PHYS 1114, Lecture 21, March 6 Contents:

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1 PHYS 1114, Lectue 21, Mach 6 Contents: 1 This class is o cially cancelled, being eplaced by the common exam Tuesday, Mach 7, 5:30 PM. A eview and Q&A session is scheduled instead duing class time. 2 Exam 2 coves chaptes 1 7: (a) All mateial of pevious test could show up in solving poblems elated to Chaptes 4 7. (b) Chapte 4: Newton s thee foce laws, Pope use of Fee-body diagams, Nomal foce, Inclined plane, Tension in ope. (c) Chapte 5: Fiction (static and kinetic), Hooke s law. (d) Chapte 6: Unifom otational motion, the vaious angula quantities, centipetal acceleation and foce, appaent weight, fiction needed to tun a cone, banked cuves, Newton s univesal gavity. (e) Chapte 7: Enegy, Kinetic and Potential Enegy, Wok, Consevative vesus Nonconsevative wok, the Consevation of Mechanical and Othe Enegy, Powe. 3 The thee examples discussed in the Q&A session have been added in the next fou pages ii v: (a) Sping poblem using enegy consevation. (b) Citical speed fo ca going ove top of hill. (c) Inclined plane poblem with pully. i

2 Sping poblem using enegy consevation v f =0 y=0 m=20g v o =0 4cm unstetched k=25n/m v m h Oiginal state: KE o = 0, x = m = 4.0 cm: E = PE g + PE s = (0.020 kg)(9.80 m/s 2 )( m) (25 N/m)( m)2 = J = J Intemediate state: x = 0, h = 0, PE s = PE g = 0: E = 1 2 mv2 = 1 2 (0.020 kg)v2 = J =) v = m/s = 1.1 m/s Maximum height: v = 0, KE = PE s = 0: E = PE g = mgh = (0.020 kg)(9.80 m/s 2 )h = J =) h = J (0.020 kg)(9.80 m/s 2 ) = m = 6.4 cm ii

3 Citical speed fo ca going ove top of hill v v>v c w=mg=f c v = v c to just stay on the oad: Data: = 30 m, g = 9.80 m/s 2 w = mg = F c = mv2 =) mg = mv2 =) v 2 = g =) v c = p g =) v c = m/s = 17 m/s Also, =) v c = 62 km/h = 38 mi/h iii

4 Inclined plane poblem with pully F N T T F N a T T m F f mg l M Mg F f mgsin mgcos m mg M Mg a Fom the two fee-body diagams we ead o : Ma = Mg T F ma = T mg sin µ k F N = mg cos N + (M + m)a = Mg mg sin µ k mg cos =) (M + m)a = M m(sin + µ k cos ) g =) a = M m(sin + µ k cos ) M + m g M = m = 1.0 kg µ k = 0.30 g = 9.80 m/s 2 = 30 l = 1.0 m a = m/s 2 = 1.2 m/s 2 iv

5 m l KE o = 0 PE o = 0 M v KE f = 1 2 mv Mv2 " " speed = ~v l lsin l PE f = mgl sin Mgl W NC = F f l = µ k (mg cos ) l (opposing) KE f + PE f = KE o + PE o + W NC 1 2 (M + m)v2 + (mgl sin Mgl) = µ k (mg cos ) l =) 1 2 (M + m)v2 = Mgl mgl sin µ k mgl cos =) v 2 = 2 Mg mg(sin µ k cos ) M + m l = 2al =) v 2 = 2( m/s 2 )(1.0 m) = m 2 /s 2 =) v = m/s = 1.5 m/s and v = v o + at with v o = 0 =) t = v a = m/s m/s 2 = 1.3 s v

6 Main Fomulae fo Second Test: Basic Mathematics: See Lectues 1, 2 and notes fo Exam 1 (Lectue 12) fo Algeba, Tigonomety, Significant Digits. Reminde of Vectos: A = (Ax, A y ) A ± B = (Ax ± B x, A y ± B y ) B = (B x, B y ) A = ( A x, A y ) (Rescaling) A A -A 3 2 A B B A B B A B A A+B (It is easie to wok with components, than with magnitudes and angles, using ules of sines and cosines epeatedly.) Kinematics: 8 < = o + = : o + t = t o + t 8 < : = o = Displacement = o = Angula Displacement t = t t o = Elapsed Time In Plane : = (x, y) = Position Vecto v = (v x, v y ) = Velocity a = (a x, a y ) = Acceleation q Speed = Magnitude of velocity = v = vx 2 + vy 2 On Cicle : = Position Angle, = Radius 1

7 Tanslational Motion (See notes fo Exam 1) = o = d = Displacement Vecto SI Unit: m v = t = o t t o = Aveage Velocity SI Unit: m/s ā = v t = v v o t t o = Aveage Acceleation SI Unit: m/s 2 Constant velocity (also setting t o = 0): v v v o = constant, a 0 = o + vt Constant acceleation (also setting t o = 0): a ā a o = constant v = v o + a t v = 1 2(v + v o ) o = vt = 1 2(v + v o ) t o = v o t + 1 2a t 2 v 2 x v 2 ox = 2a x x = 2a x (x x o ) v 2 y v 2 oy = 2a y y = 2a y (y y o ) 2

8 Newton s Laws of Motion Fist Law of Newton: An object continues in a state of est o in a state of motion at a constant speed along a staight line, unless compelled to change that state by a net foce. Second Law of Newton: F net = X F = ma X Fx = ma x, X Fy = ma y NB Fist Law follows fom second as no acceleation takes place if thee is no net foce. NB SI unit fo foce is newton = N = kg m/s 2. Thid Law of Newton: Action = Reaction Fouth Law of Newton: Univesal Gavity: F = G m 1m 2 2 whee G is Newton s constant =) F = mg with g = G M E R E 2 = 9.80 m/s2 Examples of Foces: Weight ( ~w = m~g) Nomal Foce ( ~ N) Static and Kinetic Fiction ( ~ F f,s o ~ F f,k ) Tension ( ~ T ) 3

9 Fee-Body Diagam fo a Body on an Inclined Plane: F N F f mg sin mg cos mg Static and Kinetic Fiction Foces: Static: F f apple F max f = µ s F N Kinetic: F f = µ k F N Nomal foce fom Newton s Thid Law!! Action = Reaction Sping foce: (Hooke s Law) The estoing foce towads equilibium length is linea in the length d of the deviation fom equilibium (x = 0). F s = kd Moe pecisely, the x-component of F ~ s is (F s ) x = kx, if the sping is along the x axis and the +x-diection coesponds to stetching. Thus the foce on an object attached and its displacement ae in opposite diections. 4

10 Unifom Cicula Motion v v = ~v = constant speed Howeve, ~v? adius, o, moe pecisely, ~v along instantaneous tangent =) ~v 6= constant velocity Peiod T = time fo one evolution (ev) v = 2 T = speed = 1 cicumfeence 1 peiod Angula Position = s = ac length adius measues angle in adians (ad) Aveage angula velocity! = t Instantaneous angula velocity! = t when t tends to 0! is slope of (t) gaph at t f i is aea unde!(t) gaph between t o and t f 5

11 Radial and tangential components of vecto ~ A -axis (adial) A t-axis A A A t (tangential) A = A cos A t = A sin Velocity 8 < v t = s t = t : v = v z = 0 (z is thid diection) =) v t =! v = 0 v z = 0 Centipetal foce (! in ad/s) a = v t 2 =!2 a t = 0 a z = 0 ~F net = m~a = F ~ mv 2 c =, towad cente = centipetal foce, equied to stay in cicle In components: ma = (F net ) = mv2 ma t = (F net ) t = 0 ma z = (F net ) z = 0 = m! 2 6

12 Fictitious appaent weight (nomal foce) in loop-the-loop v w y N w x top F net =m v 2 v y N bottom F net =m v 2 w w x w app = N = w + m(v bottom) 2 w app = N = m(v top) 2 Citical speed: At top N = 0 giving mv 2 c w bottom top = mg, o v c = p g. Centipetal Acceleation and Foce: a c = v2 F c = ma c with F c = ma c = mv2. 7

13 Tuning banked cone without fiction: N z Ncosq Nsinq w The ca moves in a hoizontal cicle of adius with speed v, so vetical foce components cancel, hoizontal ones give F c : X F = N sin = mv 2 = F c X Fz = N cos w = 0 w = mg =) N = w cos = mg cos =) mg cos 2 mv sin = =) v 2 = g tan v = p g tan Remak 1: At othe speeds one needs to include static fiction in the ± adial diection. (+ o depends on v < v o v > v.) Remak 2: Aiplanes also take banked cuves when they tun. Just eplace the nomal foce N with the lift and angle is now the angle of the wings with the hoizontal plane. 8

14 Wok done on an object by a constant foce F while the object is displaced ove d is W = F d cos with F = F, d= d, = 6 (F, d) Unit: newton mete = N m = J = joule Remembe: cos 0 =1, cos 90 =0, cos 180 = 1. Kinetic Enegy KE = 1 2 mv2 with v = v Wok-Enegy Theoem: W net = KE = KE 2 KE 1, Wok done = Change in KE Gavitational Potential Enegy PE gav = mgy Consevative Foces: Wok W C is independent of path. Nonconsevative Foces: Wok W NC depends on the path. In fomula: W net =W C +W NC, W C = PE = (PE f PE o ). Wok-Enegy Theoem: (Rewitten) W NC = KE + PE = E. Consevation of Mechanical Enegy: If W NC = 0 then KE f + PE f = KE o + PE o = constant E In geneal: Total enegy (mechanical and non-mechanical, like heat, electical enegy, etc.) is conseved. 9

15 Elastic Potential Enegy PE sping = 1 2 kx2 (fom foce magnitude F s = kx) Total PE = Sum of all PE contibutions. Aveage Powe: P = wok done time lapsed = W t = F d t = F v Unit: joule/second = J/s = W = watt Note: W = wok, W = watt, d= distance, d = deci. 10

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