README.md

Homework 3

Author: Igor Alentev

Telegram: m8dotpie

Email: [email protected]

Stack:

The work is done within the single Jupyter Notebook. All the simulations were printed directly to the gif files. To replicate the result you should execute the notebook and check generated files called "task1.gif" and "task2.gif".

This is the first homework when I finally started using custom class for simple and fast plotting.

• Python Notebooks
• python3
• numpy
• sympy
• ipympl
• matplotlib

Table of Contents:

• Task 1
  • Solution
    • Geometrical Intuition
    • M reaches A
    • Mechanical Intuition
  • Visual Simulation
    • Components graph
    • Total graph
    • Both graph
• Task 2
  • Solution
  • Visual Simulation
    • Components graph
    • Total graph
    • Both graph

Task 1

Solution

Geometrical intuition

This time the task does not contain any difficult steps in producing the simulation. Sinse $O_1O_2AO$ is parallelogram we can easilly find coordinates of all points. Given that $O_1$ is the origin, we know that $O_2$ is $2R$ on axis $x$. We know that $O$ is rotating around the origin, therefore $A$ is $O$ translated on $x$ axis by $2R$. $M$ is rotating on circumference of the semicircle with origin at $\frac{O + A}{2}$ with the known law of arc length.

Time when $M$ reaches $O$

Angle for the $M$ during it's movement can be calculated as follows:

$$\phi_2 = \frac{s_r}{R}$$

Therefore we have to solve equation $\phi_2 = \pi$. The answer is: $t_r \approx 1.732$

Mechanical intuition

We can split the task on the two parts.

First part (for $t < t_r$)

Point $M$ did not reach point $A$. Then we can easilly calculate relative, transport velocity, relative, corealis and transport accelerations. During my simulation I have taken point $O_s = \frac{O + A}{2}$ as a pole for $M$. It is beneficial since we know that $O_s$ is rotating with the same angle law as the $O$ and $A$ and it is the only motion. Moreover, the distance $O_sM$ is simply the radius $R$. We can easilly derive $\omega_1, \epsilon_1$ from $\phi(t)$ and $\omega_2, \epsilon_2$ from $s_r(t)$. Also we have to mention that our body make translatory motion (angle law for both $O_1O and O_2A$ is the same). There there is no rotational speed and therefore there is no coriolis acceleration:

$$ \vec{V_M^{tr}} = \vec{\omega_1} \times \vec{O_1O} $$

$$ V_M^{rel} = \left[\begin{array}{c} \dot{x}_M\\ \dot{y}_M\\ 0 \end{array}\right] $$

$$ \vec{a_M^{tr}} = \left[\begin{array}{c} \ddot{x_{O_s}}\\ \ddot{y_{O_s}}\\ 0 \end{array}\right] $$

$$ \vec{a_M^{rel}} = \left[\begin{array}{c} \ddot{x_M}\\ \ddot{y_M}\\ 0 \end{array}\right] $$

Second part (for $t \ge t_r$)

In the second part point $M$ reaches $A$ and therefore it has no relative components in velocity and acceleration. As a result there is no relative velocity, coriolis and relative acceleration. Moreover, transport velocity and acceleration is absolutely the same as $O_s$.

Visual Simulation

Components graph

Total graph

Both

Task 2

Solution

This task is relatively simple to simulate as well. Stationary coordinate system origin is at $O_1$. Moving system is at the center of the circle $O_2$. Therefore, we can simply express all the points in the stationary coordinate system. Let's take $O_2$ as the pole for the $M$ as this pole has only rotational movement. If we fix $O_2$ then $M$ is rotating around $O_2$ with a known arc length law.

$$ \vec{V}_{O_2} = \vec{\omega}1 \times \vec{R{O_2}} $$

$$ \vec{V_M^{tr}} = \vec{V_{O_2}} + \vec{\omega}_1 \times (\vec{M} - \vec{O}_2) $$

$$ V_M^{rel} = \left[\begin{array}{c} \dot{x}_M\\ \dot{y}_M\\ 0 \end{array}\right] $$

$$ \vec{a_M^{tr}} = \dot{\vec{V_{O_2}}} + \vec{\epsilon_1} \times (\vec{M} - \vec{O}_s) + \vec{\omega_1} \times (\vec{\omega_1} \times (\vec{M} - \vec{O}_2)) $$

$$ \vec{a_M^{cor}} = 2 \cdot \vec{\omega_1} \times \vec{V_M^{rel}} $$

$$ \vec{a_M^{rel}} = \left[\begin{array}{c} \ddot{x_M}\\ \ddot{y_M}\\ 0 \end{array}\right] $$

Visual Simulation

Components graph

Total graph

Both