# I Some weird circular relationship

1. Jun 17, 2016

### Runei

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. Jun 17, 2016

### haruspex

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.

3. Jun 17, 2016

### Delta²

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
4. Jun 17, 2016

### haruspex

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.

5. Jun 18, 2016

### Runei

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! :-)

6. Jun 18, 2016

### Delta²

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 $\omega(t)$.

7. Jun 18, 2016

### Runei

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.

8. Jun 18, 2016

### haruspex

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