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import numpy as np
import matplotlib.pyplot as plt

plt.style.use('fivethirtyeight')

Building Quiz 07

A bead moves along a planar sinusoidal track. It does not rotate as it moves along the path, so the generalized coordinates are its x-position y-position, $$\mathbf{q} = [x,~y]$$. There are two constraints on the motion of the bead,

• $y = \sin x$

• $x = 10 t$, where $t$ is time in seconds.

1. write the constraint equations, $\mathbf{C}(\mathbf{q},~t)$.

2. write out the Jacobian of the constraints, $\mathbf{C_q}$

3. Invert the Jacobian to solve for $\dot{x}~and~\dot{y}$, using $\mathbf{C_q \dot{q}} = -\mathbf{C}_t$

Numerical approach

def C(q, t):
    C = np.zeros(2)
    C[0] = np.array([q[1] - np.sin(q[0])])
    C[1] = np.array(q[0] - 10*t)
    return C

def Cq(q, t):
    Cq = np.zeros((2, 2))
    Cq[0, 0] = -np.cos(q[0])
    Cq[0, 1] = 1
    Cq[1, 0] = 1
    return Cq

Cq(np.array([0,0]), 1)

array([[-1.,  1.],
       [ 1.,  0.]])

SymPy approach

import sympy

sympy.var('x, y, t')
q = sympy.Matrix([x, y])
q

$$\begin{split}\displaystyle \left[\begin{matrix}x\\y\end{matrix}\right]\end{split}$$

C = sympy.Matrix([y - sympy.sin(x), x - 10*t])
C

$$\begin{split}\displaystyle \left[\begin{matrix}y - \sin{\left(x \right)}\\- 10 t + x\end{matrix}\right]\end{split}$$

C.jacobian(q)

$$\begin{split}\displaystyle \left[\begin{matrix}- \cos{\left(x \right)} & 1\\1 & 0\end{matrix}\right]\end{split}$$

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Building Quiz_06