This problem is from Introduction to Classical Mechanics by David Morin. Specifically it is problem $6.25$ (Spring on a T) in the Lagrangian Mechanics section, which is available on his webpage.

enter image description here

My solution is as follows:

Let $\theta$ be the angle between the short rod an the vertical line and let $x$ be the current length of the spring. Then we have

$r=(l\sin\theta-x\cos\theta, l\cos\theta+x\sin\theta)$ and $\dot{r}=((l\dot{{\theta}}-\dot{x})\cos\theta+x\dot{\theta}\sin\theta,(\dot{x}-l\dot{\theta})\sin\theta+x\dot{\theta}\cos\theta)$

Therefore the Lagrangian is given by $$L=\frac{1}{2}m((l\omega-\dot{x})^2+\omega^2x^2)-\frac{1}{2}kx^2$$

Then the Euler-Lagrange equation for $x$ is $$m\ddot{x}=(m\omega^2-k)x$$

Which gives that $x(t)=Ae^{s_1t}+Be^{s_2t}$ where $s_{1,2}=\pm\sqrt{\frac{m\omega^2-k}{m}}$.

Is this correct?

  • $\begingroup$ David Morin gives solution to questions, I have some of his books. $\endgroup$ – A---B Mar 18 '17 at 19:03
  • $\begingroup$ @A---B I couldn't find the solution for this specific problem in my copy of the book. $\endgroup$ – Si.0788 Mar 18 '17 at 19:04
  • $\begingroup$ Yes sorry this is an exercise question. Solutions are given for problem question. $\endgroup$ – A---B Mar 18 '17 at 19:12
  • $\begingroup$ @A---B My apologies, I should have made this clear. Would you mind checking through my solution? $\endgroup$ – Si.0788 Mar 18 '17 at 19:17
  • $\begingroup$ Although I have studied this chapter but I won't like to go through your solution because I might give a false answer. Sorry for that. $\endgroup$ – A---B Mar 18 '17 at 19:25

I've checked the maths and they seem ok. Further, you can test the solution by other way: analyzing the physical meaning of the solution:

This parameter has to be determining the behavior of the system.


With $\omega>\sqrt{k/m}$ the solution makes $x$ to diverge. It's the case when the frequency is such that the spring is unable to produce a force strong enough to make the mass veer to the center. From the rotating frame we can say that the centrifugal force overcomes the restitution force of the spring.

With $\omega<\sqrt{k/m}$ the parameter is purely imaginary and the solution is oscillatory, with maximum frequency for $\omega=0$, as expected as the force from the spring has not to overcome the centrifugal one.

With $\omega=\sqrt{k/m}$ you've found the special value that the problem says. In this case, the mass is at constant $x$.

So, the physical meaning of the solution says you did it well.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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