Physics 207: Lecture 4, Pg 1 Physics 207, Lecture 4, Sept. 15 l Goals for Chapts. 3 & 4  Perform vector algebraaddition & subtraction) graphically or by x,y & z components  Perform vector algebra (addition & subtraction) graphically or by x,y & z components  Interconvert between Cartesian and Polar coordinates  Distinguish position-time graphs from particle trajectory plots  Obtain particle velocities and accelerations from trajectory plots  Determine both magnitude and acceleration parallel and perpendicular to the trajectory path  Solve problems with multiple accelerations in 2-dimensions (including linear, projectile and circular motion)  Discern different reference frames and understand how they relate to particle motion in stationary and moving frames

Physics 207: Lecture 4, Pg 2 Physics 207, Lecture 4, Sept. 15 Assignment: Read Chapter 5 (sections 1- 4 carefully) l MP Problem Set 2 due Wednesday (should have started) l MP Problem Set 3, Chapters 4 and 5 (available today)

Physics 207: Lecture 4, Pg 3 Vector addition l The sum of two vectors is another vector. A = B + C B C A B C

Physics 207: Lecture 4, Pg 4 Vector subtraction l Vector subtraction can be defined in terms of addition. B - C B C = B + (-1)C B -C-C B - C A Different direction and magnitude ! A = B + C

Physics 207: Lecture 4, Pg 5 Unit Vectors Unit Vector l A Unit Vector is a vector having length 1 and no units l It is used to specify a direction. u U l Unit vector u points in the direction of U u  Often denoted with a “hat”: u = û U = |U| û û x y z i j k i, j, k l Useful examples are the cartesian unit vectors [ i, j, k ]  Point in the direction of the x, y and z axes. R = r x i + r y j + r z k

Physics 207: Lecture 4, Pg 6 Vector addition using components: CAB l Consider, in 2D, C = A + B. Ciji ji (a) C = (A x i + A y j ) + (B x i + B y j ) = (A x + B x )i + (A y + B y ) Cij (b) C = (C x i + C y j ) l Comparing components of (a) and (b):  C x = A x + B x  C y = A y + B y  |C| =[ (C x ) 2 + (C y ) 2 ] 1/2 C BxBx A ByBy B AxAx AyAy

Physics 207: Lecture 4, Pg 7 Exercise 1 Vector Addition A. {3,-4,2} B. {4,-2,5} C. {5,-2,4} D. None of the above l Vector A = {0,2,1} l Vector B = {3,0,2} l Vector C = {1,-4,2} What is the resultant vector, D, from adding A+B+C?

Physics 207: Lecture 4, Pg 8 Converting Coordinate Systems In polar coordinates the vector R = (r,  ) l In Cartesian the vector R = (r x,r y ) = (x,y) We can convert between the two as follows: In 3D cylindrical coordinates (r,  z), r is the same as the magnitude of the vector in the x-y plane [sqrt(x 2 +y 2 )]  tan -1 ( y / x ) y x (x,y)  r ryry rxrx

Physics 207: Lecture 4, Pg 9 Exercise: Frictionless inclined plane A block of mass m slides down a frictionless ramp that makes angle  with respect to horizontal. What is its acceleration a ?  m a

Physics 207: Lecture 4, Pg 10 Resolving vectors, little g & the inclined plane g (bold face, vector) can be resolved into its x,y or x ’,y ’ components  g = - g j  g = - g cos  j ’ + g sin  i ’  The bigger the tilt the faster the acceleration….. along the incline x’x’ y’y’ x y g  

Physics 207: Lecture 4, Pg 11 Dynamics II: Motion along a line but with a twist (2D dimensional motion, magnitude and directions) l Particle motions involve a path or trajectory l Recall instantaneous velocity and acceleration l These are vector expressions reflecting x, y & z motion r r v r a r r = r (t)v = d r / dta = d 2 r / dt 2

Physics 207: Lecture 4, Pg 12 Instantaneous Velocity l But how we think about requires knowledge of the path. l The direction of the instantaneous velocity is along a line that is tangent to the path of the particle’s direction of motion. v The magnitude of the instantaneous velocity vector is the speed, s. (Knight uses v) s = (v x 2 + v y 2 + v z ) 1/2

Physics 207: Lecture 4, Pg 13 Average Acceleration l The average acceleration of particle motion reflects changes in the instantaneous velocity vector (divided by the time interval during which that change occurs). l The average acceleration is a vector quantity directed along ∆v ( a vector! )

Physics 207: Lecture 4, Pg 14 Instantaneous Acceleration l The instantaneous acceleration is the limit of the average acceleration as ∆v/∆t approaches zero l The instantaneous acceleration is a vector with components parallel (tangential) and/or perpendicular (radial) to the tangent of the path l Changes in a particle’s path may produce an acceleration  The magnitude of the velocity vector may change  The direction of the velocity vector may change (Even if the magnitude remains constant)  Both may change simultaneously (depends: path vs time)

Physics 207: Lecture 4, Pg 15 Motion along a path ( displacement, velocity, acceleration ) l 2 or 3D Kinematics : vector equations: rrvrar r = r(  t) v = dr / dt a = d 2 r / dt 2 Velocity : r v = dr / dt r v av =  r /  t Acceleration : v a = dv / dt v a av =  v /  t vvvv y x path

Physics 207: Lecture 4, Pg 16 Generalized motion with non-zero acceleration: need both path & time Two possible options: Change in the magnitude of v Change in the direction of v a = 0 a a a a = + Animation Q1: What is the time sequence in this particle’s motion? Q2: Is the particle speeding up or slowing down?

Physics 207: Lecture 4, Pg 17 Examples of motion: Chemotaxis l Q1: How do single cell animals locate their “food” ?  Possible mechanism: Tumble and run paradigm. l Q2: How does a fly find its meal?  Possible mechanism: If you smell it fly into the wind, if you don’t fly across the wind. An example…..

Physics 207: Lecture 4, Pg 18 Kinematics l The position, velocity, and acceleration of a particle in 3-dimensions can be expressed as: r ijk r = x i + y j + z k v ijkijk v = v x i + v y j + v z k (i, j, k unit vectors ) a ijk a = a x i + a y j + a z k l All this complexity is hidden away in rrvrar r = r(  t)v = dr / dta = d 2 r / dt 2

Physics 207: Lecture 4, Pg 19 Special Case Throwing an object with x along the horizontal and y along the vertical. x and y motion both coexist and t is common to both Let g act in the – y direction, v 0x = v 0 and v 0y = 0 y t 0 4 x t = 0 4 y t x 0 4 x vs t y vs t x vs y

Physics 207: Lecture 4, Pg 20 Another trajectory x vs y t = 0 t =10 Can you identify the dynamics in this picture? How many distinct regimes are there? Are v x or v y = 0 ? Is v x >,< or = v y ? y x

Physics 207: Lecture 4, Pg 21 Another trajectory x vs y t = 0 t =10 Can you identify the dynamics in this picture? How many distinct regimes are there? 0 < t < 3 3 < t < 77 < t < 10  I.v x = constant = v 0 ; v y = 0  II.v x = v y = v 0  III. v x = 0 ; v y = constant < v 0 y x What can you say about the acceleration?

Physics 207: Lecture 4, Pg 22 Exercise 2 & 3 Trajectories with acceleration l A rocket is drifting sideways (from left to right) in deep space, with its engine off, from A to B. It is not near any stars or planets or other outside forces. l Its “constant thrust” engine (i.e., acceleration is constant) is fired at point B and left on for 2 seconds in which time the rocket travels from point B to some point C  Sketch the shape of the path from B to C. l At point C the engine is turned off.  Sketch the shape of the path after point C

Physics 207: Lecture 4, Pg 23 Exercise 2 Trajectories with acceleration A. A B. B C. C D. D E. None of these B C B C B C B C A C B D From B to C ?

Physics 207: Lecture 4, Pg 24 Exercise 3 Trajectories with acceleration A. A B. B C. C D. D E. None of these C C C C A C B D After C ?

Physics 207: Lecture 4, Pg 25 Trajectory with constant acceleration along the vertical How does the trajectory appear to different observers ? What if the observer is moving with the same x velocity (i.e. running in parallel)? x t = 0 4 y x vs y t x 0 4 x vs t

Physics 207: Lecture 4, Pg 26 Trajectory with constant acceleration along the vertical This observer will only see the y motion x t = 0 4 y x vs y x t = 0 4 y y vs t In an inertial reference frame all see the same acceleration

Physics 207: Lecture 4, Pg 27 Trajectory with constant acceleration along the vertical What do the velocity and acceleration vectors look like? Velocity vector is always tangent to the curve! Acceleration may or may not be! y t = 0 4 x x vs y

Physics 207: Lecture 4, Pg 28 Home Exercise The Pendulum 1  = 30° A) v r = 0 a r = 0 v T = 0 a T  0 B ) v r = 0 a r  0 v T = 0 a T = 0 C) v r = 0 a r  0 v T = 0 a T  0 Which statement best describes the motion of the pendulum bob at the instant of time drawn ? i.when the bob is at the top of its swing. ii.which quantities are non-zero ?

Physics 207: Lecture 4, Pg 29 Home Exercise The Pendulum Solution aTaT NOT uniform circular motion : If circular motion then a r not zero, If speed is increasing so a T not zero  = 30° a O However, at the top of the swing the bob temporarily comes to rest, so v = 0 and the net tangential force is mg sin  C) v r = 0 a r = 0 v T = 0 a T  0 mg T Everywhere else the bob has a non- zero velocity and so then (except at the bottom of the swing) v r = 0 a r  0 v T  0 a T  0

Physics 207: Lecture 4, Pg 30 Physics 207, Lecture 4, Sept. 15 Assignment: Read Chapter 5 (sections 1- 4 carefully) l MP Problem Set 2 due Wednesday (should have started) l MP Problem Set 3, Chapters 4 and 5 (available today)