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I Some weird circular relationship

  1. Jun 17, 2016 #1
    Hello there!

    I'm currently doing some mathematical modelling at my work, and I have arrived at an interesting kind of circular relationship integral - and now I'm wondering about what to do.

    The integral looks very innocent at first glance:
    $$ \theta_s = \int\limits_0^{t_1} \omega (t) dt$$
    So, it's a circular rotating body, with some time-dependent angular velocity. However, the ##\theta_s## is a constant - a parameter we can design in the system.

    But what makes the thing complex (at least in my head) is that ##t_1## is the time it takes for the body to rotate the amount ##\theta_s## - so the upper limit becomes dependent on the angular velocity also.

    Can I create another integral that relates the time ##t_1## to ##\omega (t)## and ##\theta_s##?
    Is Leibniz' rule the way to go?

    And ##\omega (t)## is an unknown function, by the way.
     
  2. jcsd
  3. Jun 17, 2016 #2

    haruspex

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    You can consider it more abstractly. You have an arbitrary function y=f(x), and you want to find the value of x2 such that the area under the curve from x=x1 to x2 is a given value. Doesn't sound like there's any way to cast that than the form of integral you quote.
     
  4. Jun 17, 2016 #3

    Delta²

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    We cant uniquely determine ##\omega(t)## neither ##t_1## just by this integral equation, need more equations to uniquely determine them.

    For example if you try ##\omega(t)=at+b## you ''ll find a ##t_1## that depends on a,b and ##\theta_s## (##at_1^2+2bt_1-2\theta_s+c=0##), if you try ##\omega(t)=sin(at)## you 'll find a totally different ##t_1=\frac{1}{a}arcos(a\theta_s+1)##.
     
    Last edited: Jun 17, 2016
  5. Jun 17, 2016 #4

    haruspex

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    I don't think that is what Runei was trying to do. Rather, he was looking to turn the equation into the form ##t_1=F(\omega, \theta_s)##, where F is some functional, maybe an integral.
     
  6. Jun 18, 2016 #5
    Thanks for the replies! The equation is coupled with another equation namely
    $$ \int\limits_0^{t_1}\tau_1(t)\omega(t)dt+\int\limits_0^{t_2}\tau_2(t)\omega(t)dt+\int\limits_0^{t_1}\tau_{in}\omega(t)dt = 0 $$
    The ##\tau_1(t)## and ##\tau_2(t)## are controllable functions - they can be chosen by design (EDIT: And they will have the opposite sign of ##\tau_{in}##). The function ##\tau_{in}## is a square wave signal (perhaps modulating another signal - but that is of lesser importance right now) with an on-time of ##t_1## and an off-time of ##t_2##. And as mentioned earlier, the ##t_1## is precisely the time it takes the system to turn an amount ##\theta_s##.

    $$ \theta_s = \int\limits_0^{t_1}\omega(t) dt $$

    What I am basically trying to do is modelling the system in steady-state, where I know that the input torque will be a square wave, with duty cycle determined by the angular velocity as mentioned above.

    I've been considering using a trapezoidal expansion of the integrals and solving the equations numerically, but I was wondering if they could be "massaged" even more.

    Thanks again for the replies! :-)
     
  7. Jun 18, 2016 #6

    Delta²

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    I might be wrong but seems to me again that we can't determine uniquely ##\omega(t)##. Can choose "quite a random " ##\omega(t)## and then just solve for ##t_1## and ##t_2##. For example if we put ##\omega(t)=C## we see how everything is simplified and easy to determine ##t_1## and ##t_2## so that the two integral equations hold. But I might be wrong.

    I think perhaps a third equation , even one not directly involving ##\omega(t)## but some equation like ##t_1+t_2=C## will narrow down our choices for ##\omega(t)##.
     
  8. Jun 18, 2016 #7
    Well there are actually some more now that I think about it.
    One thing I realized is that the integral with ##t_2## should probably have a lower bound being ##t_1## instead.
    ##\int\limits_0^{t_1}\tau_1(t)\omega(t)dt+\int\limits_0^{t_1}\tau_{in}(t)\omega(t)dt+\int\limits_{t_1}^{t_2}\tau_2(t)\omega(t)dt = 0##
    ##\theta_s = \int\limits_0^{t_1}\omega(t) dt##
    The function ##\tau_{in}## will have a period ##T_{in}## and that period will be equal to ##t_2##. Furthermore, the function ##\tau_{in}## has it's duty cycle ##D## which means that
    ##t_1 = D\cdot T_{in}##

    But there's more I see. The ##\omega(t)## is at any time related to the torque ##\tau_{net}## and moment of inertia ##I##, which means that we have in the period ##[0;t_1]##:
    ##\dot\omega(t)\cdot I = \tau_{net} = \tau_{in} + \tau_{1}##
    And in the time period ##[t_1;t_2]##
    ##\dot\omega(t)\cdot I = \tau_{net} = \tau_{2}##

    So I guess the integrals above could be rewritten as
    ##\int\limits_0^{t_1}\dot\omega(t)\omega(t) dt+\int\limits_{t_1}^{t_2}\dot\omega(t)\omega(t) dt = 0##
    The moment of inertia can be removed since it's constant and can be multiplied out.
     
  9. Jun 18, 2016 #8

    haruspex

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    Isn't ##\int \dot\omega(t)\omega(t).dt## just ##[\omega^2(t)]/2##?
     
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