# rate of change of spring potential energy $\frac{dU}{dt}$

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?

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.

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.

• 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 Mar 31, 2013 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. Mar 31, 2013 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, 2013 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? Mar 31, 2013 at 11:40

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. $$=>\dfrac{dU}{dt}=\mathbb Klv \ tan\theta;$$