make a free body diagram for a small part (single link) of the chain. What forces are acting on it and how does it move?
     
  4. Jan 5, 2016 #3
    Thank you sir.

    No, I have not leaned Lagrangian mechanics yet. This problem is part of a first semester mechanics course.

    The second approach then must be right since it is based on the FBD's for every dm.

    Could you please comment on the validity of the first approach?

    Thank you for your reply.
     
  5. Jan 5, 2016 #4

    Orodruin

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    The first method has several problems. There are several things which you cannot simply assume to be true without further justification. For example, what is the direction if the overall force on the chain and how the acceleration relates to the acceleration of the chain's center of mass (which a priori lies inside the cylinder).
     
  6. Jan 5, 2016 #5
    Thank you Orodruin! :D I will stick with the second method!

    One last question:
    In the end does the total Normal force produce any work?

    If it does, I cannot use conversation of mechanical energy to find the velocity and I will have to find the total work done in a similar way to the second approach.

    Thank you again. I appreciate your help! :)
     
  7. Jan 5, 2016 #6

    Orodruin

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    You have to be very careful here, the total normal force is an integrated force over several different points of application. You therefore run into trouble even when you want to define the work done by the "total normal force". The work done by the normal forces is zero, at each point, the chain is moving orthogonal to the direction of the force.
     
  8. Jan 5, 2016 #7
    Ok thank you! Both ways (conservation of mechanial energy and analytical integral form of work for every dm) give the same result for the velocity as a function of the angle.

    This makes me more confident about the correctness of the solution.


    Still I have not got a 100% thorough understanding of what happens with the cm. I hope in the future I will get that feeling.

    Thank you for your kind responses! :D
     
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