Having trouble controlling the size of your lasso loop? When you try to make the

loop big enough to jump through, does it fall apart? The problem may be with your equipment, and can be easily solved with the help of a little physics. You don't even need to understand the math to use the answer, which is given at the end of this article.

 

The story begins in 1975, when some friends and I started the MIT juggling club (it is now one of the oldest juggling clubs still in existence). One member from back in the 1970's was Carey Bunks. Last fall when I went to France for a few days I met up with Carey, who moved there some years ago. He gave me a small tour of Paris, which included briefly juggling in a parade we happened on in Montparnasse. Carey also showed me a monograph he's written on the lasso, describing how to get started and how to do many rope tricks. Reflecting his MIT backround, the book includes some analyses of the physics of rope spinning. If you would like to purchase a copy of Carey's lasso book, which I recommend, send $10 to Carey Bunks, Paris.

 

A lasso has a small loop called a "honda," usually made of metal, spliced into one end of the rope. The other end of the rope passes through the honda to make the large circle of rope that spins, called the "loop." The section of rope that goes from the honda to your hand is called the "spoke."

 

Now, those who don't like math should skip to the last paragraph of this article to 'find the answer to the loop size problem I described above.

 

Consider the forces experienced by the honda as the lasso spins. Call the outward (centrifugal) force on the honda F. Counteracting the outward force F is S, the tension on the spoke which prevents the honda from flying away from you. F = S to keep the loop from growing or shrinking in size. The honda is also pulled sideways by T, the tension in the loop. S = T to keep the lasso going in a circle. See Figure 1.

 

The outward force F = hrw, where h is the mass of the honda, r is the radius of the loop, and w is the angular velocity of the honda squared, which is approximately constant. Everywhere I say the term "mass," you can think of it as just the weight. Radius r just means half of the width of the loop, and angular velocity is technical term for how fast the rope is spinning.

 

Let m equal the mass of the rope comprising the loop, and imagine that it is distributed around the loop as n equally spaced points of mass. Let t equal the loop tension that would exist at anyone of these points. Think of these n points as being the corners of an n sided regular polygon. (See figure 2.)

 

Take any three adjacent mass points, and number them 1, 2, and 3 in row. Let p equal the angle made by points 1, 2 and 3. This is the angle between any two sides of a regular n sided polygon, which equals 180 ­ (360/n) degrees. The force pulling mass point 2 outward is (m/n)rw. Counteracting this outward pull is the pull of mass points 1 and 3, each of which exerts an inward pull on mass point 2 of t cos(p/2), so 2t cos(p/2) = (m/n)rw and p = 180 - (360/n). Solving for t gives t = mrw/ [2n cos(90 - (180/n)) ] = mrw / [2n sin (180/n)].

 

The real loop has mass spread equally all around it, not clumped into n points. Therefore, the real loop tension T equals the limit of the above expression for t as n goes to infinity. As n goes to infinity, n sin(180/n) goes to pi (to see this, try setting your calculator to degrees and plug n = 1000 into this expression). Therefore, T = mrw/(2pi).

 

We now have hrw = F = S = T = mrw/(2pi), so h = m/(2pi). Let d equal the linear density of yo rope (this is just its weight per foot). By geometry, m = (2pi)rd, so h = rd.

 

This final equation, h = rd, is the remarkably simple answer to the problem of controlling the size of your lasso loop. Let r be the radius of your loop in feet, that is, r is half the width you want your loop to be. Weigh your rope (without the honda), and divide its weight in ounces by its length in feet to get d. To have your loop be stable at the desired width, your honda should weigh h equals r times d ounces!

 

For example, suppose you want to do the Texas skip, like Carey is doing in the picture, and suppose you are 6 feet tall. Then you will want a loop that has about a 3 foot radius. If your rope weighs, say, 1/2 ounce per foot, then r = 3, d = 1/2, then you need your honda to weigh 3 times 1/2, which equals one and a half ounces. If your honda is too light, you can wrap some copper wire around it to make it heavier. Happy skipping!

 

"The Academic Juggler" is an occasional feature of Jugglers World, devoted to all kinds of formal analyses of juggling. Anyone with comments or suggestions for this feature are encouraged to write to Arthur Lewbel, Lexington MA.