Pencil pushers

A couple of days ago, Matt Parker posted this video to his Standup Maths YouTube channel.

In it, he and Hugh Hunt, reader in engineering dynamics and vibration at Cambridge, stand a pencil on its end at the edge of a table, give the bottom end of it a horizontal whack, and see what happens. They also calculate what happens… sort of.

See, the thing is, the title of the video is “Will a falling pencil hit the table? We do the maths!” They certainly do some math, but not enough to answer the question in the title. Kind of disappointing. So let’s do a little more math and finish the problem for them. In the following, we’ll be referring to their work, so you really should watch the video before proceeding.

Here’s the setup. A pencil, which we are modeling as a uniform rod, is standing upright at the edge of a table. We’re going to apply a sharp, short duration horizontal force at the bottom of the rod and work out its subsequent motion. The sketch below shows the layout and the coordinate system I want to use in the solution.

Rod setup

The \(g\) represents the acceleration due to gravity. The rod is characterized by its length, \(2a\), and its mass, \(m\). Because the mass is uniformly distributed along its length, the center of mass, \(C\), is halfway up the rod, and we can calculate its moment of inertia about \(C\) as

\[I_C = \frac{1}{3} m a^2\]

(If you’re looking at that formula and your first thought is “Why isn’t that a 12 in the denominator?” join the club. It took me a few minutes to remember that Hugh had defined \(a\) as half the length of the rod, and that’s why there’s a 3 instead of a 12. We’re used to seeing the formula written in terms of the full length

\[I_C = \frac{1}{12} m L^2\]

but since \(L = 2a\),

\[I_C = \frac{1}{12} m (2a)^2 = \frac{1}{12} m (4a^2) = \frac{1}{3} m a^2\]

so the 3 is correct.)

\(A\) is the end that’s initially at the top, and we’ll be tracking its position to find out if hits the table or flies past it.

\(\hat{\:F}\) is the impulse with which we strike the bottom of the rod. It’s shown as acting a little above the bottom, but I drew it that way so the arrow wouldn’t get lost in the top of the table. Impulse is defined as the integral of the force over the duration of its application. We’re going to assume the duration of the impulse is short, \(\Delta t\), and that it ends at time \(t = 0\). Thus, the impulse is

\[\hat{\:F} = \int_{-\Delta t}^0 F(t)\,dt\]

The impulse-momentum equation is essentially just Newton’s second law integrated over time. For this situation, in the horizontal direction, we have

\[\hat{\:F} = m v_0 - m v_{-}\]

where \(v_0\) is the velocity of C at \(t = 0\) (just after the impulse), and \(v_{-}\) is the velocity of C at \(t = -\Delta t\) (just before the impulse). Because the rod starts out stationary, \(v_{-} = 0\) and

\[\hat{\:F} = m v_0\]

There’s a rotational analog to the linear impulse-momentum equation

\[\hat{\:M}_C = \hat{\:F} a = I_C \,\omega_0 - I_C \,\omega_{-}\]

where \(\omega_0\) is the angular velocity of the rod at \(t = 0\) and \(\omega_{-}\) is the angular velocity at \(t = - \Delta t\). Because the rod is stationary before we hit it, \(\omega_{-} = 0\), so

\[\hat{\:F} a = I_C \omega_0 = \frac{1}{3} m a^2 \omega_0\]

There’s some simplification here that the video doesn’t mention. The full definition of angular impulse about \(C\) is

\[\hat{\:M}_C = \int_{- \Delta t}^0 F(t) \, r_{\perp}(t) \, dt\]

where \(r_{\perp}(t)\) is the perpendicular distance from the line of action of \(\hat{\:F}\) to \(C\). It’s because the impulse is assumed to act over a very short length of time that we can take \(r_{\perp}(t) = a\) throughout the impulse. The idea is that impulse is so short, the rod doesn’t have a chance to rotate significantly. And because \(a\) is not a function of \(t\), we can bring it out of the integral, leaving us with \(\hat{\:M}_C = \hat{\:F} a\).

Combining our two impulse-momentum equations to eliminate \(\hat{\:F}\) gives us

\[v_0 = \frac{1}{3} a \omega_0\]

which is exactly what the video got.

Now let’s track the motions of \(C\) and \(A\). After the impulse, the bottom of the rod will be just off the edge of the table, and the only force acting on the rod will be gravity (in the grand tradition of all elementary mechanics problems, we’re ignoring air resistance). So \(C\) will trace out a parabola, with

\[x_C = v_0 \,t = \frac{1}{3} a \,\omega_0\, t\]

and

\[y_C = a - \frac{1}{2} g \,t^2\]

being the parabola’s parametric equations.

With the only force acting through the rod’s center of mass, the angular velocity will remain constant at \(\omega_0\). So

\[\theta = \omega_0 \,t\]

and we can use that to write the equations for the coordinates of \(A\):

\[x_A = x_C - a \sin \omega_0 t = \frac{1}{3} a\, \omega_0\, t - a \sin \omega_0 t\] \[y_A = y_C + a \cos \omega_0 t = a\,(1 + \cos \omega_0 t) - \frac{1}{2} g\, t^2\]

Now, taking a hint from the video, we’re going to consider the conditions at which the tip of the rod, \(A\), is even, horizontally, with the edge of the table. In other words, when \(x_A = 0\).

\[x_A = \frac{1}{3} a\, \omega_0\, t - a \sin \omega_0 t = 0\]

That means

\[\frac{1}{3} a\, \omega_0\, t = a \sin \omega_0 t\]

or

\[\frac{1}{3} \theta = \sin \theta\]

This is the formula they got in the video, but they were looking only at the special case for which the tip of the pencil just barely hits the corner of the table. What we’ve shown is that the same formula works whenever the tip of the pencil passes a vertical line even with the edge of the table. And instead of stopping here, as the video does, we can use the solution to this formula to go further and answer the question in the video’s title: Under what conditions will the pencil hit the table?

The solution of the formula, to an unnecessarily precise seven digits, is \(\theta = 2.278863\), which is equivalent to about 130.57°. What we need to do is consider the value of \(y_A\) when \(\theta\) is equal to this value. There are three conditions of interest:

  1. If \(y_A > 0\) when \(\theta = 2.278863\), then the tip of the pencil will pass above the table and there will be no hit.
  2. If \(y_A = 0\) when \(\theta = 2.278863\), then the tip of the pencil will just graze the corner of the table. This is the boundary between hit and miss and is the condition they focused on in the video.
  3. If \(y_A < 0\) when \(\theta = 2.278863\), then the pencil hits the table.

Rod positions

Yes, \(y_A\) being less than zero for \(\theta = 2.278863\) is physically impossible because the pencil will hit the table before \(\theta\) can get to that value. But that’s OK. All we really need to know is the condition for which the pencil will hit the table. We can’t expect to track the motion of the pencil after it bounces off the table.

So let’s do some algebra on our expression for \(y_A\). We can use the \(\theta = \omega_0 t\) relationship to eliminate \(t\) from the right-hand side:

\[y_A = a\, (1 + \cos \theta) - \frac{1}{2}\frac{g}{\omega_0^2}\, \theta^2\]

Dividing through by \(a\), we get a formula with nondimensional terms, which will allow us to do some numerical work.

\[\frac{y_A}{a} = (1 + \cos \theta) - \frac{1}{2}\frac{g}{a\, \omega_0^2}\, \theta^2\]

Substituting in \(\theta = 2.278863\) and rearranging, we get these three conditions:

  1. \(y_A > 0\) when \(\frac{g}{a\, \omega_0^2} < 0.134650\) (misses table)
  2. \(y_A = 0\) when \(\frac{g}{a\, \omega_0^2} = 0.134650\) (grazes table)
  3. \(y_A < 0\) when \(\frac{g}{a\, \omega_0^2} > 0.134650\) (hits table)

So for a given rod length and a given gravitational field, it’s the angular velocity imparted to the rod by the impulse that determines whether the rod strikes the table or not. If the angular velocity is above this limit,

\[\omega_0 = \sqrt{\frac{g}{0.134650\, a}}\]

the rod will not hit the table. If the angular velocity is below this value, it will.

We can now work out what this means in terms of the magnitude of the impulse. Recall that

\[\hat{\:F} = \frac{1}{3} m\, a\, \omega_0\]

so we can solve for the value of \(\hat{\:F}\) that is the boundary between hit and miss:

\[\hat{\:F} = 0.908396\, m\, \sqrt{g\, a}\]

Impulses higher than this will be misses, impulses lower will be hits.

If you stop for a minute and think about this result, you should be able to convince yourself that it makes sense. Clearly, a larger mass will require a stronger impulse to get the rod to fly off the table. And a higher gravitational field would mean the rod drops more quickly, also requiring a stronger impulse to get it to miss the table. The relationship between \(\hat{\:F}\) and \(a\) is perhaps not as obvious, but think of it this way: The longer the rod, the more slowly it rotates for a given impulse; therefore, the only way to get it to swing around to the required angle of \(\theta = 2.278863\) before striking the table is to hit it harder.

Now let’s run some numbers for what we see in the video. We’ll assume the pencil is of length 18 cm. With \(g = 9.81\:\mathrm{m/s^2}\) and \(a = 0.09\:\mathrm{m}\), we get \(\omega_0 = 28.45\:\mathrm{rad/s}\) (about 270 rpm) as the boundary between hit and miss. That corresponds to

\[v_0 = \frac{1}{3}a\, \omega_0 = 0.85\:\mathrm{m/s}\]

as the horizontal component of velocity at \(C\). The velocity of the base of the pencil just after the impulse needs to be

\[v_0 + a\, \omega_0 = \frac{4}{3} a\, \omega_0 = 3.4\:\mathrm{m/s}\]

These values seem pretty easy to achieve, which is probably why most of the hits in the video sent their pencils flying off without hitting the table.