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Rotational work

  1. Mar 17, 2005 #1
    Q: A disk of radius 2.40 cm and mass 1 kg is pulled by a string wrapped around its circumference with a constant force of 0.36 newtons. What is the angular velocity of the disk to three decimal places after it has been turned through 0.85 of a revolution?

    So I started by obtaining my degrees from the 0.85 revolutions (S=rθ). This equals 2.225 degrees.

    Next I used ∆x=R∆θ
    (2.40)*(2.225)
    =5.340 radians

    W=F∆x
    (0.36)*(5.340)
    =1.9224

    Using conservation of work (W=∆K):
    K=.5Iω^2 (where I of a disk=.5MR^2)

    W=1.9924=.5(.5*1*2.4^2)ω^2
    ω=1.15542

    WRONG. I can't figure out where I am going wrong, please help!
     
  2. jcsd
  3. Mar 17, 2005 #2
    You have some pretty heavy duty mistakes there.

    First, do not use degrees at all whatsoever.

    ∆x=R∆θ At this point, you use degrees for ∆θ (wrong, should be radians) and solve for ∆x in radians (wrong, should be meters).

    The answer with Work and Kinetic energy looks correct, but the reason they don't match is that you didn't convert ∆x correctly.
     
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