# Spinning 4 connected springs system - Feynman exercises 14.20

189 views   b) Based on the Potential energy stability test the condition to have stable equilibrium is: second derivative < 0.

I quote from Wikipedia: ‘The potential energy is at a local minimum. This is a stable equilibrium. The response to a small perturbation is forces that tend to restore the equilibrium. If more than one stable equilibrium state is possible for a system, any equilibria whose potential energy is higher than the absolute minimum represent metastable states‘.

So the equation should be

$2K - m\omega^2 < 0$

But the answer is based on the second derivative > 0, which is unstable equilibria.

What am I misunderstanding here?

Solutions:

a) $\frac{ML\omega^2}{2K - m\omega^2}$

b) Only if $\frac{M\omega^2}{k}$ < 2

edited Oct 20, 2018
Have you solved part (a)? It is not clear from your notebook that you have. It seems that the condition in (b) is taken from the denominator in (a).
a) Yes, I just solved for $\Delta L$ from the stated equation, but did not get the solution. I know it has to be related to the elastic force as I know centripetal force is correct. I just cutted the following algebra so as to focus on the equilibrium equation. I can upload it if you want though.
b) Yes I agree with you. But my trouble here is the equilibrium equation. I think the overall issue has to be involved in elastic force
Sorry if the previous answer was not completed enough.
As you can see, I got $\sqrt {2}k$ instead of 2k

1 vote

(a) Yes, your mistake is with the elastic force, which you have written as $k\Delta L$. This is the tension $T$ in each of the springs, but it is not the force pulling each mass towards the centre.

There are 2 tension forces $T$ acting on each mass $m$. The component of each in the direction of the centre is $T\cos45^{\circ}=\frac{1}{\sqrt2}T$. Therefore the total elastic force on each mass towards the centre is $\frac{2}{\sqrt2}T=\sqrt2 T$.

Writing $F=ma$ with centripetal acceleration $a=\omega^2x$ we have $$\sqrt2 T=m\omega^2 x$$ $$\sqrt2 k\Delta L=m\omega^2(\frac{L+\Delta L}{\sqrt2})$$ $$2k\Delta L=m\omega^2(L+\Delta L)$$ $$\Delta L=\frac{m\omega^2L}{2k-m\omega^2}$$

(b) The above equation tells us that there is no equilibrium position for $2k \lt m\omega^2$, because $\Delta L$ would then be -ve, so the spring force would be outward - ie there would be no centripetal (inward) force. But it does not tell us if any equilibrium position is stable.

As you found out, to do that we have to examine how the resultant force on the mass changes as we increase/decrease the radius $x$ close to the equilibrium position $x_0$. If the resultant force becomes -ve (inward) for $x \gt x_0$ and +ve (outward) for $x \lt x_0$ then the equilibrium is stable.

Mathematically, we need to examine $\frac{dF}{dx}$ at the equilibrium position $x_0$. If it is $\lt 0$ - ie resultant force decreasing as we move outwards - then the equilibrium position is stable.

In the rotating frame of reference the resultant force $F(x)$ on each mass $m$ is the sum of the outward (+ve) centrifugal force and the elastic central force, which can be inward (-ve) or outward (+ve). Therefore : $$F(x)=m\omega^2 x-2k(x-\frac{L}{\sqrt2})$$ in which I have written the extension $\Delta L$ in terms of $x=\frac{L+\Delta L}{\sqrt2}$, and $F(x)$ is measured in the outward direction, the same as $x$.

Now the question has been somewhat ambiguous about $\omega$, so there are 2 cases to consider. We are not told whether (i) $\omega$ is fixed - eg regulated by an external torque (a governor) which keeps $\omega$ constant regardless of whether the masses move inwards or outwards, or (ii) after $\omega$ is given a fixed value at an equilibrium position the regulator is disconnected - then it is the angular momentum $m\omega x^2$ which is constant.

(i) angular velocity $\omega$ is constant
$$\frac{dF}{dx}=m\omega^2-2k$$ We already know that we must have $2k \gt m\omega^2$ for all equilibrium positions, so in this case $\frac{dF}{dx} \lt 0$ for all equilibrium positions, hence all equilibrium positions are stable.

(ii) angular momentum $m\omega x^2$ is constant

The angular velocity and radius at the equilibrium position are $\omega_0, x_0$. If the mass moves to a different radius $x$ the angular velocity $\omega$ changes but the angular momentum remains constant : $m\omega x^2 = m \omega_0 x_0^2$. Then $m\omega^2 x=\frac{m\omega_0^2 x_0^4}{x^3}$. Therefore : $$F(x)=\frac{m\omega_0^2 x_0^4}{x^3}-2k(x-\frac{L}{\sqrt2})$$ $$\frac{dF}{dx}=-\frac{3m\omega_0^2 x_0^4}{x^4}-2k$$ which is clearly $\lt 0$ for all values of $x$. So in this case also we find that all equilibrium positions are stable.

The only condition for stable equilibrium is $2k \gt m\omega^2$.

answered Sep 15, 2018 by (28,876 points)
edited Sep 18, 2018
Yes you are right, I got the formula for angular momentum wrong. Thank you for pointing that out. I have corrected my answer.
I am afraid another doubt sprang up... Imagine we are just asked for solving b). Firstly we obtain the equilibrium equation $F(dis) = 0$ where dis is distance:

$$2kdis' - M \omega^2 dis = 0$$

The derivative $\frac{d}{ddis} F (dis) = 0$ is:

$$2k - M \omega^2 = 0$$

Which is the equation we wanted. However I have just applied the first derivative and to evaluate the stable equilibrium I need to work with the second derivative. Thus the method I applied must flaw right?
I don't understand why you need the 2nd derivative. And $dis$ in your1st equation is two different quantities $\Delta L$ and $x$.
That is right, I used ' to distinguish both. Me neither, but the method technically requires the second derivative. Please see:

https://en.wikipedia.org/wiki/Mechanical_equilibrium

Section 1) stability
The 2nd derivative is used with potential energy $U$. The 1st derivative of $U$ is force $F$. You need $F=-dU/dx=0$ to find equilibrium and $dF/dx=-d^2U/dx^2<0$ to determine stability.