# Rotational Dynamics on a door knob

1. Apr 10, 2006

### 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

2. Apr 10, 2006

### 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 equasions and 4 unknows one of which would cancel out.

3. Apr 10, 2006

### Huskies213

Re

Can anyone explain more ? im still lost

4. Apr 11, 2006

### Hootenanny

Staff Emeritus
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

Last edited: Apr 11, 2006