3.2 Centripetal Acceleration

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Christopher Gilmore
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1 unifom cicula motion the motion of an object with onstant speed along a cicula path of constant adius 3.2 Centipetal Acceleation The hamme thow is a tack-and-field event in which an athlete thows a hamme a heavy metal ball attached to a wie and handle the fathest distance possible (Figue 1). To do this, the athlete fist swings the ball in icula path. Afte giving it a maximum cicula speed with thee o fou tuns, the athlete eleases the hamme, which then flies down the field. Just befoe the hamme is eleased, it has unifom cicula motion, which is motion in icula path at onstant speed. Figue 1 To give the hamme enough speed to tavel a long distance down the field, the athlete must move it apidly in icula path. centipetal acceleation (a > c) the instantaneous acceleation that is diected towad the cente of icula path By moving the ball in icle, the athlete intoduces the foce of tension in the wie. This tension keeps the ball in icula path. The geate the tension, the geate the acceleation towad the cente of the cicle and the faste the ball moves in a cicula path. When the tension is vey lage, so is the speed of the ball. When the athlete eleases the hamme, the ball tavels fa down the field. You may not always ealize it, but objects moving in cicula paths ae all aound you. Clothes in a washing machine duing the spin cycle, the dum of lothes dye, the hands of cetain electic clocks, and the spinning blades of a blende and a lawn mowe: all of these objects move with unifom cicula motion. What you may not have consideed is that these ae all among the most common non-inetial fames of efeence. These objects move in icula path, so thei velocity constantly changes diection. Theefoe they ae acceleating. Acceleation that is diected towad the cente of icula path is called centipetal acceleation, a > c. Equations fo Centipetal Acceleation Recall that the aveage acceleation, a > av, of an object equals the change in velocity, Dv >, duing an inteval of time, : a > av 5 Dv>. Fo an object moving with unifom cicula motion, the velocity changes diection continuously with time, so Dv > is definitely not zeo. Theefoe, centipetal acceleation is not zeo. To calculate centipetal acceleation, we now conside the example of a unne moving at onstant speed along icula tack. The velocity of the unne changes with time and is always tangential to the cicula path, as shown in Figue 2. Figue 2(a) shows a unne s velocity vectos at two neaby positions: A9 and A. Figue 2(b) shows the coesponding change in velocity, Dv >, ove a shot time inteval,. Recall fom Section 1.4 that the diffeence in velocity vectos is the same as adding one vecto to the negative of the othe vecto. Fist, shift vecto v > 2 down so that its head is at point A (Figue 2(b)). Then evese v > 1 so v > 1 becomes 2v > 1. Place the tail of vecto 2v > 1 at the head of vecto v > 2, so that the sum of the vectos is Dv >, which points 114 Chapte 3 Unifom Cicula Motion NEL

2 towad the cente of the cicle. It should be noted that we actually need to decease the size of the time inteval until it is vey small fo Dv > to point diectly towad the cente. Fo the sake of discussion and illustating the concept, howeve, ou model in Figue 2 is satisfactoy. As you can see in Figue 2(a), the individual velocity vectos v > 1 and v > 2 ae both tangent to the cicle, pependicula to the cicle s adius, and equal in length (magnitude). This is tue fo all of the unne s velocity vectos along the cicula path. The acceleation vecto has the same diection as Dv >, so it follows that the centipetal acceleation of an object must always point towad the cente of the cicula path. y v 2 O s A v 1 A x O 2 v 1 C A v 2 v B x (a) (b) Figue 2 (a) The velocity v > of an object (in this case, a unne) moving with unifom cicula motion is shown as v > 1 and v > 2 at two diffeent locations along the cicula path. The distance tavelled in going fom point A9 to point A is s. (b) The diffeence in the velocity vectos, Dv >, is diected towad the cente of the cicle when the time inteval is vey small. Now use the tiangle BAC in Figue 2(b) to calculate the magnitude of the acceleation. This tiangle has two sides with equal lengths, 0v > 1 0 and 0v> 20. In geneal, the velocity magnitude is the same at any point aound the cicle, so 0v > v> v> 0, o simply v. The thid side of this tiangle is the vecto Dv >, which has a length of 0Dv > 0. The tiangle BAC has the same inteio angles as tiangle AOA9 in Figue 2(a), so these tiangles ae simila. You can check this by using the elations below to show that the angle θ is the same fo both tiangles. Note in Figue 2(b) that the angle at point A between v > 2 and the adius is 908, so u 2 1 a u 1 2a u a When you add the two equal angles, a, within tiangle BAC to the thid angle, they equal 1808, so the thid angle must equal a 5 u Two sides of tiangle AOA9 ae along the adius of the cicle, so they have length, while the othe side (between points A9 and A) has length s. When is small, the ac length between points A9 and A appoaches a staight-line length that connects A9 and A. Theefoe, the distance the unne tavels fom A9 to A is appoximately equal to the distance given by s < v. The tiangles BAC and AOA9 ae simila, so the atios of thei coesponding sides ae equal. Substituting v fo the magnitude of eithe v > 1 o v > 2, and assuming a vey small : 0Dv > 0 5 s v 0Dv > 0 5 v v 0Dv > 0 5 v 2 NEL 3.2 Centipetal Acceleation 115

3 The magnitude of the aveage acceleation equals the magnitude of the diffeence in the velocities (0Dv > 0) divided by : a av 5 0Dv> 0 v 2 5 a av 5 v 2 In the above equation, a av 5 when the time inteval is vey small: 5 v 2 peiod (T ) the time equied fo a otating, evolving, o vibating object to complete one cycle whee is the magnitude of the centipetal acceleation, v is the speed of the object moving along the cicula path, and is the adius of the cicula path. Note that, although this deivation stated with the definition of aveage acceleation, the esult becomes exact fo a vey small time inteval, so the centipetal acceleation in this case is an instantaneous quantity diected towad the cente of the cicle. The equation fo centipetal acceleation indicates that, when the speed of an object moving with unifom cicula motion is lage fo onstant adius, such as in the case of the hamme in the hamme thow, the diection of the velocity changes moe apidly than it would fo a smalle speed. This means that, to poduce these apid changes in velocity, you need a lage acceleation. When the adius is lage fo onstant speed, the diection of the velocity changes moe slowly, so the object has a smalle acceleation. Sometimes you may not know the speed of an object moving with unifom cicula motion. Howeve, you may be able to measue the time it takes fo the object to move once aound the cicle, o the peiod, T. Then you can calculate the speed. Remembe that the speed is constant, and that it equals the length of the path the object tavels (the cicumfeence of the cicle, o 2π) divided by the peiod, T: v 5 Dd Dd 5 2p and 5 T, so v 5 2p T Substitute the above expession fo v into the above equation fo centipetal acceleation to obtain the acceleation in tems of the peiod and the adius: 5 v 2 a 2p 2 T b 5 4p 2 2 T p2 T Chapte 3 Unifom Cicula Motion NEL

4 Fo high otational speeds, fequency is the pefeed quantity of measuement. The fequency, f, equals the numbe of evolutions pe unit of time, o f 5 1 T The unit of fequency is hetz (Hz), o cycles pe second. In tems of fequency and adius, the equation fo centipetal acceleation takes the fom 5 4p2 T 2 5 4p2 a 1 f b 2 fequency (f ) the numbe of otations, evolutions, o vibations of an object pe unit of time; the invese of peiod; SI unit Hz 5 4p2 1 f 2 5 4p 2 f 2 You now have thee equations fo detemining the magnitude of the centipetal acceleation. When dealing with the vecto of this acceleation, emembe that centipetal acceleation always points towad the cente of the cicle. The following Tutoial models how to solve poblems involving centipetal acceleation. Unit TASK BOOKMARK You can use some of the equations fo centipetal acceleation when you complete the Unit Task on page 146. Tutoial 1 Solving Poblems with Objects Moving with Centipetal Acceleation This Tutoial shows how to calculate the centipetal acceleation fo an object undegoing unifom cicula motion using the diffeent equations fo the magnitude of centipetal acceleation. Sample Poblem 1: Calculating the Magnitude of Centipetal Acceleation A child ides aousel with a adius of 5.1 m that otates with onstant speed of 2.2 m/s. Calculate the magnitude of the centipetal acceleation of the child. Given: m; v m/s Requied: Analysis: 5 v 2 Solution: 5 v m/s m m/s 2 Statement: The magnitude of the centipetal acceleation of the child is 0.95 m/s 2. Sample Poblem 2: Calculating the Magnitude and Diection of Centipetal Acceleation A salad spinne with a adius of 9.7 cm otates clockwise with a fequency of 12 Hz. At a given instant, the lettuce in the spinne moves in the westwad diection (Figue 3). Detemine the magnitude and diection of the centipetal acceleation of the piece of lettuce in the salad spinne at the moment shown in Figue 3. diection of otation v lettuce N Figue 3 NEL 3.2 Centipetal Acceleation 117

5 Given: cm m; f 5 12 Hz Requied: a > c Analysis: Fist, detemine the diection of the acceleation fom Figue 3. Then calculate the magnitude of the acceleation using the equation 5 4p 2 f 2. Solution: The westwad velocity vecto is at the south end of the spinne, as Figue 3 indicates. The diection of the centipetal acceleation is noth. 5 4p 2 f 2 5 4p m2 112 Hz m/s 2 Statement: The centipetal acceleation of the lettuce at the moment shown in Figue 3 is m/s 2 [N]. Sample Poblem 3: Calculating Fequency and Peiod of Rotation fo a Spinning Object The centipetal acceleation at the end of an electic fan blade has a magnitude of m/s 2. The distance between the tip of the fan blade and the cente is 12 cm. Calculate the fequency and the peiod of otation of the fan. Given: m/s 2 ; 5 12 cm m Requied: f ; T Analysis: Use the equation fo centipetal acceleation that includes fequency and adius: 5 4p 2 f 2 ; eaange and solve fo f. Then use the equation elating fequency and peiod to calculate the peiod of otation: T 5 1 f. 5 4p 2 f 2 4p 2 5 f 2 f 5 Å 4p 2 Solution: f 5 Å 4p 2 5 Å m/s 2 4p m Hz Choose the positive oot because fequency cannot be negative. f Hz 1one exta digit caied2 T 5 1 f Hz T s Statement: The fequency of the fan is 19 Hz, and the peiod of otation is s. Pactice 1. At a distance of 25 km fom the eye (cente) of a huicane, the wind moves at nealy 50.0 m/s. Assume that the wind moves in icula path. Calculate the magnitude of the centipetal acceleation of the paticles in the wind at this distance. T/I A [ans: 0.10 m/s 2 ] 2. An athlete in a hamme thow competition swings the hamme with unifom cicula motion clockwise as viewed fom above at a speed of 4.24 m/s and a distance of 1.2 m fom the cente of the cicle. At a given instant, the hamme s velocity is diected southwad. Detemine the centipetal acceleation at this instant. T/I [ans: 15 m/s 2 [W]] 3. A ball on a sting moves in a hoizontal cicle of adius 1.4 m. The centipetal acceleation of the ball has a magnitude of 12 m/s 2. Calculate the speed of the ball. T/I A [ans: 4.1 m/s] 4. The planet Venus moves in a nealy cicula obit aound the Sun. The aveage adius of its obit is m. The centipetal acceleation of Venus has a magnitude of m/s 2. Calculate Venus s peiod of evolution aound the Sun (a) in seconds and (b) in Eath days. T/I A [ans: (a) s; (b) 226 days] 5. Suppose a satellite evolves aound Eath in icula obit. The speed of the satellite is m/s, and the adius of its obit, with espect to Eath s cente, is m. Calculate the magnitude of the satellite s centipetal acceleation. T/I A [ans: 7.01 m/s 2 ] 6. A eseach appaatus called entifuge undegoes centipetal acceleation with a magnitude of m/s 2. The centifuge has a adius of 8.4 cm. Calculate the fequency of the centifuge (a) in hetz and (b) in evolutions pe minute (pm). T/I A [ans: (a) Hz; (b) pm] 118 Chapte 3 Unifom Cicula Motion NEL

6 3.2 Review Summay Unifom cicula motion is the motion of any body that follows icula path at onstant speed. Centipetal acceleation is the instantaneous acceleation of an object towad the cente of icula path. Thee ae thee equations to detemine centipetal acceleation: 5 v 2, 5 4p2 T 2, and 5 4p 2 f 2. Questions 1. You have a puck on a sting, and you twil the puck with unifom cicula motion in a hoizontal cicle along vitually fictionless ice. K/U T/I A (a) What causes the centipetal acceleation of the puck? (b) How does doubling the adius of the cicle and leaving the speed unchanged affect the centipetal acceleation? (c) How does doubling the speed and leaving the adius unchanged affect the centipetal acceleation? 2. Two athletes compete in the hamme thow. One athlete can spin the hamme twice as fast as the second athlete. Compae the magnitudes of the two centipetal acceleations fo the two hamme thows. Explain you answe. T/I C A 3. In a odeo, a pefome twils a lasso (ope) at a constant speed, and the lasso tuns in icle of adius 0.42 m. The lasso has a peiod of otation of 1.5 s. Calculate the magnitude of the centipetal acceleation of the lasso. T/I A 4. A motocyclist maintains onstant speed of 28 m/s while acing on icula tack with onstant adius of 135 m. Calculate the magnitude of the centipetal acceleation of the motocyclist. T/I A 5. The centipetal acceleation of an object at Eath s equato esults fom the daily otation of Eath. Calculate the object s centipetal acceleation, given that the adius of Eath at the equato is m. T/I A 6. An amusement pak ide consists of a otating cylinde with oase fabic on the walls, fo fiction. Paticipants on this ide stand against the wall as the cylinde otates. Afte the cylinde eaches onstant speed, the floo of the ide dops away beneath the occupants. They emain against the wall because of the centipetal acceleation, which must be geate than about 25 m/s 2. This ide has a adius of 2.0 m. Detemine the minimum fequency of otation of the cylinde. T/I A 7. The centipetal acceleation of a moving aound iculuve at onstant speed of 22 m/s has a magnitude of 7.8 m/s 2. Calculate the adius of the cuve. T/I A 8. A jogge is unning aound icula tack that has icumfeence of 478 m. The magnitude of the centipetal acceleation of the jogge is m/s 2. Calculate the jogge s speed in kilometes pe hou. T/I A 9. A bicycle wheel with a adius of m is spinning clockwise at a ate of 60.0 pm. T/I A (a) Calculate the peiod of the wheel s motion. (b) Calculate the centipetal acceleation of a point on the edge of the wheel if at that instant it moves westwad. 10. The Moon s peiod of evolution is 27.3 days, and the magnitude of its centipetal acceleation is about m/s 2. T/I A (a) Calculate the distance between the cente of the Moon and the cente of Eath. Assume that the obit of the Moon is cicula and that its speed is constant. (b) Compae you answe with the value povided in Appendix B. If diffeent, suggest easons why. 11. The ecod distance fo the hamme thow is about 87 m. To achieve this distance, an athlete must poduce entipetal acceleation of nealy 711 m/s 2. K/U T/I A (a) Given a adius of 1.21 m, calculate the speed of the ball when it is eleased. (b) The athlete lets go when the ball is 2.0 m above the gound and moving at an angle of 428 above the hoizontal. Detemine the ange. Ignoe any ai fiction. NEL 3.2 Centipetal Acceleation 119

( ) ( ) 1.4 m ( ) Section 3.2: Centripetal Acceleration Tutorial 1 Practice, page Given: r = 25 km = m; v = 50.0 m/s. Required: a!

( ) ( ) 1.4 m ( ) Section 3.2: Centripetal Acceleration Tutorial 1 Practice, page Given: r = 25 km = m; v = 50.0 m/s. Required: a! Section 3.2: Centipetal Acceleation Tutoial 1 Pactice, page 118 1. Given: 25 km 2.5 10 4 m; v 50.0 m/s Requied: Analysis: Solution: ( 50.0 m/s) 2 2.5!10 4 m 0.10 m/s 2 Statement: The magnitude of the centipetal