# Damped pendulum and sliding rod

1. Feb 26, 2006

### Frobar

Problem 1:

I have a mathematical pendulum with a mass m connected to a string of length l. The pendulum is damped by air resistance that is proportional to the velocity, Ffric = -k*v. I need to derive the damping effect the air resistance has on the pendulum - that is, the decrease of the total energy of the energy it causes per unit time. The damping effect should be expressed as a function of k, l and a, the angular velocity ("omega-dot" in common notation).

I know that the speed of the point mass must be a*l, because the angular velocity is expressed in radians. The force on it from air resistance must then be -k*a*l. I thought that, because work is force*displacement, and effect is work/time, maybe I will get the effect if I take the force times the speed (displacement/time). In that case, the effect from air friction would be -k*(a*l)^2. Is this correct?

What worries me is that I may not have understood the ramifications of potential and kinetic energy being constantly swapped in this system.

Is there some better form I could get the solution in given that the next problem is to derive the movement equation for the pendulum (a differential equation in theta) by comparing dE/dt to the result? E is the total energy.

Problem 2:

A rod is lying still on a frictionless surface when a force is applied to one end of it. As a result of this the rod will start to slide and rotate. However, one point will not get displaced for small angles of rotation. I need to calculate this point.

If I knew how to calculate the slide speed, I could probably derive the point with a little trig, but I'm uncertain how to get at it. The rotation can probably be had using tau=I*alpha.

Any help appreciated!

2. Feb 27, 2006

### gandharva_23

problem 2: are you applying a force at the end of the rod ? if it is a force then please mention whether its a constant force (direction ?) . i am sending a solution assuming you are applying an impulse at the end . Let J be the impuse applied at the end of the rod . then i have
v = J/m
w = 6J/mL
where v is the speed of com of the rod and w is the angular speed of the rod .
If J is applied at the bottom end of the rod let us us assume that the IAOR is at a distance x from the centre of the rod .
this implies v - wx = 0
this gives x = L/6 .

3. Feb 27, 2006

### qtp

if you look at the new component it ends up adding a term which is dependant on the angular velocity. i found that the net force is
$$F_{net} = -m g \sin{\theta} -k l \frac{d\theta}{dt}$$
then u use small angle and angular acceleration to get it down to
$$\frac{d^2 \theta}{dt^2} + \frac{k}{m} \frac{d \theta}{dt} + \frac{g}{l} \theta = 0$$

4. Feb 27, 2006

### Frobar

It is a constant force that is perpendicular to the rod along the surface its lying on, applied at one of its ends. The rod is very thin by the way, if that matters. Could you elaborate on the w = 6J/mL step? I'm still very shaky when it comes to angular momentum.

5. Feb 28, 2006

### Frobar

I guess what I'm really having trouble with is the "translational" (as opposed to the rotation they cause) effect of forces not applied at the center of mass. Maybe problem 2 could be solved without dealing explicitly with this motion, but I would still like to get a better understanding of how it acts with forces at different lengths from the center of mass, and exactly where it comes from.