# Homework Help: Damped pendulum

1. Feb 6, 2004

### Norman

Problem:
A piece of wire of mass m and resistance R forms a circular ring of area A. It is suspended by an insulatd, massless rod of length l to form a frictionless pendulum in the presence of a uniform, vertical magnetic field B. The normal to the ring is parallel to its velocity. Find the time required for the energy of a small oscillation of the pendulum to decay to 1/e of its original value.

What I have so far:
I have the differential equation for the damped harmonic oscillator:
$$m\ddot{x}+\frac{mg}{l} sin[\theta]+F_B = 0$$
where $$F_B$$ is the damping force. Basically I know that I want to find the torque, not the force at first then relate this back to the force in the end. I am having a really hard time finding $$T_B$$. The problem I am having with the torque is finding the correct differential length to integrate so that my L x B term actually damps out the motion. Please help.
Thanks

2. Feb 7, 2004

### HallsofIvy

An annoyingly worded problem! They tell you the ring "has area A" but it's not the area that's important. Since A= &pi;r2, the important information is that the radius of the ring is &radic;(A/&pi;).

Your equation, $$m\ddot{x}+\frac{mg}{l} sin[\theta]+F_B = 0$$ has both "x" and "&theta;" which are not independent (I think. You didn't actually say what they were). If I am understanding these correctly knowing l lets you replace one by the other.

At first I thought this was a ring rotating on its axis but "normal to the ring is parallel to its velocity" implies that the ring is at the bottom of the pendulum, "facing" the direction of swing.

Finally, in order for the magnetic field to affect the swing at all, you have to have some current in the ring. Are you given that?

3. Feb 8, 2004

### Norman

You are correct about "x" and "&theta". They are the linear and rotational position, so they are relate by the length of the pendulum.

Correct. The ring is at the bottom of the pendulum and "facing" the direction of motion.

I can find the EMF by taking the time derivitive of the flux through the loop. Once I have the EMF, I am given the resistance and I can find the current.

Once again my biggest problem is the integration over the $$dl$$ components. Because of the orientation of the loop and the $$B$$ field, using the right hand rule you can see that the torque on the ring is perpendicular to the motion ($$\tau = A \times B$$ approx). So there must be some component of the $$dl$$ that allows for their to be a damping force. I am killing myself trying to determine how to do the correct integral which gives me the contribution that actually damps out the motion.

4. Feb 8, 2004

### Norman

what I have so far:
$$m\ddot{x}+\frac{mg}{l} sin[\theta]+F_B = 0$$
the torque on the wire loop due to the B-field $T_B$ is:
$$T_B=IAB = \frac{EAB}{R}$$
where $E$ is the EMF in the loop
$$E=\frac{\partial f}{\partial t}$$
where $f$ is the flux through the loop.
$$f=BASin[\theta]=BA\theta$$ (small angle approx}
$$E=BA\dot{\theta}$$
this implies:
$$T_B=\frac{A^2 B^2}{R}\dot{\theta}$$
$$F_B=\frac{T_B}{l}=\frac{A^2 B^2}{l R} \dot{\theta}, \dot{\theta} l = \dot{x}$$
finally:
$$F_B = \frac{A^2 B^2}{l^2 R} \dot{x}$$

from here I can solve the differential equation for $x$ and find the decay parameter $\tau$ which charecterizes the time in which the amplitude function falls to 1/e of its initial value. Obviously my original torque equation $T_B=IAB$is incorrect because there is no $l$ showing up in it. Any help getting the correct torque on this loop would be appreciated.
Thanks,
Norm

Last edited: Feb 8, 2004
5. Feb 9, 2004

### Norman

Help? Anyone?

6. Feb 9, 2004

### jamesrc

Hi,

I'll try to help, but I make no guarantees on accuracy. I'd stay away from x altogether, write your equations in terms of &theta;.

Here's what the equation of motion of the pendulum should look like:

$$ml^2\ddot{\theta} + \tau_b + mgl\sin\theta = 0$$

where &tau;b is the damping torque due to the magnetic field interacting with the induced current in the loop.

$$\tau_b = \vec{\mu}\times\vec{B}$$

with &mu; being the magnetic moment of the current loop, IA. I comes from Lentz + Faraday, as you have worked on:

$$|I| = \frac{d}{dt}\frac{BA\sin\theta}{R} = \frac{BA\cos\theta\dot{\theta}}{R}$$

so that:

$$\tau_b = \frac{B^2A^2(\cos^2\theta)\dot{\theta}}{R}$$

If we do the small angle approximation where sin(&theta;) ~ &theta; and cos(&theta;) ~ 1 - &theta;2 and drop all but the first order terms, we should get:

$$ml^2\ddot{\theta} + \frac{B^2A^2}{R}\dot{\theta} + mgl\theta = 0$$

The time constant for this system is

$$\tau = \frac{2Rml^2}{B^2A^2}$$

(compare with standard form of a second order differential eqiuation to check that. The units work out so that's a good sign.)

7. Feb 9, 2004

### Norman

James,

Thanks a lot, I actually got this answer the way I did it above. I believe your equation for $\tau_B$ is incorrect. I am trying it a very different way right now and will let you know if I get the same solution or not. Once again thanks a lot for taking the time to post that solution.
Cheers,
Norm