1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Object sliding on inclined plane whose angle increases

  1. Nov 3, 2016 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    ##\frac{d}{dt}\frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} = 0##

    3. 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?
     
  2. jcsd
  3. Nov 4, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In your first part, ##r## only is the generalized coordinate.
    In the second part you phantasize: 'LE in terms of ##\theta##' means ##r## and ##\dot r## are givens ?
    And you don't solve for ##\theta## but you impose ##\theta= \omega t## ?
     
  4. Nov 6, 2016 #3
    Thanks! That's the confirmation I was looking for!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted