Spring Energy System: Finding Total Energy with User-Selected Position (x,y)

[member 428835]

Homework Statement

The picture is too tough to draw online, so I've attached a picture of it. The illustrated mechanism shows two springs both with a known spring constant $k$ and rest length of $2R$. One end of the top spring is fixed above a wheel and the other end is attached to the wheel that may rotate but not move. The other spring also has one end attached to the wheel, while the other end of this spring is at some user-selected distance $(x,y)$.

I am trying to find the total energy given some location $(x,y)$.

Homework Equations

Spring energy $k(L-2R)^2/2$ where $L$ is the length of the spring.

The Attempt at a Solution

Energy $E$ is $$E = k(L_1-2R)^2/2+k(L_2-2R)^2/2$$ So all I need to do to express energy as a function of $x,y$ is write $L_1$ and $L_2$ in terms of $x,y$. To find $L_1$ let's use law of cosines:
$$L_1^2=(3R)^3+R^2-2\cdot3R\cdot R\cos\theta$$ I could find the length of $L_2$ the same way: instead of using $\theta$ I would use $\pi-\theta$. One side of the triangle has $L_2$, another has length $R$, and the last has length $\sqrt{x^2+y^2}$. But this is where I run into problems. Specifically, how to I express $\theta$ as a function of $x,y$?

Mathematically, $\theta$ does not depend on $x,y$. But physically it does, so there must be some physical law I'm not accounting for. Any ideas?

I should say, $\theta$ is the state variable and that $x,y$ are the independent parameters; does that imply it is independent on $x,y$ and I need not bother finding a relation?

 

[TSny]

Yes, $x$ and $y$ are parameters that are independent of the angle $\theta$ shown in your picture. Express $L_2$ in terms of $x, y, R$ and $\theta$.

Your expressions for $E$ and $L_1^2$ look correct.

 

[member 428835]

Thank you very much!

 

[member 428835]

Actually, TSny are you sure I don't need to express $\theta$ as a function of $x,y$? $\theta$ is definitely a function of $x,y$; do you know how I could find out what this function is?

I ask because I'm then asked to find the surface of equilibrium solutions and then plot a few bifurcation diagrams.

 

[TSny]

Uh oh. Maybe I'm not interpreting the problem correctly. I thought that values of $x$ and $y$ are first selected and then you are asked to find the PE of the system as a function of the angle $\theta$, where you could imagine reaching in and rotating the wheel to select any value of $\theta$ that you want (while keeping $x$ and $y$ fixed). In that case, $\theta$ is an independent variable and the potential energy is a function of $\theta, x$ and $y$: $U(\theta, x, y)$.

But, maybe you are meant to find the PE only for the equilibrium position of the wheel. In that case, the equilibrium value of $\theta$ can be expressed in terms of $x$ and $y$. So, the equilibrium potential energy would be a function only of $x$ and $y$: $U_{eq}(x, y)$. (But to find the equilibrium value of $\theta$ for chosen $x$ and $y$, you could use the PE function as defined in the previous paragraph, in principle anyway.)

So I could be misleading you. And I'm not sure how the surface of equilibrium solutions is defined. Finally, I know almost nothing about bifurcation. If I had known that this is where this problem was heading, I would probably not have replied. But I look forward to learning something by seeing what develops here.

Hopefully, others will chime in and set you on the right track.

 

[Nidum]

There is :

a balance of forces at the point of joining of the two springs .

a relationship between spring extension and applied force .

 

[member 428835]

Thanks for both you responses. I think TSny actually has the right idea. So moving down the path of choosing an $x,y$ and letting $\theta$ vary and seeing how potential energy reacts, I wrote the potential energy $U$ as
$$U=\frac{k}{2}(L_1-2R)^2+\frac{k}{2}(L_2-2R)^2\\
L_1=R\sqrt{10-6\cos\theta}\\
L_2=\sqrt{R^2+x^2+(y+R)^2+2R\sqrt{x^2+(y+R)^2}\cos\theta}$$ but when I input this into mathematica

k=1
R=1
U[x_, y_, t_] := k ((R^2 + x^2 + (y + R)^2 + 2 R (x^2 + (y + R)^2)^(1/2) Cos[t])^(1/2) - 2 R)^2/2 + k (R (10 - 6 Cos[t])^(1/2) - 2 R)^2/2

and look at values, say $x,y=1,1$ I get an energy equilibrium always at $\theta=\pi$, which I know is only true when $x=0$. I've uploaded a picture for your consideration. Any idea why this graph is coming out wrong (symmetric about $\theta=\pi$)?

 

[TSny]

The expression for L₂ doesn't look correct. For example, if you let x = R , y = -R, and θ = π/2, what should you get for L₂?

There will probably be positions of θ corresponding to stable equilibrium as well as unstable equilibrium for given x and y.

 

[member 428835]

The expression for L₂ doesn't look correct. For example, if you let x = R , y = -R, and θ = π/2, what should you get for L₂?

There will probably be positions of θ corresponding to stable equilibrium as well as unstable equilibrium for given x and y.

Good catch! So I'm thinking to find $L_2$ we still use law of cosines:
$$L^2=R^2+x^2+(y+R)^2+R\sqrt{x^2+(y+R)^2}\cos(\gamma)$$ where $\gamma$ is the angle opposite of $L_2$. Since I know everything but $\gamma$, I'll find it by making its two adjacent sides vectors, and invoking the definition of the dot product. The first vector would be (length doesn't matter) $\langle\sin\theta,-\cos\theta\rangle$ and the second would be $\langle x,y+R \rangle$. Now use the definition of the dot product:
$$\langle x,y+R \rangle\cdot\langle\sin\theta,-\cos\theta\rangle = \sqrt{x^2+(y+R)^2}\cos\gamma\implies\\
\cos\gamma = \frac{x\sin\theta-(y+R)\cos\theta}{\sqrt{x^2+(y+R)^2}}$$
This implies $$L^2=R^2+x^2+(y+R)^2+R\sqrt{x^2+(y+R)^2}\left[ \frac{x\sin\theta-(y+R)\cos\theta}{\sqrt{x^2+(y+R)^2}} \right]$$
Then I would substitute this into the potential energy equation. Do you agree?

 

[TSny]

I don't think that's right. (x, y) are the coordinates of one end of the spring. What are the coordinates of the other end of the spring (in terms of R and θ)?

Then use the usual formula for the distance between two points in terms of the coordinates of the points.

 

[member 428835]

Sorry, I made a mistake in post 9! Specifically, I accidentally computed the dot product wrong, but I just now edited it to a correct form. It simplifies below:
$$L^2=R^2+x^2+(y+R)^2+R(x\sin\theta-(y+R)\cos\theta)$$
Do you now agree?

 

[member 428835]

What are the coordinates of the other end of the spring (in terms of R and θ)?

This is much easier, huh? Ok, so we have $(x,y)$ and $(R\sin\theta,R(1-\cos\theta))$. Then $$L=\sqrt{(R(1-\cos\theta)-y)^2+(R\sin\theta-x)^2}$$
Do you agree?

 

[TSny]

Ok, so we have $(x,y)$ and $(R\sin\theta,R(1-\cos\theta))$. Then $$L=\sqrt{(R(1-\cos\theta)-y)^2+(R\sin\theta-x)^2}$$
Do you agree?

Yes, almost. I think you have a sign error in the y coordinate of $(R\sin\theta,R(1-\cos\theta))$.

 

[member 428835]

Shoot, dumb algebra mistakes. Ok, so we have $(x,y)$ and $(R\sin\theta,R(-1-\cos\theta))$. Then $$L=\sqrt{(R(1+\cos\theta)+y)^2+(R\sin\theta-x)^2}$$
hahaha took waaaaaay too long, but thanks for all your help!

 

[TSny]

Shoot, dumb algebra mistakes. Ok, so we have $(x,y)$ and $(R\sin\theta,R(-1-\cos\theta))$. Then $$L=\sqrt{(R(1+\cos\theta)+y)^2+(R\sin\theta-x)^2}$$
hahaha took waaaaaay too long, but thanks for all your help!

That looks good.

 

[Chestermiller]

Shouldn't the elastic strain energy pass through a minimum at the equilibrium value of $\theta$?

 

[member 428835]

Shouldn't the elastic strain energy pass through a minimum at the equilibrium value of $\theta$?

Yep, so $dp/d\theta = 0$ gives the optimum $\theta$ values to substitute back into $P$ for a surface of equilibria, right?

 

[Chestermiller]

Yep, so $dp/d\theta = 0$ gives the optimum $\theta$ values to substitute back into $P$ for a surface of equilibria, right?

I think so. Try solving the problem this way, and see if it matches the value of theta obtained from the balance of moments. It would certainly be easier to do the problem this way.

 

[member 428835]

I think so. Try solving the problem this way, and see if it matches the value of theta obtained from the balance of moments. It would certainly be easier to do the problem this way.

But $dP/d\theta$ gives an implicit solution for $\theta(x,y)$. Then I'm unsure how to explicitly express $P(x,y)$.

 

[Nidum]

The solution to the bifurcation part of this problem may be load path dependent .

Referring to the lower end of the lower spring : The final position coordinates do not give enough information . You need to start with coordinates at a position where the tension of both springs is zero and then define a path through successive later positions .

The solution to the basic equilibrium problem with lower spring end at specific coordinates is just simple statics .