每日彙整: 2018-04-22

樹莓派、樹莓派之學習、樹莓派之教育

STEM 隨筆︰古典力學︰運動學【四‧E】

俗話說︰千里來龍，在此結穴。

且讓我們借 A 參考系，用 \theta 角為變數，將

Nonminimal Coordinates Pendulum

In this example we demonstrate the use of the functionality provided in mechanics for deriving the equations of motion (EOM) for a pendulum with a nonminimal set of coordinates. As the pendulum is a one degree of freedom system, it can be described using one coordinate and one speed (the pendulum angle, and the angular velocity respectively). Choosing instead to describe the system using the x and y coordinates of the mass results in a need for constraints. The system is shown below:

../../../../_images/pendulum_nonmin.svg

The system will be modeled using both Kane’s and Lagrange’s methods, and the resulting EOM linearized. While this is a simple problem, it should illustrate the use of the linearization methods in the presence of constraints.

 

文本改寫一番

 

如是當能品味異同也！

撞日趁興何不一鼓作氣玩轉

Pendulum on a movable support

Sketch of the situation with definition of the coordinates (click to enlarge)

Consider a pendulum of mass m and length ℓ, which is attached to a support with mass M, which can move along a line in the x-direction. Let x be the coordinate along the line of the support, and let us denote the position of the pendulum by the angle θ from the vertical. The coordinates and velocity components of the pendulum bob are

\displaystyle {\begin{array}{rll}&x_{\mathrm {pend} }=x+\ell \sin \theta &\quad \Rightarrow \quad {\dot {x}}_{\mathrm {pend} }={\dot {x}}+\ell {\dot {\theta }}\cos \theta \\&y_{\mathrm {pend} }=-\ell \cos \theta &\quad \Rightarrow \quad {\dot {y}}_{\mathrm {pend} }=\ell {\dot {\theta }}\sin \theta \end{array}}

The generalized coordinates can be taken to be x and θ. The kinetic energy of the system is then
\displaystyle T={\frac {1}{2}}M{\dot {x}}^{2}+{\frac {1}{2}}m\left({\dot {x}}_{\mathrm {pend} }^{2}+{\dot {y}}_{\mathrm {pend} }^{2}\right)
and the potential energy is
\displaystyle V=mgy_{\mathrm {pend} }
giving the Lagrangian
\displaystyle {\begin{array}{rcl}L&=&T-V\\&=&{\frac {1}{2}}M{\dot {x}}^{2}+{\frac {1}{2}}m\left[\left({\dot {x}}+\ell {\dot {\theta }}\cos \theta \right)^{2}+\left(\ell {\dot {\theta }}\sin \theta \right)^{2}\right]+mg\ell \cos \theta \\&=&{\frac {1}{2}}\left(M+m\right){\dot {x}}^{2}+m{\dot {x}}\ell {\dot {\theta }}\cos \theta +{\frac {1}{2}}m\ell ^{2}{\dot {\theta }}^{2}+mg\ell \cos \theta \end{array}}
Since x is absent from the Lagrangian, it is a cyclic coordinate. The conserved momentum is
\displaystyle p_{x}={\frac {\partial L}{\partial {\dot {x}}}}=(M+m){\dot {x}}+m\ell {\dot {\theta }}\cos \theta \,.
and the Lagrange equation for the support coordinate x is
\displaystyle (M+m){\ddot {x}}+m\ell {\ddot {\theta }}\cos \theta -m\ell {\dot {\theta }}^{2}\sin \theta =0
The Lagrange equation for the angle θ is
\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left[m({\dot {x}}\ell \cos \theta +\ell ^{2}{\dot {\theta }})\right]+m\ell ({\dot {x}}{\dot {\theta }}+g)\sin \theta =0;
and simplifying
\displaystyle {\ddot {\theta }}+{\frac {\ddot {x}}{\ell }}\cos \theta +{\frac {g}{\ell }}\sin \theta =0.
These equations may look quite complicated, but finding them with Newton’s laws would have required carefully identifying all forces, which would have been much more laborious and prone to errors. By considering limit cases, the correctness of this system can be verified: For example, \displaystyle {\ddot {x}}\to 0 should give the equations of motion for a simple pendulum that is at rest in some inertial frame, while \displaystyle {\ddot {\theta } \to 0 should give the equations for a pendulum in a constantly accelerating system, etc. Furthermore, it is trivial to obtain the results numerically, given suitable starting conditions and a chosen time step, by stepping through the results iteratively.

 

乎？