# Circular motion

## A simple one-link manipulator
Consider the simple manipulator below, consisting of a homogenous rod of mass $m=0.1$kg and length $l=1$m. The rod is rotating about the single joint with a constant angular velocity of $\omega=2$rad/s. 

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### What is the centripetal acceleration at the endpoint?



```python
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
```


```python
# Parameters. SI units, of course
m = 0.1
l = 1.0
omega = 2.0

# The formular for centripetal acceleration is
a_c = omega**2 * l
print "The centripetal acceleration at the endpoint is %1.2f m/s^2" % a_c
```

    The centripetal acceleration at the endpoint is 4.00 m/s^2


### What is the centripetal acceleration at the center of mass?


```python
# %load com-acceleration.py
# The center of mass for a homogeneous rod is of course in the middle of the rod
from IPython.display import display, Math, Latex
display(Latex(r"The centripetal acceleration at the center of mass is %1.2f $m/s^2$" % (omega**2*l/2)))
```


The centripetal acceleration at the center of mass is 2.00 $m/s^2$


## Derivation of the acceleration for uniform circular motion
Define the x- and y axis as in the figure below. The angle $\theta$ to be the angle to the x-axis

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We have defined unit vectors in a polar coordinate system attached to the moving link
\begin{align}
e_r &= \begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}\\
e_\theta &= \begin{bmatrix}-\sin\theta\\\cos\theta\end{bmatrix}
\end{align}
With these unit vectors it is easy to express the position $p$ of the endpoint with respect to the origin, which is located at the joint:
$$ p(t) = l e_r(t) = l \begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}. $$

In order to find the velocity of the endpoint we take the time-derivative of the position
\begin{align}
   v(t) &= \frac{d}{dt} p(t) = \left( \frac{d}{dt} l \right) e_r + l \frac{d}{dt} e_r\\
        &= l \frac{d}{dt} \begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}\\
        &= l \frac{d}{d\theta}\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix} \frac{d\theta}{dt}\\
        &= l \omega \begin{bmatrix}-\sin\theta\\\cos\theta\end{bmatrix} = l \omega e_\theta.
\end{align}
We see that the *speed*, the magnitude of the velocity, $|v| = l \omega $ is proportional to both the distance from the center of rotation and to the angular velocity. The velocity vector is tangential to the path of the endpoint, and if $\omega$ is positive, then the direction of the velocity is in the positive direction of $e_\theta$.  

### Exercise: derive the centripetal acceleration by differentiating the velocity
Also: Discuss the magnitude and direction of the centripetal acceleration.


```python
# %load centripetal-acceleration-solution.py
display(Latex("""
The acceleration becomes
\\begin{align}
a(t) &= \\frac{d}{dt} v(t) = \\frac{d}{dt} l \\omega e_\\theta = \\frac{d}{dt} l \\omega \\begin{bmatrix} -\\sin\\theta \\\\ \\cos\\theta \\end{bmatrix} \\\\
     &= \\frac{d}{dt} \left( l \\ omega \\right) e_\\theta + l \\omega \\frac{d}{dt} \\begin{bmatrix} -\\sin\\theta \\\\ \\cos\\theta \\end{bmatrix} \\\\
     &= 0 + l \\omega^2 \\begin{bmatrix} -\\cos\\theta \\\\ -\\sin\\theta \\end{bmatrix} = - l \omega^2 e_r.  
\\end{align}
"""))
```



The acceleration becomes
\begin{align}
a(t) &= \frac{d}{dt} v(t) = \frac{d}{dt} l \omega e_\theta = \frac{d}{dt} l \omega \begin{bmatrix} -\sin\theta \\ \cos\theta \end{bmatrix} \\
     &= \frac{d}{dt} \left( l \ omega \right) e_\theta + l \omega \frac{d}{dt} \begin{bmatrix} -\sin\theta \\ \cos\theta \end{bmatrix} \\
     &= 0 + l \omega^2 \begin{bmatrix} -\cos\theta \\ -\sin\theta \end{bmatrix} = - l \omega^2 e_r.  
\end{align}




```python

```


```python

```

f952bd671399cfca99a9adb8379f121e11bba48f
ipynb
Jupyter Notebook
introduction/notebooks/circular-motion.ipynb
alfkjartan/robotica-aplicada
515265059792d0e8586e2f450bd1af34a47fb963
MIT
introduction/notebooks/circular-motion.ipynb
alfkjartan/robotica-aplicada
515265059792d0e8586e2f450bd1af34a47fb963
MIT
introduction/notebooks/circular-motion.ipynb
alfkjartan/robotica-aplicada
515265059792d0e8586e2f450bd1af34a47fb963
MIT
Qwen/Qwen-72B
1. YES
2. YES