Kinetic terms of the double pendulum

Calculus Level pending

The kinectic part of the Lagrangian for a double pendulum is

\(T=\quad \frac { 1 }{ 2 } ({ m }_{ 1 }+{ m }_{ 2 }){ l }_{ 1 }^{ 2 }{ \dot { \theta } }^{ 2 }+\quad \frac { 1 }{ 2 } { m }_{ 2 }({ l }_{ 2 }^{ 2 }(\dot { \theta } +\dot { \varphi } { ) }^{ 2 } +\quad 2{ l }_{ 1 }{ l }_{ 2 }\dot { \theta } (\dot { \theta } +\dot { \varphi } )\cos { \varphi } )\)

What is \(\frac { d }{ dt } \frac { \partial T }{ \partial \dot { \theta } } \) ?

1)

\(\quad { m }_{ 1 }{ l }_{ 1 }^{ 2 }\dot { \theta } +\quad { m }_{ 2 }{ l }_{ 1 }^{ 2 }\dot { \theta } +\quad { m }_{ 2 }{ l }_{ 2 }^{ 2 }(\dot { \theta } +\dot { \varphi } )+\quad { m }_{ 2 }{ l }_{ 1 }{ l }_{ 2 }(2\dot { \theta } +\dot { \varphi } )\cos { \varphi }\)

2) \(\quad { m }_{ 1 }{ l }_{ 1 }^{ 2 }\ddot { \theta } + { m }_{ 2 }{ l }_{ 1 }^{ 2 }\ddot { \theta } + { m }_{ 2 }{ l }_{ 2 }^{ 2 }(\ddot { \theta } +\ddot { \varphi } )+{ m }_{ 2 }{ l }_{ 1 }{ l }_{ 2 }((2\ddot { \theta } +\ddot { \varphi } )\cos { \varphi } -\dot { \varphi } (2\dot { \theta } +\dot { \varphi } )\sin { \varphi } )\)

3)

\(\quad { m }_{ 1 }{ l }_{ 1 }^{ 2 }\ddot { \theta } +\quad { m }_{ 2 }{ l }_{ 2 }^{ 2 }(\ddot { \theta } +\ddot { \varphi } )+\quad { m }_{ 2 }{ l }_{ 1 }{ l }_{ 2 }(2\dot { \theta } +\dot { \varphi } )\cos { \dot { \varphi } } \)

4)

\(\quad ({ m }_{ 1 }+{ m }_{ 2 }){ l }_{ 1 }^{ 2 }\ddot { \theta } +\quad { m }_{ 2 }{ l }_{ 2 }^{ 2 }(\ddot { \theta } +\ddot { \varphi } { ) }+\quad { m }_{ 2 }{ l }_{ 1 }{ l }_{ 2 }(2\ddot { \theta } +\ddot { \varphi } )\cos { \dot { \varphi } } \)

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