Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
    
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 –  Bernhard Mar 31 '13 at 7:52
    
@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. –  Ashish Gaurav Mar 31 '13 at 8:20
    
@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 $dt$] –  ABC Mar 31 '13 at 11:06
    
@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? –  Ashish Gaurav Mar 31 '13 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.

share|improve this answer
    
thanks for the confirmation, but now that the discussion has turned to Hamiltonian mechanics, I'll wait for an answer in that way. –  Ashish Gaurav Mar 31 '13 at 11:33
1  
@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? –  ABC Mar 31 '13 at 11:37
    
Yes, that's true, and I have given the link above in comments. You might want to go over and have a look. –  Ashish Gaurav Mar 31 '13 at 11:39
    
@AshishGaurav Still Newton's Laws are best. Without much of mathematics involved. Pure physics at its best..... –  ABC Mar 31 '13 at 11:42
1  
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. –  Ashish Gaurav Mar 31 '13 at 16:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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