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Rotational Dynamics on a door knob

  1. Apr 10, 2006 #1
    Can anyone help me with where to begin ? ? ?

    Turning a doorknob through 1/6 of a revolution requires 0.1 J of work. What is the torque required to turn the doorknob?

    i know that T = F x r
    what is the force and radius ?(is radius 3/6=1/2??)
  2. jcsd
  3. Apr 10, 2006 #2
    Energy equals force times distance. 0.1 = F*ds where ds = (pi/3)*R, pi/3 is 1/6 of revolution.
    As you said T = F*R, So now you have 3 equasions and 4 unknows one of which would cancel out.
    I hope it is helpful.
  4. Apr 10, 2006 #3

    Can anyone explain more ? im still lost
  5. Apr 11, 2006 #4


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    As you correctly say torque is given by;

    [tex]\tau = F r[/tex]

    I would think you also know that work done is given by;

    [tex]Wd = Fs[/tex]

    If we have a circle of radius [itex]r[/itex] and an angle of [itex]\theta[/itex] then the arc length subtended by that angle is given by;

    [tex]s = r\theta[/tex]

    Now substituting [itex]s = r\theta[/itex] into the work equation (because the force acts tangentally) gives;

    [tex]\fbox{wd = F r\theta = \tau\theta}[/tex]

    This is basically what Dmitri said, but simplyfying before plugging in the numbers. Remember that a full revolution is [itex]2\pi[/itex] radians.

    Hope this helps

    Last edited: Apr 11, 2006
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