Rotational Dynamics on a door knob

[Huskies213]

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??)

 

[Dmitri]

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 equations and 4 unknows one of which would cancel out.
I hope it is helpful.

 

[Huskies213]

Re

Can anyone explain more ? I am still lost

 

[Hootenanny]

As you correctly say torque is given by;

\tau = F r

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

Wd = Fs

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

s = r\theta

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

\fbox{wd = F r\theta = \tau\theta}

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

Hope this helps

-Hoot