rt () is constant. We know how to find the length of the radius vector by r( t) r( t) r( t)

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1 Cicula Motion Fom ancient times cicula tajectoies hae occupied a special place in ou model of the Uniese. Although these obits hae been eplaced by the moe geneal elliptical geomety, cicula motion is still an impotant special motion to undestand. The simplicity of the shape sometimes hides the complexity of the motion. We stat by ecognizing that if you choose the simplest oigin fo the coodinate system, the cente of the cicle, that the motion can constained by the condition const. This is a tue statement at all times, but it is not a full desciption of the tajectoy. Fo that we need to know moe about the motion. The geneal case is that the speed of the paticle is not constant. Howee, let s stat by consideing the special case of unifom cicula motion, i.e. the speed of the paticle does not change. It is impotant to notice that we said the speed and not the elocity, fo the fome is constant, while the latte is not. The motion made by the hands of a clock is one example of unifom cicula motion. If we conside the paticle at the tip of the hand, then that paticle epeatedly taces out a cicle with a constant peiod. Its speed is constant, but it is impotant to note that the speed does depend of the adius, so othe paticles in the clock s hand will also hae constant speeds, but the speeds will diffe fom each othe depending on the paticle s adius. The quantity that is constant fo all paticles is the angula elocity i.e. the ate of angula change of the paticles. A diagam will be helpful.

2 is the angle that the adius ecto makes with some efeence axis. As the paticle otates aound the cente this angle changes. The ate of change of this angle is the angula elocity,. Mathematically this is witten as d and is constant fo unifom cicula motion. dt Now with a well defined complete desciption of this special motion we can wite down the tajectoy by inspection: ( t) cos( t ) x sin( t ) y whee xyae, othogonal unit ectos, and sets the initial position at time t 0. The fist thing that you might like to check is that the magnitude of t () is constant. We know how to find the length of the adius ecto by ( t) ( t) ( t) which indeed equals the constant. Since we hae an explicit expession fo the position as a function of time, it is easy to find the elocity and acceleation by taking deiaties with espect to time. The elocity is ( t) sin( t ) x cos( t ) y. The magnitude of the elocity is the speed and it is a constant () t adius has a popotionately lage speed. Taking the deiatie we find the acceleation Again, this is a ecto with a constant magnitude. Notice that as mentioned aboe, a paticle with a lage a( t) cos( t ) x sin( t ) y. a() t. We hae discussed the magnitudes of the elocity and acceleation, but not the diections. It is ey impotant to undestand that these ae ecto quantities. Just because the speed of the paticle is constant does NOT imply that the acceleation is zeo. So what diection is the elocity in? Well, it must be tangent to the cicle. If it wee not, then thee would be a adial component to the elocity and the adius would change. How can we check this? We just noted that t () must be pependicula to tat () all times. Can we check this? Yes, by taking the dot poduct and seeing if it is zeo at all times. Please find ( t) ( t ) and show that it is indeed always zeo. So let s assume that the otation is counteclockwise ( >0) and then the diagam looks like this.

3 The elocity is tangent to the tajectoy. What about the acceleation, what diection is that in? This can be seen by noting the elationship between the adius and acceleation: ( t) cos( t ) x sin( t ) y a( t) cos( t ) x sin( t ) y See the elationship? a t ( ) ( t) ecto. The constant of popotionality is The acceleation is in the opposite diection to the adial. With this the diagam becomes: a

4 This is a complete desciption of the kinematics of unifom cicula motion, as we hae found the tajectoy, elocity and acceleation as a function of time. What about the dynamics, in othe wods what about the foces inoled? We hae teated the poblem ey geneally and hae not specified any paticula foces, howee we can aleady make a ey stong and impotant statement: whatee the foces, thei sum, the net foce, must be equal to the mass times the acceleation o F t ma t m t. This is a net ( ) ( ) ( ) ey poweful statement, fo an object to be in unifom cicula motion thee must be a ey paticula foce on the paticle. This puts temendous constaints on paticles in cicula motion. Tuning the statement aound if the net foce on a paticle is not equal to paticle is not moing in a unifom cicula tajectoy about the oigin. m () t, then the Fo some of you this fomulation of the foce fo cicula motion will look diffeent fom what you leaned in high school. You may hae leaned that This is simply an equialent statement because F net m diected adially inwad. ( ) so the net foce can be ewitten as Fnet m o m. Whatee fom you pefe, thee must be a net foce on the paticle equal to this magnitude and diected adially inwad. Tuning to nonunifom cicula motion, this means that thee is a component of the acceleation paallel to the elocity and theefoe the paticle speed is not constant. An appopiate diagam looks like: a

5 The acceleation can be boken up into two components. One component will be adial o pependicula to the elocity and the othe will be tangential o paallel to the elocity. The elocity itself must still be tangential o else the adius would not be constant. a a pependicula a a paallel The paallel acceleation changes the otational speed of the paticle. In the diagam aboe the paticle is slowing its otational speed. What about the pependicula o adial component of the acceleation? It must be the same magnitude as befoe a pependicula. What is diffeent is that this is no longe a constant. Since the instantaneous and ae changing then the adial acceleation must also change to maintain cicula motion. Of couse this means that the adial component of the foce on the paticle must change to poduce this esult. The classic example of nonunifom cicula motion is an object otating in a etical cicle in a gaitational field. Two examples ae a bucket being swung aound etically on a ope, o a toy ca doing a loop-to-loop. Let s daw the fee-body diagam fo an example such as this, specifically the swinging bucket. T ope F g

6 Thee ae two foces on the bucket, gaity and the tension due to the ope. We can esole these foces into adial and tangential components. T ope F g,tangential F g,adial Algebaically F,tangential mgcos( ) and F,adial mg sin( ). The tangential foce component g g changes the angula elocity, while the adial foce component changes the diection of the elocity just so that a cicula tajectoy is followed. This means that the total adial foce must hae the coect magnitude fo cicula motion o Fnet, adial Tope Fg,adial must equal m o equialently m. This is a ey impotant point: the sum of the adial foces on an object in cicula motion must be equal to m. As the object otates aound the obit its elocity will change, being smallest at the top and highest at the bottom. This then demands that the adial foce must change as the object is otating. This is accomplished by the tension in the ope aying as the bucket goes aound. The tension is highest as the bucket swings though the bottom of its ac and lowest at the top. You should note that the fee-body diagam will look diffeent as the angle changes. If would be good pactice to daw the fee-body diagam fo the bucket at seeal positions aound the cicle. Then calculate what the tension in the ope must at that time. I hae witten a code to simulate the motion of the swinging bucket. The fist paametes that I use ae: mass of bucket = 1kg, length of ope = 1 m, initial position was down( 0 ), initial elocity = 1m/s ( 1 ad/sec ). The image to the ight shows the tajectoy in white and the tension ecto in blue.

7 Hee is the angle eses time cue: This looks a lot like a sine function with a peiod of ~ sec and an amplitude of 0.3 adians (~18 ). The angula elocity cue looks simila: Note the phase diffeence between the two. When the angle is zeo the magnitude of the angula elocity is geatest. This last plot showed the tension in the ope as a function of angle.

8 The positie diection is defined as adially inwad. If the bucket was just hanging and not swinging the tension would be mg 9.8( N). Howee, due to the need to acceleate the bucket adially inwad to hae it execute cicula motion, the tension is lage than the esting alue. A numbe of things change as the bucket is gien a lage initial elocity, 4m/s, see figue on ight. Clealy the amplitude of oscillation is lage (1.3 adians ~78 ) and you can see fom the gaph below that the peiod has inceased to.5 sec. The angula elocity is beginning to look less sinusoidal and a bit moe like a tiangle wae.

9 The most obious change is that the tension now aies oe a much lage ange. Thee ae times ( 0) when the tension is almost tiple the weight and othe times ( 75 ) when the tension is less than one thid the weight. Inceasing the initial elocity to 6 m/sec yields a completely diffeent type of behaio. Now clealy no longe is the motion sinusoidal. The peiod has inceased to oe 3 seconds

10 The angula elocity is stating to hae inflection points. But the most impotant diffeence is obseed in the tension of the ope: it goes negatie! The image on the ight shows a positie tension, i.e. adially inwad. The image on the left shows a negatie tension, pointing adially outwad. This cannot happen with a ope. The bucket cannot hae this tajectoy, it would hae fallen.

11 Does this mean that such motion can nee occu? No, but it cannot happen with a ope. You need to use a od o othe tethe that can apply both positie and negatie foces on the bucket. If you use a stiff od you can find a stating condition that allows the mass to just each the top and almost stop. The tajectoy looks like this; howee thee ae two VERY diffeent situations. The fist is that the mass does not quite make it to the top and the second is that the mass goes oe the top and aound in a complete obit. The next six plots show the two diffeent conditions. Fo the fist thee plots the initial elocity was 6.45 m/s, while fo the last thee gaphs the initial elocity was 6.76 m/s. This small diffeence (~1%) in the initial elocities poduced a ey lage change in the tajectoies. With the lowe initial elocity the angle s. time cue shows a ey asymmetic shape. The flat tops ae whee the mass emains nea the top of the tajectoy fo a long time. You can see that the angula displacement is bounded by. The angula elocity cosses zeo and changes sign as the paticle appoaches the top and eeses diection to fall back down oe the same side.

12 The tension still displays the negatie egions and so this behaiou cannot be obseed by using a ope. Tuning to the next initial condition with a 1% lage stating elocity, below ae the esults fom that simulation. The fist thing to notice is that the angle is no longe bounded by, but will now ee incease as the mass otates counteclockwise aound and aound. The low slope egions ae whee the mass is nea the top and the angle changes ey slowly. This is easily seen in the next gaph of angula elocity s. time.

13 In the angula elocity s. time cue you can see whee it eaches a low alue and stays thee fo some time. Howee in contast to befoe the elocity nee eaches zeo o changes sign. Although the tension gaph looks quite diffeent fom befoe, actually little has changed, only the ange of the abscissa. The tension alues ae essentially identical fo the two situations.

14 These two cases ae impotant to undestand. They show a lage change in behaio with a ey small change in initial conditions. This is not the typical cicumstance that we hae inestigated befoe in class, but we will see it again by the end of the tem. Thee is just one last condition that we need to discuss. In the last two situations we still had a negatie tension. Howee, we know that if we swing a bucket fast enough we can get it to go aound. Thee is a citical initial condition that must be met fo this to occu. You can see it fom the fee-body diagam of the bucket at the top of the obit. The Tope must be zeo o geate, the limiting condition is that the tension is zeo. F g T ope Theefoe the minimum adially inwad foce would equal mg, but we know that this must equal m m so this set a minimum elocity fo the paticle to hae at the top. To obtain this elocity at the top, the mass must hae sufficient initial elocity. Stating the mass at the bottom, 5g the citical initial alue fo o 5g 7 m / s. (We will late see how to deie this analytically when we examine enegy conseation.) This is when the tension will be zeo and the top. Fo lage initial angula elocities the tension will be geate than zeo. But this is fine fo a ope. Below ae the obital gaphs fo this condition. The displacement angle is steadily inceasing and the angula elocity shows oscillations as the mass acceleates towads the bottom and deceleates towads the top.

15 The tension s. angle gaph shows that indeed the tension is always positie. When the bucket is at the top of the obit the tension goes to zeo, thee is a slack ope. Howee, when the mass ounds the bottom the tension is six times the weight of the bucket. Top 180 Cicula motion is ey common. It is impotant that you undestand how it aises and what is special about the foces that poduce it.

16 Impotant Note You may hae noticed that in these notes the wods centipetal foce o centipetal acceleation hae not yet appeaed. This is because I eally don t want you to use them. The foces that poduce cicula motion ae eal foces, i.e. gaitational, sping, tension, contact, magnetic. Thee is no foce in natue named the centipetal foce, no paticle geneates it, no any field caies it. This is just a name gien to the adial component of the net sum of the foces on an object that is moing in a cicle, theefoe its magnitude must be m o m. Many students want to put a centipetal foce on the mass to make it moe in a cicle. Please do not do this! Put a gaitational foce o a tension o a nomal foce, put a eal foce on the object. If you hae to use the wod centipetal make sue you know what you ae doing and just don t let me hea it. Appendix, poof of acceleation components fo nonunifom cicula motion Let () t descibe the angula aiation of the paticle with time. In the case of unifom cicula motion this would be () t twith being a constant. We want to conside the moe geneal d case with an abitay () t. Now () t. The position of the paticle is gien by dt ( t) cos( ( t)) x sin( ( t)) y Diffeentiating this with espect to time yields the elocity d ( t) ( sin( ( t)) x cos( ( t)) y) ( t) Taking anothe time deiatie dt d d a( t) ( sin( ( t)) x cos( ( t)) y) (cos( ( t)) x sin( ( t)) y) dt dt o d d () t. It is customay to denote the angula acceleation () t dt dt a() t And so the total acceleation equals a( t) ( t) ( t) This is a simple expession to undestand, the fist tem gies the acceleation paallel to the elocity and the second tem gie the adially inwad acceleation needed to moe in a cicle. Defining the paallel and pependicula acceleations yields the total acceleation a a a as the sum of the paallel a and pependicula a acceleations.

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