Dot Physics

Friction Demo with a meterstick

This is one of my favorite demos. Easy to do and doesn’t really require any equipment. The basic idea is that you support a meter stick horizontally with two fingers. Slide your fingers in and they will both meet at the center of mass. Here is a video.


Meterstick friction demo from Rhett Allain on Vimeo

So, how does it work? To explain this, you need to understand friction and equilibrium. Well, nobody really understands friction – but anyway..

First, what is equilibrium? Equilibrium means that an object’s motion does not change. For this case that means that both it’s velocity and it’s angular momentum are constant or zero (constant zero is a constant). If the velocity is constant then:

i-a4c20c82f48d22df57a8e6e6b968c1a4-equil-1.jpg

The meter stick is also constant in it’s angular motion. A key point is that it is not rotating. This means that it is not rotating about ANY point. For any point on the meter stick the following is true:

i-02013247ea449c9090ba5201ca03821c-nettorque.jpg

Where ? is the torque defined as:

i-4d96f7e6b4054e1facab97d823d3c63f-torquedef.jpg

Didn’t I talk about torque before? Yes, I did. It was for the torque by rotating balls in fantastic contraption. I am surprised I never made “basics” out of this. Essentially, the torque is the “rotational force”. It is what causes it to change an objects rotational motion. To calculate the torque, you need to pick a point about which to calculate. For this case it doesn’t matter what point you use since it is not rotating about all points. Also, torque IS a vector. However, in this case you can just deal with it’s magnitude. That makes it easier for intro students to calculate (don’t need the cross product). Here is a good illustration for toque. Suppose I push on a door (top view).

i-4c1cc7d7042c284f68d74719bce95c56-doortorque.jpg

In this diagram, imagine you were trying to open this door by pushing at the location and direction of the three forces (A, B, and C). Which one would be the easiest to open the door with? I think your experience would say B. Force C would be the next hardest and force A wouldn’t do anything, right? The torque about the hinge is going to be the force times the distance from the point about which we are calculating the torque (the hinge) time the sine of the angle between the vector from the hinge to the force. Let me draw these r vectors on the diagram.

i-ca2f7c9dce8d1c98c86064943b92f031-rarrows-1.jpg

So, comparing forces B and C (which have the same magnitude), force B has a much longer r (sometimes called the arm length) than C. Both of these forces have the same angle (about 90 degrees) between the force and the arm. This means that force B will have a much larger torque. For force A, the angle between the force and the arm is 180 degrees. The sin(180) = 0, so this force produces no torque about the hinge.

This whole torque thing gets much more complicated, but this is enough for us to look at the meter stick again. In this following diagram, I will show the forces acting on the meter stick for the case where the two fingers are not evenly distributed and not moving in. Note also that the gravitational force on the meter stick acts on all parts, however this can be modeled as a single gravitational force acting at the center of mass (which happens to be the center of the meter stick.)

i-bfd433464b43401bedf0270913483f08-meterstickfbd.jpg

What is important in this diagram? First, if you can’t remember anything about free body diagrams (that is what it’s called) check out my free body diagram post. In this diagram, the two forces from the fingers add up to the gravitational force. If this weren’t true, the total force in the vertical direction would not be zero and the thing would not be in equilibrium. Also, the forces from the fingers are not equal because they are not equally spaced. If I label the forces from the fingers to be F1 and F2 then the following must be true for the y-components of force:

i-977d459fcbadf0e70687d0319c557df8-sumforcesmeterstick-1.jpg

Again – remember I said that this is for the y-components so they are NOT vectors. However, this expression is not enough to find out how much force the two fingers push on the meter stick. If there are two variables (the two fingers) then I would need two equations to solve the problem. I can do this by writing down the expression for the total torque about the center (I could do this about any point, but I will choose the center). Let me call x1 and x2 the distances the fingers are from the center. One thing I did not mention earlier is the sign of torques. If I want the torques to add up to zero, then at least one of them will have to be negative. If I used the full vector nature of torques, this would come out automatically, however here I will call torques that want to twist the thing counterclockwise positive and torques that twist clockwise as negative. This would give the following expression for torques:

i-f53744c072f449ff145829ff488b6b87-sumtorques-1.jpg

Here I explicitly put the distance between mg and the point about which the torques are calculated as (0). Removing the gravity term, you can see that there is a relationship between the forces the two fingers exert on the meter stick.

i-89019a23fa122cd376e898be5c0215e4-finger-1finger-1.jpg

Where am I going with all this? Well, the force the finger pushes on the stick is related to the frictional force between the finger and the stick. If I use the following model for friction:

i-8f08d0eafcde59b6699f6c8669b3ef9b-friction-123.jpg

Here N is the force the two surfaces (finger and meter stick) exert on each other. The harder they are pushed together, the greater the frictional force. Maybe you can see where this is going. The finger that is closer to the center will have a greater frictional force on it. This means that if I push the two fingers in together, one will stick and one will slip. The one that is further from the center of mass (where ever that center of mass is) will slide and the one that is closer will stick.

I guess if you were awesome you could make it so that both fingers slide together, but you would have to keep everything completely even. Also, note that this ‘trick’ does not depend on the mass distribution of the object, but it DOES depend a uniform coefficient of friction.

Ok, that should be enough for you to have an understanding of the ‘trick’. Of course, you don’t really need to understand this to do the trick.

Comments

  1. #1 Uncle Al
    February 21, 2009

    The phenomenon is entertaining. Its automatic self-correction through negative feedback is awesome. Compare with ongoing multi-$trillion green hemmorrhages in Washington and the effects of positive feedback (No oversight! No restrictions! No penalties!) upon extrapolated results. Ragnarok.

  2. #2 Fran
    February 21, 2009

    I love this demo, too!

  3. #3 Neil
    February 22, 2009

    Interesting, but not really as cool as an attack on photons. Just sayin.

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