Balancing a Pencil

Does quantum mechanics matter at everyday scales? It would be very surprising if quantum effects were to be manifest in a macroscopic system. This has been claimed for the problem of balancing a pencil on its tip. But the behaviour of a tipping pencil can be explained in purely classical terms.

A pencil balanced on its point. Is there a trick? Yes: see below.

A pencil balanced on its point. Is there a trick? See below.

Modelling a balanced pencil

We have all tried to balance a pencil on its sharp end. Although this is essentially impossible, you may often notice someone at a tedious committee meeting trying to do it.

We can model the pencil as an inverted simple pendulum. For equilibrium, the bob must be positioned exactly over the fixed point, and the angular momentum must vanish. For a small swing angle, the pendulum is approximately isochronous, but for large swings the period increases with amplitude. Denoting the deflection from the downward vertical by ${\theta}$ , the dynamical equation is

$\displaystyle m\ell\ddot\theta = - mg\sin\theta$

where ${m}$ is the mass of the bob, ${\ell}$ is the length of the rod and ${g}$ is the acceleration due to gravity. Defining the frequency ${\omega = \sqrt{g/\ell}}$ , this is

$\displaystyle \ddot\theta + \omega^2\sin\theta = 0 \,.$

For small amplitude motion, we can replace the sine by its argument and the solution is simple harmonic motion with period ${T = 2\pi/\omega}$ , independent of the amplitude.

For finite motions, the equation is harder to solve but it is a standard problem in classical dynamics. The solution may be expressed as

$\displaystyle \sin\theta/2 = \sin\theta_0/2\ \mathrm{sn}[\omega(t-t_0),k]$

where ${\theta_0}$ is the amplitude, ${k=\sin\theta_0/2}$ and ${\mathrm{sn}}$ is the Jacobian elliptic function (see a text on classical mechanics, e.g. [3], for details). The period is

$\displaystyle T = \frac{4K}{\omega}$

where ${K=K(k)}$ is the complete elliptic integral of the first kind,

$\displaystyle K(k) = \int_0^{\pi/2}\frac{d\phi}{\sqrt{1-k^2\sin\phi}} \,.$

The function ${K}$ varies very slowly with ${k}$ so that, for moderate amplitudes, the period depends only weakly on the value of ${\theta_0}$ . However, ${K}$ is unbounded as ${\theta_0 \rightarrow \pi}$ and the period becomes infinitely long in this limit.

The table below shows the time to drop, $T/4 = \frac{K}{\omega}$ , for a range of initial angles. The time increases as ${\theta_0 \rightarrow \pi}$ , but the growth is very slow.

Drop time for a bencil initially stationary with angle theta. The frequency is omega=10.

Drop time for a pencil initially stationary with angle theta. The frequency is omega=10.

Asymptotic Estimate of the Period

There is a detailed analysis in [2], available online. We give only the main results here.

We can use an asympototic expression for the integral ${K(k)}$ . The Digital Library of Mathematical Functions gives an expansion (DLMF, Eqn. 19.12.1.) valid for ${0 < k < 1}$ :

$\displaystyle K(k) \approx \log\left(\frac{4}{k^\prime}\right) = \log\left(4\sec\theta_0/2\right) \,.$

Here ${k^\prime}$ is defined by ${k^2+k^{\prime 2}=1}$ . Given that $k=\sin\theta_0/2$ , we have ${k^\prime=\cos\theta_0/2}$ .

Since ${K(0)=\pi/2}$ we can improve the approximation by adding a simple factor linear in ${k}$ . Defining $\kappa=(\pi/2)/\log 4$ , we write

$\displaystyle K(k) \approx [\kappa+(1-\kappa)\sin\theta_0/2]\log\left(4\sec\theta_0/2\right) \,. \ \ \ \ \ \ \ \ \ (1)$

This makes the approximation exact at ${\theta_0=0}$ . The values of ${K(k)}$ and the approximate expressions are given in the table below. The expression (1) is within about 3% of the true value throughout the range ${0 < k < 1}$ .

${Complete elliptic integral and approximations for a range of values of {\theta}$

Exact period and approximations for a range of values of theta (omega=10) .

Using typical values for the parameters, we find that, even with a tiny deviation of the initial position from vertical — less than the width of an atom — the bob will reach the bottom point within just a few seconds (see [2] for details).

We can also get an estimate of the time-scale without using elliptic integrals. If ${\psi}$ is the angle between the rod and the upward vertical, then the motion is governed by

$\displaystyle m\ell\ddot\psi = mg\sin\psi \,.$

We find that, even with disturbances on an atomic scale, the time for the pendulum to swing to the bottom is just a few seconds.

Conclusion

In a classical system, there are no constraints on the initial conditions. There is an equilibrium solution, corresponding to a perfectly balanced pencil. It is effectively unattainable, but theoretically possible. In reality, there are always small errors in setting the initial conditions. What the analysis of the inverted pendulum shows is that, however tiny the initial displacement, the pencil will drop within a few seconds. This is a consequence of the asymptotic properties of the complete elliptic function, which tends logarithmically to infinity as ${k\rightarrow 1}$ . Persistence of balance is impossible in practice.

Easy to balance a brush handle. Very hard to balance a pencil (this one is held up by a paper clip).

Sources

[1] DLMF: NIST {Digital Library of Mathematical Functions}. Release 1.0.8 of 2014-04-25. http://dlmf.nist.gov/19.12.E1.

[2] Lynch, Peter, 2014: The Not-so-simple Pendulum: Balancing a Pencil on its Point. ArXiV: Online, 5 June 2014: http://arxiv.org/abs/1406.1125

[3] Synge, J. L. and B. A. Griffith, 1959: {Principles of Mechanics}. McGraw-Hill, Third Edition, 552pp.

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