Suppose we have a setup like this. In orange are two wooden sticks sort of things, and they are attached to the block of mass $m$(as usual) at a joint which is hinge type something. A similar connection is with spring and with the ground.The block is moving off to the right with velocity $v$. The spring is already extended by a length $l$. I want to find the rate of change of spring potential energy at the instant.

If the block moves towards right then I think the spring is going to move a little bit to the right, not staying perfectly vertical and in that case there would be some sort of angle $\phi$ between force and displacement. But that case is difficult to imagine. Any other ideas/hints?

Thanks in advance.

EDIT: I've decided to solve this thing the Hamiltonian way.(as suggested by Bernhard). Even though I don't know much of it, this is my attempt.enter image description here

I've read here basic intro(I didn't understand much of Wikipedia, as it was too technical). So now I know I have to make a variable $H=T+V$(the hamiltonian) and do something. Now the question is about $\frac{dU}{dt}$ at any time $t$ where at $t=0$, $\gamma=45^o$ (for ease of coordinates).

$$H=T+V=\frac{1}{2}m\dot x^2+\frac{k}{2}(\sqrt{(Rsin\gamma-\frac{R}{\sqrt{2}}-l-l_0)^2+(Rcos\gamma - \frac{R}{\sqrt{2}})^2}-l_0)^2$$

I've assumed sticks, spring as massless. $l_0$ is the natural length of spring.

What should I do ahead? Please try this problem in simplest language, as I don't know if this involves something difficult like tensors etc. I don't know if this is the place to ask something like this, or if this problem can be solved in Hamiltonian at all. Thanks a lot.

  • $\begingroup$ For these kind of problems, I think it is good to first think about the number of degrees of freedom, before continuing. Is the left block fixed? Then I think you have. You can either take two angles, or two position coordinates of the joint. Maybe you want to have a look at: en.wikipedia.org/wiki/Lagrangian_mechanics en.wikipedia.org/wiki/Hamiltonian_mechanics $\endgroup$
    – Bernhard
    Mar 31, 2013 at 7:52
  • $\begingroup$ @Bernhard: I've just passed high school, so I don't think I know those things. If you could elaborate in a simple language using any mechanics, I'll be able to use that in future problems. Thanks. $\endgroup$ Mar 31, 2013 at 8:20
  • $\begingroup$ @AshishGaurav Just ignore the movement of spring towards right, as you have get this problem solved on this instant itself.[When spring is vertical , for a small time $dt$] $\endgroup$
    – ABC
    Mar 31, 2013 at 11:06
  • $\begingroup$ @Bernhard: OK; I've read this. I'm thinking of a solution in the Hamiltonian terms. I thought it this way: the left wooden stick is fixed at one end to the ground, so let that angle be $\gamma$, and the position of block be $x$. Let the origin for this question(in xy plane) be at the fixing point of the left stick. How do I proceed? $\endgroup$ Mar 31, 2013 at 11:40

1 Answer 1


You are very right in your approach. Just ignore the movement of the spring towards right as we have to work at this very instant, where spring is vertical. $$ {F}=\mathbb Ky =\dfrac{dU}{dy}$$ $$\dfrac{dU}{dt}=\dfrac{dU}{dy}\times \dfrac{dy}{dt};$$ $$\dfrac{dx}{dt}=v$$ Constraint in motion:$$dy=dx \times tan\theta$$ And so the answer. $$=>\dfrac{dU}{dt}=\mathbb Klv \ tan\theta;$$

There is no problem in your answer.

  • $\begingroup$ thanks for the confirmation, but now that the discussion has turned to Hamiltonian mechanics, I'll wait for an answer in that way. $\endgroup$ Mar 31, 2013 at 11:33
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    $\begingroup$ @AshishGaurav Well it's very correct and even i would be pleased to see a answer in other variant. :p . You are an Indian , right ? I'm too an Indian , but we are not taught Lagrangian and Hamiltonian Mechanics till high school....... Have you studied them for yourself? $\endgroup$
    – ABC
    Mar 31, 2013 at 11:37
  • $\begingroup$ Yes, that's true, and I have given the link above in comments. You might want to go over and have a look. $\endgroup$ Mar 31, 2013 at 11:39
  • $\begingroup$ @AshishGaurav Still Newton's Laws are best. Without much of mathematics involved. Pure physics at its best..... $\endgroup$
    – ABC
    Mar 31, 2013 at 11:42
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    $\begingroup$ This is not for JEE: I mean the question was a part of JEE assignment, but I am not interested in a classical way solution. Please stop panicking that this thing is going to come into some competitive exam. $\endgroup$ Mar 31, 2013 at 16:50

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