Maybe you have seen this trick. Basically, you hold by supporting it with two fingers from the bottom. You then move your hand around to keep it balanced while the stick is vertical. It is really not as hard as it looks. Also, there are two things that can make your job easier. Use a longer stick, or add an extra mass at the end of the stick. Here is a video of me demonstrating this. (I forgot, this also an event in the show Unbeatable Banzuke)

Balancing a Stick Demo from Rhett Allain on Vimeo.

So, how does this work? Let me start with a stick that is mostly vertical and supported by the bottom.

This is a rather simple diagram – meaning it could be a lot more complicated. However, I just want to look at the angular motion of this stick (of length *l*). To do this, I will use the angular motion equation:

Where ? (the greek letter tau) represents the torque about some point (it’s a vector), I represents the moment of inertia for rotation about an axis through that point and ? is the angular acceleration. Rotation of rigid objects can get crazy complicated, but in this case everything will behave. In fact, I can even drop the vector notation because the torque and the angular acceleration are in the same direction.

Notice that this looks a like like the equation:

So, torque is like the rotational force and I is like the rotational mass. Maybe I will show a demo on the moment of inertia sometime. Back to this case. If I look at the torque about the bottom point, I could use the following for the magnitude of the torque:

Here, *r* is a vector from the pivot point to the point where the force is applied. F is the force, and ? is the angle between r and F. For this case, the torque due to the gravitational force will be: (note that the gravitational froce acts on all of the stick. However, this can be modeled as if it were all acting at a single point known as the center of mass).

So, the magnitude of the torque would be: (note that the hand exerts no torque about the end point because *r* = 0 )

And this would be equal to the product of the moment of inertia and the angular acceleration. For a stick held at one end, the moment of inertia (about a fixed axis) would be: (again, maybe I will derive this someday for you)

So, putting it all together:

The important thing to notice is first – the mass cancels as you would expect. However, the length does NOT cancel. A longer stick would have a lower angular acceleration. What about the case with the extra mass at the end of the stick like in the video? Let me draw a free body diagram for this:

For this example, I will assume that the mass on the end is twice the mass of the stick (I just randomly picked that). First, I need to find the new center of mass. The system can be treated as two masses, the stick with a mass *m* located at the center (l/2) and another mass (2m) located at l. In this case, the center of mass (from the end) would be:

Reality check – the center of mass is still inside the stick. It would be crazy if I got a center of mass greater than *l*. Ok, now what about the new moment of inertia? I can think of this object as being two objects put together, the stick plus the extra mass. This would give a moment of inertia of:

The torque will be the same as before except there is a new mass and a new center of mass. Putting this all together, I get:

The angular acceleration with the extra mass is less than without. This means you would have more time to move your hand to bring it back in the vertical position.

Here are some plots of the angular position of the three above rods. The one that is going to fall over the fastest is the plain rod of length *l*. The rod with length 2*l* takes the longest to fall over. I made these plots with vpython. If you want to code, let me know.