Ex 1.21 The dumbbell, 2

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Structure and Interpretation of Classical Mechanics

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Notice that these forces of constraint are proportional to the normal to the constraint surface at each instant and thus do no work for motions that obey the constraint.

b. Write the formal Lagrangian

\displaystyle{L(t;x_0, y_0, x_1, y_1, F; \dot x_0, \dot y_0, \dot x_1, \dot y_1, \dot F)}

such that Lagrange’s equations will yield the Newton’s equations you derived in part a.

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[guess]

Eq (1.99):

\displaystyle{L(t;x,\lambda; \dot x, \dot \lambda) = \sum_\alpha \frac{1}{2} m_\alpha \dot{\mathbf{x}}_\alpha^2 - V(t,x) + \lambda \phi(t,x)}

\displaystyle{ \begin{aligned} L &= \frac{1}{2} m_0 (\dot x_0^2 + \dot y_0^2) + \frac{1}{2} m_1 (\dot x_1^2 + \dot y_1^2) + \lambda \left[ (x_1(t) - x_0(t))^2 + (y_1(t) - y_0(t))^2 - l^2 \right] \\ \end{aligned}}

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The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} \frac{d}{dt} \frac{\partial L}{\partial \dot x_0} - \frac{\partial L}{\partial x_0} &= 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot x_1} - \frac{\partial L}{\partial x_1} &= 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot y_0} - \frac{\partial L}{\partial y_0} &= 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot y_1} - \frac{\partial L}{\partial y_1} &= 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot \lambda} - \frac{\partial L}{\partial \lambda} &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} \frac{d}{dt} \left( m_0 \dot x_0 \right) + \left( \lambda 2 (x_1 - x_0)\right) &= 0 \\ \frac{d}{dt} \left( m_0 \dot x_1 \right) - \left( \lambda 2 (x_1 - x_0)\right) &= 0 \\ \frac{d}{dt} \left( m_0 \dot y_0 \right) + \left( \lambda 2 (y_1 - y_0)\right) &= 0 \\ \frac{d}{dt} \left( m_0 \dot y_1 \right) - \left( \lambda 2 (y_1 - y_0)\right) &= 0 \\ \frac{d}{dt} \left( 0 \right) - \left[ (x_1(t) - x_0(t))^2 + (y_1(t) - y_0(t))^2 - l^2 \right] &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} m_0 \ddot x_0 &= - 2 \lambda (x_1 - x_0) \\ m_0 \ddot x_1 &= 2 \lambda (x_1 - x_0) \\ m_0 \ddot y_0 &= - 2 \lambda (y_1 - y_0) \\ m_0 \ddot y_1 &= 2 \lambda (y_1 - y_0) \\ (x_1(t) - x_0(t))^2 + (y_1(t) - y_0(t))^2 - l^2 &= 0 \\ \end{aligned}}

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Let \displaystyle{\lambda = - \frac{F}{2l}}:

\displaystyle{ \begin{aligned} m_0 \ddot x_0 &= F \frac{(x_1 - x_0)}{l} \\ m_0 \ddot x_1 &= -F \frac{(x_1 - x_0)}{l} \\ m_0 \ddot y_0 &= F \frac{(y_1 - y_0)}{l} \\ m_0 \ddot y_1 &= -F \frac{(y_1 - y_0)}{l} \\ \end{aligned}}

\displaystyle{ \begin{aligned} m_0 \ddot y_0 &= F \sin \theta \\ m_0 \ddot x_0 &= F \cos \theta \\ m_0 \ddot y_1 &= -F \sin \theta \\ m_0 \ddot x_1 &= -F \cos \theta \\ \end{aligned}}

[guess]

— Me@2021-04-09 11:18:07 PM

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2021.04.10 Saturday (c) All rights reserved by ACHK