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CAF123
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Homework Statement
Write down the Lagrangian of a simple pendulum in terms of it's angle θ to the vertical suspended from a pivot attached to a moving carriage at constant velocity ##v##. Suppose that the carriage is now moving at a velocity ##v(t)=at## so it is accelerating uniformly. Show that the Lagrangian is equivalent to that of a pendulum in a modified gravitational field and deduce the strength and direction of this field.
Homework Equations
The lagrangian for the pendulum with carriage at constant speed v (1)
The lagrangian for the pendulum with carriage at variable speed v(t). (2)
The Attempt at a Solution
(1) is $$L = \frac{m}{2}\ell^2 \dot{\theta}^2 + mg\ell \cos \theta + \frac{m}{2}v^2 + mv\ell \cos \theta \dot{\theta}$$
For purposes of finding the E.O.M the third term here can be neglected.
(2) is $$L' = \frac{m}{2}\ell^2 \dot{\theta}^2 + mg\ell \cos \theta + \frac{m}{2}(at)^2 + m (at)\ell \cos \theta \dot{\theta}$$
Since ##\theta## is the generalized coordinate, I think I can neglect the third term again. (It will vanish when I take derivatives), but I do not see why physically. The quantity ##at## will have a different value at each point in time so it affects the Lagrangian numerically. Is it because the 'information' that is useful is encoded in the functional form of the Lagrangian through it's dependence on ##\theta##, ##\dot{\theta}## and ##t##? So, in general, does that mean if there is a term in the Lagrangian that is independent of all the generalized coords/ velocities and time, we can neglect it?
As for the question, there is a hint to use the derivative ##\frac{d}{dt} (t \sin \theta)## in the calculation and I can see where that is going, however, I was wondering if there was another way to solve this without requiring this 'trick'.
Many thanks.