I’ve searched my old calculus text, my differential equations text and the web and I can’t find how to solve the following problem, or for that matter how to even approach the solution. The SD is my last hope.
a = constant = length of pendulum anchored to the top of a disk.
R = constant = radius of disk
x = theta = angle at which pendulum intersects disk.
x ’ = theta dot = time rate of change of theta
x ‘’ – theta dot dot
g = constant = gravitational acceleration
Eps = small angle
My solutions manual says, “Lagrange’s equation for x is”
(a - Rx)x ‘’ – R(x ')[sup]2[/sup] – g cos(x) = 0
“Now let us expand about some angle x[sub]0[/sub], and assume the deviations are small. Defining eps = x - x[sub]0[/sub], we obtain”
eps ‘’ + [g sin (x[sub]0[/sub])] / (a - R x[sub]0[/sub]) = (g cos x[sub]0[/sub]) / (a - R x[sub]0[/sub])
I know how to get a Taylor’s expansion for sines and cosigns and functions etc. but I haven’t the foggiest how to do this. I could really use some help here as I think this is something I’m going to need to know in the future.
Substitue x[sub]0[/sub]+eps for x, giving :
(a - R(x[sub]0[/sub]+eps))((x[sub]0[/sub]+eps)’’) – R((x[sub]0[/sub]+eps)’)2 – g cos(eps+x[sub]0[/sub]) = 0
Don’t use Taylor series, use the identity for cos(a+b) (which I always look up to make sure I get it right, but it’s something like cos(a+b) = cos(a)cos(b) - sin(a)sin(b)) and drop terms proportional to eps[sup]2[/sup]. Remember that x[sub]0[/sub]’ = x[sub]0[/sub]’’ = 0.
A non-obvious part is that eps’ and eps’’ are also being assumed small. Anyway, you get
(a - Rx[sub]0[/sub])eps ‘’ – g cos(eps+x[sub]0[/sub]) = 0
except that cos(eps+x[sub]0[/sub]) should be something like cos(x[sub]0[/sub]) * cos(eps) - sin(x[sub]0[/sub]) * sin(eps) from the identity. For small eps, sin(eps) becomes eps, and cos(eps) becomes 1, and you should be able to get there.
I suspect there is supposed to be a factor of eps multiplying either sin(x[sub]0[/sub]) or cos(x[sub]0[/sub]) in your last equation.
The important point to remember in all this is to get everything to be linear in eps, eps’ and eps’’ by using whatever identities you need, and by dropping quadratic (and higher) powers of eps.
Thanks ZenBeam that’s some pretty good stuff. I guess I shouldn’t always think “expand” means to use the Taylor series.
I don’t think I would ever have realized to “Remember that x0’ = x0’’ = 0.” so I probably would never have figured it out.