# Disclaimer: The following material is being kept online for archival purposes.

## Although accurate at the time of publication, it is no longer being updated. The page may contain broken links or outdated information, and parts may not function in current web browsers.

Lesson Plan #29         http://www.phy6.org/stargaze/Lcircul.htm

# (19) Motion in a Circle

A short but important section, deriving centripetal acceleration for motion at constant speed around a circle.

Part of a high school course on astronomy, Newtonian mechanics and spaceflight
by David P. Stern

 This lesson plan supplements: "Motion in a Circle," section #19   http://www.phy6.org/stargaze/Scircul.htm "From Stargazers to Starships" home page: ....stargaze/Sintro.htm Lesson plan home page and index:             ....stargaze/Lintro.htm

 Goals: The student will learn About uniform circular motion, and the relation of its frequency of N revolutions/sec with the peripheral velocity v and with the rotation period T. That uniform circular motion is a type of accelerated motion. That the "centripetal acceleration" of an object going around a circle of radius R with constant speed v equals v2/r and is directed towards the center of the motion. An elementary proof of the preceding result, using the theorem of Pythagoras. The proof depends on neglecting a small quantity x, but because x can be made as small as we please, it holds exactly. That intuition can deceive. Even though an object whirling around a circle strains to pull away from the center, if it is cut loose, it will not move in that direction, but will continue along a straight line tangential to the circle. Terms: uniform circular motion, frequency, peripheral velocity, centripetal acceleration and force. Starting out:  Today we will learn about the simplest case of motion not in a straight line, uniform motion around a circle. Of course, that is an accelerated motion, because by Newton's first law (or at least, by what that law implies), any motion not in a straight line is accelerated, and requires a force for maintaining it. (Illustrate the following by a drawing on the board, to which details are added as the discussion progresses.) When we whirl around a weight tied to a string, we feel it pull our hand, which means (because forces occur in equal and opposite pairs), our hand is pulling that string inwards. Why does this happen? [If a students says "because of the centrifugal (or centripetal) force," say "that is just a word, a technical term. What is actually happening?"] What is happening, is that the weight, left to itself, would continue moving along a straight line, further and further from your hand (show on the board).   The string does not allow it to do so, but pulls it back towards your hand, to keep it in its circle. We will show today that motion in a circle can be viewed as the combination of two motions taking place at the same time--like the motion of the airplane, flying in a cross-wind (Section #14). One motion is the continuation of the existing velocity along a straight line (show on the drawing)--the way the weight would move by Newton's first law, if no outside force acted on it.   The other is a motion towards the center of the circle (draw it, too), returning the weight to its circular path. (Figuratively returning it; in actuality, both motions are simultaneous and the weight never leaves the circle.) That second motion, it will be shown, is an accelerated one...   Now for the details..... (Continue on the board with the derivations, while the students copy into their notebooks.) Guiding questions and additional tidbits (Suggested answers, brackets for comments by the teacher or "optional") --Why is motion at a constant speed around a circle an accelerated motion? The speed, the magnitude of the velocity, is indeed constant. But the direction changes all the time. Velocity is a vector quantity, and any change of its direction also involves acceleration. --When you whirl a stone at the end of a string and let go, how does the stone move? It continues along a straight line, tangent to the circle in which it moved. -- Why doesn't the released stone move outwards, in the direction in which it pulled? Because from the moment when it is released, no forces exist any more in the direction of the string. The stone strains against the centripetal force only as long as it moves in a circle. --Riding over a dirty road, the wheels of your car acquire a coating of mud, which soon flies off again. How does it fly off? Along a line tangent to the wheel. If, as is likely, the mud flies off soon after the wheel has picked it up from the road, it will fly backwards and upwards from the rim of the wheel--in the direction of the mudguards which truckers hang behind their wheels to intercept it. [Draw schematic on the blackboard]. --What is the acceleration of a stone rotating with speed v around a circle of radius r? v2/r, directed to the center of the circle. --In the derivation of the formula a = v2/r we neglected a small quantity x. Does that mean that the formula is only approximate, not exact? No, it is exact, because x can be made as small as we please. --If a stone makes N circuits per second around the center, what is its rotation period, T? T = 1/N. The total time spent in those N circuits is found by multiplying the number of circuits (=N) by the length of each one (=T). But that time, by definition, is one second, so NT = 1 from which T = 1/N. --If a stone makes N circuits per second around a circle of radius R, what is its centripetal acceleration? The distance covered in each revolution is 2 πR The distance covered in one second is (2 πR)N--which by definition is also its peripheral speed. Hence a = (2πRN)2/R = (2πN)2 R2/R = (2πN)2 R Additional discussion: Astronauts are subjected to large accelerations during launch and re-entry. The forces associated with such accelerations are often called "g forces" because they are measured in gravities, i.e. the acceleration is measured in units of g, the acceleration due to gravity for which we will use the approximate value 10 m/sec2. In another lesson in "'Stargazers" it was noted that the V2 rocket of World War II started with an acceleration of 1 g and ended at "burn-out", with its mass (mostly fuel) greatly reduced, at about 7 g. The space shuttle (I believe) pulls 3g before burnout, which is quite uncomfortable, even for someone lying flat on the back on a contoured surface. Re-entry has comparable (negative) accelerations. To get astronaut used to taking such forces, they are whirled around during their ground training in a centrifuge, inside a small cabin mounted at the end of a horizontally rotating arm. (Anyone seen such centrifuges on TV?) They are a bit like some amusement park rides, but can create greater stresses, and have TV cameras that monitor the rider. -- A "human-rated centrifuge" has a radius of 6 meters and creates a 1-g acceleration. How many revolutions per second must it make, and what is its rotation period? Taking g = 10 m/s2, 10 = (2πN)2 R = (6.28)2 N2 (6) = 236.63 N2 N2 = 0.04226 N = 0.206 rev/sec (larger than N2, of course, since N<1 !) T = 1/N ~ 5 seconds --If you double the rotation speed, so that each revolution only takes 2.5 seconds, what will the g-force due to the rotation be? It will be 4g, since it grows like the square power of v (or of N). How is the centripetal force transmitted to the astronaut? That is the force keeping the astronaut moving in a circle. It is transmitted to the astronaut's body by the chair or couch that supports it.

Guides to teachers...       A newer one           An older one             Timeline         Glossary

Author and Curator:   Dr. David P. Stern
Mail to Dr.Stern:   audavstern("at" symbol)erols.com .

Last updated: 12.17.2001      Kept unchanged 10-20-04