High School Math-Physics SMILE Meeting
23 November 2004
Notes Prepared by Porter Johnson

Bill Blunk [Joliet Central HS, retired]           How Smooth is Smooth? 
Bill  passed around the following information, which was excerpted from instructions for Brownells Flex Hone System provided by Brownell Corporation http://www.brownells.com/ .

"How Smooth is Smooth?
We often use the phrase 'as smooth as glass', yet glass is not really very smooth! In fact, it's quite rough. To prove this point, perform the following simple experiment. Select a piece of glass such as a window pane glass top or even a mirror. Using the forefinger and middle finger, very lightly slide your finger across the surface of the glass. it will feel smooth. Now, remove the cellophane wrapper and repeat the performance."
Bill felt that they meant for you to remove a cellophane wrapper from a package and place it between your fingers and the glass as you lightly slide your fingers across the surface of the glass. But, just why does it feel rough in that case?  On the other hand, without the cellophane the glass feels smooth.  But when we slid our fingers on the mirror of a small telescope, the mirror felt quite smooth, both with and without the cellophane  Why?  For additional information on smoothness of glass see Chapter 37 of a book  by Eric Mazur [Prentice-Hall 2003]. We will discuss this matter in detail after we all have had a chance to "stew over it". Thanks for the puzzler, Bill.

Don Kanner [Lane Tech HS, physics]           Helium 
Don first showed us how to get Helium balloons off the ceiling of a room, using a 2 meter stick with double-sided masking tape wrapped around it.  In addition, he showed how to do the same thing using ordinary masking tape -- just wrap the tape around an end of the stick, giving it a twist so that some of the sticky side would be outward. Very useful tricks with a clever twist, Don!

Don went on to measure the time for a Mylar® balloon filled with Helium gas to rise H = 2 meters to the ceiling, when released from rest.  The time was measured to be t = 2.7 seconds.  If we assume that the motion corresponded to a uniform acceleration, that acceleration is

a = 2 H /t2 = 0.74 meters/second2.
He placed the balloon on an electronic balance, and recorded an "apparent mass" of - 3.4 gm = - 0.0034 kg. The net upward force on the balloon is computed to be  0.0034 kg ´ 9.8 m/s2 = 0.033 Nt.  Don then estimated the volume of the balloon by approximating it as a spheroid of principal radii a = b = 20 cm = 0.20 m, and c = 35 cm = 0.35 m. The volume is
V = 4 p a2 c / 3 = 4 p 0.2 ´ 0.2 ´ 0.35 / 3 = 0.008 m3 = 8 liters.
This volume of Helium (near STP, with a density of 0.179 grams per liter) should correspond to a mass of 1.6 grams of Helium.  Then, he released the helium from the balloon, and recorded the mass of the empty balloon to be + 10.6 gm = + 0.0106 kg.  The mass of the balloon filled with Helium should be the sum, about 12.2 gm = 0.0122 kg. The upward acceleration of the balloon (uniform acceleration assumed) is given by
a = Fnet / m = 0.033 Nt/ 0.0122 kg  = 2.8 meters/second2
This number is in serious disagreement with the value 0.74 m/s2 , which was determined above. This indicates that the rising of the helium balloon does not correspond to our assumption of uniform acceleration. Presumably, air resistance is very important in the rise of the balloon.

Don finished by taking several breaths from the Helium balloon, after which he said -- in a high-pitched "Helium voice" -- the very familiar-sounding words: Tha- tha- thats all, folks! Fascinating, Don!

Fred Farnell [Lane Tech HS, physics]           The Physics of Sign Hanging 
Fred analyzed the forces involved in hanging a sign, which are represented in the diagram:

Hanging Sign

If T1 and T2 are the tension forces in the cords suspending the sign with weight W, then for equilibrium the vector sum of these forces must be zero; i.e.:
T1 + T2 + W = 0
We may write this equation in component form as follows:
Vertical:       T1 sin j + T2 sin q = W
Horizontal:    T1 cos j = T2 cos q 
Each end of a string was held by a volunteer, who each had a spring scale to determine the tensions T1 and T2 when a weight W was suspended from a point along the string. We also measured the angles, and obtained the following data:
Quantity            Value
T1  8.9 Nt
T2 7.5 Nt
W 8.4 Nt
j 45°
q 18°
We then checked to see how well the equilibrium conditions were satisfied:
Vertical Check  Horizontal Check
  T1 sin j + T2 sin q = W    T1 cos j = T2 cos q 
8.9 sin 45°+ 7.5 sin 18° =? 8.40     8.9 cos 45° =?  7.5 cos 18°
6.29 + 2.32 = 8.61 Nt =?  8.40 Nt    6.29  Nt =?  7.13 Nt
 0.21 Nt discrepancy (about 2%)   0.84 Nt discrepancy (about 18%)

Various explanations for these discrepancies were proposed, such as friction between the string and the hook on the weight at the hanging location, as well as the discrepancy in setting the zero location of the spring scales when these scales were, in fact, tilted.  Still, the agreement was fairly good.  Nicely done, Fred!

Charlotte Wood-Harrington [Brooks College Preparatory School,  physics]           Newton's Third Law -- a 'Tom Senior' Demo 
Charlotte placed some dowel rods (each about 30 cm long) on the table, and on top of them placed a sheet of pink foam insulating board, which was about 30 cm wide, 100 cm long, and 2 cm thick. Then she put a small self-propelled toy car on the top of the foam board, which served as a racetrack for the car.  When the car was turned on and then placed on the foam board to travel in the long direction, it went forward, the dowel rods rotated, and the foam board went backwards --- as required by Newton's Third Law.  The arrangement worked very well, except that Charlotte had inadvertently gotten a 'hot' car, which traveled only at top speed. The car went so fast that the foam board almost immediately shot off the table in the opposite direction.  We need to find a car that isn't such a speed demon!

A very nice illustration of physics in action, Charlotte!

Bill Shanks [Joliet Central, happily retired]               $1.00 for 3.5 meters
showed us a tape measure -- with a metric scale of more than 3.5 meters! He recently obtained it at an "ultra-cheap tool bin" at Menards for only 99¢Bill used the tape measure to determine the size of wooden cubes that were being given away.  He obtained 1.27 cm [or 1/2 inch], and calculated a volume of just over 2.0 cubic centimeters. Those cubes looked really small!  Bill also showed us some ratchet clamps, which he also obtained from Menards for about $4.50.  A good, inexpensive method of clamping and holding things.  Useful stuff, Bill!

A general discussion occurred as to why we speak of centripetal and centrifugal forces, despite the fact that forces outside the nuclear domain are either caused by contact, gravity, or electromagnetic fields.  Don Kanner indicated that every centripetal force has a centrifugal reaction force, as required by Newton's Third Law.  If you twirl a slingshot over your head, there is a force pulling the sling inward and the reaction force pulls your hand outward.  For the moon revolving around the earth (ignoring the effect of the sun), the earth and moon actually rotate about their common center of mass, which actually lies inside the earth.  Thus, the earth and the moon each experience a centripetal acceleration, caused by their mutual gravitational attraction.  Porter Johnson suggested that we use the word "centripetal" to describe the direction of the force, rather than as a source of the force itself.  For example, we could say that the gravitational force produced by the earth on the moon lies in a centripetal direction.

We did not have time for Ann Brandon to make her presentation, Non-scrambled EggsAnn will be scheduled for our next meeting, Tuesday 07 December 2004.  See you there!

Notes taken by Porter Johnson