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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.

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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?

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  • $\begingroup$ 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
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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.

$s_{1,2}=\pm\sqrt{\frac{m\omega^2-k}{m}}$

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.

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