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Ok then since I am wrong then formula for torque is wrong

  1. Apr 1, 2012 #1
    Force F is applied to the rim of a uniform disk (M=3.13kg, R=.193m). The disk is mounted on a fixed frictionless axis through its center, and the force is applied at an angle β=45.8° to the radius. The disk starts at rest, and reaches frequency f = 61.1 revolutions per second after rotating through an angle θ = 609 radians.

    What is the magnitude of force F?

    Torque = FRsinβ

    Torque = I∂

    ∂ = ωf - ωi / t

    ωf = 2∏f = 383.9026 rad/s

    ωf = θ/t; 609 rad / 383.9026 rad/s = 1.586, so t = 1.586 s

    ∂ = 383.9026 / 1.586 = 242.005 rad/s^2

    Torque = (1/2)(MR^2)(∂)
    Torque = (1/2)(3.13)(.193)^2(242.005) = 14.1076

    14.1076 = F(.193)(sin 45.8)
    F = 102 N

    But 102 N is wrong so I guess Torque really does not = FRsinβ


    Can anyone help here....cause I have no clue why or how this could be wrong...
     
  2. jcsd
  3. Apr 1, 2012 #2

    Doc Al

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    Staff: Mentor

    Good. That's the final angular velocity.
    Careful. Do not assume that the angular velocity is constant. Redo this. (What's the average angular velocity?)
     
  4. Apr 1, 2012 #3

    tms

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    So [itex]\omega = 2\pi[/itex]?
     
  5. Apr 1, 2012 #4

    Doc Al

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    I think he meant ωf = 2∏f.
     
  6. Apr 1, 2012 #5
    Thanks Doc Al!!

    You're right!

    ωavg = 0 + 383.9026 / 2 = 191.9513 rad/s so t= 609 rad / 191.9513 rad/s = 3.17267 s and ∂ = 383.9026 / 3.17267 = 121.002 rad/s^2 so....

    Torque = (1/2)(3.13)(.193)^2(121.002) = 7.0538

    and F = 7.0538 / (.193)(sin 45.8) = 51.0N......right????
     
  7. Apr 1, 2012 #6

    Doc Al

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    Staff: Mentor

    Looks good!
     
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