Uniform Circular Motion and Newton s Law. Physics 201, Lecture 10

Transcription

1 Physics 201, Lectue 10 Today s Topics n Cicula Motion and Newton s Law (Sect. 6.1,6.2) n Centipetal Foce in Unifom Cicula Motion n Examples n n Motion in Acceleated Fame (sec. 6.3, conceptual undestanding) Motion with Resistance (sec. 6.4, slides at the end, self eading) n Hope You e Peiewed Chapte 6 (and also Chapte 5) Unifom Cicula Motion and Newton s Law q Recall: Centipetal Acceleation = ω = 2π/T, always in tangential diection = ω 2 = 2 /, always pointing to the cente q Now, pe Newton s 2 nd law, thee must be a net-foce that is esponsible fo. F = m this F is called Centipetal Foce (F c ) q F c can be in the fom of tension, fiction, gaitation, o combination of them Example: Ball on a Sting in Hoizontal Cicula Motion. q Execise: A ball attached on a sting of length is in unifom cicula motion, if the ball is moing at a (constant) linea speed, what is the tension T in the sting? Quick Quiz: What If the Centipetal Foce is Lost Solution: the only foce in the hoizontal plane is the tension (which sees as the centipetal foce) T = m = m 2 / demo: what if the sting is cut? iew in hoizontal plane (top iew) q Fo the aboe cicula motion, how will the ball continue to fly if the sting is cut off? Ø answe path #3. 1

2 Example/Demo: Conical Pendulum What is the peiod of the conical pendulum? y x Solution Daw FBD as shown y diection: ΣF y = Tcosθ-mg =0 à T= mg/cosθ Example: Ca at a Tun (Leel Road) q When a is tuning along a hoizontal cue, the static fiction between the tie and the oad suface supplies the equied centipetal foce. f s = m = m 2 / T x diectioon: ΣF x = Tsinθ = ma x =m 2 / tigonomety: = Lsinθ à (g/cosθ) sinθ = 2 /Lsinθ =sqt(lg sinθtanθ) Peiod T= 2π/ = 2π sqt(lcosθ/g) let µ s be the coefficient of static fiction f s < µ s n = µ s mg à 2 /< µ s g (can you see n=mg?) < µ g Quiz: why is static fiction used hee? answe: thee is no elatie motion in adial diection s Example: Ca Tuning on a Banked Cue q In cases of low fiction oad suface, (o when speed is high), oad tuns ae designed to be banked. In such cases, nomal foce poides the equied centipetal foce. Demo/Execise: Rolle Coaste q What is the minimum speed at the top of a olle coaste? A De-compose nomal foce n: F c = n x = nsinθ. Execise: show = g tanθ q at top point A: F c = mg + T top = m 2 /R à 2 = (mg + T top ) R/m > mgr/m =gr (note: T top >0) (see boad) à > gr 2

3 Rolle Coaste Quiz q In this olle coaste design that the cat is olling aboe the tack, at top point B, the cat s speed Non-Unifom Cicula Motion q In a geneic (non-unifom) cicula motion, acceleation usually has both centipetal and tangential components can not be too high, can not be too low, no limit. Ø answe: at top point B: F c = mg - N top = m 2 /R < mg a = + a t ΣF = ΣF + ΣF t a t ΣF = m, ΣF t = ma t Conceptual undestanding only fo this couse Example of non-unifom Cicula Motion q Conside a mass in etical cicula motion with aying speed At any point, the centipetal foce is poided by ombination of tension T and a component of gaitation mgcosθ F c = T- mgcosθ = m = m 2 /R 2 T = m( + g cosθ ) R Quiz: Block in Acceleating Ca q A block on the fictionless floo of an acceleating tain. To a bystande on the gound, what is the blocks acceleation? 0, +a 0, -a 0, othe Newton s 2 nd Law (in eath fame) : F = ma, F=0, a=0 To the obsee standing inside the tain, what is the block s acceleation? (standing=no elatie motion) 0, +a 0, -a 0, othe a 0 Conceptual undestanding only fo this couse Newton s 2 nd Law (in tain fame) : F=0, a = - a 0, F=ma? 3

4 Fictitious Foce q Newton s 2 nd Law is alid only in the inetia efeence fame i.e. IF a is measue in an inetia efeence fame F eal = ma q In an acceleating fame (a 0 ), the 2 nd is not alid. To foce the fom of 2 nd law, one has to add an fictitious foce F fictitious = -Ma 0 into the equation F = F eal + F fictitious = ma Motion In Acceleated Fame q Newton s 2 nd Law Applies only in inetial efeence fame q One can deie the 2 nd Law in acceleated (non-inetial) efeence fame: Refeence Fame A: inetial, F = ma =m d/dt Refeence Fame B: Moing w..t to Fame A with a 0 =d 0 /dt In Fame B: ʹ = - 0 aʹ = dʹ /dt = d/dt d 0 /dt maʹ = F ma 0 =Fʹ è a fictitious foce F fictitious =-ma 0 has to be intoduced to, atificially, keep the same fom of the 2 nd Law Conceptual Undestanding only Newton s 2 nd Law: Two Pactical Appoaches q Fist Pinciple: Newton s 2 nd Law Applies only in inetial efeence fame Quiz: Test You Imagination Quiz: A lady is sitting on a otating table watch a wood bock which is also fixed on the table. The distance between the lady and the block is. To the lady s iew, the motion of the block is: q Appoach 1: Woking in inetia efeence fame: F = m a Staightfowad, but may need to do Galilean tansfomation. No motion, Cicula motion with adius, Cicula motion with adius R, Motion in moe complicated cue q Appoach 2: Woking in a non-inetia fame of acceleation a 0 Intoduce a fictitious foce F fictitious = - ma 0 Add the fictitious foce to the eal foce: F = F + F fictitious è So we can, atificially, keep the same fom of the 2 nd Law R maʹ =F (= F ma 0 ) 4

5 Example: Fictitious Foce In Cicula Motion One moe Example: Fictitious Foce In Linea Motion Inetial Obsee: Obsee In the Ca y: ΣF y = Tcosθ - mg = 0 (Noninetial) T=ma T= m 2 / Centifugal Foce F fictitious =-T = -ma x: ΣF x = Tsinθ = mgtanθ = ma tanθ = a/g y: ΣF y =Tcosθ-mg =0 x: ΣF x =Tsinθ ma =0 (a=0, in otating fame) F fictitious = - ma in x diection Study afte class Motion with Resistance Foce q So fa, we hae consideed on fee fall fo pojectile motion Teminate Speed q If we only cae about the maximum speed the falling object can each, the math is quite simple: At maximum speed: mg=r (quiz: why?) This maximum speed is called teminate speed ( T ) Eg. if R = b T = mg/b o if R = 1/2DρA 2 Fee Fall m d/dt = mg Falling with Resistance R m d/dt = mg R examples Model 1: R=-b Model 2: R=1/2 DρA 2 T = 2mg D ρ A Self eading to know the meaning of quantities in the equation. 5

6 Some Teminal Speeds 6