A pendulum with viscous friction

[wrobel]

TL;DR Summary

Not a challenge merely unexpectedly solvable by hands problem

Consider the standard pendulum with a weightless rod of length b and a mass point m and mg is applied. In the hinge there is a torque of viscous friction which is proportional $\omega^2$.
Now release the pendulum from the horizontal position. What biggest height does the point m attain after its first passing through the vertical position?

 

[anuttarasammyak]

Let me know whether I got you. Is the equation of motion
$$L=ml^2\dot{\theta}$$
$$\dot{L}=ml^2\ddot{\theta}=-mgl\sin\theta-k\dot{\theta}^2 sgn\ \dot{\theta}$$?

 

[erobz]

I get the following eq of motion:

$$ mb^2 \ddot \theta + \beta \dot \theta^2 - mg b \cos \theta = 0 $$

You are saying this is analytically solvable?

 

[vanhees71]

Note the important additional factor $\text{sign} \dot{\theta}$ written in #2. That's also what I'm puzzled about. I don't see, how to solve the question posed in #1, because I don't see how to calculate the energy loss due to friction without solving for the equation of motion first, and that's for sure not easily done in analytic form. Even the frictionless case leads to elliptic functions! Maybe @wrobel gives the solution, he has in mind soon?

Let me see, how far a I get. Using the EoM given in #2, with $\theta$ being the angle wrt. the direction of $\vec{g}$ as usual. Then the energy of the pendulum is
$$E=\frac{m b^2}{2} \dot{\theta}^2 - m g b \cos \theta,$$
and
$$\mathrm{d}_t E=\dot{\theta} (m \ddot{\theta} + m g b \sin \theta)=-k |\dot{\theta}|^3.$$
From this I don't see, how to calculate the energy loss without knowing the solution of the EoM :-(.

 

[erobz]

Note the important additional factor $\text{sign} \dot{\theta}$ written in #2. That's also what I'm puzzled about. I don't see, how to solve the question posed in #1, because I don't see how to calculate the energy loss due to friction without solving for the equation of motion first, and that's for sure not easily done in analytic form. Even the frictionless case leads to elliptic functions! Maybe @wrobel gives the solution, he has in mind soon?

Would we have to have the sign factor if we are just trying to get to the first turning point?

 

[vanhees71]

That's true, because then $\dot{\theta}>0$ during this time.

 

[wrobel]

I get the following eq of motion:

View attachment 329418

$$ mb^2 \ddot \theta + \beta \dot \theta^2 - mg b \cos \theta = 0 $$

You are saying this is analytically solvable?

Yes,
$$\dot \theta=\omega(\theta),\quad mb^2\omega'\omega+\beta\omega^2-mgb\cos\theta=0$$
and $$\omega'\omega=\frac{1}{2}\frac{d}{d\theta}\omega^2$$
a linear equation with respect to $\omega^2(\theta)$

 

[erobz]

Yes,
$$\dot \theta=\omega(\theta),\quad mb^2\omega'\omega+\beta\omega^2-mgb\cos\theta=0$$
and $$\omega'\omega=\frac{1}{2}\frac{d}{d\theta}\omega^2$$
a linear equation with respect to $\omega^2(\theta)$

If that term that has $\cos \theta$ was a constant $k$ its solvable with the substitution $u = \beta \omega^2 + k$ , but it's not constant?

 

[vanhees71]

Yes,
$$\dot \theta=\omega(\theta),\quad mb^2\omega'\omega+\beta\omega^2-mgb\cos\theta=0$$
and $$\omega'\omega=\frac{1}{2}\frac{d}{d\theta}\omega^2$$
a linear equation with respect to $\omega^2(\theta)$

How do you get to this equation? I still don't see, how to answer your problem, about the height after "one swing" analytically.

 

[wrobel]

I rewrote the equation from post 3. The standard trick:$$\dot\theta=\omega(\theta)\Longrightarrow \ddot\theta=\omega'\dot\theta=\omega'\omega$$
gives a first order linear ODE with respect to $\omega^2(\theta)$

 

[erobz]

I rewrote the equation from post 3. The standard trick:$$\dot\theta=\omega(\theta)\Longrightarrow \ddot\theta=\omega'\dot\theta=\omega'\omega$$
gives a first order linear ODE with respect to $\omega^2(\theta)$

$$ mb^2 \frac{d \omega}{dt} + \beta \omega^2 - mgb \cos \theta = 0 $$

$$ mb^2 \frac{d \omega}{d \theta} \frac{d \theta}{ dt} + \beta \omega^2 - mgb \cos \theta = 0$$

$$ mb^2 \frac{d \omega}{d \theta} \omega + \beta \omega^2 - mgb \cos \theta = 0 $$

While it is true that

$$ u = \omega^2 $$

$$ \implies \frac{du}{d \theta} = 2 \omega \frac{d \omega}{d \theta} $$

would convert that to a linear equation, it is not true that $u =\omega^2$ is a viable substitution given the non-constant term $mgb \cos \theta$ that remains in the equation.

 

[vanhees71]

I rewrote the equation from post 3. The standard trick:$$\dot\theta=\omega(\theta)\Longrightarrow \ddot\theta=\omega'\dot\theta=\omega'\omega$$
gives a first order linear ODE with respect to $\omega^2(\theta)$

Great, but shouldn't it be the EoM in #2?

 

[wrobel]

I do not know what EoM is and I do not understand what is written in 2.
Introduce a notation $y(\theta)=\omega^2(\theta)$ then the equation from 3
takes the form $$mb^2y'/2+\beta y=mgb\cos\theta.$$
To solve it use the method of variation of a constant
https://personal.math.vt.edu/renardym/class_home/firstorder/node1.html

 

[anuttarasammyak]

Even in no friction case elliptic integral is tough for me. I am afraid there is no analytical solution to find at least for the beginners like me.

 

[erobz]

Even in no friction case elliptic integral is tough for me. I am afraid there is no analytical solution to find at least for the beginners like me.

It would be nice to see what level of a “hands” solution this is. I suspect it will be entirely lost on me, but I’m always curious about stuff I can’t comprehend.

 

[vanhees71]

I do not know what EoM is and I do not understand what is written in 2.
Introduce a notation $y(\theta)=\omega^2(\theta)$ then the equation from 3
takes the form $$mb^2y'/2+\beta y=mgb\cos\theta.$$
To solve it use the method of variation of a constant
https://personal.math.vt.edu/renardym/class_home/firstorder/node1.html

EoM=equation of motion, and I think the correct equation of motion, with $\theta$ counted relative to the direction of $\vec{g}$ is
$$mb^2 \ddot{\theta} + k \dot{\theta}^2 \text{sign} \dot{\theta} + \frac{m g b} \sin \theta.$$

 

[wrobel]

It would be nice to see what level of a “hands” solution this is. I suspect it will be entirely lost on me, but I’m always curious about stuff I can’t comprehend.

I do not have any other words to explain this standard technique. But I think if you ask it in the topic "Differential Equations" then maybe there is more gifted and more patient educator than me.

 

[anuttarasammyak]

@wrobel I am curious to know how you treat $cos\theta$ term as g(t) in the variation constants method and the final result of maximum angle to reach.

 

[vanhees71]

$\theta$ is now the independent variable!