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Cylinder rolling thanks to external torque.

  1. Jun 5, 2016 #1
    1. The problem statement, all variables and given/known data

    A cylinder of mass m and radius R is set on a plane, with large enough friction coefficient to ensure at any moment rolling without slipping. A constant torque is applied along the axis passing through the center of mass (G) of the cylinder and perpendicular to the basis of the cylinder.
    What is the acceleration of the center of mass G?

    2. Relevant equations

    The central equation in this problem is
    I dw / dt = M
    This equation should be valid if w (angular velocity) and I are referred to either an axis passing through the center of mass of the system (I_G) or through a point that is fixed along the motion.

    I will also need to compute I for the cylinder,
    I_G=1/2 mR^2,
    while remembering that, if I need I for a different axis,
    I_O=I_G+ m (GO)^2

    3. The attempt at a solution

    To find a solution is easy, I have two! I can set myself either in the center of mass, or in the contact point between the cylinder and the plane, since it is fixed.

    In the second case, I have I_O = 3/2 mR^2 and therefore the angular acceleration is 2 M/(3 mR^2), and the linear acceleration should be 2M/(3 mR)
    In the first case, I have I_G=1/2 mR^2, and therefore the angular acceleration is 2 M/(mR^2) and the linear acceleration would be 2 M/(mR).

    I am pretty sure the proper solution is the first, but I am not able to understand why I can't solve it easily in the center of mass... I think I am missing in this case the additional torque caused by the attrition force, but I don't understand how that would bring the total to 2M/(3 mR)
     
  2. jcsd
  3. Jun 5, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Welcome to PF!
    Yes.
    Try setting up the torque equation relative to the center of mass with the torque due to friction included. Don't forget that you also have Fnet = M aG.
     
  4. Jun 5, 2016 #3
    Thanks a lot! I think I am there:

    In the center of mass, I have
    1) M_T = M - F_a R
    2) F_a = m a_G
    3) M_T= I_G dw/dt
    where M_T is the total torque and the friction is the only force parallel to the plane. I also have
    4) a_G = dw / dt R,
    since the cylinder is rolling without slipping.

    Therefore, from 1) and 2),
    5) M_T=M-m a_G R

    While from 3) and 4),
    6) M_T=1/2 m R a_g

    Therefore, from 5) and 6),
    7) M-m R a_G = 1/2 m R a_g,
    from which a_g = 2 M / (3 m R)

    Perfect! Finally the two solutions are only one!
     
  5. Jun 5, 2016 #4

    TSny

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    Homework Helper
    Gold Member

    Looks great. Good work!
     
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