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Transcript
Chapter 6: Circular Motion and Other Applications of Newton’s Laws Chapter 6 Goals: • to deepen our understanding of the causes of circular motion in various scenarios • to gain skill with applying the FBD method to such problems • to learn how to handle the case of velocity-dependent resistive forces • note: we will mix things up in dealing with this chapter, and this lecture will include material from chapter 4, on relative motion Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example of curvy motion A bicycle/rider of mass 100 kg proceeds around a curve at 25 mi/hr, and experiences a centripetal acceleration of g/2 while doing so -- any faster and the bicycle skids!! a) Find the radius of the curve. b) If static friction is what keeps the bike from skidding, what is the value of ms? Use FBD method. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. +y x+ • v = (25 mi/hr)(1/3600 hr/s) (1609 m/mi) = 11.2 m/s • ac = v2/r so r = v2/ac = (11.2 m/s)2/(5 m/s2) = 25.1 m • x: fs = mac msn = mv2/r ms = mv2/nr • y: n – Fg = 0 n = mg • combining, the m cancels, and ms = v2/gr • but recall that v2/r = g/2 so ms = .50 fs n ac Fg Example of curvy motion A Ferris wheel begins to rotate and ‘tangentially’ accelerates until the speed at the rim is v = 3.20 m/s. Its radius is 8 m, and it gets ‘up to speed’ in 5.0 seconds. A child of weight 200 N is riding. (a) At what instant of time (if any) is ac = at? (b) What is the child’s apparent weight when at the top? • v = vi + at t so v = 0 + at +y 2 at = v/t = (3.2 m/s)/(5 s)= .65 m/s n • at end, ac = v2/r = (3.2) 2(8) = 1.28 m/s2 • ac = (at t)2/r = at t = (r/at )1/2 = (8/.65) 1/2 = 3.5 s ac • y: n – Fg = – mac n = – mac + mg • so the weight is ‘reduced’ to 87%: n = 174 N Fg • at the bottom the weight is ‘enhanced’ to 113% Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The simple pendulum I at +t +r • simple means that mass is the point-like bob mass m • length of pendulum is L • operates in gravity g = –jhatg • FBD is at left: ‘position’ is q • introduce radial and tangential coordinates r and t (t in direction of increasing q) • at = dv/dt ar = –v 2/L • t: –mg sinq = – mat = at = g sinq = 0 • r: mg cosq - T = –v 2/L • T = mg cosq + mv 2/L {show Active Figure 15.16} Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The simple pendulum II • a challenging non-linear differential equation for angle position: since at = L d2q/dt2 d2q/dt2 + (g/L) sinq = 0 • if the amplitude is small, then sinq ≈ q R • in radians, q := s/R h • but sin q := o/h s o q • s ≈ o; h ≈ R sin q ≈ s/R = q • cos q := a/h ≈ 1 –q2/2 tan q := o/h ≈ s/R = q a, R Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The simple pendulum III • under this approximation, we get d2q/dt2 + (g/L) q = 0 • this is a well-studied equation for q (t) and it is known that the motion is simple harmonic oscillations • for small amplitudes, a pendulum executes SHM in its variable q, the angle away from the vertical!! Period is amplitude-independent!! Makes a nice clock!! • What about the equation for the tension? • T = mg cosq + mw 2L • an interesting combination of terms, that allow T to ‘balance’ the bob’s weight [mg cosq] and also to provide the needed radial (centripetal) force [mw 2L] • our other equation shows that mg sinq provides the need tangential (Hooke-like) force to cause the SHM Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Relative Motion as seen from two inertial frames of reference • recall that an inertial frame of reference is an arena for doing physics in which Newton’s first law is correct: no net force no acceleration • [if acceleration is observed then the frame is noninertial (accelerating) and humans invent fictitious forces to explain the weird motions observed] • back to section 4.6… • two observers measuring some kinematics: the motion of a body at point P • observer A is not moving, and A measures position rPA, velocity vPA, and acceleration aPA • observer B is moving as seen by A (at vBA) and it measures rPB, vPB, and acceleration aPB Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. How do we relate what A and B observe? • basic kinematical fact:s • rPA = rPB + rBA ° rPA A P• take one time derivative: • vPA = vPB + vBA rBA • in English: the velocity of a point with rPB respect to frame A is the velocity of the B point with respect to frame B plus the velocity of B with respect to A • we are assuming B is moving at constant velocity vBA • another time derivative yields aPA = aPB so accelerations are the same when observed from either inertial frame!! • N2 is correct in either frame Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. An alternative way of writing the same stuff • vPA = vP/A vPB = vP/B vBA = vB/A • we have the Galilean transformation: vP/A = vP/B + vB/A • note how the algebra works: P/A = (P/B)(B/A) • addition of vectors multiply subscripts • thus, vP/B = vP/A – vB/A or P/B = (P/A)/(B/A) • subtraction of vectors divide subscripts • the challenge is to identify the frames and the object Simple 1d example: An empty water bottle rolls toward the back of a bus at 2.5 m/s. The bus is traveling at 55 mph ≈ 25 m/s. What is the velocity of the bottle as seen from the ground? Ground is frame A; bus is frame B; bottle is object P vP/A = vP/B + v B/A vP/A = – 2.5 + 25 = 22.5 m/s Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Trickier 2d example: An empty water bottle rolls sideways across the back of a bus at 5 m/s. The bus is traveling at 55 mph ≈ 25 m/s. What is the velocity of the bottle as seen from the ground? What is the speed of the bottle as seen from the ground? In what direction is the bottle moving as seen from the ground? Ground is frame A; bus is frame B; bottle is object P Take x direction to be along bus’s motion as seen from ground Take y direction to be perpendicular to that: across the road vP/A = vP/B + v B/A vP/A = 5 m/s j+ 25 m/s i Speed is |vP/A | = √(52 + 252) = 25.5 m/s Direction is Arctan(5/25) = 11.3° away from the x direction y y x vP/B As seen from B x vP/A As seen from A Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Trickier 2d example: You are canoeing across a river that is 75 m wide, and in which the current moves at 2 m/s. You can paddle at 2.5 m/s. How should you paddle to go directly across the river, and how much time will it take [hint: if you aim straight across, you will reach the other side in time t = 75 m/2.5 m/s = 30 s but will have drifted downstream a distance d = 2 m/s(30 s) = 60 m!!!] Ground is frame A; river is frame B; canoe is object P Take x direction to be along river’s motion as seen from ground Take y direction to be perpendicular to that: across the river vP/A = vP/B+ v B/A vP/A := vP/A j = vP/B + 2 m/s i vP/B= vP/A j – 2 m/s i But we know vP/B = 2.5 m/s so we can solve: 2 2.5 = vP/A 2 2 vP/A = 2.52 - 2 2 = 1.5 m/s [3 : 4 : 5 triangle] vP/B= 1.5 m/s j – 2 m/s i and vP/A = 1.5 m/s j Speed is 1.5 m/s with respect to the shore Canoe is aimed at Arctan(1.5/2) = 36.9° away from UPSTREAM! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. y y x x vP/B As seen from B vP/A As seen from A It will require a time t = 75 m/1.5 m/s = 50 s to get across, but at least you will be where you intended to be!! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Extremely tricky 2d example: High-altitude winds are blowing at 450 km/hr at an angle of 35° S of E. A plane with airspeed 1100 km/hr desires to fly from New York to Los Angeles, a distance of 5000 km at a bearing of 25°S of W . How much time for the flight? How should the pilot aim the plane? What will the plane’s speed be with respect to the ground? Ground is frame A; air is frame B; plane is object P Take x direction to be E and y direction to be N LA's position is rP/A = -5000 cos 25 ˆi - 5000 sin 25 ˆj = -4532 ˆi - 2113 ˆj ˆj rP/A -4532 ˆ plane' s velocity is (where t is the time of flight) v P/A = = i t t wind velocity is v = 450 cos 35 ˆi - 450 sin 35 ˆj = 368 .6ˆi - 258 .1ˆj - 2113 t B/A v P/B = vP/B,E ˆi vP/V,N ˆj 2 2 airspeed is | v P/B |= vP/B, E vP/B,N = 1100 km/hr v P/B = v P/A - v B/A = ˆi ˆj - 368.6ˆi - 258.1ˆj -4532 t -2113 t 2 1100 t 2 = - 4532 - 368 .6t - 2113 258 .1t 2 Solve for t and substitute back in… quadratic equation!! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 1.210 x10 6 t 2 = 20.54 x10 6 3.341x10 6 t .1359 x10 6 t 2 4.465 x10 6 - 1.091x10 6 t .06662 x10 6 t 2 [cancel the 10 6 ] 2.165t 2 - 2.250t - 25.00 = 0 t = -2.2502 -4 2.165 -25.00 = 21.007 2.250 v P/A = 2.25014.885 = 8.51 hr 2.014 ˆi ˆj = -533 km/hr ˆi - 248 km/hrˆj -4532 8.51 -2110 8.51 plane' s speed is vP/A = - 533 2 - 248 2 v P/B = v P/A - v B/A = - 4532 ˆ i - 2110 ˆj - 368 .6ˆi - 258 .1ˆj 8.51 8.51 = -901 km/hr ˆi - 10 km/hr ˆj plane is ' aimed' at angle = .63 Arctan -10 -901 180 = 180 .63 = .63 N of W Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. = 635 km/hr Resistive Forces and Fluid Drag I • a different type of ‘friction’ in which the friction force depends on how fast the object is moving • friction we know and love (‘kinetic’) is also dubbed Coulomb friction • linear drag obeys R = – b v where b is a constant • resistance force R is opposed to v and proportional to v • taking + = ↓ and assuming object released from rest in gravity FBD dv dv b m = -bv mg =- vg dt dt m • if dv/dt = 0, v is terminal speed vT • vT = mg/b Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. R a Fg +y Resistive Forces and Fluid Drag II dv dv b m = -bv mg =- vg dt dt m • this is a differential equation for v(t) • we think about what functions have a first derivative that ‘looks like them’… accounting for the g is issue too • thought reveals exponential functions of time Try v(t ) = Ae t B and stick it into the differenti al equation b -b mg t t Ae = - ( Ae B) g works if = ,B= = vT m m b • no information about value of A as yet • we assumed that the object was released from rest, so the velocity at t = 0 is 0 [v(0) = 0] • this is called implementation of initial conditions Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Resistive Forces and Fluid Drag III v(0) = 0 = Ae ( 0) B 0 = A B mg A = -B = = -vT b - bt / m v (t ) = vT (1 - e ) dv(t ) a(t ) = = ge -bt / m dt • here is the graph for v(t) • exponential decay ‘up to a constant’ • the acceleration is simple exponential decay Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.