enter image description hereTwo particles, $P_1$ and $P_2$, of equal masses $m$ are linked by a spring of stiffness $k$ and natural length $a$. They are sliding freely without friction along a horizontal fixed hoop of radius $R$.

I need some help in writing Lagrangian for generalized coordinates $\theta_1$ and $\theta_2$ given as $\theta_1=\frac{\phi_1+\phi_2}{2}$, $\theta_2=\frac{\phi_1-\phi_2}{2}$.

I wrote that position vectors for $P_1$ and $P_2$ are $\vec r_1$ and $vec r_2$, respectively.
$\vec r_1=-R\cos\phi_1 \vec i+R\sin\phi_1\vec j$
$\vec r_2=R\cos\phi_2\vec i+R\sin\phi_2\vec j$
From this
$\dot{\vec r_1}=R\dot\phi_1\sin\phi_1\vec i+R\dot\phi_1\cos\phi_1\vec j$
$\dot{\vec r_2}=-R\dot\phi_2\sin\phi_2\vec i+R\dot\phi_2\cos\phi_2\vec j$

Kinetic energy is

$T=\frac{1}{2}m(\dot{\vec r_1})^2+\frac{1}{2}m(\dot{\vec r_2})^2=\frac{1}{2}mR^2(\dot\phi_1^2+\dot\phi_2^2)=mR^2(\dot\theta_1^2+\dot\theta_2^2)$

I need help about forces that are doing work.
I know that there is a force in the string, $F=ke$ where $e$ is extension, but I don't know how to determine its direction.

I wrote that
$\vec {P_1P_2}=\vec r_2-\vec r_1$
$|\vec{P_1P_2}|=R\sqrt{2(1+cos(\phi_1+\phi_2))}$ and $|\vec {P_1P_2}|=a+e$

This is all I did. I'm not sure if normal reaction does any work.

Any help would be appreciated.

Left particle is $P_1$, right one is $P_2$, I forgot to write it on the picture.


1 Answer 1


In this case, the lagrangian is the difference between kinetic and potential energy, $T-V$. You have found $T$ already, so it remains to find $V$. The ring is horizontal, so the only potential energy comes from the spring. If you can write this spring stretching energy in terms of your generalized coordinates you are done. Finding this energy should be a straightforward geometry problem, as it only depends on the distance between the particles.

  • $\begingroup$ Is potential energy is this case $E_p=\frac{1}{2}ke^2$ where $e=|\vec {P_1P_2}|-a=R\sqrt{2(1+\cos(\phi_1+\phi_2))}-a$? $\endgroup$
    – gov
    Commented Aug 16, 2013 at 15:01
  • 1
    $\begingroup$ yes that is right. $\endgroup$ Commented Aug 19, 2013 at 2:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.