HOW MUCH? The heck you say.

We're studying circular motion in regular physics, and I'm preparing my laboratory activity for the week after Thanksgiving break.  (Not sooner -- we have our trimester exams next week.)  I want to do the "swing a stopper on a string in a horizontal circle above your head" experiment, a classic developed in the PSSC era.  I discovered my setups for this experiment to be a tangled mess, with numerous missing pieces, broken strings, and not enough hollow tubes for the string to go through in any case.  Now you can tell why I haven't done this experiment with my class in about five years.

I looked on the PASCO site, hoping to find a reasonably priced set of hollow tubes with stoppers and the light, low friction thread that leads to quality data.  And what to my eyes did appear:  $39 big ones for a set of five stoppers, two tubes, ten zip ties, and some regular old string. 

Okay, my department's budget is nearly unlimited.  I can -- and do -- order any equipment I want or need for my class.  Nevertheless, there's something to be said for intelligent use of resources.  $39 for items available in the storeroom?  Neither frugal nor intelligent.  This is why a former Chief Reader for the AP exam defined "Pasco" as a Latin verb meaning "to rob."*

* Before the Pasco Police come my way, please note that I am a HUGE customer, and a huge supporter of the company in general.  They sent me two loaner heat engines for my summer institutes -- no charge, no hassle, no problem.  (Of course, I probably garnered them 5-10 orders for said heat engines, so they got their money's worth.)  I tell anyone who will listen how reliable PASCO's products are, and how good their technical support is.  But the downside:  they're expensive.  And in this case, obnoxiously expensive.

I had no trouble finding stoppers, thread, and zip ties.  I'm going to use thread from Burrito Girl's* sewing kit rather than regular string; the chemistry department has stoppers of all sizes.  The trick was finding the hollow tubes without a trip to the hardware store.

* Burrito Girl is my wife and sidekick. 

My classroommate Alex Tisch looked at the picture, and offered up a suggestion that would make the editors of the Tightwad Gazette croon:  what about a BIC pen with the ink part removed?  Two decades ago I used to take apart these pens when I was bored in class... now I could use that experience to save my department some dough. 

In the event, I used a papermate brand pen.  The pen-tip is connected to a thin tube of ink, all of which can be removed from the pen casing easily; the cap on the other side took some wedging, but I got it out with a fingernail in less than one minute.  Voila, a "hollow tube," at a cost of about a quarter.

(Oh, you want to know about the actual experiment?  Attach the stopper to the string with the zip-tie, thread the string through the tube, and hang a mass from the bottom end of the string.  Hold the tube and swing the stopper in a horizontal circle at constant speed such that the hanging mass hangs in equilibrium.  The radius of circular motion can be measured with a ruler.  The speed of the mass can be determined with a stopwatch, knowing that speed is circumference divided by the time for one revolution.  A graph of speed squared on the vertical axis and radius on the horizontal axis yields a line whose slope is the centripetal acceleration of the stopper.  This acceleration can be shown to be equal to g times the ratio of the hanging mass to the stopper mass.)

GCJ

Circular motion demonstration: turntable

When I arrived at Woodberry Forest eleven years ago, I discovered several 1970s-vintage turntables in the storeroom.  I turned one into a nice, quantitative, circular motion demonstration.

In the picture you see the turntable, with five identical brass masses placed on top.  Before I start the demonstration, I show the class that a brass mass placed right near the center stays put when the table rotates; but the same mass placed near the outer edge flies off.  The goal of the demonstration is to predict the maximum radius for which the mass will not fly off.

The friction force on the brass mass acts toward the center of the rotation, and is a centripetal force equal to mv²/r.  The variable r represents the radius of the circle -- that's what we're looking for here. 

Problem is, the speed v itself depends on r, because a mass near the edge covers a larger distance in the same time as a mass near the center.  So we write v in terms of the period of revolution, T:  v = 2πr/T.  The period of revolution is measured with a stopwatch.

The friction force can be set equal to μFn.  The normal force on the mass is simply its weight.  The coefficient of (static) friction must be measured... but I can do that with a spring scale or a force probe.  I pull a stack of brass masses with the scale.  I divide the reading in the scale just as the masses start to move by the weight of the stack of masses. 

(Here the class can be asked whether the coefficient of friction for a single mass will be greater, less than, or the same as that for the stack of masses.  Answer:  the same, because we still have brass in contact with the same surface.  I only use a stack of masses to make the spring scale reading easier to obtain.)

So now we know everthing in the relevant equations necessary to solve for the radius r.  We find that radius to be in the neighborhood of 4-5 cm.  That's in the middle of the second mass, as measured from the center.

What does this mean experimentally?  When I turn the turntable on, the outer masses should fly off, but the first one or two should stay put.  And sure enough, that's what happens -- physics works.

Follow-up:  Next I replace the 20 g brass masses with 10 g brass masses from the same set.  Which ones should fly off now?  Answer: still, the masses more than 4-5 cm from the center will fly off, the others will stay put.  The coefficient of friction still hasn't changed, because we're using the same surfaces.  Mass cancels out when solving in variables, so the mass doesn't matter.  Sure enough, the 10 g masses fly off outside of 4-5 cm, as predicted.

GCJ

Centripetal vs. Centrifugal Force: Golf Cart

A golf cart is moving in a straight line. I want the cart to move in a circle. Should I push or pull the cart TOWARD the center of the circle, or AWAY FROM the center of the circle?

Of course, this is the central (ha!) question of the circular motion unit. Students have preconceived notions of "centrifugal force," as well as mistaken ideas about force in the direction of motion. It's nice to begin the circular motion unit with this central question, followed by a demonstration that shows unambiguously and memorably that force toward the center of the circle is required.

Since I live on campus, about 0.5 miles from my classroom, I drive a golf cart to work. This morning I blocked off about 10 spaces in the little parking lot next to the science dungeon. I tied a sturdy rope to the corner of my cart. With the class watching, I drove the cart forward. A physically strong student pulled on the rope in a direction perpendicular to the cart's velocity. Sure enough, the cart's path arced slightly.

Next, I had THREE students tug on the rope. This time the cart's path described a "tighter" circle. We will use this qualitative observation on Monday, when we write and use the equation for centripetal acceleration.

And finally, I turned to a student who originally answered that we should pull the cart AWAY from the center of the circle. I asked him to do so, but he smiled and politely declined. Woo-hoo -- he gets it.

GCJ