3

The system is placed on a vertical plane and composed of a homogeneous bar of mass m and length l which can only rotate about its baricenter at the origin. A particle P of mass m can move along the bar. Two springs are shown with constant $k>0$ The spring at B stays always vertical. The other spring connects the particle with the origin $O$. Using as lagrangian coordinates the angle $\theta$ and the distance $OP=s$ along the bar. $-l/2 \leq s \leq l/2$ Determine the Lagrange equations enter image description here

MY SOLUTION:

The potential of the system is: $ U(\theta, s) = mgs \cos \theta − \frac 18kl^2 \cos^2 \theta − \frac 12 ks^2 + C$

(Note :in my course the relation between potential and force is with a + sign, for instance $\dfrac {\partial U}{\partial \theta} =F_{\theta}$, so the Langrangian is L=T+U)

The kinetic energy of the system is

$T= \dfrac 1{24}ml^2 \dot\theta^2 +\dfrac 12m(\dot s^2+s^2\dot \theta^2)$

The Lagrange equations I found are:

$\dfrac 1{12}ml^2 \ddot \theta+2ms\dot\theta\dot s +m s^2\ddot \theta=-mgs\sin \theta -\dfrac 14 kl^2\sin\theta\cos \theta$

$m\ddot s -ms\dot\theta^2=mg\cos\theta-ks$

Instead the answer from my lecture notes is the one in the picture below. Did I make some mistake or is the lecture notes' answer wrong? ( I found the same answer as the notes for $U $ and $T$ )

enter image description here

some_math_guy
  • 3,230
  • I see two points of disagreement. The first is the coefficient of $1/6$ in their Lagrange equations, and on that one I'm inclined to agree with you (assuming that their statement of $T$ agrees with yours). However, the term $-ms\dot{\theta}^2$ in your Lagrange equations seems spurious: there's no minus signs in your $T$, after all. – Semiclassical Jan 18 '21 at 00:22
  • @Semiclassical $\dfrac{d}{dt}(\dfrac{\partial L}{\partial \dot s})-\dfrac{\partial L}{\partial s}=\dfrac{d}{dt}(\dfrac{\partial T}{\partial \dot s})-\dfrac{\partial (T+U)}{\partial s}=\dfrac{d}{dt}(\dfrac{\partial T}{\partial \dot s})-\dfrac{\partial T}{\partial s}-\dfrac{\partial U}{\partial s}$ .The - sign comes from the term $-\dfrac{\partial T}{\partial s}=-ms\dot\theta^2$ – some_math_guy Jan 18 '21 at 00:36
  • You're quite right. I'd neglected that this term arises from the $\partial_s$ rather than $\partial_{\dot s}$. – Semiclassical Jan 18 '21 at 00:40
  • @Semiclassical There is a third point of disagreement on the third term of the first Lagrange equation: a dot – some_math_guy Jan 18 '21 at 00:41
  • On that one I strongly agree with yours, on the grounds that $m\dot{s}^2\ddot{\theta}$ would have units of kg.m^2/s^4 rather than joules = kg.m^2/s^2. – Semiclassical Jan 18 '21 at 00:43

0 Answers0