# Ex 1.21 The dumbbell, 3

Structure and Interpretation of Classical Mechanics

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c. Make a change of coordinates to a coordinate system with center of mass coordinates $x_{cm}$, $y_{cm}$, angle $\theta$, distance between the particles $c$, and tension force $F$. Write the Lagrangian in these coordinates, and write the Lagrange equations.

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[guess] \displaystyle{ \begin{aligned} y_{cm} &= \frac{m_0 y_0 + m_1 y_1}{m_0 + m_1} \\ x_{cm} &= \frac{m_0 x_0 + m_1 x_1}{m_0 + m_1} \\ \end{aligned}} \displaystyle{ \begin{aligned} y_{cm} &= \frac{m_0 y_0 + m_1 (y_0 + c(t) \sin \theta)}{m_0 + m_1} \\ x_{cm} &= \frac{m_0 x_0 + m_1 (x_0 + c(t) \cos \theta)}{m_0 + m_1} \\ \end{aligned}}

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Let $\displaystyle{M = m_0 + m_1}$. \displaystyle{ \begin{aligned} y_{cm} &= \frac{M y_0 + m_1 c(t) \sin \theta}{M} \\ x_{cm} &= \frac{M x_0 + m_1 c(t) \cos \theta}{M} \\ \end{aligned}} \displaystyle{ \begin{aligned} y_0 &= y_{cm} - \frac{m_1}{M} c(t) \sin \theta \\ x_0 &= x_{cm} - \frac{m_1}{M} c(t) \cos \theta \\ \end{aligned}} \displaystyle{ \begin{aligned} y_1 &= y_{cm} + \frac{m_0}{M} c(t) \sin \theta \\ x_1 &= x_{cm} + \frac{m_0}{M} c(t) \cos \theta \\ \end{aligned}}

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The Lagrangian \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] \\ &= \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) - \frac{F}{2l} \left[ (c(t))^2 - l^2 \right] \\ \end{aligned}}

. \displaystyle{ \begin{aligned} \dot y_0 &= \dot y_{cm} - \frac{m_1}{M} \dot c(t) \sin \theta - \frac{m_1}{M} c(t) \dot \theta \cos \theta\\ \dot x_0 &= \dot x_{cm} - \frac{m_1}{M} \dot c(t) \cos \theta - \frac{m_1}{M} c(t) \dot \theta \sin \theta \\ \end{aligned}}

[guess]

— Me@2021-04-17 05:40:46 PM

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