# A conical pendulum. A textbook claims one can start the circular motion in an unintuitive way

A textbook claims that if you hang a mass from a string that's attached to a rod (so that the string is parallel with the vertical axis) and you spin the rod fast enough, you can put the mass into circular motion by slightly perturbing it. I don't believe this. If it's true, please help me believe it. If it's false, please help me fight the textbook. (I think the mass will swing like a simple pendulum no matter how fast you spin the rod.)

Here's a description of the situation, given by the textbook: Here's a diagram of the situation, given by the textbook: Here's the claim the textbook makes (note 1: "g" represents the local gravitational acceleration; note 2: the term "solution" is with respect to the task of finding the angle alpha): Here's the plot related to the claim: http://i.stack.imgur.com/ALJ08.png

A way to get the mass into circular motion, that I find intuitive, is to make an angle between the string and the vertical axis (which is what the textbook means by "slightly perturb") and then throw the mass (such that some of the velocity is tangential to the circular path). The claim by the textbook is that if you spin the rod fast enough, you only have to do the first part of the way I find intuitive ("slightly perturb the mass") and don't have to do the second part of the way I find intuitive ("throw the mass").

• That's the nice thing about energy arguments. They work where forces are not intuitive or not known. To work with force, you need the path the mass will follow. All you need for energy is the beginning and end states. The energy of the circular solution is lower at high rotation rates. Atomic orbitals are solved with energy because the path of an electron is not even well defined. – mmesser314 Aug 14 '15 at 14:01

This type is problem is best analyzed using Lagrangian mechanics, since the effect of the rotating pendulum can be absorbed easily using rotating coordinates. The solution boils down to fixed point analysis.

First start with $$z=\ell(1-\cos\theta)\, ,\qquad y=\ell\sin\theta\cos(\omega t)\, ,\qquad x=\ell\sin\theta \sin(\omega t)\, .$$ With the kinetic energy $T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)$ and the potential $V=mgz$ we get, after obvious manipulations: $$L=T-V=\frac{m}{2}(\ell^2\dot{\theta}^2+\omega^2\ell^2\sin^2\theta)+mg\ell \cos\theta$$ with a constant term ignored. Although the coordinates depend explicitly on time, the Lagrangian itself does not, so the resulting Hamiltonian will be conserved. However, this is not a natural system so the Hamiltonian will not be the total energy of the system, but just some constant without any obvious physical interpretation.

The exact equation of motion is thus found to be $$m\ell\ddot{\theta}= m\omega^2\ell^2\sin\theta\cos\theta-mg\ell\sin\theta\, .\tag{1}$$

Using the momentum $$p=\frac{\partial L}{\partial \dot{\theta}}-m\ell^2\dot{\theta}\qquad \Rightarrow \qquad \dot{\theta}=\frac{p}{m\ell^2}\, ,$$ we find the Hamiltonian $$H=\dot{\theta}p-L= \frac{p^2}{m\ell^2}-\frac{1}{2}m\omega^2\ell^2\sin^2\theta+mg\ell(1-\cos\theta)\, .$$ The fixed points of the system are given by $$\left(\frac{\partial H}{\partial p},\frac{\partial H}{\partial\theta} \right)=(0,0)$$ which immediately implies $p=0$. With this, we can investigate $$\frac{\partial H}{\partial\theta}\vert_{p=0}= mg\ell\sin\theta \left(1-\frac{\omega^2}{g/\ell}\cos\theta\right)=0\, .$$ If $g/\ell > \omega^2$, then only $\theta=0,\pi$ are solutions. If, on the other hand, $g/\ell < \omega^2$ there is an additional solution given by $\cos\theta= \frac{g/\ell}{\omega^2}$. Thus there exists a critical frequency $\omega_c$ defined by $g/\ell=\omega_c^2$ above which there is a simple bifurcation with two new fixed points developing.

• Near the fixed point $(p,\theta)=(0,0)$ we have $\theta=\epsilon \Delta \theta$, $\dot{\theta}=\epsilon \Delta \dot{\theta}$ so, inserting this into the exact equation of motion (1), expanding in powers of $\epsilon$ and keeping only terms of order ${\cal O}(\epsilon)$ gives the linearized equation of motion $$\epsilon\ell\Delta\ddot{\theta}=\epsilon\frac{g}{\ell}\left(\frac{\omega^2}{g/\ell}-1\right)\, .$$ If $\omega<\omega_c$, then $\frac{\omega^2}{g/\ell}-1$ is negative and we recover the EOM of a simple harmonic oscillator. The origin is then a stable fixed point If $\omega>\omega_c$ then the sign of $\frac{\omega^2}{g/\ell}-1$ is positive and the solutions are in terms of hyperbolic functions: the fixed point is unstable.
• One easily shows by expanding about $(p,\theta)=(0,\pi)$ that this is always an unstable fixed point.
• Lastly, near $(p,\theta)=(0,\arccos(\frac{g/\ell}{\omega^2}))$. This can only occur when $\omega>\omega_c$. We must now expand the exact EOM of Eq.(1) using $\theta = \theta_0 +\epsilon\Delta \theta$ with $\theta_0=\arccos(\frac{g/\ell}{\omega^2})$. After straightforward manipulations one obtains, to ${\cal O}(\epsilon)$: $$\epsilon\ell \Delta\ddot{\theta}=-\epsilon\Delta\theta \omega^2\left(1-\frac{g^2\ell^2}{\omega^4}\right)$$ (with a sweet cancellation of the term of order ${\cal O}(1)$.) This is the equation of motion of a harmonic oscillator with frequency $\Omega^2=\omega^2\left(1-\frac{g^2\ell^2}{\omega^4}\right)$.

This clearly shows that, if you rotate you pendulum fast enough, i.e so that $\omega>\omega_c$, the pendulum can have small oscillation about a fixed point at some angle given by $\arccos(\frac{g/\ell}{\omega^2})$.

The analysis is supported by looking at level curves of $H$ (it is conserved, even if it is not the total energy).

Below the critical frequency we have the normal phase curves of a harmonic oscillator with stable fixed point at the origin and unstable in the inverted position. Above the critical frequency, the stable fixed point at $\theta_0$ is clearly visible, as is the unstable fixed point at the origin. So to answer the question highlighted by the OP: if you spin the pendulum fast enough, and start it near $\theta=0$, i.e. near the downward vertical, it will NOT oscillate near this position.

Edit: We set one up at work, using a good quality skateboard bearing with very little damping in the plane of the pendulum (with no rotation the amplitude dampens by 90% over 400 oscillations). The stiff string (actually a piece of aluminum wire) is required for otherwise the pendulum will not remain in the same (rotating) plane and the motion becomes closer to a spinning spherical pendulum.

Even with a good bearing, there is always a bit of sideways wiggle of the bearing during rotation (similar to the wiggle of trains on a track). This dampens the amplitude of small oscillations about the fixed point (at least this is what we think is happening as qualitatively the $Q$ value for the pure oscillation motion of the pendulum is not enough to account for the rapid damping of the oscillation at high $\omega$).

Also for short penduli length the frequency $\omega$ has to be large, which makes it hard to "see" small oscillations during the fast rotation about the axis. Presumably one would need to keep track of the position of the bob using some sensors.

• Nice theoretical work. I think it assumes a 'stiff' string, so isnt it a bit doubtful that this Hamiltonian applies as a good approximation to the real world problem? (But most likely the book author had this model in mind) – lalala Mar 23 '17 at 9:48
• @lalala see edit. – ZeroTheHero Mar 23 '17 at 12:42
• Can you provide a video of your experiment? – Matt Kleinsmith Mar 24 '17 at 0:34
• @MatthewKleinsmith Good question. Not right now but I'll look into it. I know we had plan to "refine" the setup over the summer and I'll inquire. The most recent setup involved attaching the pivot at the end of a motor normally used to control the speed of a drill. (as an aside I think someone still have the Mathematica code used to simulate the trajectories.) – ZeroTheHero Mar 24 '17 at 0:39

I think you may be forgetting that unless the bob undergoes circular motion, the string will swiftly become twisted. The string must either then break under its own torsion, or, through its torsion, impart a torque that makes the bob begin to spin. The bob's angular speed must "catch" up to the string, meaning that its time averaged angular speed must equal that of the string, otherwise, again, the torsion in the string will break it.

From this argument it follows that the only steady state behaviors are those where the ball's angular velocity in the horizontal plane (about the vertical axis) is the same as that of the string.

If everything hung down vertically, the bob must begin to spin about the vertical axis, with the string hanging straight downwards. This is the meaning of the solution $\cos\alpha = 0$.

The point that the book is making is that this steady state behavior is unstable: we must think about what happens if a small draught kicks the ball sideways, by an arbitrarily small amount. Now this is a highly nontrivial problem.

Actually, I don't thing that the method of Klein Gordon's Answer, where we are asked to put ourselves in a noninertial frame fixed to the ball, will answer this question definitively. There is nothing wrong with this method: what KleinGordon is saying is that if the ball shifts a little off-axis, there is a centrifugal force that tends to amplify the shift, hence the instability. This statement does establish a plausibility for the instability, but the centrifugal inertial force is not the only effect at work here. A sideways kick sends a torsonal / helical wave up the string, and a complicated set of wave reflexions on the string will follow. These would need to be analysed in detail to find the linearized equations of motion for the system and thus establish instability (through the positive real part frequency eigenvalues).

Having said this, in the face of KleinGordon's argument I find it intuitively hard to accept anything other than instability. Experimentally one finds that the equilibrium is unstable. But I seriously wonder whether the book author has rigorously established this fact other than through KleinGordon's argument which, although physically compelling, is not mathematically rigorous. A full system analysis would be an interesting addition to the answers to this question.

Once positive real part system eigenvalues are established, we don't even need a draught to set off an instability. Nor do we even need so much as the sideways light pressure from even a lone photon. Indeed we only need the Heisenberg uncertainty principle to show that the bob's ultimate quantum nature will set it on a path away from the unstable equilibrium: see Floris's wonderful answer here.

• I will have to think about this. I believe the string is itself rigid so the bob is rotating even when the string is pointing straight down. But it's an unstable equilibrium, which should not be too hard to prove. Not the stuff of cell phones though. – Floris Aug 11 '15 at 0:52
• @Floris Yes, of course, a two piece system of rigid rod and ball would be much easier to analyse. Brain asleep this morning! – Selene Routley Aug 11 '15 at 0:58
• Rather belatedly I have attempted to bring some mathematics to the problem. See if my recent answer gives some new insights. – Floris Aug 12 '15 at 22:45

When you consider a rotating frame of reference, you can assign a "rotational potential" to the centrifugal force.

If centrifugal force is

$$F_c = m\omega^2 r$$

then the centrifugal potential is

$$V_c = -\int F_c dr = -\frac12 m \omega^2 r^2\tag1$$

The potential energy as a function of deflection $\theta$ is given by

$$V_g = - m\;g\;\ell\;(1-\cos\theta)\tag2$$

For small angles $\theta$, we have that $r = \ell\sin\theta \approx \ell\theta$. This means that (1) can be written as

$$V_c=-\frac12 m \omega^2 \ell^2 \theta^2$$

Further, since $\cos\theta\approx1-\frac12\theta^2$ (2) can be written as

$$V_g = \frac12 m\;g\ell\;\theta^2$$

The sum of these is the effective potential well that the bob is moving in:

$$V_{eff} = \frac12m\ell(g-\omega^2\ell)\theta^2$$ If this potential well has a minimum, the bob is stable; if it has a maximum, the bob is unstable. This follows from the sign of the expression in front of $\theta^2$, and thus we know the motion is stable if

$$g-\omega^2\ell \lt 0\\ \omega \gt \sqrt{\frac{g}{\ell}}$$

Note that in order to find the new steady state, you can't use the small angle approximation - you need to use the full $\cos\theta$ term to find the angle at which the conical pendulum will settle.

The assumption in the above is that the rotation of the rod will translate to the mass - in other words, that if there is a slight displacement of the mass it will rotate at $\omega$. Now if the bob is rotating while it is hanging vertically, that must mean that the rotation of the rod is being transmitted to the bob - which can only happen if the string does indeed transfer torque.

Consider the situation in the rotating frame of the rod, which is a non-inertial reference frame. Then the textbook's comment, which I believe is correct, might become more intuitively clear.

You might also try setting up the experiment yourself!

• I tried with a shoe lace (1.4 meters long) as the string, a wrench socket as the mass, and a power drill as the spinning rod. I didn't observe the phenomenon. Here's a picture of the equipment: i.stack.imgur.com/rQ01L.jpg – Matt Kleinsmith Feb 20 '15 at 3:10
• omega = sqrt(g / l) = sqrt(9.8 / 1.4) ~= 2.6/s. 2.6 radians per second? (I'm trying to see find out if the power drill was spinning fast enough, with respect to the omega = sqrt(g / l) equation). – Matt Kleinsmith Feb 20 '15 at 3:17
• Neat! Your calculation seems correct to me. I would have thought the drill would be spinning faster than that, so I am a bit surprised. I'll have to ponder this some more. Also, ideally, your experiment should use an angular velocity that is not too much different from the pendulum frequency, so that the equilibrium angle is not so large that you enter the nonlinear regime. – kleingordon Feb 20 '15 at 3:30
• I don't have experience with considering situations from non-inertial reference frames. I'll give it a go: Let the mass be hanging down and spinning, with respect to an inertial reference frame. From the rod's perspective, the mass looks like it's not spinning. When the mass is perturbed..., I don't know what to visualize. If the textbook is incorrect, and we assume the mass is swinging like a simple pendulum from an inertial frame, the mass will move in a spiral path? If the textbook is correct, the mass will either look stationary or the mass will move in a circular path? – Matt Kleinsmith Feb 20 '15 at 3:35
• What I had in mind was the centrifugal force, the notorious "fictitious" force that appears in equations of motion when written down in a rotating frame. My thought was that if the centrifugal force were greater than the gravitational restoring force, then the slightly perturbed rotating pendulum would continue to move out. It's possible I'm overlooking something simple, though, especially considering that your experiment isn't confirming this idea. Maybe try again with a shorter string, and slightly larger initial perturbation? – kleingordon Feb 20 '15 at 3:55

The Hamiltonian of this system, as given by @ZeroTheHero, is $$H(\theta,p)=\frac{p^2}{ml^2}-\frac{1}{2}m\omega^2l^2sin^2\theta+mgl(1-cos\theta)$$ The process how to determine the system stays stable or not qualitatively is that write down the fixed ponts or critical points firstly and then substitute these points into Eigenvalues of the coefficient matrix of the system and finally the behavior of the system will be determined depending on the attribute of Eigenvalues.

The fixed ponts or critical points are given by $$(\frac{\partial H}{\partial p},-\frac{\partial H}{\partial \theta})=(0,0)$$ which corresponds to two equations $$\frac{2p}{ml^2}=0,\quad -mglsin\theta+m\omega^2l^2cos\theta sin\theta=0$$ The first set of solutions is $$(\theta_0,p_0)=(k\pi,0),\quad with\quad k\in Z$$ and if $\theta\neq k\pi$, we obtain the second set of solutions $$(\theta_0,p_0)=(arccos\frac{g/l}{\omega^2},0)$$ The coefficient matrix of the system is $$M=\begin{bmatrix} \frac{\partial^2H}{\partial p\partial\theta}&\frac{\partial^2H}{\partial^2 p}\\ -\frac{\partial^2H}{\partial^2 \theta}&-\frac{\partial^2H}{\partial\theta\partial p} \end{bmatrix}=\begin{bmatrix} 0&\frac{2}{ml^2}\\ -mglcos\theta+m\omega^2l^2cos2\theta&0 \end{bmatrix}$$ The Eigenvalues are $$\lambda_1=-\sqrt{2}\sqrt{-\omega_c^2cos\theta+\omega^2cos2\theta},\quad\lambda_2=-\lambda_1$$ in which $\omega_c^2=\frac{g}{l}$.

Case1. investigating the first set of fixed points $(\theta_0,p_0)=(k\pi,0),\quad with\quad k\in Z$:

The corresponding Eigenvalues are $\lambda_1=-\sqrt{2}\sqrt{-\omega_c^2(-1)^k+\omega^2}=-\lambda_2$

1. If $k=2n,n\in Z$, then $\lambda_1=-\sqrt{2}\sqrt{-\omega_c^2+\omega^2}$
1. If $\omega>\omega_c$, then $\lambda_1=-\lambda_2\in R$, the fixed point is a saddle (unstable).
2. If $\omega<\omega_c$, then $\lambda_1=-\lambda_2\in C$, the fixed point is a center (stable).
2. if $k=2n+1,n\in Z$, then $\lambda_1=-\sqrt{2}\sqrt{\omega_c^2+\omega^2}$, which is always a $Real$ number, so the fixed point is a saddle (unstable).

Case2. investigating the second set of fixed points $(\theta_0,p_0)=(arccos\frac{g/l}{\omega^2},0)$:

The corresponding Eigenvalues are $\lambda_1=-\sqrt{2}\omega\sqrt{\frac{\omega_c^4}{\omega^4}-1}=-\lambda_2$

1. If $\omega>\omega_c$, then $\lambda_1=-\lambda_2\in C$, the fixed point is a center (stable).
2. If $\omega<\omega_c$, then $\lambda_1=-\lambda_2\in R$, the fixed point is a saddle (unstable).

Conclusion: if $\omega>\omega_c$, only $(\theta,p)=(n\pi,0),n\in Z$ is the unstable fixed point; and if $\omega<\omega_c$, only $(\theta,p)=(2n\pi,0),n\in Z$ is the stable fixed point; 