How can I calculate the torque on a damped torsion pendulum?

[JolleJ]

If I damp a torsion pendulum, a force will work on it given by F = -k*v, where k is some constant and v is the velocity. My question is, how can I from this calculate the torque, which this affects the torsion pendulum with?

I've tried myself, however, I'm sure there's something wrong:
For a particle in the pendulem, since v = r*w, where r is the distance from axis to point in which the force works and w is the angular speed, I get: < EDIT
F = -k*r*w
Then I multiply both sides with r, getting the torque in this point:
F*r = -k*r^2*w <=> T = -k*r^2*w, where T is the torque.

For the entire pendulum the torque is then:
T = \sum(-k*r^2*w)
-k and w are constant, so:
T = -k*w\sum(r^2)

However, I don't get this, since the sum of the distance of all particles is either infinit or zero. Could someone explain to me what I've done wrong, or perhaps just show me, with what torque the damping really works.

Thank you very much.

 

[maverick280857]

Can you list all the forces acting on the body and also write the equation of motion?

Also, if you write v = rw where r has dimensions of length, then w has to have dimensions of angular speed and not angular momentum.

 

[JolleJ]

Sorry; changed that.

I'm not sure, which forced act on it. On the pendulum I've put a cardboard, so this creates the damping. I'm pretty sure that I'm doing something wrong, by insetting v = w*r, since v have different directions all over the cardboard.
And then of course the torque from the string works: T = -D*phi. Where phi is the angel and D is the torsionconstant of the string.

 

[maverick280857]

Well, the velocity is given by

\vec{v} = \vec{\omega}\times\vec{r}

Taking \vec{\omega} = \omega\hat{e_{z}} and \vec{r} = r\hat{e_{r}}, I get \vec{v} = r\omega\hat{e_{\theta}}. I am using cylindrical coordinates here.

So the damping force is \vec{F_{D}} = -br\omega\hat{e_{\theta}}.

The other force, as you have correctly stated, is \vec{T} = -D\theta\hat{e_{\theta}}. The equation of motion is thus

-D\theta - br\dot{\theta} = I\ddot{\theta}

As you can see, \omega = \dot{\theta}.

By the way, what are you summing over anyway?

 

[JolleJ]

Thank you. However I can't quite follow you. I've never seen cylindrical coordinates before.
I'm trying to get to this:
-D\theta - br\dot{\theta} = I\ddot{\theta}
Simply don't know how to explain it in my text. I know you just did it, but can't it be done using normal coordiantes (so that I can understand it)?

Again thanks.

 

[maverick280857]

Ok, forget about the polar (cylindrical) coordinates. Just consider this: I am using \theta to denote the angular coordinate and the left hand side of that equation is the sum of the torques (with correct signs). The right hand side is just the moment of inertia times the angular acceleration. So, basically, I have just written the equation

\sum_{i}\vec{N_{i}} = I\vec{\alpha}

 

[maverick280857]

As for doing it with "normal" coordinates, well, this is the easiest way to do it. This is just a differential equation in \theta. If you convert this back into cartesian coordinates using

x = r\cos\theta
y = r\sin\theta

so that (here x and y denote the coordinates of some point on the rim of the disk relative to its center taken as the origin)

\theta = \tan^{-1}\frac{y}{x}

then you'll get a messy equation. Thats the whole idea of preferring description in certain coordinates over others.