Overdamping limit

[Math Jeans]

1. The problem statement, all variables and given/known data

Find the conditions under which it is valid to approximate the equation

mL^2\ddot{\theta}+b\dot{\theta}+mgLsin\theta=\Gamm a

by its overdamped limit
b\dot{\theta}+mgLsin\theta=\Gamma

2. Relevant equations

The formula is for an overdamped pendulum where b is the damping coefficient, and Gamma is a torque on the pendulum.

3. The attempt at a solution

I know from past experience that the conditions are b^2>>m^2gL^3, but I'm not sure how to obtain that from the equation.

Thanks,
Jeans

[Mark44]

If \theta is "small," sin(\theta)~\approx~\theta. Under this condition you can replace sin(\theta) in your differential equation to make it a linear d.e with constant coefficients. Is that enough of a start?

[Math Jeans]

I had considered doing that and then equalizing the equations, but the problem is implying that it is done for all theta.

Here's an email from my teacher regarding it. I don't really know how much of it I understand (I just got the email this morning):

"No you don't need to solve for any ODE. Rescale the problem with time scale T, the coefficient preceding theta' should be 1, the coefficient preceding theta'' should be much less than 1. Check the overdamped bead on the hoop problem for similar procedure."

[Math Jeans]

Ok! I think I get it! Essentially I had to non-dimensionalize the equation by getting rid of the double-dot term.

mL^2\ddot{\theta}+b\dot{\theta}+mgLsin\theta=\Gamm a

mL^2\frac{d^2\theta}{dt^2}+b\frac{d\theta}{dt}+mgL sin\theta=\Gamma

\frac{L}{g}\frac{d^2\theta}{dt^2}+\frac{b}{mgL}\fr ac{d\theta}{dt}+sin\theta=\frac{\Gamma}{mgL}

Define: \tau=\frac{mgL}{b}t --> t=\frac{b}{mgL}\tau

\frac{d\theta}{dt}=\frac{d\theta}{\frac{b}{mgL}d\t au}=\theta'
\frac{d^2\theta}{dt^2}=\frac{d^2\theta}{\frac{b^2} {m^2g^2L^2}d\tau}=\theta''

Apply the new definition:

\frac{L*m^2g^2L^2}{gb^2}\theta''+\theta'+sin\theta =\frac{\Gamma}{mgL} --> \frac{m^2gL^3}{b^2}\theta''+\theta'+sin\theta=\fra c{\Gamma}{mgL}

And there is my \frac{m^2gL^3}{b^2} which agrees with what I thought was the answer.

Then if we multiply out the function under Gamma on the right side of the equation and substitute t back in, we are left with the original equation minus the double-dot term. Yay!

QATC?

thanks a bunch,
Jeans