Flipping Phyic Lecture Nte: Intrductin t Circular Mtin and Arc Length Circular Mtin imply take what yu have learned befre and applie it t bject which are mving alng a circular path. Let begin with a drawing f an bject which i mving alng a circle with a cntant iu. We have paued the bject at ne pint in time t dicu hw we identify the lcatin f an bject which i mving alng a circle. The x-y crdinate ytem which lcate an bject in tw r three dimeninal pace relative t an rigin wa intrduced by René Decarte in the 1600. When we identify the lcatin f an bject which i mving alng a circle uing Carteian crdinate, ntice that bth the x and y pitin value f the bject change a a functin f time. We can al identify the bject lcatin uing plar crdinate which ue the iu and angular pitin t identify the lcatin f the bject. Ntice when we ue iu and angular pitin t identify the lcatin f an bject mving alng a circle, the angular pitin change, hwever, the iu tay cntant. Having nly ne variable change a a functin f time while decribing the lcatin f an bject i much eaier t wrk with than when tw variable change. We can relate Carteian and plar crdinate uing trig functin: in θ = O y A x = y = r in θ & c θ = = x = r c θ H r H r Nw let dicu hw we decribe an bject mving frm ne lcatin t anther when it i mving alng a circle with a cntant iu. In the drawing the bject ha mved frm the initial pitin t the final pitin while mving thrugh an angular diplacement: Δθ = θ f θ i Arc Length: The linear ditance traveled when an bject i mving alng a curve. The ymbl fr arc length i. The equatin fr arc length i = rδθ Yu mut ue ian fr the angular diplacement in the arc length equatin. We will demntrate why n in a future vide. The arc length when an bject mve thrugh a full circle i called Circumference. The angular diplacement when an bject mve thrugh a full circle i 2! ian. The equatin fr circumference i ( ) C = 2π r = r 2π, which yu can ee i a pecial cae f the equatin fr arc length when the angular diplacement i ne revlutin. 0206 Lecture Nte - Intrductin t Circular Mtin and Arc Length.dcx page 1 f 1
Defining Pi fr Phyic Cmmn tudent anwer t the quetin, What i!? A number 3 3.1 3.14 ~3.141592653589793238462643383279502884197169399375105820974944592307816406286 An irratinal number Smething gd t eat By definitin pi i the rati f a circle circumference t it diameter: π = C D = 3.14159... Which we can rearrange C = π D = π ( 2r) C = 2π r t get the equatin fr circumference The equatin fr circumference i jut a retatement f the definitin f! Fribee example: π = C D = 86.9cm = 3.16 3.14159... 27.5cm The unit fr! are π = C D meter meter =1 In ther wrd! ha n unit, it i dimeninle We give thi rati a pecific name, it i called ian C D = π ian! i in ian and ian are dimeninle.! ian repreent the rati f the circumference t the diameter f every circle. Radian are a placehlder and we will ue thi fact repeatedly in phyic. 1 revlutin = 360 = 2π ian Knw thi!! Nte: 1 revlutin 2 ian Fr me rean tudent ften imply leave the! ut, dn t be that tudent. Abbreviatin: r = iu = ian d NOT ue r fr ian, r i fr iu, i fr ian. = rδθ = 1.5m ( )( 2π r) lead t r cnfuin, = rδθ = ( 1.5m) ( 2π ) de nt. 0207 Lecture Nte - Defining Pi fr Phyic.dcx page 1 f 1
Flipping Phyic Lecture Nte: Intrductry Arc Length Prblem Gum n a Bike Tire Example Prblem: Hw far de a piece f gum tuck t the utide f a 67 cm diameter wheel travel while the wheel rtate thrugh 149? Knwn: D = 67cm; r = 2π ian D 67cm = = 33.5cm; Δθ = 149 = 2.6005 ian; =? 2 2 360 Suggetin: Whenever the diameter i given in a phyic prblem, immediately determine the iu a well. T ften I have een tudent ue the diameter a the iu. ( )( ) = rδθ = 33.5 2.6005 = 87.118 87cm Unit: = rδθ cm i = cm Radian have n unit and are jut a placehlder. The ian drp ut becaue we n lnger need them a a placehlder. If we had left the angular diplacement in degree, the unit fr arc length wuld wrk ut t be in cm i which make n ene. 0208 Lecture Nte - Intrductry Arc Length Prblem Gum n a Bike Tire.dcx page 1 f 1
Angular Velcity Intrductin The equatin fr average linear velcity i: v! avg = Δ! x Δt Average linear velcity equal change in linear pitin ver change in time. Therefre the equatin fr average angular velcity i: ω! avg = Δ! θ Δt Average angular velcity equal change in angular pitin ver change in time. The ymbl fr angular velcity i the lwercae Greek letter mega,ω. The tw mt cmmn unit fr angular velcity are: ued mt ften phyic. revlutin minunte = rev = rpm ued mt ften in the real wrld. min Nte: Fr quite a while we will nt dicu the directin f angular quantitie like angular diplacement and angular velcity. In my experience it i much eaier fr tudent t get t knw the angular variable withut directin firt and then intrduce directin. Have patience, directin will cme n enugh. Fr the f yu wh are nt atified with that, undertand that clckwie and cunterclckwie are berver dependent and therefre we will be uing the Right Hand Rule intead f clckwie and cunterclckwie, but nt yet. Example f bject with angular velcity: 0209 Lecture Nte - Angular Velcity Intrductin.dcx page 1 f 1
Intrductry Angular Velcity Prblem A Turning Bike Tire Example: The wheel f a bike rtate exactly 3 time in 12.2 ecnd. What i the average angular velcity f the wheel in (a) ian per ecnd and (b) revlutin per minute? Knwn: Δθ = "exactly"3 rev; Δt =12.2ec; ω avg =? a ( ) & b ( ) rev min Nte: Unfrtunately the wrd exactly i metime ued in phyic prblem and it mean the number referred t ha an infinite number f ignificant digit. Hpefully yu recgnize thi i impible. (a) ω avg = Δθ Δt = 3 rev 12.2ec (b) ω avg =1.54505 = 0.24590 rev 60 1rev 1min 2π 2π 1rev =1.54505 1.55 =14.7541 14.8 rev min Typical mitake with thi cnverin: 1) Frget t include the parenthei arund 2! in yur calculatr and therefre actually divide by tw and then multiply by! reult in 145.617 which i nt crrect. 2) Frget t include the number! in yur calculatin which reult in 46.351 which i al nt crrect. 3) Add! t yur anwer even thugh yu typed it in t yur calculatr and therefre already ued it value reult in 14.8! which i, yu gueed it, al nt crrect. 0210 Lecture Nte - Intrductry Angular Velcity Prblem - A Turning Bike Tire.dcx page 1 f 1
Angular Acceleratin Intrductin The equatin fr average linear acceleratin i: a! avg = Δ! v Δt Average linear acceleratin equal change in linear velcity ver change in time. The equatin fr average angular acceleratin i: α! avg = Δ ω! Δt Average angular acceleratin equal change in angular velcity ver change in time. The ymbl fr angular acceleratin i the lwercae Greek letter alpha, α. We ften call it fihy thing intead. Unit fr angular acceleratin are: 2 Jut like I mentined when learning abut angular velcity, we are nt ging t dicu directin until later. 0211 Lecture Nte - Angular Acceleratin Intrductin.dcx page 1 f 1
Angular Acceleratin f a Recrd Player Example Prblem: A recrd player i plugged in, unifrmly accelerate t 45 revlutin per minute, and then i unplugged. The recrd player (a) take 0.85 ecnd t get up t peed, (b) pend 3.37 ecnd at 45 rpm, and then (c) take 2.32 ecnd t lw dwn t a tp. What i the average angular acceleratin f the recrd player during all three part? Part (a) knwn: ω ai = 0; ω af = 45 rev 1min min 60ec α a = Δω a Δt a 2π 1rev = ω ω af ai = 1.5π 0 = 5.54399 5.5 Δt a 0.85 2 Part (b) knwn: ω b = cntant Δω b = 0 α b = Δω b Δt b = 0 Δt b = 0 Part (c) knwn: ω ci =1.5π ec ; ω cf = 0; Δt c = 2.32ec; α c =? α c = Δω c Δt c = ω cf ω ci Δt c = 0 1.5π = 2.03120 2.0 2.32 2 =1.5π ec ; Δt a = 0.85ec; α a =? 0212 Lecture Nte - Angular Acceleratin f a Recrd Player.dcx page 1 f 1
Intrductin t Unifrmly Angularly Accelerated Mtin Jut like an bject can have Unifrmly Accelerated Mtin r UAM and bject can have Unifrmly Angularly Accelerated Mtin, UαM. Thi table cmpare the tw: If the fllwing i cntant linear acceleratin, a angular acceleratin, α we can ue the equatin UAM UαM There are 5 variable v f, v i, a, Δx, Δt ω f, ω i, α, Δθ, Δt There are 4 equatin v f = v i + aδt Δx = v i Δt + 1 2 aδt 2 v f 2 = v i 2 + 2aΔx Δx = 1 ( 2 v + v f i )Δt ω f = ω i + αδt Δθ = ω i Δt + 1 2 αδt 2 ω f 2 = ω i 2 + 2αΔθ Δθ = 1 ( 2 ω + ω f i )Δt Ue Radian! If we knw 3 f the variable we can find the ther 2. Which leave u with 1 Happy Phyic Student! When yu ue the Unifrmly Angularly Accelerated Mtin equatin pleae ue ian fr yur angular quantitie. Mt f the time yu have t ue ian in the UαM equatin and yu alway can ue ian in the UαM equatin. Therefre, pleae, alway ue ian in the UαM equatin. 0213 Lecture Nte - Intrductin t Unifrmly Angularly Accelerated Mtin.dcx page 1 f 1
Intrductry Unifrmly Angularly Accelerated Mtin Prblem Example Prblem: What i the angular acceleratin f a cmpact dic that turn thrugh 3.25 revlutin while it unifrmly lw t a tp in 2.27 ecnd? 2π Knwn: α =?; Δθ = 3.25rev 1rev = 6.5π ; ω = 0; Δt = 2.27ec f A cmpact dic will lw with a cntant angular acceleratin we can ue the Unifrmly Angularly Accelerated Mtin (UαM) equatin. There i n UαM equatin that ha all fur f ur knwn variable in it, we firt need t lve fr angular velcity initial. Δθ = 1 ( 2 ω + ω f i )Δt 6.5π = 1 ( 2 0 + ω i )2.27 ω i = 2 ( )( 6.5π ) 2.27 =17.9915 And nw that we have the initial angular velcity, we can lve fr the angular acceleratin. α = ω f ω i Δt = 0 17.9915 2.27 = 7.9258 7.93 2 0214 Lecture Nte - Intrductry Unifrmly Angularly Accelerated Mtin Prblem.dcx page 1 f 1
Flipping Phyic Lecture Nte: Human Tangential Velcity Demntratin In the demntratin: Each pern ha the ame angular velcity. A larger iu mean a larger arc length r linear ditance travelled. Therefre a larger iu mean a larger linear velcity called tangential velcity. 0215 Lecture Nte - Human Tangential Velcity Demntratin.dcx page 1 f 1
Intrductry Tangential Velcity Prblem Example Prblem: Three mint are itting 3.0 cm, 8.0 cm, and 13.0 cm frm the center f a recrd player that i pinning at 45 revlutin per minute. What are the tangential velcitie f each mint? Knwn: r 1 = 3.0cm; r 2 = 8.0cm; r 3 =13.0cm; ω = 45 rev min cm v t1 = r 1 ω = ( 3) ( 1.5π ) =14.1372 14 cm v t2 = r 2 ω = ( 8) ( 1.5π ) = 37.6991 38 cm v t3 = r 3 ω = ( 13) ( 1.5π ) = 61.2611 61 cm 2π 1rev 1min 60ec =1.5π ; v =? each t ( ) The tangential velcity f an bject i, by definitin, tangent t the circle the bject i decribing. Thi mean tangential velcity i, by definitin, at a 90 angle t the iu f the circle. 0216 Lecture Nte - Intrductry Tangential Velcity Prblem.dcx page 1 f 1
Intrductin t Tangential Acceleratin with Recrd Player Example Prblem When an bject i mving in a circle it travel a linear ditance which i called arc length. = rδθ it ha a linear velcity which i called tangential velcity. v t = rω it ha a linear acceleratin which i called tangential acceleratin. a t = rα Yu mut ue ian in all three f thee equatin! Example Prblem: A recrd player i plugged in and unifrmly accelerate t 45 revlutin per minute in 0.85 ecnd. Mint are lcated 3.0 cm, 8.0 cm, and 13.0 cm frm the center f the recrd. What i the magnitude f the tangential acceleratin f each mint? Knwn: ω i = 0; ω f = 45 rev min 2π 1rev 1min 60ec ( ) r 1 = 3.0cm; r 2 = 8.0cm; r 3 =13.0cm; a t =? each =1.5π ; Δt = 0.85ec; Firt we need t lve fr the angular acceleratin f the recrd player and therefre each f the mint: α = Δω Δt = ω ω f i Δt = 1.5π 0 0.85 = 5.54399 2 Nw we can lve fr the tangential acceleratin f each mint: a t1 = r 1 α = 3 ( )( 5.54399) =16.63196 cm a t2 = r 2 α = ( 8) ( 5.54399) = 44.35190 44 cm a t3 = r 3 α = ( 13) ( 5.54399) = 72.07183 72 cm 2 2 2 17 cm 2 Tangential velcity and tangential acceleratin are by definitin tangent t the circle thrugh which the bject i mving, that i what the wrd tangential mean. Thi al mean the tangential velcity and acceleratin are perpendicular t the iu f the circle. If yu undertand the derivative, yu can ee the relatinhip between the arc length, tangential velcity, and tangential acceleratin equatin: = rδθ d ( = rδθ ) d dt dt = r dθ dt v = rω t v t = rω d ( v dt t = rω ) dv t dt = r dω dt a = rα t 0217 Lecture Nte - Intrductin t Tangential Acceleratin with Recrd Player Example Prblem.dcx page 1 f 1
Demntrating the Directin f Tangential Velcity and Acceleratin There are three different part t the demntratin. 1. The turntable i plugged in and angularly accelerate at 4.2 / 2 up t 33 rev/min in le than ne ecnd. 2. The turntable rtate at 33 rev/min fr arund ne and a half ecnd. 3. The turntable i unplugged and angularly accelerate at -1.5 / 2 t a tp in lightly mre than tw ecnd. We can viualize all the tangential velcitie and acceleratin: It i al imprtant t undertand tangential velcity and acceleratin directin relative t ne anther: And Δθ, ω, α refer t the whle bject, hwever,, v t, a t refer t a pecific pint n the bject. 0218 Lecture Nte - Demntrating the Directin f Tangential Velcity and Acceleratin.dcx page 1 f 1
Centripetal Acceleratin Intrductin When an bject i rtating at a cntant angular velcity, the whle bject ha a cntant angular velcity. Therefre, every mint n the turntable ha the ame, cntant angular velcity. Lking at a ingle mint n the turntable: ω = cntant Becaue the angular velcity i cntant, there i n angular acceleratin. α = Δω Δt = 0 Δt = 0 Becaue the angular acceleratin i zer, the tangential acceleratin f the mint i zer. a t = rα = r 0 ( ) = 0 The angular velcity f the mint i cntant, hwever, the tangential velcity f the mint i nt cntant. Remember tangential velcity i a vectr. The magnitude f the tangential velcity f the mint i cntant. The directin f the tangential velcity f the mint i nt cntant. Becaue the tangential velcity f the mint i changing, the mint mut have a linear acceleratin.! a = Δ! v Δt (If velcity i changing, there mut be a linear acceleratin.) A hwn abve, thi line acceleratin i nt a tangential acceleratin. It al i nt an angular acceleratin. Angular acceleratin i angular, nt linear. Al, it zer anyway. The acceleratin which caue the tangential velcity t change directin i called Centripetal Acceleratin. Centripetal Acceleratin: The acceleratin that caue circular mtin. Centripetal mean Center Seeking. Centripetal acceleratin i alway in tward the center f the circle. Cined by Sir Iaac Newtn. Cmbinatin f the Latin wrd centrum which mean center and petere which mean t eek. I a linear acceleratin. ( ) 2 a c = v 2 t r = rω r = r 2 ω 2 r = rω 2 a c = v 2 t r = rω 2 Bae S.I. unit fr centripetal acceleratin are m 2 ( ) a c = rω 2 m 2 = mi 2 2 = m 2 0219 Lecture Nte - Centripetal Acceleratin Intrductin.dcx page 1 f 1
Intrductry Centripetal Acceleratin Prblem: Cylindrical Space Statin Example: A cylindrical pace tatin with a iu f 115 m i rtating at 0.292 /. A ladder ge frm the rim t the center. What i the magnitude f the centripetal acceleratin at (1) the tp f the ladder, (2) the middle f the ladder, and (3) the bae f the ladder? ω = 0.292 ; r = 0m; r = 115m = 57.5m; r 1 2 2 3 =115m; a c =? ( each) (1) a c = v 2 t r = rω 2 a c1 = r 1 ω 2 = ( 0)ω 2 = 0 (2) a 2c = r 2 ω 2 = ( 57.5) ( 0.292) 2 2 m = 4.90268 4.90 m 2 2 (3) a 3c = r 3 ω 2 = ( 115) ( 0.292) 2 = 9.80536 9.81 m 2 Rim tangential velcity calculatin: v t3 = r 3 ω = 115 ( )( 0.292) = 33.58 m 3600 1hr 1km 1000m =120.888 km hr v t3 =120.888 km hr 1mile 1.609km = 75.13238 75 mi hr 0220 Lecture Nte - Intrductry Centripetal Acceleratin Prblem - Cylindrical Space Statin.dcx page 1 f 1
Newtn Secnd Law tate: Flipping Phyic Lecture Nte: Centripetal Frce Intrductin and Demntratin! F = m! a Previuly we hwed that an bject mving alng a curved path mut have a centripetal acceleratin that i inward. Which mean: The net frce in the in-directin r! F in = m a! c! F in i called the Centripetal Frce. There are three thing I need yu t remember abut the Centripetal Frce: 1) Centripetal Frce i nt a new frce. 2) Centripetal Frce i never in a free bdy diagram. 3) When umming the frce in the in-directin, the in-directin i pitive and the ut-directin i negative. Full Free Bdy Diagram with mint t the left and right f the center f the recrd player: Free Bdy Diagram with jut the Frce f Static Frictin t hw it i alway inward: 0221 Lecture Nte - Centripetal Frce Intrductin and Demntratin.dcx page 1 f 1
Intrductry Centripetal Frce Prblem Car ver a Hill Example: A 453 g ty car mving at 1.05 m/ i ging ver a emi-circular hill with a iu f 1.8 m. When the car i at the tp f the hill, what i the magnitude f the frce frm the grund n the car? 1kg Knwn: m = 453g 1000g = 0.453kg; v =1.05 m t ; r =1.8m; F =? n F in = F g F N = ma c mg F N = m v 2 t r F N = mg + m v 2 t r F = mg m v t N r F n = ( 0.453) ( 9.81) ( 0.453) 1.052 1.8 2 = 4.1665 4.2N Nte: The frce cauing the circular mtin, the Centripetal Frce, r the net frce in the in-directin, in thi cae i the Frce f Gravity minu the Frce Nrmal. Al nte: F g = mg = 0.453 F in = F g F N ( )( 9.81) = 4.444 4.4N F N < F g In ther wrd, a yu g ver a hill in a car, yu feel a if yu weigh le. And the fater yu mve, the maller the frce nrmal, and the lighter yu feel. 0222 Lecture Nte - Intrductry Centripetal Frce Prblem - Car ver a Hill.dcx page 1 f 1
What i the Maximum Speed f a Car at the Tp f a Hill? Example: What i the maximum linear peed a car can mve ver the tp f a emi-circular hill withut it tire lifting ff the grund? The iu f the hill i 1.8 meter. The fater the car mve, the maller the Frce Nrmal. When we increae the peed f the car t the pint where the Frce Nrmal i zer, then the nly frce cauing circular mtin i the Frce f Gravity acting n the car. Any fater and the wheel f the car will leave the. Knwn: r =1.8m; v t max =? F in = F g F N = ma c mg 0 = m v 2 t r g = v 2 t r v 2 = gr v t t = gr v t = ( 9.81) ( 1.8) = 4.20214 4.2 m 0223 Lecture Nte - What i the Maximum Speed f a Car at the Tp f a Hill.dcx page 1 f 1
Flipping Phyic Lecture Nte: Mint n a Rtating Turntable Determining the Static Cefficient f Frictin Example: A turntable i turning 45 revlutin per minute. Mint are lcated 0.030 m, 0.080 m, and 0.130 m frm the center f the recrd. Determine what yu can abut the cefficient f tatic frictin between the turntable and the mint. Knwn: ω = 45 rev 1min 2π = 1.5π ; r1 = 0.030m; r2 = 0.080m; r3 = 0.130m; µ =? min 60ec 1rev The free bdy diagram fr all three mint i the ame: F F () y = FN Fg = may = m 0 = 0 FN = Fg = mg in = Ff max = mac µ FN = mrω 2 µ mg = mrω 2 µ g = rω 2 µ 1 = µ 2 = µ 3 = r1ω 2 g r2ω 2 g r3ω 2 g (0.03)(1.5π ) = 2 = 0.067910 0.068 9.81 (0.08)(1.5π ) = 2 = 0.18109 0.18 9.81 (0.13)(1.5π ) = 2 9.81 = 0.29428 0.29 Becaue the frce f tatic frictin i actually le than r equal t the cefficient f tatic frictin time frce nrmal and we ued the maximum frce f tatic frictin, we actually determined the minimum cefficient f frictin neceary t keep each mint n the turntable. Therefre the anwer i that the cefficient f tatic frictin mut be greater than r equal t 0.29. µ 0.29 0224 Lecture Nte - Mint n a Rtating Turntable - Determining the Static Cefficient f Frictin.dcx page 1 f 1
Flipping Phyic Lecture Nte: Determining the Frce Nrmal n a Ty Car mving up a Curved Hill Example: A 0.453 kg ty car mving at 1.15 m/ i ging up a emi-circular hill with a iu f 0.89 m. When the hill make an angle f 32 with the hrizntal, what i the magnitude f the frce nrmal n the car? Knwn: m = 0.453kg; v t = 1.15 m ; r = 0.89m; θ = 32 ; FN =? Draw FBD: Break frce int cmpnent (in-directin and parallel t in-directin) Fg = mg c θ & Fg = mg in θ! Re-draw FBD: Fin = FN Fg = mac FN = Fg + mac = mg cθ + m ( )( ) ( ) ( FN = 0.453 9.81 c 32 + 0.453 vt 2 r 2 = 4.3804 4.4N )1.15 0.89 0225 Lecture Nte - Determining the Frce Nrmal n a Ty Car mving up a Curved Hill.dcx page 1 f 1
Flipping Phyic Lecture Nte: Demntrating Why Water Stay in a Bucket Revlving in a Vertical Circle Inertia i the tendency f an bject t maintain it tate f mtin. Lking at the tp pitin: The inertia f the water trie t keep the water mving t the left and the frce f gravity pull the water dwn. Here are tw different ituatin with tw different reult: The inertia f the water, r the tendency f the water t maintain it tate f mtin, i what keep water in the bucket. In ther wrd, the water i mving and want t keep mving in a traight line, hwever, the bucket keep getting in the way, which i why water tay in the bucket. 0226 Lecture Nte - Demntrating Why Water Stay in a Bucket Revlving in a Vertical Circle.dcx page 1 f 1
Flipping Phyic Lecture Nte: Analyzing Water in a Bucket Revlving in a Vertical Circle Free Bdy Diagram: Ntice hw all free bdy diagram are different at different bucket lcatin. Hwever, frce f gravity i alway dwn and frce f tenin i alway in. Let tart by analyzing when the bucket i at the bttm. F in = FT Fg = mac FT mg = mrω 2 FT = mg + mrω 2 A cuple f thing t ntice: 1) The centripetal frce i Frce f Tenin minu Frce f Gravity. 2) The fater the bucket pin the larger the frce f tenin and therefre 3) The fater the bucket pin the larger the centripetal frce neceary t keep the bucket mving in circular mtin. If we knew the iu f the path, the ma and angular peed f the bucket and water we culd lve fr the frce f tenin. We culd d thi at any bucket lcatin: Tp, Bttm, Side, etc. Ntice at a lcatin which i nt the bttm, tp, r ide, we wuld need t break the frce f gravity int it cmpnent in the in-directin and the tangential-directin, jut like we did with the ty car and the curved hill. (http://www.flippingphyic.cm/car-hill-frce-nrmal.html) But, we dn t have a number dependency, let tp here and nt have any number in thi len. J Fr the f yu taking AP Phyic 1, be aware I have yet t ee a prblem n the AP Phyic 1 exam that ha the bucket in a pitin ther than the tp, bttm, r ide. 0227 Lecture Nte - Analyzing Water in a Bucket Revlving in a Vertical Circle.dcx page 1 f 1
Flipping Phyic Lecture Nte: Analyzing Water in a Bucket Revlving in a Vertical Circle Example: What i the minimum angular peed neceary t keep water in a vertically revlving bucket? The rpe iu i 0.77 m. F in = Fg + FT = mac mg + 0 = mrω 2 g = rω 2 At the minimum angular peed neceary t keep water in a vertically revlving bucket the tenin in the rpe i reduced t zer. At thi angular peed the centripetal frce i jut the frce f gravity. ω2 = g ω = r g 9.81 = = 3.5694 3.6 r 0.77 0228 Lecture Nte - Minimum Speed fr Water in a Bucket Revlving in a Vertical Circle.dcx page 1 f 1
The Right Hand Rule fr Angular Velcity and Angular Diplacement Step #1) Dn t be t cl fr the Right Hand Rule. Step #2) Limber Up! (Nt kidding. Yu have t rtate yur hip and hulder metime when ding the right hand rule) Step #3-1) Take the finger f yur right hand. Step #3-2) Curl yur finger in the directin f the turning mtin f the bject. Step #3-3) Stick ut yur thumb. Step #3-4) Yur thumb indicate the directin f the bject angular velcity and angular diplacement. Tw example: Nte: Clckwie and cunterclckwie are berver dependent directin. The directin that reult frm the Right Hand Rule i nt berver dependent. Thi i ne f the rean we ue the Right Hand Rule in phyic t determine the directin f angular diplacement and angular velcity. Al nte: The directin f the angular diplacement and angular velcity f a turning bject i perpendicular t the plane in which the bject i lcated. 0229 Lecture Nte - The Right Hand Rule fr Angular Velcity and Angular Diplacement.dcx page 1 f 1
A Tale f Three Acceleratin r The Difference between Angular, Tangential, and Centripetal Acceleratin An bject mving in a circle can have three different type f acceleratin: Angular Acceleratin: α = Δω Δt in 2 i an angular quantity. Tangential Acceleratin: a t = rα m in i a linear quantity. 2 Centripetal Acceleratin: a c = v 2 t r = rω m 2 in 2 i a linear quantity. Angular acceleratin eparate itelf frm the ther: 1) Becaue it i an angular quantity, wherea the ther tw are linear quantitie. 2) Becaue angular acceleratin applie t the whle rigid bject, hwever, tangential acceleratin and centripetal acceleratin are fr a pecific iu. A majr difference between tangential acceleratin and centripetal acceleratin i their directin. Centripetal mean center eeking. Centripetal acceleratin i alway directed inward. Tangential acceleratin i alway directed tangent t the circle. By definitin, tangential acceleratin and centripetal acceleratin are perpendicular t ne anther. Anther majr difference between tangential acceleratin and centripetal acceleratin i that circular mtin cannt exit withut centripetal acceleratin. N centripetal acceleratin mean the bject i nt mving in a circle. Centripetal acceleratin reult frm the change in directin f the tangential velcity. If the tangential velcity i nt changing directin, then the bject i nt mving in a circle. Tangential acceleratin reult frm the change in magnitude f the tangential velcity f an bject. An bject can mve in a circle and nt have any tangential acceleratin. N tangential acceleratin imply mean the angular acceleratin f the bject i zer and the bject i mving with a cntant angular velcity. α = Δω Δt = 0 Δt = 0 a t = rα = r 0 ( ) = 0 0230 Lecture Nte - A Tale f Three Acceleratin r The Difference between Angular, Tangential, and Centripetal Acceleratin.dcx page 1 f 1
Flipping Phyic Lecture Nte: Cnical Pendulum Demntratin and Prblem Example: Tw phere attached t a hrizntal upprt are rtating with a cntant angular velcity. Determine the angular velcity. A hwn in the figure: L = 9.7 cm, x = 3.4 cm and θ = 43. Knwn: 1m 1m ω =?; x = 3.4cm = 0.034m; L = 9.7cm = 0.097m; θ = 43 100cm 100cm in θ = F O FTin A FT = FT = FT in θ & c θ = = y FT = FT c θ in y H FT H FT () y = FT Fg = may FT c θ mg = m 0 = 0 FT c θ = mg FT = in = FT = mac FT in θ = mrω 2 F y in mg c θ mg in θ = mrω 2 g tan θ = rω 2 c θ g tan θ g tan θ ω = r r Need r: r = L in + x therefre we need Lin ω2 = in θ = ω = O L in = L in = L in θ & r = L in + x = L in θ + x H L g tan θ = L in θ + x (9.81) tan ( 43) = ±9.55716 9.6 (0.097) in ( 43) + 0.034 Accrding t the right hand rule, the directin f the angular velcity and angular diplacement f the phere i dwn, which i negative. ω meaured = Δθ 2π = = 3π = 9.51998 Δt 0.66 Ue the meaured value a ur berved value and the predicted value a ur accepted value: Er = ( ) 9.55716 9.51998 O A 100 = 100 = 0.39052 0.39% A 9.51998 0231 Lecture Nte - Cnical Pendulum Demntratin and Prblem.dcx page 1 f 1