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UW-Madison PHYSICS 207 - Lecture 6 Notes

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Lecture 6Uniform Circular Motion (UCM) has only radial accelerationAgainUniform Circular Motion (UCM) is common so we have specialized termsMass-based separation with a centrifugeExampleSlide 12Slide 13Slide 14Sketch of q vs. timeSlide 16Slide 17Non-uniform CM: What if w is linearly increasing …Slide 19Tangential acceleration?Non-uniform Circular MotionSlide 22Chaps. 5, 6 & 7 What causes motion? (What is special about acceleration?) What are forces ? What kinds of forces are there ? How are forces and changes in motion related ?Newton’s First Law and IRFsNo Net Force, No acceleration…a demo exerciseSlide 26Physics 207: Lecture 6, Pg 1Lecture 6GoalsGoals Discuss uniform and non-uniform circular motion  Recognize different types of forces and know how they act on an object in a particle representation Identify forces and draw a Free Body Diagram Solve problems with forces in equilibrium (a=0) and non-equilibrium (a≠0) using Newton’s 1st & 2nd laws.Physics 207: Lecture 6, Pg 3Uniform Circular Motion (UCM) has only radial acceleration UCM changes only in the direction of v1. Particle doesn’t speed up or slow down!2. Velocity is always tangential, acceleration perpendicular ! path radial aa0Taavv|v||v|vvvTvvrrd tddtdaTTTTTr/v /vvvvv2Physics 207: Lecture 6, Pg 4AgainCentripetal Acceleration ar = vT2/r = 2rCircular motion involves continuous radial accelerationvTrarUniform circular motion involves only changes in the direction of the velocity vectorAcceleration is perpendicular to the trajectory at any point, acceleration is only in the radial direction.Physics 207: Lecture 6, Pg 5Uniform Circular Motion (UCM) is common so we have specialized termsNew definitions1. Period (T): The time required to do one full revolution, 360° or 2 radians2. Frequency (f): f ≡ 1/T, number of cycles per unit timeAngular velocity or speed in UCM = t = 2/ T = 2f (radians per unit time)rvTsPhysics 207: Lecture 6, Pg 6Mass-based separation with a centrifugeHow many g’s (1 g is ~10 m/s2)?ar = vT2 / r = 2 r f = 6000 rpm = 100 rev. per second is typical with r = 0.10 mar = (2 102)2 x 0.10 m/s2 Before Afterbb5ar = 4 x 104 m/s2 or ca. 4000 g’s !!!but a neutron star surface is at 1012 m/s2Physics 207: Lecture 6, Pg 11Example A horizontally mounted disk 2.0 meters in diameter (1.0 m in radius) spins at constant angular speed such that it first undergoes (1) 10 counter clockwise revolutions in 5.0 seconds and then, again at constant angular speed, (2) 2 counter clockwise revolutions in 5.0 seconds. 1 What is T the period of the initial rotation? T = time for 1 revolution = 5 sec / 10 rev = 0.5 sPhysics 207: Lecture 6, Pg 12Example A horizontally mounted disk 2 meters in diameter spins at constant angular speed such that it first undergoes 10 counter clockwise revolutions in 5 seconds and then, again at constant angular speed, 2 counter clockwise revolutions in 5 seconds.2 What is  the initial angular velocity?= d /dt = /t = 10 • 2π radians / 5 seconds = 12.6 rad / s ( also 2  f = 2  / T )Physics 207: Lecture 6, Pg 13Example A horizontally mounted disk 2 meters in diameter spins at constant angular speed such that it first undergoes 10 counter clockwise revolutions in 5 seconds and then, again at constant angular speed, 2 counter clockwise revolutions in 5 seconds.3 What is the tangential speed of a point on the rim during this initial period?| vT | = ds/dt = (r d /dt = r | vT | = r = 1 m • 12.6 rad/ s = 12.6 m/sPhysics 207: Lecture 6, Pg 14Example A horizontally mounted disk 1 meter in radius spins at constant angular speed such that it first undergoes 10 counter clockwise revolutions in 5 seconds and then, again at constant angular speed, 2 counter clockwise revolutions in 5 seconds.4 Sketch the  (angular displacement) versus time plot. =  + t rvtsPhysics 207: Lecture 6, Pg 15Sketch of vs. timetime (seconds) (radians) =  + t =  +  5rad =  + t =  rad + (x 2/5) 5 rad = 24 radPhysics 207: Lecture 6, Pg 16Example5What is the average angular velocity over the 1st 10 seconds?Physics 207: Lecture 6, Pg 17Sketch of vs. timetime (seconds) (radians) =  + t =  +  5rad =  + t =  rad + ( 5) 5 rad =  rad5 Avg. angular velocity =  / t = 24  /10 rad/sPhysics 207: Lecture 6, Pg 18Then angular velocity is no longer constant so d /dt ≠ 0Define tangential acceleration as aT ≡ dvT/dt = r d/dtDefine angular acceleration ≡ d dtLet  be constant & integrating:= 0 +  t Integrating again:= 0 + 0 t + ½  t2 Multiply by rr = r 0 + r 0 t + ½ r  t2 s= s0 + vT t + ½ aT t2 Many analogies to linear motion but it isn’t one-to-oneNon-uniform CM: What if  is linearly increasing …Physics 207: Lecture 6, Pg 19Example6 If now the turntable starts from rest and uniformly accelerates throughout and reaches the same angular displacement in the same time, what must be the angular acceleration  ?Key point ….. is associated with tangential acceleration (aT).Physics 207: Lecture 6, Pg 20 Tangential acceleration? = o + o t + t2 (from plot, after 10 seconds)rvts12aT r 6 If now the turntable starts from rest and uniformly accelerates throughout and reaches the same angular displacement in the same time, what must the “tangential acceleration” be?7 What is the magnitude and direction of the acceleration after 10 seconds? 24  rad = 0 rad + 0 rad/s t + ½ (aT/r) t2 48 rad 1m / 100 s2 = aT =0.48  m/s2Physics 207: Lecture 6, Pg 21Non-uniform Circular MotionFor an object moving along a curved trajectory, with varying speedVector addition: a = ar + aT (radial and tangential)araT2Tangential2radial|a| aa r2Tradialva dtd |v|aTTangentialaPhysics 207: Lecture 6, Pg 22 Tangential acceleration? aT = 0.48 m / s2 and vT = 0 + aT t = 4.8 


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UW-Madison PHYSICS 207 - Lecture 6 Notes

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