ME 5180: Advanced Dynamics | Project 01 | Group 06

Group Members:

Christian DiPietrantonio

Project Overview:

In this project, a pendulum is attached to a spinning frame. The frame has dimensions,

$h_1 = 0.2~m$

$w_1 = 0.1~m$

and the pendulum length is $L=0.15$ m with a $m=0.1$ kg point mass at the end of the system. The pendulum swings in the $x'$ - $z'$ plane as it rotates at a constant speed, $\Omega$

Your team's goal is to

• build the equations of motion using Lagrange and least action $L=T-V$
• solve for the motion for a slow rotation speed and a fast rotation speed
• visualize the solution with plots and animations

Running the Program:

From the project directory, run:

julia main.jl 

All source code is in the src/ directory.

Project Structure

TBD

Results

TBD (Include Plots, Animations, Comparison of hand derrived eqns to those derived using Julia)

Analysis:

To solve this problem, we can define a reference frame that rotates with the pivot about the z-axis. At time t, the rotating reference frame rotates by $\phi(t) = \Omega t$ relative to the inertial frame. Let's say we have some vector $\vec{v} = x' \widehat{i'} + y' \widehat{j'} = x \widehat{i} + y \widehat{j}$. We know that $\widehat{i'} = \cos(\phi) \widehat{i} + \sin(\phi) \widehat{j}$ and $\widehat{j'} = \cos(90^\circ + \phi) \widehat{i} + \sin(90^\circ + \phi) \widehat{j} = -\sin(\phi) \widehat{i} + \cos(\phi) \widehat{j}$. We also know that $z = z'$. Now, since the trajectory of the pendulum is defined in the inertial frame, we must introduce a method to transform coordinates from the rotational frame back to the inertial frame. Using our original expression for $\vec{v}$, we can substitute the expressions for $\widehat{i'}$ and $\widehat{j'}$ to find a rotation matrix about the z-axis, $R_{x}(\phi)$, such that

$$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = R_z(\phi) \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix} $$

Doing so yields:

$$ R_z(\phi) = \begin{bmatrix} \cos \phi & -\sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

We can now convert the position of the pivot and the mass from the rotating frame to the inertial frame. For the pivot (in the rotating frame):

$$ \vec{r'}_{Pivot} = \begin{bmatrix} w_1 \\ 0 \\ h_1 \end{bmatrix} $$

Therefore, the pivot position in the inertial frame can be expressed as: $$ \vec{r}_{Pivot} = \begin{bmatrix} w_1 \cos(\Omega t) \ w_1 \sin(\Omega t) \ h_1 \end{bmatrix} $$

The position of the mass the rotating frame relative to the pivot can be expressed as:

$$ \vec{r'}_{Mass, rel} = \begin{bmatrix} L \sin(\theta) \\ 0 \\ -L \cos(\theta) \end{bmatrix} $$

Where $\theta$ is the angle in the $x'- z'$ plane between the pendulum and the verticle $z'$ axis. Therefore, the position of the mass can be expressed in the inertial frame as:

$$ \vec{r}_{Mass, rel} = \begin{bmatrix} L \sin(\theta) \cos(\Omega t) \\ L \sin(\theta) \sin(\Omega t) \\ -L \cos(\theta) \end{bmatrix} $$

The absolute position of the mass in the inertial frame can then be expressed as:

$$ \vec{r}_{Mass} = \vec{r}_{Pivot} + \vec{r}_{Mass, rel} = \begin{bmatrix} \cos(\Omega t)\left[w_1 + L \sin \theta (t)\right] \\ \sin(\Omega t)\left[w_1 + L \sin \theta (t)\right] \\ h_1 - L \cos \theta (t) \end{bmatrix} $$

Therefore:

$$ x(t) = \cos(\Omega t)\left[w_1 + L \sin \theta (t)\right] $$

$$ y(t) = \sin(\Omega t)\left[w_1 + L \sin \theta (t)\right] $$

$$ z(t) = h_1 - L \cos \theta (t) $$

Taking the time-derrivative of each allows us to find the speed of the mass, which can then be used to find the kinetic energy of the system:

$$ T = \frac{1}{2} m \left[L^2 \dot{\theta}(t)^2 + \Omega^2 (w_1 + L \sin \theta (t)^2) \right] $$

The potential energy of the system can also be expressed as:

$$ V = mg \left(h_1 - L \cos \theta (t) \right) $$

We can now write the Lagrangian as:

$$ L = T - V = \frac{1}{2} m \left[L^2 \dot{\theta}(t)^2 + \Omega^2 (w_1 + L \sin \theta (t)^2) \right] - mg \left(h_1 - L \cos \theta (t) \right) $$

Finally we can used the Euler-Lagrange equation, $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0 $, to find the equation of motion for the system:

$$ \ddot{\theta}(t) = \frac{\Omega^2}{L} \left(w_1 + L \sin \theta (t) \right) \cos \theta (t) - \frac{g}{L} \sin \theta (t) $$

Please note, a full derrivation of the equation above can be found in the project01_team06_notes.pdf file in the notes directory of the repository.

Now, in order to program this system in Julia, we must decompose the second-order ODE above into two first-order ODEs. We can do that by defining:

$$ \begin{aligned} u_1 = \theta \\ u_2 = \dot{\theta} \end{aligned} $$

Therefore, we have:

$$ \begin{aligned} \dot{u}_1 &= u_2 \\ \dot{u}_2 &= \frac{\Omega^2}{L} \left(w_1 + L \sin \theta (t) \right) \cos \theta (t) - \frac{g}{L} \sin \theta (t) \end{aligned} $$

Finally, we must determine a low and high speed for $\Omega$. Given that the natural frequency of the pendulum is

$$ \omega_o = \sqrt{\frac{g}{L}} = \sqrt{\frac{9.81}{0.15}} \approx 8.08 : \text{rad/s} $$

Therefore, we can choose a low speed of $ \Omega_s = 1 : \text{rad/s}$ and a high speed of $ \Omega_f = 12 : \text{rad/s}$. We can also evaluate the cases where the pendulum starts at rest, $\theta_o = 0$, and where the pendulum stars with a small displacement, $\theta_o = 0.05 : \text{rad}$.