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I've got complex dynamics problem I'm trying to solve, but the crux of my problem at the moment is describing the path a pulley would take on a rope of fixed length.

Assume the rope is fixed at points $A$ and $B$ and that the rope length is longer than the distance from $A$ to $B$.

I know how long the rope is, the pulley radius, where the fixed points are, and I know the angle the pulley makes to the center of the rope field.

My initial approach is to start with a diagram like the following:

Overall layout

This is a lot to take in, so I'll go over each part in particular to approach the solution I'm currently using, which utilizes a numeric solver to find the answer. I'm attempting to write the into a dynamics model, though, with the ultimate goal of building a state feedback controller around the system, so what I really need is an equation of motion.

First, the pulley location:

Pulley Location

This is straightforward enough. I don't know what the distance from the pulley center is to the center of the rope field ($L$), but I know the angle the pulley makes to the center of the rope field, $\theta$. I then know that the horizontal and vertical positions should be:

$$ L_x = L\sin{\theta} \tag{1}\\ $$ $$ L_y = -L\cos{\theta} \tag{2}\\ $$

Now, moving on to the length from the $A$ fixed point to the tangent of the pulley:

A-side setup

The tangent side of the triangle formed, $T_a$, forms a right triangle with the pulley radius $r$. The hypotenuse, $L_a$, is the distance from the pulley center to the fixed point $A$. Finally, the angle of the triangle that occupies a portion of the pulley is $\alpha_{\mbox{minor}}$, so-named because it's the a-side angle of the smaller triangle.

I know that $L_a$ is the distance between the pulley center and $A$:

$$ L_a = \sqrt{(L_x-A_x)^2 + (L_y-A_y)^2} \tag{3} \\ $$

Since I know the pulley radius $r$, I can calculate:

$$ \alpha_{\mbox{minor}} = \mbox{acos}{(r/L_a)} \tag{4} \\ $$

And then the tangent length to be:

$$ T_a = L_a \sin{\alpha_{\mbox{minor}}} \tag{5} \\ $$

The process is the same for the b-side:

B-side setup

Following the same steps:

$$ L_b = \sqrt{(L_x-B_x)^2 + (L_y-B_y)^2} \tag{6} \\ $$

$$ \beta_{\mbox{minor}} = \mbox{acos}{(r/L_b)} \tag{7} \\ $$

$$ T_b = L_b \sin{\beta_{\mbox{minor}}} \tag{8} \\ $$

Then I need to find the wrap angle, but in order to do so I first find the "major" angles, to find out what portion of the interior angle of the pulley isn't being used for rope wrap:

Major angle setup

This is straightforward, like calculating the minor angles, but instead here the "short" leg of the right triangle is the distance $-Ly$ (taking negative such that the net length is a positive). The hypotenuse for the a-side and b-side is still $L_a$ and $L_b$, respectively, so:

$$ \alpha_{\mbox{major}} = \mbox{acos}{(-L_y/L_a)} \tag{9} \\ $$ $$ \beta_{\mbox{major}} = \mbox{acos}{(-L_y/L_b)} \tag{10} \\ $$

Finally, looking at the wrap angle:

Wrap angle

Now it's clear that the wrap angle is what's left over after taking away the a- and b-side major and minor angles. The rope wrapped around the pulley is then just the arc lengths through that angle:

$$ \mbox{wrapAngle} = 2\pi - \left(\alpha_{\mbox{major}} + \alpha_{\mbox{minor}} + \beta_{\mbox{major}} + \beta_{\mbox{minor}}\right) \tag{11} \\ $$ $$ \mbox{wrap} = r\left(\mbox{wrapAngle}\right) \tag{12} \\ $$

Finally, I solve for the pulley center distance $L$ in the equation:

$$ \mbox{ropeLength} = T_a + \mbox{wrap} + T_b \tag{13} \\ $$

When I try to solve for $L$ with Matlab, I get:

Warning: Unable to find explicit solution. For options, see help. 
> In solve (line 317) 

ans =

Empty sym: 0-by-1

My real hope here was that I could start with the basic pendulum equation:

$$ \frac{d^2\theta}{dt^2} + \frac{g}{L}\sin{\theta} = 0 \tag{14}\\ $$

And have the "pendulum length" $L$ as a function of $\theta$, such that I wind up with a more complex pendulum equation, and then can substitute that in my control equations, but I can't get past the non-solution presented here.

Am I expressing the problem poorly? Is there a way to formulate these equations such that there is an analytic solution?

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closed as off-topic by Aaron Stevens, John Rennie, Yashas, JMac, GiorgioP May 10 at 19:49

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Here I'm exploring the 'simplification' of looking at an ellipse only:

A point mass is moving without friction inside an ellipse with focal points $A$ and $B$:

Ellipse

Due to inertia the mass will oscillate. We want to find the Equation of Motion (EoM).

Assume the particle starts at $y_0$ height and $v=0$ then the energy equation is:

$$mg(y-y_0)+\frac12 mv^2=0$$

Or:

$$g(y-y_0)+\frac12(v_x^2+v_y^2)=0$$

$$g(y-y_0)+\frac12(\dot{x}^2+\dot{y}^2)=0$$

To get the EoM, derive to time, so:

$$g\dot{y}+\ddot{x}\dot{x}+\ddot{y}\dot{y}=0\tag{1}$$

In the case where $|AP|+|BP|\gg|AB|$ the ellipse then approximates a circle and a simple EoM is then obtained (I think). But we're stuck with an ellipse:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

where $a$ is the semi-major axis and $b$ the semi-minor axis. Also:

$$x=\pm\frac{a}{b}\sqrt{b^2-y^2}$$

$$\dot{x}=\pm\frac{a}{b}(b^2-y^2)^{-1/2}y\dot{y}$$

Now we could certainly still obtain $\ddot{x}$ and then inserting $\dot{x}$ and $\ddot{x}$ into $(1)$ would yield an expression in $y$, $\dot{y}$ and $\ddot{y}$. But it's very messy.

And there's no way it could be made explicit in $\ddot{y}$ like in:

$$\ddot{y}=F(y,\dot{y})$$

as is possible for the SHO.

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  • $\begingroup$ I'm going to go ahead and select this as the answer, because I think showing the intractability of the "simplified" case is sufficient to show there is no solution to the more general case. Thanks for all your time and effort! $\endgroup$ – Chuck May 10 at 15:05
  • $\begingroup$ You're welcome. It's a tough nut, obviously. $\endgroup$ – Gert May 10 at 15:10
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Am I expressing the problem poorly? Is there a way to formulate these equations such that there is an analytic solution?

The best way to tackle this problem and obtain an Equation of Motion (EoM) that can be solved analytically or numerically is by the principle of Conservation of Energy. Derived to time, the energy equation yields a Newtonian EoM.

Pulley

Assuming no friction or drag, the energy equation simply is: $$\Delta U+\Delta K=0$$ So: $$mg(y-y_0)+\frac12 mv^2+\frac12 I\omega^2=0$$ where $y_0$ is the initial ($t=0$) position in $y$. The second RH term is the rotational kinetic energy of the pulley wheel, where: $$I=\gamma mR^2$$ with $\gamma$ a shape factor and $R$ the pulley wheel radius. Assuming no slipping of the pulley wheel: $$v=\omega R \Rightarrow \omega=\frac{v}{R}$$ $$\frac12 I\omega^2=\frac12 \gamma mv^2$$ $$mg(y-y_0)+\frac12 mv^2+\frac12 \gamma mv^2=0$$ $$g(y-y_0)+\frac{1+\gamma}{2}v^2=0$$

The difficulty now is that the instantaneous velocity vector $\vec{v}$ is the resultant of two vectors, $\vec{v_x}$ and $\vec{v_y}$.

Starting from:

$$v^2=v_x^2+v_y^2$$

your task is therefore 'eliminating' $\vec{v_x}$, so that an expression in $\vec{v_y}$ and $y$ only is obtained. This is done by means of the type of tedious trigonometry used in your question, deploying relations between the angles, $x$, $y$ and the total length of rope.

You will obtain something like:

$$g(y-y_0)+\frac{1+\gamma}{2}f(v_y^2)=0$$

with $f$ a function.

Now derive both sides to time $t$:

$$\frac{\mathbf{d}}{\mathbf{dt}}(g(y-y_0))+\frac{\mathbf{d}}{\mathbf{dt}}\Big(\frac{1+\gamma}{2}f(v_y^2)\Big)=0 \Rightarrow gv_y+\frac{\mathbf{d}}{\mathbf{dt}}\Big(\frac{1+\gamma}{2}f(v_y^2)\Big)=0$$

For the RHS, using the chain rule we get:

$$(1+\gamma)v_y a_y\frac{\mathbf{d}}{\mathbf{dt}}f(v_y^2)$$

The $v_y$ drop out, so we get:

$$g+(1+\gamma)a_y \frac{\mathbf{d}}{\mathbf{dt}}f(v_y^2)=0$$

This will give you an expression for the instantaneous acceleration $a_y$:

$$a_y=\frac{\mathbf{d}v_y}{\mathbf{dt}}$$

That is the Newtonian EoM. Solve it and will likely be an oscillator (but not a simple harmonic one).

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  • $\begingroup$ I appreciate the help. My approach with the simple pendulum was the Lagrange equation, $\frac{\partial{L}}{\partial{\theta}} - \frac{d}{dt}\frac{\partial{L}}{\partial{\dot{L}}} = 0$ and solving for $\ddot{\theta}$. I'm comfortable with moving through those steps but, as I mention at the end of the problem statement, I can't come up with an expression for the path - the only way I've managed to get any results so far is with numeric solvers. If this is the case, then my $f(v_y^2)$ stays as a black-box term and I can't do any further analysis because I can't solve the differential equation. $\endgroup$ – Chuck May 9 at 13:22
  • $\begingroup$ Oh, but you can always solve numerically. But it might not even be necessary: just fully develop the EoM and see what it looks like. $\endgroup$ – Gert May 9 at 13:49
  • $\begingroup$ But the end goal here for me is to generate a differential equation that I can linearize and put into state feedback form to serve as the basis for a state feedback controller. I can't wrap a controller around numerically-solved equations :( $\endgroup$ – Chuck May 9 at 14:39
  • $\begingroup$ I might develop a full solution, just to see... $\endgroup$ – Gert May 9 at 14:48
  • $\begingroup$ Turns out it's not possible to eliminate $x$ from the expression of $v$. The point $P$ moves on an ellipse, with foci $A$ and $B$. This problem probably requires two coupled ODEs. Not at all simple. $\endgroup$ – Gert May 9 at 22:08

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