If someone doesn't get back to you before me I'll take a look at it later.
-Dan
Hi!
I'm trying to derive the periodic time of a pendulum motion. I have started with writing the height of the object as a function of the angle . Then I have calculated . After that I have written the gravitational potential energy as , so I got an expression for as a function of , but since I'm only working with small angles I have set the second derivative to a constant, so . Then I have written an expression for the kinetic energy as a proportion of the angular velocity in square, .
So, I have:
Since the total amount of energy, is always the same, I obtained , so I got the new system
and by substitution I could remove as well from the system, so I could minimize it to a single equation:
where
Now, this is where I got stuck. I don't know how to make it from here. Am I on the right track, and in that case can someone explain to me how to continue? Besides, can I make the assumption that for all alpha, since I am only woring with small angles? Thanks in advance!
So far so good. We then have the Taylor expansion:
How? I'm not sure how you can do this. Are you thinking of a specific starting point for the motion?
I am having serious trouble with this last statement. The only way I can make sense of this is that you are setting giving:
( because we are at the bottom of the swing. From your potential energy expression I can see that you have set h = 0 here.)
For small this becomes
which is not constant as you have supposed. I think this is where your problem lies.
The typical (that is to say, non-Lagrangian) derivation of the pendulum equation is to resolve the tension and weight into components along the direction of motion and radially. Then we know that there is no radial acceleration, so the acceleration along the direction of motion is given by
(The negative sign comes from the fact that the radial direction, which I have taken as the +y direction, points toward the axis of rotation. Since we are using a right handed coordinate system (as we always do in Physics) the direction of the acceleration is always opposite to the direction of the displacement.)
And we can easily see that
(where L is the length of the pendulum)
since the path is circular.
Thus
Thus
Now,
So for small angular displacements we have
and this is the SHO differential equation.
You can do this by using energy methods and, as I recall, this equation comes out by considering constants of the motion. But that way is a great deal more involved, Mathematically speaking.
-Dan
Yes, I thought that . Is that rash?How? I'm not sure how you can do this. Are you thinking of a specific starting point for the motion?
I am having serious trouble with this last statement. The only way I can make sense of this is that you are setting giving:
( because we are at the bottom of the swing. From your potential energy expression I can see that you have set h = 0 here.)Well, I don't know if you mean that would be constant, I ment that would be constant, that cannot make it harder, right? Maybe it's bad to make that assumption, cause it's not the real case, but that's another thing I guess.For small this becomes
which is not constant as you have supposed. I think this is where your problem lies.
Besides, , and what is does not matter, the result will be the same anyway.
I don't know if that's so easy to do, my pendulum motion does not come from an ordinary pendulum, but from an swing seat like thing that twists and get higher up because of that. is zero when the swing is as far down as possible.The typical (that is to say, non-Lagrangian) derivation of the pendulum equation is to resolve the tension and weight into components along the direction of motion and radially.
Ok, but it would be cool to do it that wayYou can do this by using energy methods and, as I recall, this equation comes out by considering constants of the motion. But that way is a great deal more involved, Mathematically speaking.
I can try to continue from where I last stopped:
That means that
Since
And means that , so we have that
So we have
If we integrate that expression, we get
, where is not a single function of , but can be many functions of , and is a constant (or many constants).
So we get
Now we must remember that , so . But since can be many functions (with different signs and starting values), we could choose different values for and stitch some intervals together, so we can form a function that is defined for all . We need to know that and is continuous functions. Therefore, as soon as we have choosen a start value for , we must use that whole interval, not changing the sign during the interval. Also, in the middle of two intervals, can't just stop, since a pendulum does not just stop at it's highest possition. No, it's turning down again. So we must start a new interval immediately, and that is with the oppisite sign ( ). So we get
, hence, the periodic time is .
Does that make sense? Can I reason like this? Or is there some simpler solution? :P
You are right, my bad!
You are now dealing with a "physical pendulum" (one where we need to account for the construction of the pendulum, not just the pendulum bob.) There appears to be a torsion in the problem, too if I am reading it right.
I'm sorry to say that you are a bit beyond me in what you are trying to do. Again, I'd choose the Lagrangian method to simplify the work, but it looks to me like you're going to run into complexities no matter how you derive it. I agree with you, given how simple the answer you derived looks, there is likely a shortcut in here somewhere.
-Dan
Hm, let's find that shortcut.
What we want to have is , where K is a constant. If we can get there, we know we are dealing with a sinus like movement of .
Starting with an early equation from my last attempt:
, where k is either 1 or -1
Let's derivate:
And multiplicate with to get rid of that in the denominator
And since is either 1 or -1,
So,
And our constant .
Since means that the periodic time is , this means that the periodic time is , which the previous result also showed.
So, I had to integrate, take the square root, then derivate again... Rather complicated, even though it's easier than the other way :/
Wow. I have started to discuss all types of mathematical problems on this forum (if not with someone else then with myself), partly because of the response and tips I get, and partly because that the equations display so nicely ^^ I am currently trying to get latex to work on my computer to, so I can imbed nicer equations into word...