# 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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