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Rolling without slipping magnitude

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

    A thin light string is wrapped around a solid uniform disk of mass M and radius r, mounted as shown. The loose end of the string is attached to the axle of a solid uniform disc of mass m and the same radius r which can roll without slipping down an inclined plane that makes angle θ with the horizontal. Find the magnitude of the acceleration of the center of mass of the rolling disc, a. Neglect friction in the axles of the pulley and the rolling disk.

    3. The attempt at a solution
    1- I would use ##F=ma## for the disc and ##τ=Iα## for the pulley to write to equations and use the condition ##a=αr##.

    2-friction b/w the incline causes torque in the disc


    4-the net torque on the pulley will have a cos(θ) term (to have only perpendicular elements).

    what I don't know-
    will only the force of static friction cause net torque in the pulley (or T will cause torque in
    the pulley)

    assuming T causes torque
    ##Tr cos(θ)=0.5 M r^2 α##
    using ##α r= a##

    I solved the equations but μ and cos(θ) terms don't get cancelled
    the answer is just in terms of M,m and g and θ

    plzz help!!
    Last edited: Jan 1, 2016
  2. jcsd
  3. Jan 1, 2016 #2


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    You'll need to use more than just F = ma for the disk on the incline. It is rotating as well as translating.

    Which disc?
    The string comes off of the pulley in a direction that is tangent to the rim of the pulley.

    The static friction force acts on the disc on the incline, not on the pulley. T is the only force applying a torque to the pulley.

    Also, the static friction is not necessarily at its maximum possible value. You cannot assume ##f = \mu mg\cos \theta##.
    Last edited: Jan 1, 2016
  4. Jan 1, 2016 #3
    the one on the incline
    will static friction point down the incline?
  5. Jan 1, 2016 #4


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    You should be able to decide this based on the direction of the angular acceleration of the disc on the incline.
  6. Jan 1, 2016 #5
    the disc is rolling down the incline therefore static friction should also point down the incline (because it is the one causing the rotation)
    but the answer turns out to be correct if I consider the static friction up the incline (like T)
    angular acceleration is pointing out of the screen.(counter clockwise rotation)
    what is wrong?
  7. Jan 1, 2016 #6


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    Yes, angular acceleration is pointing out of the screen. So, which way does the torque have to act on the disc?
  8. Jan 1, 2016 #7
    so stupid of me obviously up the incline!! I am confusing net F (linear acceleration) with net torque(angular)
    thanks a lot
  9. Jan 1, 2016 #8


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    OK. Good work.
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