Lagrange mechanics: Pendulum attached to a massless support

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SUMMARY

The discussion focuses on analyzing the motion of a simple pendulum of length b and mass m, which is attached to a massless support moving vertically upward with constant acceleration a. The key equations derived include the potential energy U = mgh and kinetic energy T = (1/2)mv^2. The participant successfully identified the degrees of freedom as vertical and circular, utilizing polar coordinates for analysis. The final position equation for the pendulum is y = (1/2)at^2 + v_0t - bcos(\theta), indicating the integration of the support's acceleration into the pendulum's motion.

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Homework Statement


A simple pendulum of length ##b## and bob with mass ##m## is attached to a massless support
moving vertically upward with constant acceleration ##a##. Determine (a) the
equations of motion and (b) the period for small oscillations.

2. Formulas

##U = mgh##

##T = (1/2)mv^2 ##

##L= T-U##

The Attempt at a Solution



I need help finding the Kinetic and Potential energy in this problem.
My understanding is that the pendulum is attached to something that its moving upwards. With this motion and the oscillation of the pendulum, I only find 2 degrees of freedom: vertical and circular (this one is in polar coordinates).

So far I have:

##U= -(mgy + mgbcos(\theta)## and ##K = (1/2)m\dot y^2 + (1/2)mb^2\dot\theta^2##

I think there is something that I am missing because ##\dot y = at +k## (where ##k## is a constant) if one is to integrate the acceleration of the support(##a##). Also, I am not sure if I should add the velocity vectors to find the total velocity of the pendulum.

Thank you for the help.
 
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If the support has constant acceleration, is its position a degree of freedom?
 
I thought it was until you mentioned... I revised and I think I was able to figure it out. Another thing that I realized is that ##y = (1/2)a t^2 + v_0 t -bcos\theta##

Thank you for pointing that out!
 

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