Lagrangian mechanics: pivoted bar with a moving support
My first step in the Lagrangian approach is to determine the kinetic energy T and potential energy V of the system.
The potential energy V is due to gravity. With \theta measured from the downward vertical, the vertical position of the center of mass G, relative to its height when \theta=\pi/2, is -L\cos(\theta). Thus, the potential energy is:
V = -mgL\cos(\theta)
To find the kinetic energy, I need the velocity of the center of mass G and the angular velocity of the bar. The horizontal position of G is x_G = s(t) + L\sin(\theta), and its vertical position (measured downwards from the pivot A) is y_G = L\cos(\theta).
By differentiating these with respect to time, I find the velocity components of G:
\begin{aligned} v_{Gx} & = \dot{s} + L\dot{\theta}\cos(\theta) \\ v_{Gy} & = -L\dot{\theta}\sin(\theta) \end{aligned}
The squared speed of G is:
v_G^2 = v_{Gx}^2 + v_{Gy}^2 = \dot{s}^2 + 2\dot{s}L\dot{\theta}\cos(\theta) + L^2\dot{\theta}^2
The kinetic energy T includes the translational energy of the center of mass and the rotational energy about the center of mass:
T = \frac{1}{2}m v_G^2 + \frac{1}{2}I_G \dot{\theta}^2
Substituting v_G^2, I get:
T = \frac{1}{2}m\left(\dot{s}^2 + 2\dot{s}L\dot{\theta}\cos(\theta) + L^2\dot{\theta}^2\right) + \frac{1}{2}I_G \dot{\theta}^2
Which I can rewrite as:
T = \frac{1}{2}m\dot{s}^2 + mL\dot{s}\dot{\theta}\cos(\theta) + \frac{1}{2}\left(mL^2 + I_G\right)\dot{\theta}^2
The Lagrangian L is T - V:
\mathcal L = \frac{1}{2}m\dot{s}^2 + mL\dot{s}\dot{\theta}\cos(\theta) + \frac{1}{2}(mL^2 + I_G)\dot{\theta}^2 + mgL\cos(\theta)
Since s(t) and thus \dot{s}(t) and \ddot{s}(t) are known functions of time, the only generalized coordinate describing the system’s configuration is \theta(t). I use the Euler-Lagrange equation for \theta:
\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial \mathcal L}{\partial \dot{\theta}}\right) - \frac{\partial \mathcal L}{\partial \theta} = 0
I calculate the necessary partial derivatives:
\frac{\partial \mathcal L}{\partial \dot{\theta}} = (mL^2 + I_G)\dot{\theta} + mL\dot{s}\cos(\theta)
Its time derivative is:
\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial \mathcal L}{\partial \dot{\theta}}\right) = (mL^2 + I_G)\ddot{\theta} + mL\ddot{s}\cos(\theta) - mL\dot{s}\dot{\theta}\sin(\theta)
And the other partial derivative:
\frac{\partial \mathcal L}{\partial \theta} = -mL\dot{s}\dot{\theta}\sin(\theta) - mgL\sin(\theta)
Substituting these into the Euler-Lagrange equation, the terms -mL\dot{s}\dot{\theta}\sin(\theta) cancel out, leading to the equation of motion:
(mL^2 + I_G)\ddot{\theta} + mL\ddot{s}\cos(\theta) + mgL\sin(\theta) = 0
This equation describes the angular motion \theta(t) of the bar. The term mL\ddot{s}\cos(\theta) represents the effect of the horizontal acceleration of the pivot point A on the bar’s rotation. The term (mL^2 + I_G) is the moment of inertia of the bar about the pivot A, often denoted I_A. This example shows how systems with moving constraints or externally driven components can be handled systematically using the Lagrangian method.
For more insights into this topic, you can find the details here.