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Cylinders rolling down without slipping

  1. Feb 7, 2015 #1
    1. The problem statement, all variables and given/known data

    Two solid cylinders are placed on an inclined plane with inclined angle Θ. Both mass of cylinders are m, but the radius bigger cylinder is two times the radius of small cylinder. A string links the big cylinder's center to small cylinder's top (see picture). Both cylinders are freed from static condition and then roll down the inclined plane without slipping.
    Determine the relation between the big cylinder tangential acceleration and its center of mass acceleration
    http://www.sumoware.com/images/temp/xzlkhnoflaqjkbbp.png [Broken]

    2. Relevant equations
    atangential = α r

    3. The attempt at a solution

    I think that rolling without slipping means the tangential acceleration equals the center of mass acceleration.
    I think that the acceleration of big cylinder center of mass equals the tangential acceleration of the small cylinder.
    It rolls without slipping means that the big cylinder center of mass acceleration equals its tangential acceleration.
    So, I think atan = acm

    But, my book says that
    atan = acm + α r
    acm = α r
    atan = 2 acm

    But, I don't understand how atan = acm + α r comes from

    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Feb 7, 2015 #2

    Edit : The question is to determine the relation between the top point of small cylinder and its acceleration of center of mass
    Last edited by a moderator: May 7, 2017
  4. Feb 8, 2015 #3


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    And what is your question?
  5. Feb 8, 2015 #4
    My book says that
    atop-point = acm + α r
    acm = α r
    atop-point = 2 acm

    But, I don't understand how atop-point = acm + α r comes from
    What I know is that atop-point = atan = α r (since it's rolling without slipping)
    If the top point of the small cylinder accelerates more than its center point, it will roll with slipping, right ?
  6. Feb 8, 2015 #5


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    The points at the perimeter of the cylinder move with respect to the centre, so their relative velocities add to that of the CM.
    At the top, this velocity has the same direction as that of the CM. The bottom point has relative velocity opposite to the CM. The speed of the perimeter is ωR, so the tangential speed of the bottom point is Vcm-ωR, that of the top point is Vcm+ωR.
    In case of rolling without slipping, the bottom point is in rest with respect to the ground. So Vcm=ωR. At the top, the velocity is Vcm+ωR=2Vcm.
    The same is valid for the tangential acceleration.
  7. Feb 8, 2015 #6
    So, for a cylinder or ball or anything that's rolling, the tangential acceleration is not just α r, but acm+α r , right ?
    And if it's rolling without slipping, the tangential acceleration is 2 α r ?
    Or the formula is just relevant for the cylinder whose top point is linked by a string to a center of another cylinder ?
  8. Feb 8, 2015 #7


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    The tangential acceleration of a point on the perimeter depends on the position of that point. On the top, it is acm+α r. At the bottom, it is acm-α r, and both are horizontal. The tangential acceleration at a general position is the vector sum of the horizontal acceleration of the CM and the tangential acceleration with respect to the CM .
    The blue vector is the acceleration of the CM. The green vectors are the accelerations with respect to the CM. The magnitudes of the green vectors are the same αr. The red vectors are the sum of the blue and green ones, the tangential acceleration of a point of the perimeter with respect to the ground.

  9. Feb 8, 2015 #8
    Thanks for your help :)
  10. Feb 8, 2015 #9


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    You are welcome :)
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