Page 1Physics 207 – Lecture 5Physics 207: Lecture 5, Pg 1Physics 207, Physics 207, Lecture 5, Sept. 17Lecture 5, Sept. 17Goals:Goals:Solve problems with multiple accelerations in 2Solve problems with multiple accelerations in 2--dimensions dimensions (including linear, projectile and circular (including linear, projectile and circular motion)motion)Discern different reference frames and understand Discern different reference frames and understand how they relate to particle motion in stationary and how they relate to particle motion in stationary and moving framesmoving framesRecognize different types of forces and know how they Recognize different types of forces and know how they act on an object in a particle representationact on an object in a particle representationIdentify forces and draw a Identify forces and draw a Free Body DiagramFree Body DiagramAssignment: HW3, (Chapters 4 & 5, due 9/25, Wednesday)Read through Chapter 6, Sections 1-4Physics 207: Lecture 5, Pg 2Exercise (2D motion with acceleration)Relative Trajectories: Monkey and HunterA. go over the monkey?B. hit the monkey?C. go under the monkey?A hunter sees a monkey in a tree, aims his gun at the monkey and fires. At the same instant the monkey lets go. Does the bullet …Page 2Physics 207 – Lecture 5Physics 207: Lecture 5, Pg 3Schematic of the problem xB(∆t) = d = v0cos θ ∆t yB(∆t) = hf= v0sin θ ∆t – ½ g ∆t2 xM(∆t) = d yM(∆t) = h – ½ g ∆t2 Does yM(∆t) = yB(∆t) = hf? Does anyone want to change their answer ?(x0,y0) = (0 ,0)(vx,vy) = (v0cos θ, v0sin θ)Bulletθv0g(x,y) = (d,h)hfWhat happens if g=0 ?How does introducing g change things?MonkeyPhysics 207: Lecture 5, Pg 4Uniform Circular Motion (UCM) is common so we have specialized terms Arc traversed s = θθθθ r Tangential velocity vt Period, T, and frequency, f Angular position, θθθθ Angular velocity, ωPeriod (T): The time required to do one full revolution, 360° or 2π radiansFrequency (f): 1/T, number of cycles per unit timeAngular velocity or speed ω = 2πf = 2π/T, number of radians traced out per unit time (in UCM average and instantaneous will be the same)rθvtsPage 3Physics 207 – Lecture 5Physics 207: Lecture 5, Pg 5Example Question (note the commonality with linear motion) 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. 1 What is the period of the initial rotation? 2 What is initial angular velocity? 3 What is the tangential speed of a point on the rim during thisinitial period? 4 Sketch the angular displacement versus time plot. 5 What is the average angular velocity? 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 angular acceleration be? 7 What is the magnitude and direction of the acceleration after 10 seconds? Physics 207: Lecture 5, Pg 6Example QuestionExample Question A horizontal turntable 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. 1 What is the period of the turntable during the initial rotationT (time for one revolution) = ∆t /# of revolutions/ time = 5 sec / 10 rev = 0.5 s 2 What is initial angular velocity?ω = angular displacement / time = 2 π f = 2 π / T = 12.6 rad / s 3 What is the tangential speed of a point on the rim during this initial period? We need more…..Page 4Physics 207 – Lecture 5Physics 207: Lecture 5, Pg 7Relating rotation motion to linear velocity In UCM a particle moves at constant tangential speed vtaround a circle of radius r (only direction changes). Distance = tangential velocity · time Once around… 2πr = vtTor, rearranging (2π/T) r = vtω r = vtDefinition: If UCM then ω = ω = ω = ω = constant3 So vT= ω r = 4 π rad/s 1 m = 12.6 m/srθvts4 A graph of angular displacement (θ) vs. timePhysics 207: Lecture 5, Pg 8Angular displacement and velocity Arc traversed s = θθθθ rin time ∆t then ∆s = ∆θ rso ∆s / ∆t = (∆θ / ∆t) rin the limit ∆t 0one getsds / dt = dθ / dt rvt= ω rω = dθ / dtif ω is constant, integrating ω = dθ / dt, we obtain: θ = θο+ ω ∆tCounter-clockwise is positive, clockwise is negativerθvtsPage 5Physics 207 – Lecture 5Physics 207: Lecture 5, Pg 9Sketch of Sketch of θ θ vs. timevs. timetime (seconds)020π10π51030πθ (radians)θ = θο+ ω ∆tθ = 0 + 4π 5 radθ = θο+ ω ∆tθ = 20π rad + 4π rad5 Avg. angular velocity = ∆θ / ∆t = 24 π /10 rad/sPhysics 207: Lecture 5, Pg 10Next partNext part…….... 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?Page 6Physics 207 – Lecture 5Physics 207: Lecture 5, Pg 11 Then angular velocity is no longer constant so dω/dt ≠ 0 Define tangential acceleration as at = dvt/dt = r dω/dt So s = s0+ (ds/dt)0∆t + ½ at∆t2 and s = θ r We can relate at to dω/dtθ = θo+ ωo∆t + ∆t2ω = ωo+ ∆t Many analogies to linear motion but it isn’t one-to-one Note: Even if the angular velocity is constant, there is always a radial acceleration.Well, if ω is linearly increasing …atr12atrPhysics 207: Lecture 5, Pg 12Tangential acceleration? Tangential acceleration? θ = θo+ ωo∆t + ∆t2(from plot, after 10 seconds)24 π rad = 0 rad + 0 rad/s ∆t + ½ (at/r) ∆t248 π rad 1m / 100 s2= atrθvts12atr 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?Page 7Physics 207 – Lecture 5Physics 207: Lecture 5, Pg 13Circular motion also has a radial (perpendicular) componentCircular motion also has a radial (perpendicular) componentCentripetal Accelerationar= vt2/rCircular motion involves continuous radial accelerationvtrarUniform circular motion involves only changes
View Full Document
Unlocking...