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Rotational mechanics

  1. Nov 25, 2013 #1
    1. The problem statement, all variables and given/known data

    A plank having mass 3.7 kg rides on top of two identical solid cylindrical rollers each having radius 5.5 cm and mass 2.9 kg. The plank is pulled by a constant horizontal force of 6 N applied to its end 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.

    1.) Find the acceleration(linear) of the plank(m/s^2).

    2.) Find the frictional force acting acting on the plank(N).

    M=3.7 kg
    m=2.9 kg
    r=.055 m
    F=6 N
    α = angular acceleration
    a=linear acceleration


    2. Relevant equations

    τ=Fr
    I=(1/2)m(r^2)
    Ʃτ=Iα
    F=ma
    Frictional force(static) = force applied


    3. The attempt at a solution

    torque of one wheel: τ=Fr = 6*.055 = .33 N

    Ʃτ=Iα = (1/2)*(2.9)*(.055^2)*α →
    .33=.00438625α → α(of one wheel)=75.2351907 rad/s^2

    a=αr = 75.2351907*.055 = 4.13793103 m/s^2

    Ʃa(both wheels and board)= (4.13793103*2) + a of board

    F=ma→ 6=(3.7 + 2.9 +2.9)a → a of board = .631578947 m/s^2

    ∴ Ʃa = 8.27586207 + .631578947 = 8.90744102 m/s^2
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 25, 2013 #2

    haruspex

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    That's wrong in two ways. First, the inertia of the plank and other cylinder come into it. The 6N is in some way shared between them. (If each component had a separate 6N it would go a lot faster.)
    Secondly, there are two ways of viewing the rotation of a rolling wheel. You can think of it as a rotation about its centre (plus a linear movement) or as rotating about the point of contact with the ground. With the first view, there would also be a torque from the friction on the ground; with the second view, the distance is 2r. Both views lead to the same result.
    The safest approach is to consider the forces and accelerations of each of the three components separately (but two are identical, so only two lots of equations). Make sure to use unique symbols to represent all the forces.
     
  4. Nov 25, 2013 #3
    Your solution starts out with this equation: torque of one wheel: τ=Fr = 6*.055 = .33 N

    This equation is incorrect. The force acting on one wheel is not 6 N, or even 3 N, and the units of torque are also incorrect. Start out by doing a force balance on the plank. Draw a FBD. Let F1 be the tangential (frictional) force exerted by each of the wheels on the plank. What direction is this force pointing (in the direction the plank is moving, or the opposite direction)? What other horizontal forces are acting on the plank? Do a force balance on the plank, including ma.

    Let F2 represent the tangential force exerted by the flat surface on each wheel. Do a FBD on each wheel. What are the horizontal forces acting on each wheel? Write a horizontal force balance on each wheel, including the ma term for the center of mass of the wheel. What are the torques around the center of mass of each wheel in terms of F1 and F2? How is the angular acceleration of each wheel related to these torques? How is the angular acceleration of the wheels related kinematically to the horizontal acceleration of the center of mass of the wheels, if the wheels do not slip relative to flat surface?

    Chet
     
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