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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