Miscellaneous Pendulum Pivot Motions

Tag: LnL_1.5.3

Source:

• Textbook: Landau and Lifshitz, Mechanics, 3rd Edition
• Chapter 1 The Equations of motion
• Section 5 The Lagrangian for a system of particles
• Problem 3: Pendulum with a moving suspension point

In all three cases described below, the point of support moves in the same plane as the pendulum as per a given function of time. The lagrangian of the pendulum is to be found.

Part a)

Description: The pendulum moves uniformally in a vertical circle of radius $a$ and angular velocity $\gamma$

Solution: The coordinates of m can be expressed as: $$ \begin{aligned} x &= a\cos(\gamma t) + l\sin\phi\\ y &= a\sin(\gamma t) - l\cos\phi \end{aligned} $$ (where generalized coordinate, $\phi$, is the angle the string makes with vertical, $\gamma$ is angle with horizontal)

$$\mathcal{L} = T-V$$ where Kinetic and potential energies: $$ \begin{aligned} T &= \tfrac{1}{2}m(\dot{x}^2 + \dot{y}^2) = \tfrac{1}{2}m\bigl(l^2\dot{\phi}^2 + 2al\gamma\sin(\phi-\gamma t)\dot{\phi} + a^2\gamma^2\bigr)\\ V &= mg y = mg\bigl(a\sin(\gamma t) - l\cos\phi\bigr) \end{aligned} $$

We can ignore the constant term $\tfrac{1}{2}ma^2\gamma^2$ in $T$ as it does not affect the equations of motion. Like wise, $mg a\sin(\gamma t)$ in $V$ can be ignored since its contribution to action is constant Let us note that the total derivative of $a l \cos(\phi - \gamma t)$ is: $$ \frac{d}{dt}\bigl[al\cos(\phi-\gamma t)\bigr] = -al\sin(\phi-\gamma t)\dot{\phi} + al\gamma\sin(\phi-\gamma t) $$

Rearraging the terms, the lagrangian can be written as: $$ L = \tfrac{1}{2}m l^{2}\dot{\phi}^{2} + m a l \gamma^{2}\sin(\phi-\gamma t) + m g l \cos\phi- m\gamma\frac{d}{dt}\bigl[al\cos(\phi-\gamma t)\bigr] $$ Since lagrangians differing by a total time derivate lead to the same equations of motion we can ignore the last term. So that the final lagrangian becomes: $$ L = \tfrac{1}{2}m l^2\dot{\phi}^2 + m a l \gamma^2 \sin(\phi-\gamma t) + m g l \cos\phi $$

Part b)

Description: The point of support oscillates horizontally according to the law $x = a\cos(\gamma t)$

Solution: The coordinates of $m$ can be expressed as: $$ \begin{aligned} x &= a\cos(\gamma t) + l\sin\phi\\ y &= -l\cos\phi \end{aligned} $$ (where generalized coordinate, $\phi$, is the angle the string makes with vertical) so that $\dot{x} = -a\gamma\sin(\gamma t) + l\dot{\phi}\cos\phi$ and $\dot{y} = l\dot{\phi}\sin\phi$ $\mathcal{L} = T - V$ where

$$ \begin{aligned} T &= \tfrac{1}{2}m(\dot{x}^2 + \dot{y}^2) = \tfrac{1}{2}m\bigl(l^2\dot{\phi}^2 + a^2\gamma^2\sin^2(\gamma t) - 2 a l \gamma\sin(\gamma t)\dot{\phi}\cos\phi\bigr)\\ V &= m g y = - m g l \cos\phi \end{aligned} $$ We can ignore the constant term $\tfrac{1}{2}m a^2\gamma^2\sin^2(\gamma t)$ in $T$ again. The derivative of $\sin\phi\sin(\gamma t)$ is $$ \frac{d}{dt}\bigl[\sin\phi\sin(\gamma t)\bigr] = \dot{\phi}\cos\phi\sin(\gamma t) + \gamma\sin\phi\cos(\gamma t) $$ substituting this in the lagrangian, we get: $$ L = \tfrac{1}{2} m l^2\dot{\phi}^2 + m a l \gamma^2 \sin\phi\cos(\gamma t) + m g l \cos\phi- m a l \gamma\frac{d}{dt}\bigl[\sin\phi\sin(\gamma t)\bigr] $$

Ignoring the total time derivate term, the final lagrangian becomes: $$ L = \tfrac{1}{2} m l^2\dot{\phi}^2 + m a l \gamma^2 \sin\phi\cos(\gamma t) + m g l \cos\phi $$

Part c)

Description: The point of support oscillates vertically according to the law $y = a\cos(\gamma t)$

Solution: The coordinates of m can be expressed as: $$ \begin{aligned} x &= l\sin\phi\\ y &= a\cos(\gamma t) - l\cos\phi \end{aligned} $$ (where generalized coordinate, $\phi$, is angle string makes with vertical) $L = T - V$ where $$ \begin{aligned} T &= \tfrac{1}{2}m(\dot{x}^2 + \dot{y}^2) = \tfrac{1}{2}m\bigl(l^2\dot{\phi}^2 + a^2\gamma^2\sin^2(\gamma t) - 2 a l \gamma\sin(\gamma t)\dot{\phi}\sin\phi\bigr)\\ V &= m g y = m g\bigl(a\cos(\gamma t) - l\cos\phi\bigr) \end{aligned} $$

We can ignore the constant term $\tfrac{1}{2}m a^2\gamma^2\sin^2(\gamma t)$ in $T$ as it does not lead to variation. Likewise, $m g a\cos(\gamma t)$ in $V$ can be ignored since its contribution to action is constant. The time derivative of $\cos\phi\sin(\gamma t)$ is: $$ \frac{d}{dt}\bigl[\cos\phi\sin(\gamma t)\bigr] = -\dot{\phi}\sin\phi\sin(\gamma t) + \gamma\cos\phi\cos(\gamma t) $$ substituting this in the lagrangian, we get: $$ L = \tfrac{1}{2} m l^2\dot{\phi}^2 + m g l \cos\phi - m a l \gamma^2 \cos\phi\cos(\gamma t) + m a l \gamma\frac{d}{dt}\bigl[\cos\phi\cos(\gamma t)\bigr] $$ Ignoring the total time derivate term, the final lagrangian becomes: $$ L = \tfrac{1}{2} m l^2\dot{\phi}^2 + m g l \cos\phi - m a l \gamma^2 \cos\phi\cos(\gamma t) $$