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Homework Statement
I have a question that came up during a mechanics problem. (I put my question towards the end of the post)
A particle slides on a smooth inclined plane whose inclination ##\theta## is increasing at a constant rate ##\omega##. If ##\theta(t=0)=0## at which time the particle starts from rest, find the motion of the particle.
Homework Equations
##\frac{d}{dt}\frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} = 0##
The Attempt at a Solution
Using polar coordinates for the position of the particle the Lagrangian is
##L = \frac{m}{2}\left(\dot r^2 +r^2\dot \theta^2 \right) -mgr\sin \theta##
Inserting this in L.E. we get
##m\ddot r -mr\dot \theta^2 + mg \sin \theta = 0##
Since ##\theta = \omega t## this describes the motion of the particle.
Solving this gives
##r(t) = \frac{g\sin (\omega t)}{2 \omega^2} + r_0\coth \omega t - \frac{g}{2\omega^2}\sinh \omega t##
Now my question is that what if we instead take L.E. in terms of ##\theta## then
##2mr \dot \theta \dot r - mr g \cos \theta = 0 \Longrightarrow \dot r =\frac{g}{2 \omega} \cos \omega t##
with the solution ##r(t) = \frac{g \sin \omega t}{2 \omega^2} + r_0##
This solution is pretty close to the other one but not identical. Is the problem here that ##\theta## isn't a generalised coordinate for this problem so doing this doesn't make any sense and it's just coincidence?