Combination of Forces: Lecture 7 Net Force Centripetal Motion http://web.njit.edu/~sirenko/ F net = F1 + F2 F net = ma Physics 105; Fall 2009 Lecture 7 Andrei Sirenko, NJIT 1 Lecture 7 Andrei Sirenko, NJIT 2 Uniform Circular Motion Net Force and Centripetal Force Centripetal Force Centripetal acceleration Top view: Period Centripetal force : F = ma Lecture 7 Andrei Sirenko, NJIT 3 Lecture 7 Andrei Sirenko, NJIT 4 Centripetal Force is a Centripetal Force and combination of: Tension Force: ¾ Gravitational Force: mg down to the ground ma = T ¾ Tension Force: T 2 along the string ma = mv /R = T ma = N – mg ¾Normal Force: N 2 perpendicular to the support ma = mv /R ¾Static Friction Force max maximum value Ffr = µstN Lecture 7 Andrei Sirenko, NJIT 5 Lecture 7 Andrei Sirenko, NJIT 6 Centripetal Force and Net Force and Centripetal Force Kineti c Fri cti on Force: How fast can we turn? v2 ffldfor a flat road: m = f r st Kinetic friction does not affect v 2 max speed: mfNmgvgrmax ===max µµ; = µ Centripetal acceleration directlly r st st stmax st Lecture 7 Andrei Sirenko, NJIT 7 Lecture 7 Andrei Sirenko, NJIT 8 Net Force and Centripetal Force Uniform Circular Motion What is the origin of the Centripetal acceleration centripetal force ? Period Wha t is the direc tion of centripetal force ? Lecture 7 Andrei Sirenko, NJIT 9 Lecture 7 Andrei Sirenko, NJIT 10 Uniform Circular Motion Problem #1 m = 5 kg L = 5 m Centripetal acceleration R = 2 m Find v, T, and a L Period: R v = 100 m /12 s = 8. 33 s; R = 100/π = 31.8 m a = (8.33)2/31.8 m/s 2 = 2.2 m/s 2 Lecture 7 Andrei Sirenko, NJIT 11 Lecture 7 Andrei Sirenko, NJIT 12 Problem solving tactics: m = 5 kg Problem solving tactics: m = 5 kg L = 5 m L = 5 m R = 2 m L R = 2 m θ Find v,T, and a L Find v,T, and a R R sin θ = R/L = 0.4; tan θ = (()(R/L)/ (1-(()R/L)2) ½ = 0.44 X: ma = T * sin θ Y: ma = 0 = - mg + T * cos θ Lecture 7 Andrei Sirenko, NJIT 13 Lecture 7 Andrei Sirenko, NJIT 14 Problem solving tactics: Problem is solved: m = 5 kg m = 5 kg L L = 5 m L L = 5 m θ R = 2 m θ R = 2 m Find v,T, and a Find v,T, and a R R Centripetal Force originates from the tension force! sin θ = R/L = 0. 4; tan θ = (R/L)/ (1-(R/L)2) ½ = 0440.44 X: ma = T * sin θ 2 ½ sin θ = R/L = 0.4; tan θ = (R/L)/ (1-(R/L) ) = 0.44 Circular motion: Circular motion: Y: ma = 0 = - mg + T * cos θ X: ma = T * sin θ ma = mv2/R ma = mv2/R ma = mg*i*sin θ / cos θ = mg*tan θ Y: ma = 0 = - mg + T * cos θ a = v2/R a = v2/R T = mg/ cos θ a = g*tan θ θ θ θ ½ ma = mg*sin / cos = mg*tan T = 5 kg *9.8 m/s 2 /(1-(2m/5m) 2) ½ = 53 N v = (aR) ½ θ θ v = (aR) ½ T = mg/ cos a = g*tan a = 4.3 m/s 2; ma = 5 kg * 4.3 m/s^2 = 21 N; v = (4.3*2) m/s = Lecture 7 Andrei Sirenko, NJIT 15 Lecture 7 Andrei Sirenko, NJIT2.9 m/s 16 µ µ Problem #2 R = 20 m; st= 0.5 Problem #2 R = 20 m; st= 0.5 Angle 10° Angle 10° Find vmax Find vmax R θ R θ Lecture 7 Andrei Sirenko, NJIT 17 Lecture 7 Andrei Sirenko, NJIT 18 µ µ Problem #2 R = 20 m; st= 0.5 Problem #2 R = 20 m; st= 0.5 Angle 10° Angle 10° Find vmax Find vmax X: ma = N sin θ + µst N cos θ X: ma = N *(sin θ + µst cos θ) Y: 0 = N cos θ -mg -µst N sin θ Y: N = mg / (cos θ - µst sin θ) ma = mv2/R R θ θ R θ θ Lecture 7 Andrei Sirenko, NJIT 19 Lecture 7 Andrei Sirenko, NJIT 20 µ Problem #2 R = 20 m; st= 0.5 QZ # 7 Analyze the previous problem Angle 10° ma = N *(sin θ + µst cos θ) N = mg / (cos θ - µst sin θ) Find vmax ma = m v 2 /R max 1. R = 20 m; µst= 0.5; Angle 10° m v 2 /R = mg *(s in θ + µ cos θ)/ (cos θ - µ sin θ) max st st What is going to happen to θ 2 θ µ θ θ µ θ vmax = g*R *(sin + st cos )/ (cos - st sin ) the static friction force for Mass m disappeared !!! the case when the velocity of the R vmax ≈ 12. 8 m/s ≈ 29 ml/h track is doubled: v = vmax * 2 What is going to happen to the truck? (describe) 2. R = 20 m; µ = 0.5, v=5 m/s; m=3000 kg, θ st R and θ = 0 What is the value and direction of the static friction force ? Lecture 7 Andrei Sirenko, NJIT 21 Lecture 7 Andrei Sirenko, NJIT 22.