# Ex 1.21 The dumbbell, 3.2

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 $\displaystyle{x_{cm}}$, $\displaystyle{y_{cm}}$, angle $\displaystyle{\theta}$, distance between the particles $\displaystyle{c}$, and tension force $\displaystyle{F}$. Write the Lagrangian in these coordinates, and write the Lagrange equations.

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[guess]

\displaystyle{ \begin{aligned} m_0 \ddot y_0 &= F \sin \theta \\ m_0 \ddot x_0 &= F \cos \theta \\ m_1 \ddot y_1 &= -F \sin \theta \\ m_1 \ddot x_1 &= -F \cos \theta \\ \end{aligned}}

\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} \ddot y_{cm} &= \frac{F \sin \theta - F \sin \theta}{m_0 + m_1} = 0 \\ \ddot x_{cm} &= \frac{F \cos \theta - F \cos \theta}{m_0 + m_1} = 0 \\ \end{aligned}}

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\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 \\ 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}}

\displaystyle{ \begin{aligned} x_1 - x_0 &= \frac{m_0}{M} c(t) \cos \theta + \frac{m_1}{M} c(t) \cos \theta \\ &= c(t) \cos \theta \\ \end{aligned}}

\displaystyle{ \begin{aligned} \dot x_1 - \dot x_0 &= \dot c(t) \cos \theta - c(t) \dot \theta \sin \theta \\ \end{aligned}}

\displaystyle{ \begin{aligned} \ddot x_1 - \ddot x_0 &= \ddot c(t) \cos \theta - \dot c(t) \dot \theta \sin \theta - \dot c(t) \dot \theta \sin \theta - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}

\displaystyle{ \begin{aligned} y_1 - y_0 &= c(t) \sin \theta \\ \dot y_1 - \dot y_0 &= \dot c(t) \sin \theta + c(t) \dot \theta \cos \theta \\ \ddot y_1 - \ddot y_0 &= \ddot c(t) \sin \theta + \dot c(t) \dot \theta \cos \theta + \dot c(t) \dot \theta \cos \theta + c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta \\ \end{aligned}}

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\displaystyle{ \begin{aligned} m_0 \ddot y_0 &= F \sin \theta \\ m_0 \ddot x_0 &= F \cos \theta \\ m_1 \ddot y_1 &= -F \sin \theta \\ m_1 \ddot x_1 &= -F \cos \theta \\ \end{aligned}}

When $\displaystyle{\dot c(t) = 0}$ and $\displaystyle{\ddot c(t) = 0}$,

\displaystyle{ \begin{aligned} \ddot x_1 - \ddot x_0 &= - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta \\ - \left( \frac{1}{m_1} + \frac{1}{m_1} \right) F \cos \theta &= - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}

\displaystyle{ \begin{aligned} \ddot y_1 - \ddot y_0 &= ... \\ - \left( \frac{1}{m_1} + \frac{1}{m_0} \right) F \sin \theta &= c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta \\ \end{aligned}}

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\displaystyle{ \begin{aligned} \tan \theta &= \frac{ c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta }{- c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta} \\ &= \frac{ \ddot \theta - \dot \theta^2 \tan \theta }{- \ddot \theta \tan \theta - \dot \theta^2} \\ \end{aligned}}

\displaystyle{ \begin{aligned} \tan \theta \left( - \ddot \theta \tan \theta - \dot \theta^2 \right) &= \ddot \theta - \dot \theta^2 \tan \theta \\ &... \\ 0 &= \ddot \theta (1 + \tan^2 \theta) \\ \ddot \theta &= 0 \\ \end{aligned}}

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\displaystyle{ \begin{aligned} - \left( \frac{1}{m_1} + \frac{1}{m_0} \right) F \sin \theta &= c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta \\ - \left( \frac{1}{m_1} + \frac{1}{m_1} \right) F \cos \theta &= - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}

Let $\displaystyle{\frac{1}{\mu} = \left( \frac{1}{m_1} + \frac{1}{m_1} \right)}$ and since $\displaystyle{\ddot \theta = 0}$,

\displaystyle{ \begin{aligned} - \frac{1}{\mu} F \sin \theta &= - c(t) \dot \theta^2 \sin \theta \\ - \frac{1}{\mu} F \cos \theta &= - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}

Since $\displaystyle{\sin \theta}$ and $\displaystyle{\cos \theta}$ cannot be both zero at the same time,

\displaystyle{ \begin{aligned} - \frac{1}{\mu} F &= - c(t) \dot \theta^2 \\ \end{aligned}}

Put $\displaystyle{c(t) = l}$,

\displaystyle{ \begin{aligned} \frac{1}{\mu} F &= l \dot \theta^2 \\ \dot \theta^2 &= \frac{1}{l \mu} F \\ \end{aligned}}

[guess]

— Me@2021-08-08 05:41:21 PM

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