Dot Physics

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.

i-484d04d193449f78cdc7f7bb0ac5d1e6-balance-stick-fbd.jpg

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:

i-c83881d322a8188124018df5af9665b6-torque-0-1.jpg

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:

i-c1aaa950b7f25edfc5b69315315074ef-newton-similar.jpg

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:

i-32fafcdc30c904b0e1f8f09e3e508cc2-torque-def.jpg

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).

i-9cf883671c233ff52771c14d4da607e9-torque-calc.jpg

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

i-667db8e83380ed2298d57f2d96db138b-torque-for-stick.jpg

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)

i-031127af58cef04fe11ec8b0a3f320a1-i-stick.jpg

So, putting it all together:

i-cdfb2953704e3873232a14563b81c950-alpha-1.jpg

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:

i-6dcfe7efcfb61060aaa2328742ea41bd-hammer-stick-1.jpg

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:

i-0822050c842d7b18c9d6d97400ea7d5d-ycm.jpg

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:

i-6dcc53fca7b3b84f77943c0c46c2a6c5-new-i.jpg

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:

i-28bb7b02b2f8cc0f5cad21bd76d8bc8e-alpha-with-mass.jpg

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 2l takes the longest to fall over. I made these plots with vpython. If you want to code, let me know.

i-82e2fd04620f7c74f67e06b1851c1b81-numerical-rod-plots.jpg

Comments

  1. #1 Ben
    May 12, 2009

    This was fun. I would love to see the physics of balancing a meter stick horizontally. Particularly my question is: does the support need to be of some width for the meter stick to maintain balance? On your hand, if you balance a meter stick, it will return to its horizontal resting position. However, narrow the edge that it rests on and it becomes increasingly hard for the meter stick to balance. What is going on here?

  2. #2 Dan Meyer
    May 13, 2009

    Yeah, fun stuff. I dig the illustrations, though I understand half of them. All in Keynote?

  3. #3 Rhett
    May 13, 2009

    @Dan,

    Keynote is my favorite drawing program. Makes stuff pretty. You can even just use the trial version which does not let you save after 30 days – but hey, that’s ok because I just use skitch to make a screen capture. (Shadows are what makes it look good).

  4. #4 Alfredo Louro
    May 13, 2009

    Very nice example. As an anecdote, I can tell you that I ride our local train almost every morning and afternoon, and I like to remain standing. It’s actually quite easy to respond to the occasional lurches of the train, and I have found that a heavy backpack actually helps! This is consistent with your adding mass to the top of the stick, which raises the center of mass even more.

  5. #5 AJD
    May 14, 2009

    For an in-class demo, I bring two meter sticks. Each one has a large clamp attached about 20 cm from one end. I ask the class whether it would be easier to balance one of the sticks with the mass closer or farther from your hand. Two students who gave opposite answers are asked to compete to see who can balance longer. I give a simplified explanation which ignores the moment of inertia of the stick because the clamps are pretty heavy.

  6. #6 Rhett
    May 14, 2009

    @Ben,

    Basically, if you hold the meterstick horizontally your finger (though small) can exert torques on the stick. This is enough to “push” it back towards equilibrium if it is going off. Maybe I can put together a more detailed post on this at some point.

  7. #7 Jari
    May 26, 2009

    How could this be implemented? Assume there is a stick of length l, held at an angle theta. What would the angle be at the next time step, considering gravity? How about when the hand is moved?

  8. #8 Rhett
    May 26, 2009

    @Jari,

    This is a much more complicated question. Can it be modeled? Absolutely. I think the best way to model this is with Lagrangian mechanics, which is a bit more complicated that newtonian (well, not really always more complicated).

  9. #9 Jari
    May 26, 2009

    Say you have a computer model with a position vector (the point where you hold the stick) and a “stick vector. I am trying to make a balancing game just like this, but I don’t know how to implement the physics. Great blog btw :)

  10. #10 Rhett
    May 26, 2009

    @Jari,

    This is a good question, but not really explainable in this comment. Give me a couple of days and I will see if I can write a blog post about it.

  11. #11 Jari
    May 27, 2009

    I did some searching. The dynamics for the cart-pole problem (as it is apparently called) can be found here:
    http://www.coneural.org/florian/papers/05_cart_pole.pdf