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Help please! cylinder, torque, rolling w/o slipping problem!

  1. Dec 17, 2012 #1
    A massless string is wrapped around a 10 kg cylinder. the picture will look like the letter b. you pull upward on the string with a force of P = 49N. The bottom of the cylinder s on horizontal table and will roll w/o slipping. using dynamics involving center of mass, find the magnitude and the direction of frictional force (f). (no slippage between string and cylinder)


    Solution:

    Torque about cm: RP - Rf = Iα ---> R(P-f) = Iα

    I = 1/2MR^(2)
    α = a/R

    R(P-f) = 1/2MR^(2) (a/R) ---> P-f = 1/2Ma

    Forces Horz: f = ma

    Why do we need to consider the forces horizontal?? Isn't it already included in the torque calculation?

    P-ma = 1/2Ma
    P = 3/2Ma
    Ma = 2/3P
    a = 2/3 (49N) / 10kg ---> 3.2667 N

    My teacher got 32.667N? but how?
     
  2. jcsd
  3. Dec 17, 2012 #2

    SammyS

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    You have one equation with two unknowns. You need another equation in the same two unknowns.

    f = Ma gives you that.
    You solved for a and got 3.2667 m/s2 . (You had the wrong units.)

    What does that give you for f ?
     
  4. Dec 18, 2012 #3
    ooh ok i got it. thanks!
     
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