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Velocity of center of mass of spinning disk

  1. Oct 31, 2011 #1
    A solid uniform disk of mass 19.0 kg and radius 70.0 cm (.7 m) is at rest FLAT on a frictionless surface. A string is wrapped around the rim of the disk and a constant force of 35.0 N is applied to the string. The string does not slip on the rim. The string is being pulled. When the disk has moved a distance of 6 m, what is the speed of the center of mass?



    Relevant equations (I think...):
    I=1/2MR^2
    FR=Iα
    α=a/R


    So far, I have:
    FR=Iα
    FR=1/2MR^2(a/R)
    F=1/2Ma
    a=2*F/M=3.33333 m/s^2

    and then,
    v^2=2*a*6m, since initial velocity is 0.
    v=7.26 m/s

    But this is wrong, and I'm not sure why. Please help!
     
  2. jcsd
  3. Oct 31, 2011 #2

    gneill

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    Staff: Mentor

    Newton's second law holds for translational motion, even if the net force results in a torque. The angular and linear components can be dealt with separately when they are not "coupled" via some friction or other mechanical mechanism (like a rolling object or a pulley and string).
     
  4. Nov 1, 2011 #3
    Thank you. I'm not sure how I would use Newton's Second Law, due to the fact that the normal force and mg are perpendicular to the movement and therefore not contributing to the velocity.

    I'm so frustrated :)
     
  5. Nov 1, 2011 #4

    gneill

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    Staff: Mentor

    mg and the normal force are equal and opposite, so their contribution vanishes. You only have to deal with the force applied via the string.
     
  6. Nov 1, 2011 #5
    OH, thank you gneill!!!

    Answer is:

    F=ma
    35N=(21kg)a
    a=1.667 m/s^2

    v^2=0 + 2(1.667)(7.9), 0 being the initial velocity
    v=5.1 m/s
     
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