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Hard Integral

  1. Jul 6, 2011 #1
    I've tried to ask this on the physics section but nobody know how to do it

    [itex]a = g(sen\alpha - cos\alpha u) -(\frac{v² u}{R})[/itex]
    [itex]V = \sqrt{Vo² + \int 2a.ds}[/itex]

    [itex]\alpha = \frac{s}{R}[/itex]


    Then, find the function v
     
  2. jcsd
  3. Jul 6, 2011 #2
    What is the question and where is your attempt? I'm new here, but I don't think I'm out of line by saying that this is a "help" community and not a "do" community... If you just want the answer to your problem, plug it into wolfram... If you want some "help", state specifically what you've attempted and where you're stuck...

    Cheers!

    Ken
     
    Last edited: Jul 6, 2011
  4. Jul 6, 2011 #3

    Ray Vickson

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    I cannot figure out your question. What is [itex]sen\alpha[/itex]? Do you mean [itex]\sin \alpha[/itex] or [itex] \sin(\alpha)[/itex]? Is [itex] cos\alpha u[/itex] supposed to be [itex] \cos(\alpha u)[/itex], or is it [itex] u \cos \alpha [/itex]? I will assume you want to integrate 2a with respect to s, where
    [tex] a = g \displaystyle \sin\left(\frac{s}{R}\right) - g \cos\left(\frac{u s}{R}\right) - \frac{uv^2}{R}.
    [/tex]
    That is an easy, elementary integral of the type you saw in Calculus 101 (assuming, or course, that v and u do not involve s in some unstated way).

    RGV
     
  5. Jul 6, 2011 #4
    [itex]a = g(sin(\alpha) - cos(\alpha ) u) -(\frac{v² u}{R})[/itex]
    [itex]V = \sqrt{Vo² + \int 2a.ds}[/itex]

    [itex]\alpha = \frac{s}{R}[/itex]


    Its a problem to determinate the velocity function of a car that is
    describing a loop trajectory with friction
    a is momentary acceleration, \alpha is the momentary angle that the
    car is in the loop, u is the friction coefficient, v is the momentary
    velocity, vo is initial v and s is the distance already traveled, R is
    the radius of the loop

    I know it's a help community, but in the hole problem, I balked just here, I reay dont now what to do. I have wolfram mathemat ica here and I already tried to solve this integral, but I don't know the function that does that. Besides, it's a problem for trainning in the Ipho, the original problem had no friction, so I solved it easily, but then I tried to solve same problem with "friction" (that btw, I created) and I've gotten in this integral. I don't know actually if this type of integral is needed for the test, but I want to know how to solve it, the problem is that I've never solved a integral that is in function of 3 things. V is in function of a and s, a is in function os v and s

    I'm sorry, I'm a high school student, I 've not learned integral with teachers, I actually did that by the internet. Brazil don't have teachers that teach this in the secondary school (I hoped it had), but you actually have to learn everything alone, then if you are selected to the IphO, you have some "classes" that is more " question solvng classes" and nobody explain things like solving an integral or other math things, they suppose you already know all those things, they only teach the physics stuff.

    Could you tell me at least what to search for?

    []'s
    John
     
  6. Jul 6, 2011 #5

    SammyS

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    The physics post has more information. The link is https://www.physicsforums.com/showthread.php?t=511697".
    I've edited the above quoted post to make it a bit more readable.

    Added in Edit:
    The above looks somewhat different than the original post in the physics section.

    There jaumzaum had:

    Fc=N+g.cos alpha

    [tex]A(\alpha ) = g.\sin\alpha - (v^2/R + g.\cos \alpha)u
    [/tex]

    [tex]V = \sqrt{Vo^2 + \int a.ds}[/tex]
     
    Last edited by a moderator: Apr 26, 2017
  7. Jul 6, 2011 #6
    To be honest, and I may have the wrong idea of the problem you're describing, it seems like your equation of motion is wrong. You'll have tangential and radial components for the acceleration. Radial should be more or less what you have, the v2/r term, but tangential should be something like [itex]\frac{dv(\theta)}{dt}[/itex] which must be evaluated using the chain rule.
     
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