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[MagmaMcFry]

How is the m.o.i found when the pivot is in the center, though?

The moment of inertia of any object is found by integrating the squared pivot distance over the mass of the object. If you don't want to do that, here's a list of preintegrated moments of inertia for commonly used objects.

[da_nang]

Double pendulums are suffering. Never ask about them, never be told about them, save everybody some time.

Unless of course, you're majoring Control Theory.

[MagmaMcFry]

Double pendulums are suffering. Never ask about them, never be told about them, save everybody some time.

Unless of course, you're majoring Control Theory.

In which case double pendulums are easy.

[Vector]

.

[Sinistar]

So, I guess it has come to this.

u_xy + u_x - 2y + 4y = 0

u = ?

(where u_x = du/dx)

My best guess would be substituting x ( v = u_x ) but the final result before changing back ( v = 2xy - 2y² - vy ) isn't quite to my liking.

I would be very appreciated if someone could explain solving this thing step by step.

EDIT:
Stupid me. Can't even copy it right. Here:

u_xy + u_x - 2x + 4y = 0

Thanks for reply though, MagmaMcFry, I'll check it in detail later in about half an hour.

[MagmaMcFry]

Sure why not. First step is as you said, let's let v = du/dx. Our new equation is dv/dy + v + 2y = 0, or dv/dy = -2y-v. Now you can solve this equation via guessing, or via this handy formula. In either case you'll get this:

v = c*exp(-y) - 2y + 2

Now substitute u_x back for v, integrate for x, and you'll get u = x*(c*exp(-y) - 2y + 2) + c₂.

[Sinistar]

Hmmm, still can't wrap my head about this one. The jump from

dv/dy = - v - 2x + 4y
 
to

v = c*exp(-y) + 2x - 4y + 4

isn't quite clear to me. The thing is I have to write this by hand and WolframAlpha requires subscription to see step-by-step explanation. 

[MagmaMcFry]

Okay, we have two differential equations that interest us:

(a) dv/dy = -v-2x+4y
(b) dv/dy = -v

(b) is called a homogeneous part of (a).
Okay, here's the trick: If v_1 is a solution to (a) and v_0 is a solution to (b), then v_0+v_1 is also a solution to (a). Conversely, if v_1 and v_2 are solutions to (a), then v_1-v_2 is a solution to (b).

What this means is that if we find one solution of (a) and add it to each solution of (b), we get each solution of (a).

Okay, let's solve (a) first. Since (a) is of the form v'=c*v+p(y), there is a polynomial solution for v (of the same degree as p). You can find that however you want, for example by solving for its coefficients. Example: If you have f'=2f-4x², then a polynomial solution f=ax²+bx+c satisfies 2ax+b=2ax²+2bx+2c+4x², so 0=2a-4, 2a=2b and b=2c, which is a solvable linear equation system in (a, b, c) that gives you f=2x²+2x+1.

In your particular case, we get a solution v=-4y+2x+4.

Now let's solve (b). That's an easy one, I'm sure you learned that f'=af is solved by all c*exp(ax). More generally, f'=g(f) is solved by all h^-1(x+c), where h'=1/g.

In your particular case, we have v'=-v, so we get the solutions v=c(x)*exp(-y).

Now if we add the solution of (a) to all solutions of (b), we get v=c(x)*exp(-y)-4y+2x+4.

[da_nang]

So, I guess it has come to this.

u_xy + u_x - 2y + 4y = 0

u = ?

(where u_x = du/dx)

My best guess would be substituting x ( v = u_x ) but the final result before changing back ( v = 2xy - 2y² - vy ) isn't quite to my liking.

I would be very appreciated if someone could explain solving this thing step by step.

EDIT:
Stupid me. Can't even copy it right. Here:

u_xy + u_x - 2x + 4y = 0

Thanks for reply though, MagmaMcFry, I'll check it in detail later in about half an hour.

I'm working under the assumption you meant the partial derivative since you've got mixed derivatives.



Clearly, this is a second-order linear non-homogenous PDE with constant coefficients. So we split the solution into two parts, the homogenous solution and the particular solution.

Solving for the homogenous solution:


Here, the Psi functions are arbitrary.

To solve the particular solution, we want to assume the solution is a function such that the left hand side becomes identical to the right hand side (there may be infinitely many):


Let's start with the x term by assuming it's a polynomial of degree 1 which clearly must be multiplied by x lest one of its terms be swallowed up by the homogenous solution:


Do the same for the y term (make sure none of its terms are linearly dependent on either the homogenous solution or the first particular solution):


Adding these to the homogenous solution gives us a general solution:


EDIT: Fixed typo

[Helgoland]

Topology question:
Prove or disprove: For all covering maps p: E \rightarrow X, where both E and X are path-connected, all continuous functions f: E \rightarrow E with p°f=p are already homeomorphisms.
I've been wrecking ym brain about this all weekend - does anyone have an idea?

EDIT: I think I can construct a counter-example using the Koch curve. Let's see what my tutor will say...

[Sinistar]

Ok, I've been going through what's been posted and I've made some connections, things are starting to make sense and even where they don't I still get the nagging feeling in the back of my head telling me this does look similar to something I should know.

In the end it seems like it all boils down to me not making right connections AND forgetting stuff I learned in the past. Dammit.

So, your help was very much appreciated, da_nang and MagmaMcFry and I thank you both from the bottom of my heart.

[Skyrunner]

I made a single pendulum! Thanks, da_nang and McFry and everyone else who helped me. Now I just need to figure out how to make a double pendulum.

[Shakerag]

Ok, I tried this in the small questions thread, but let's give this another go here:

Okay, geometry question.  Note that I am not a math person, so this may be worded awkwardly.

Using Schläfli symbols:
{3,3} = Tetrahedron (or a d4)
{3,4} = Octahedron (or a d8)
{3,5} = Icosahedron (or a d20)

So what the hell is a {3,6} if such a creature exists?  How many sides does it have?

[frostshotgg]

Not possible, to fit 6 regular triangles around the same vertex you'd take up all 360 degrees, so it'd be flat and thus 2d.

[Shakerag]

Not possible, to fit 6 regular triangles around the same vertex you'd take up all 360 degrees, so it'd be flat and thus 2d.

Ah.  That makes sense, thanks.