# Object sliding on inclined plane whose angle increases

1. Nov 3, 2016

### Incand

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. Nov 4, 2016

### BvU

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 $\theta= \omega t$ ?

3. Nov 6, 2016

### Incand

Thanks! That's the confirmation I was looking for!