Slide 1 / 71 Slide 2 / 71 P Physics 1 irculr Motion 2015-12-02 www.njctl.org Topics of Uniform irculr Motion (UM) Slide 3 / 71 Kinemtics of UM lick on the topic to go to tht section Period, Frequency, nd Rottionl Velocity ynmics of UM Verticl UM uckets of Wter Rollercosters rs going over hills nd through vlleys Horizontl UM Unbnked urves nked urves onicl Pendulum
Slide 4 / 71 Kinemtics of UM Return to Tble of ontents Kinemtics of Uniform irculr Motion Slide 5 / 71 Uniform circulr motion: motion in circle of constnt rdius t constnt speed Instntneous velocity is lwys tngent to circle. Kinemtics of Uniform irculr Motion Slide 6 / 71 This ccelertion is clled the centripetl, or rdil, ccelertion, nd it points towrds the center of the circle.
Kinemtics of Uniform irculr Motion Slide 7 / 71 Looking t the chnge in velocity in the limit tht the time intervl becomes infinitesimlly smll, we see tht we get two similr tringles. # l r ## r r r ## Kinemtics of Uniform irculr Motion Slide 8 / 71 If displcement is equl to velocity multiplied by time, then vt is the displcement covered in time t. uring tht sme time, velocity chnged by n mount, Δv. r # vt v 1 r v 2 # v 1 v 2 # v Kinemtics of Uniform irculr Motion These re similr tringles becuse the ngles re ll congruent, so the sides must be in proportion. # v vt = v r Δv vt Slide 9 / 71 # v = v 2 t r = v 2 r Tht's the mgnitude of the ccelertion. v 1 v 2 θ r θ r
v1 Kinemtics of Uniform irculr Motion Trnspose v 2 to see the vector ddition. Slide 10 / 71 v1 v2 v2 # v θ θ The chnge in velocity, # v, shows the direction of the ccelertion. In the picture, you cn see # v points towrds the center of the circle. Kinemtics of Uniform irculr Motion Slide 11 / 71 This ccelertion is clled the centripetl, or rdil, ccelertion. It direction is towrds the center of the circle. It's mgnitude is given by 1 Is it possible for n object moving with constnt speed to ccelerte? Explin. Slide 12 / 71 No, if the speed is constnt then the ccelertion is equl to zero. No, n object cn ccelerte only if there is net force cting on it. Yes, lthough the speed is constnt, the direction of the velocity cn be chnging. Yes, if n object is moving it is experiencing ccelertion.
1 Is it possible for n object moving with constnt speed to ccelerte? Explin. Slide 12 () / 71 No, if the speed is constnt then the ccelertion is equl to zero. No, n object cn ccelerte only if there is net force cting on it. Yes, lthough the speed is constnt, the direction of the velocity cn be chnging. Yes, if n object is moving it is experiencing ccelertion. 2 onsider prticle moving with constnt speed such tht its ccelertion of constnt mgnitude is lwys perpendiculr to its velocity. Slide 13 / 71 It is moving in stright line. It is moving in circle. It is moving in prbol. None of the bove is definitely true ll of the time. 2 onsider prticle moving with constnt speed such tht its ccelertion of constnt mgnitude is lwys perpendiculr to its velocity. Slide 13 () / 71 It is moving in stright line. It is moving in circle. It is moving in prbol. None of the bove is definitely true ll of the time.
3 n object moves in circulr pth t constnt speed. ompre the direction of the object's velocity nd ccelertion vectors. Slide 14 / 71 oth vectors point in the sme direction. The vectors point in opposite directions. The vectors re perpendiculr. The question is meningless, since the ccelertion is zero. 3 n object moves in circulr pth t constnt speed. ompre the direction of the object's velocity nd ccelertion vectors. Slide 14 () / 71 oth vectors point in the sme direction. The vectors point in opposite directions. The vectors re perpendiculr. The question is meningless, since the ccelertion is zero. Slide 15 / 71 4 Two crs go round the sme circulr trck t the sme speed. The first cr is closer to the inside of the circle. Which of the following is true bout their centripetl ccelertion? oth crs hve the sme centripetl motion since they both hve the sme speed. The centripetl ccelertion of the first cr is greter since its rdius is smller. The centripetl ccelertion of the second cr is greter since its rdius is lrger. The centripetl ccelertion of the first cr is less since its rdius is smller.
4 Two crs go round the sme circulr trck t the sme speed. The first cr is closer to the inside of the circle. Which of the following is true bout their centripetl ccelertion? oth crs hve the sme centripetl motion since they both hve the sme speed. The centripetl ccelertion of the first cr is greter since its rdius is smller. The centripetl ccelertion of the second [This object cr is pull tb] is greter since its rdius is lrger. The centripetl ccelertion of the first cr is less since its rdius is smller. Slide 15 () / 71 Slide 16 / 71 Period, Frequency, nd Rottionl Velocity Return to Tble of ontents Period Slide 17 / 71 The time it tkes for n object to complete one trip round circulr pth is clled its Period. The symbol for Period is "T" Periods re mesured in units of time; we will usully use seconds (s). Often we re given the time (t) it tkes for n object to mke number of trips (n) round circulr pth. In tht cse,
Frequency The number of revolutions tht n object completes in given mount of time is clled the frequency of its motion. Slide 18 / 71 The symbol for frequency is "f" Frequency re mesured in units of revolutions per unit time; we will usully use 1/seconds (s -1 ). nother nme for s -1 is Hertz (Hz). Frequency cn lso be mesured in revolutions per minute (rpm), etc. Often we re given the time (t) it tkes for n object to mke number of revolutions (n). In tht cse, Period nd Frequency Slide 19 / 71 Period Frequency These two equtions look similr. In fct, they re exctly opposite one nother. nother wy to sy this is tht they re inverses. Period nd Frequency Slide 20 / 71 We cn relte them mthemticlly: Period is the inverse of Frequency Frequency is the inverse of Period
Rottionl Velocity Slide 21 / 71 In kinemtics, we defined the speed of n object s For n object moving in circle, insted of using t (time), we will mesure the speed with T (period), the time it tkes to trvel round circle. To find the speed, then, we need to know the distnce round the circle. nother nme for the distnce round circle is the circumference. Rottionl Velocity Slide 22 / 71 Ech trip round circle, n object trvels length equl to the circle's circumference. The circumference of circle is given by The time it tkes to go round once is the period nd the object's speed is given by So the mgnitude of its velocity must be: Rottionl Velocity Slide 23 / 71 velocity must hve mgnitude nd direction. The mgnitude of n object's instntneous velocity is its speed. So for n object in uniform circulr motion, the mgnitude of its velocity is: If n object is in uniform circulr motion, the direction of its velocity is lwys chnging! We sy the velocity is tngent to its circulr motion.
* Rottionl Velocity Slide 24 / 71 Since, we cn lso determine the velocity of n object in uniform circulr motion by the rdius nd frequency of its motion. nd so Of course the direction of its velocity is still tngent to its circulr motion. 5 girl whirls toy t the end of string round her hed. The string mkes one complete revolution every second. She keeps the rdius constnt but increses the speed so tht the string mkes two complete revolutions per second. Wht hppens to the centripetl ccelertion? The centripetl ccelertion remins the sme. The centripetl ccelertion doubles. The centripetl ccelertion qudruples. The centripetl ccelertion is cut to hlf. Slide 25 / 71 5 girl whirls toy t the end of string round her hed. The string mkes one complete revolution every second. She keeps the rdius constnt but increses the speed so tht the string mkes two complete revolutions per second. Wht hppens to the centripetl ccelertion? The centripetl ccelertion remins the sme. The centripetl ccelertion doubles. The centripetl ccelertion qudruples. The centripetl ccelertion is cut to hlf. Slide 25 () / 71
6 bll is swung in circle. The frequency of the bll is doubled. y wht fctor does the period chnge? Slide 26 / 71 It remins the sme. It is cut to one hlf. It doubles. It is cut to one fourth. 6 bll is swung in circle. The frequency of the bll is doubled. y wht fctor does the period chnge? Slide 26 () / 71 It remins the sme. It is cut to one hlf. It doubles. It is cut to one fourth. Slide 27 / 71 ynmics of UM Return to Tble of ontents
ynmics of Uniform irculr Motion Slide 28 / 71 For n object to be in uniform circulr motion, there must be net force cting on it. We lredy know the ccelertion, so we cn write the force: ynmics of Uniform irculr Motion Slide 29 / 71 We cn see tht the force must be inwrd by thinking bout bll on string: Force on hnd exerted by string Force on bll exerted by string ynmics of Uniform irculr Motion There is no centrifugl force pointing outwrd; wht hppens is tht the nturl tendency of the object to move in stright line must be overcome. If the centripetl force vnishes, the object flies off tngent to the circle. This hppens. Slide 30 / 71 This does NOT hppen.
urved Pths Slide 31 / 71 This concept cn be used for n object moving long ny curved pth, s smll segment of the pth will be pproximtely circulr. entrifugtion Slide 32 / 71 centrifuge works by spinning very fst. This mens there must be very lrge centripetl force. The object t would go in stright line but for this force; s it is, it winds up t. 7 When n object experiences uniform circulr motion, the direction of the net force is: Slide 33 / 71 in the sme direction s the motion of the object. in the opposite direction of the motion of the object. is directed towrd the center of the circulr pth. is directed wy from the center of the circulr pth.
7 When n object experiences uniform circulr motion, the direction of the net force is: Slide 33 () / 71 in the sme direction s the motion of the object. in the opposite direction of the motion of the object. is directed towrd the center of the circulr pth. is directed wy from the center of the circulr pth. 8 boy whirls toy t the end of string round his hed. The string mkes one complete revolution every second. He keeps the rdius constnt but decreses the speed so tht the string mkes one revolution every two seconds. Wht hppens to the tension in the string? The tension remins the sme. The tension doubles. The tension is cut to hlf. The tension is cut to one fourth. Slide 34 / 71 8 boy whirls toy t the end of string round his hed. The string mkes one complete revolution every second. He keeps the rdius constnt but decreses the speed so tht the string mkes one revolution every two seconds. Wht hppens to the tension in the string? The tension remins the sme. The tension doubles. The tension is cut to hlf. The tension is cut to one fourth. Slide 34 () / 71
9 (Multi-correct irections: For ech of the following, two of the suggested nswers will be correct. Select the best two choices to ern credit. No prtil credit will be erned if only one correct choice is selected.) n object is moving in uniform circulr motion with mss m, speed v, nd rdius r. Which of the following will qudruple the centripetl force on the object? oubling the speed. utting the speed to one hlf. utting the rdius to one hlf. utting the rdius to one fourth. Slide 35 / 71 9 (Multi-correct irections: For ech of the following, two of the suggested nswers will be correct. Select the best two choices to ern credit. No prtil credit will be erned if only one correct choice is selected.) n object is moving in uniform circulr motion with mss m, speed v, nd rdius r. Which of the following will qudruple the centripetl force on the object? oubling the speed. utting the speed to one hlf. utting the rdius to one hlf. utting the rdius to one fourth. nd Slide 35 () / 71 Slide 36 / 71 Verticl UM Return to Tble of ontents
r on hilly rod Slide 37 / 71 When cr goes though dip, we cn consider it to be in circulr motion. Its ccelertion is towrds the center of the circle, which is up. We cn use free body digrm nd Newton's second lw to derive n eqution for the norml force on the cr. F N r on hilly rod Slide 38 / 71 When cr goes over hill, we cn consider it to be in circulr motion. Its ccelertion is towrds the center of the circle, which is down. We cn use free body digrm nd Newton's second lw to derive n eqution for the norml force on the cr. F N uckets nd Rollercosters Slide 39 / 71 bucket on string moving in verticl circle is lso in circulr motion. When it is t the bottom of the circle, it is in the sme sitution t cr going through dip. F T
uckets nd Rollercosters Slide 40 / 71 bucket on string moving in verticl circle is lso in circulr motion. When it is t the top of the circle, there is no force upwrd. The tension nd weight re both down. F T uckets nd Rollercosters Slide 41 / 71 The minimum velocity for bucket to mke it round the circle is chieved when the tension in the string becomes zero t the top of the circle. uckets nd Rollercosters Slide 42 / 71 roller coster cr going round loop works exctly like bucket on string. The only difference is tht insted of tension, there is norml force exerted on the cr. F T F T
10 roller coster cr is on trck tht forms circulr loop in the verticl plne. If the cr is to just mintin contct with trck t the top of the loop, wht is the minimum vlue for its centripetl ccelertion t this point? g downwrd 0.5g downwrd g upwrd 2g upwrd Slide 43 / 71 10 roller coster cr is on trck tht forms circulr loop in the verticl plne. If the cr is to just mintin contct with trck t the top of the loop, wht is the minimum vlue for its centripetl ccelertion t this point? g downwrd 0.5g downwrd g upwrd 2g upwrd Slide 43 () / 71 11 roller coster cr (mss = M) is on trck tht forms circulr loop (rdius = r) in the verticl plne. If the cr is to just mintin contct with the trck t the top of the loop, wht is the minimum vlue for its speed t tht point? Slide 44 / 71
11 roller coster cr (mss = M) is on trck tht forms circulr loop (rdius = r) in the verticl plne. If the cr is to just mintin contct with the trck t the top of the loop, wht is the minimum vlue for its speed t tht point? Slide 44 () / 71 12 pilot executes verticl dive then follows semicirculr rc until it is going stright up. Just s the plne is t its lowest point, the force of the set on him is: Slide 45 / 71 less thn nd pointing up. less thn nd pointing down. more thn nd pointing up. more thn nd pointing down. 12 pilot executes verticl dive then follows semicirculr rc until it is going stright up. Just s the plne is t its lowest point, the force of the set on him is: Slide 45 () / 71 less thn nd pointing up. less thn nd pointing down. more thn nd pointing up. more thn nd pointing down.
13 (Short nswer) bll is whirled on string s shown in the digrm. The string breks t the point shown. rw the pth the bll tkes nd explin wht this hppens. Slide 46 / 71 13 (Short nswer) bll is whirled on string s shown in the digrm. The string breks t the point shown. The initil velocity is tngent to the circle rw the pth the bll tkes nd explin wht this but there is lso grvittionl force hppens. which cuses the bll to move in prbol. The horizontl component of the velocity is constnt but the verticl component is subject to n ccelertion of 9.8 m/s 2 downwrd. Slide 46 () / 71 Slide 47 / 71 Horizontl UM Return to Tble of ontents
nked nd Unbnked urves Slide 48 / 71 When cr goes round curve, there must be net force towrds the center of the circle of which the curve is n rc. If the rod is flt, tht force is supplied by friction. nked nd Unbnked urves Slide 49 / 71 If the frictionl force is insufficient, the cr will tend to move more nerly in stright line, s the skid mrks show. nked nd Unbnked urves Slide 50 / 71 s long s the tires do not slip, the friction is sttic. If the tires do strt to slip, the friction is kinetic, which is bd in two wys: 1. The kinetic frictionl force is smller thn the sttic. 2. The sttic frictionl force cn point towrds the center of the circle, but the kinetic frictionl force opposes the direction of motion, mking it very difficult to regin control of the cr nd continue round the curve.
r Unbnked urves Slide 51 / 71 Top View Front View (the cr heding towrds you) F N f s v 14 cr goes round curve of rdius r t constnt speed v. Then it goes round the sme curve t hlf of the originl speed. Wht is the centripetl force on the cr s it goes round the curve for the second time, compred to the first time? Slide 52 / 71 twice s big four times s big hlf s big one-fourth s big 14 cr goes round curve of rdius r t constnt speed v. Then it goes round the sme curve t hlf of the originl speed. Wht is the centripetl force on the cr s it goes round the curve for the second time, compred to the first time? Slide 52 () / 71 twice s big four times s big hlf s big one-fourth s big
15 The top speed cr cn go round n unbnked curve sfely (without slipping) depends on ll of the following except: The coefficient of friction. The mss of the cr. The rdius of the curve. The ccelertion due to grvity. Slide 53 / 71 15 The top speed cr cn go round n unbnked curve sfely (without slipping) depends on ll of the following except: The coefficient of friction. The mss of the cr. The rdius of the curve. The ccelertion due to grvity. Slide 53 () / 71 16 You re sitting in the pssenger set of cr going round turn. Which of the following is responsible for keeping you moving in circle? Slide 54 / 71 The grvittionl force. The outwrd force of the cr set nd set belt. The inwrd force of the cr set nd set belt. The centrifugl force.
r 16 You re sitting in the pssenger set of cr going round turn. Which of the following is responsible for keeping you moving in circle? Slide 54 () / 71 The grvittionl force. The outwrd force of the cr set nd set belt. The inwrd force of the cr set nd set belt. The centrifugl force. nked urves Slide 55 / 71 nking curves cn help keep crs from skidding. In fct, for every bnked curve, there is one speed where the entire centripetl force is supplied by the horizontl component of the norml force, nd no friction is required. Let's figure out wht tht speed is for given ngle nd rdius of curvture. nked urves Slide 56 / 71 Top View Front View (the cr heding towrds you) y v x We know the direction of our ccelertion, so now we hve to crete xes. # Note tht these xes will be different thn for n Inclined Plne, becuse the cr is not expected to slide down the plne. Insted, it must hve horizontl ccelertion to go in horizontl circle.
nked urves Slide 57 / 71 verticl Next, do the free body digrm. y rdil Let's ssume tht no friction is necessry t the speed we re solving for. x # nked urves Slide 58 / 71 verticl Next, decompose the forces tht don't lign with n xis, F N. # F N y rdil x # nked urves Slide 59 / 71 F Ncos# verticl # F N Now let's solve for the velocity such tht no friction is necessry to keep cr on the rod while going round curve of rdius r nd bnking ngle θ. F Nsin# rdil
nked urves Slide 60 / 71 verticl verticl direction rdil direction F Ncos# # F N F Nsin# rdil 17 Which of the following is responsible for how cr stys in plce on frictionless bnked curve? Slide 61 / 71 The verticl component of the cr's weight. The horizontl component of the cr's weight. The verticl component of the norml force. The horizontl component of the norml force. 17 Which of the following is responsible for how cr stys in plce on frictionless bnked curve? Slide 61 () / 71 The verticl component of the cr's weight. The horizontl component of the cr's weight. The verticl component of the norml force. The horizontl component of the norml force.
18 etermine the velocity tht cr should hve while trveling round frictionless curve with rdius 250 m is bnked t n ngle of 15 degrees. Slide 62 / 71 18 etermine the velocity tht cr should hve while trveling round frictionless curve with rdius 250 m is bnked t n ngle of 15 degrees. Slide 62 () / 71 19 Two bnked curves hve the sme rdius. urve is bnked t n ngle of 37 degrees, nd curve is bnked t n ngle of 53 degrees. cr cn trvel round curve without relying on friction t speed of 30 m/s. t wht speed cn this cr trvel round curve without relying on friction? 20 m/s 30 m/s 40 m/s 60 m/s Slide 63 / 71
19 Two bnked curves hve the sme rdius. urve is bnked t n ngle of 37 degrees, nd curve is bnked t n ngle of 53 degrees. cr cn trvel round curve without relying on friction t speed of 30 m/s. t wht speed cn this cr trvel round curve without relying on friction? 20 m/s 30 m/s 40 m/s 60 m/s Slide 63 () / 71 onicl Pendulum Slide 64 / 71 onicl Pendulum is pendulum tht sweeps out circle, rther thn just bck nd forth pth. Since the pendulum bob is moving in horizontl circle, we cn study it s nother exmple of uniform circulr motion. However, it will prove necessry to decompose the forces. onicl Pendulum Slide 65 / 71 # rw sketch of the problem, unless one is provided. l Then drw free body digrm nd indicte the direction of the ccelertion.
onicl Pendulum Slide 66 / 71 Then drw xes with one xis prllel to ccelertion Tcos# # Then decompose forces so tht ll components lie on n xis...in this cse, decompose T. Tsin# Then solve for the ccelertion of the bob, bsed on the ngle #, by pplying Newton's Second Lw long ech xis. onicl Pendulum x - direction y - direction Slide 67 / 71 Tcos# # Tsin# onicl Pendulum Slide 68 / 71 # T n lterntive pproch is to solve this s vector eqution using #F=m. (This will work whenever only two forces re present.) Just trnslte the originl vectors (before decomposing them) to form right tringle so tht the sum of the two forces equls the new vector "m".
onicl Pendulum Slide 69 / 71 # m T nd to find tension... 20 0.5 kg object is whirled t the end of rope in conicl pendulum with rdius of 2 m t speed of 4 m/s. Wht is the tension in the rope? Slide 70 / 71 20 0.5 kg object is whirled t the end of rope in conicl pendulum with rdius of 2 m t speed of 4 m/s. Wht is the tension in the rope? Slide 70 () / 71
21 (Short nswer) Explin why it is impossible to whirl n object t the end of string in horizontl circle. Slide 71 / 71 21 (Short nswer) Explin why it is impossible to whirl n object t the end of string in horizontl circle. Slide 71 () / 71 If the string is horizontl, there is no verticl component to support the weight of the object. The string must be t n ngle, however slight, in order to hve verticl component to support the weight of the object.