Ex 1.21 The dumbbell, 3.4

Structure and Interpretation of Classical Mechanics

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e. Make a Lagrangian (\displaystyle{= T - V}) for the system described with the irredundant generalized coordinates \displaystyle{x_{cm}}, \displaystyle{y_{cm}}, \displaystyle{\theta} and compute the Lagrange equations from this Lagrangian. They should be the same equations as you derived for the same coordinates in part d.

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Eq. 1.95:

\displaystyle{ L'(t;q,c,F; \dot q, \dot c, \dot F)}

\displaystyle{ = \sum_\alpha \frac{1}{2} m_\alpha \left( \partial_0 f_\alpha (t, q, c)   + \partial_1 f_\alpha (t, q,c) \dot q + \partial_2 f_\alpha (t, q, c) \dot c \right)^2   }

\displaystyle{   - V(t, f(t,q,c)) - \sum_{\{ \alpha, \beta | \alpha < \beta, \alpha \leftrightarrow \beta \}}   \frac{F_{\alpha \beta}}{2 l_{\alpha \beta}}   \left[   c_{\alpha \beta}^2 - l_{\alpha \beta}^2  \right]  }

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Eq. 1.97:

\displaystyle{ L''(t,q,\dot q)}

\displaystyle{ = \sum_\alpha \frac{1}{2} m_\alpha \left( \partial_0 f_\alpha (t, q)   + \partial_1 f_\alpha (t, q) \dot q \right)^2 - V(t, f(t,q,l))  }

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71

Consider a function \displaystyle{g} of, say, three arguments, and let \displaystyle{g_0} be a function of two arguments satisfying \displaystyle{g_0(x, y) = g(x, y, 0)}.

Then \displaystyle{(\partial_0 g_0)(x, y) = (\partial_0 g)(x, y, 0)}.

The substitution of a value in an argument commutes with the taking of the partial derivative with respect to a different argument. In deriving the Lagrange equations for \displaystyle{q} we can set \displaystyle{c = l} and \displaystyle{  \dot c = 0} in the Lagrangian, but we cannot do this in deriving the Lagrange equations associated with \displaystyle{c}, because we have to take derivatives with respect to those arguments.

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[guess]

\displaystyle{ M = m_1 + m_2 }

\displaystyle{ \frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2} }

\displaystyle{ V = 0 }

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\displaystyle{ T }

\displaystyle{ = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 }

\displaystyle{ = \frac{1}{2} (m_0 + m_1) (\dot x_{CM}^2 + \dot y_{CM}^2 ) + \frac{1}{2} \left( m_1 r_1^2 + m_2 r_2^2 \right) \dot \theta^2 }

\displaystyle{ = \frac{1}{2} (m_0 + m_1) (\dot x_{CM}^2 + \dot y_{CM}^2 ) + \frac{1}{2} \left[ m_1 \left( \frac{m_2 l}{m_1 + m_2} \right)^2 + m_2 \left( \frac{m_1 l}{m_1 + m_2} \right)^2 \right] \dot \theta^2 }

\displaystyle{ = \frac{1}{2} (m_0 + m_1) (\dot x_{CM}^2 + \dot y_{CM}^2 ) + \frac{1}{2} m_1 m_2 \left( \frac{1}{m_1 + m_2} \right)^2   ( m_2 + m_1 ) l^2 \dot \theta^2 }

\displaystyle{ = \frac{1}{2} (m_0 + m_1) (\dot x_{CM}^2 + \dot y_{CM}^2 ) + \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} l^2 \dot \theta^2 }

\displaystyle{ = \frac{1}{2} (m_0 + m_1) (\dot x_{CM}^2 + \dot y_{CM}^2 ) + \frac{1}{2} \mu l^2 \dot \theta^2 }

[guess]

— Me@2021-11-24 09:58:31 PM

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2021.11.26 Friday (c) All rights reserved by ACHK