lecture3 - Chapter 3 Kinematics-2D Posi%on and ...

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Unformatted text preview: Chapter 3 Kinematics-2D Posi%on and Displacement •  The displacement of the object is defined as the change in its posi-on •  Δr = rf - ri •  Now r and Δr are vectors, which have both magnitude and direc%on Vectors and Scalars A vector has magnitude as well as direction. Some vector quantities: displacement, velocity, force, momentum A scalar has only a magnitude. Some scalar quantities: mass, time, temperature Vector Nota%on A •  When handwriCen, use an arrow: •  When printed, in bold: A •  When dealing with just the magnitude of a vector in print, an italic leCer is used: |A| Proper%es of Vectors •  Equality of Two Vectors –  Two vectors are equal if they have the same magnitude and the same direc%on •  Movement of vectors in a diagram –  Any vector can be moved parallel to itself without being affected Addition of Vectors—Graphical Methods We can add two vectors by using the Pythagorean Theorem. When adding vectors, their directions must be taken into account Units must be the same V R = V1 + V 2 Addition of multiple Vectors—Graphical Methods We can add many vectors graphically by using the tail-to-tip method. u r uuu rr r V R = V1 + V 2 + V 3 Continue drawing the vectors “tip-to-tail” until all are included The resultant is drawn from the origin of the first vector to the end of the last vector Example •  Three displacements are A = 200 m, due south; B = 250 m, due west; C = 150 m, 30.0° east of north. Construct a separate diagram for each of the following possible ways of adding these vectors: R1 = A + B + C; R2 = B + C + A; R3 = C + B + A. Subtraction of Vectors, and Multiplication of a Vector by a Scalar In order to subtract vectors, we define the negative of a vector, which has the same magnitude but points in the opposite direction. Then we add the negative vector. A vector can be multiplied by a scalar c; the result is a vector that hasrthe same u direction but a magnitude cV If c is negative, the resultant vector points in the opposite direction. Example Find graphically •  (a) A+B, •  (b) A- B , •  (c) B- A , •  and (d) A- 2B . Project Vectors to Components Any vector can be expressed as the sum of two other vectors, which are called its components. Usually the other vectors are chosen so that they are perpendicular to each other. If the components are perpendicular, they can be found using trigonometric functions. Adding Vectors by Components The components are effectively one-dimensional, so they can be added arithmetically. Adding vectors: 1.  Draw a diagram; add the vectors graphically. 2.  Choose x and y axes. 3.  Resolve each vector into x and y components. 4.  Calculate each component using sines and cosines. 5.  Add the components in each direction. 6.  To find the length and direction of the vector, use: Example •  While exploring a cave, a spelunker starts at the entrance and moves the following distances. She goes 75.0 m north, 250 m east, 200 m at an angle 30.0o north of east, and 150 m south. Find the resultant displacement from the cave entrance. Unit Vectors Unit vectors have magnitude 1. Using unit vectors, any vector can be written in terms of its components: Average Velocity •  The average velocity is the ra%o of the displacement to the %me interval for the displacement Δr v= Δt –  The direc%on of the average velocity is the direc%on of the displacement vector, Δr Instantaneous Velocity •  The instantaneous velocity is the limit of the average velocity as Δt approaches zero –  The direc%on of the instantaneous velocity is along a line that is tangent to the path of the par%cle’s direc%on of mo%on –  The magnitude of the instantaneous velocity vector is the speed Accelera%on vf − vi Δv a= = t f − ti Δt Δ v dv a ≡ lim = dt Δt → 0 Δ t Change in either speed or direction of a particle’s motion may produce an acceleration Kinema%c Equa%ons •  Posi%on vector r = xˆ + yˆ i j •  Velocity v = dr = v ˆ + v ˆ xi yj dt –  Since accelera%on is constant, we can also find an expression for the velocity as a func%on of %me: vf = vi + at Kinema%c Equa%ons •  The posi%on vector can also be expressed as a func%on of %me: –  rf = ri + vit + ½ at2 –  This indicates that the posi%on vector is the sum of three other vectors: •  The ini%al posi%on vector •  The displacement resul%ng from vi t •  The displacement resul%ng from ½ at2 Kinema%c Equa%ons, Component Equa%ons •  Two- dimensional mo%on at constant accelera%on is equivalent to two independent mo%ons •  vf = vi + at becomes •  vxf = vxi + axt and •  vyf = vyi + ayt •  rf = ri + vi t + ½ at2 becomes •  xf = xi + vxi t + ½ axt2 and •  yf = yi + vyi t + ½ ayt2 Example •  At t = 0, a par%cle moving in the xy plane with constant accelera%on has a velocity of and is v i (3 . 00 ı − 2 . 00 ıj) m / s = i at the origin. At t = 3.00 s, the par%cle's velocity ı j is v = ( . 00 i + . 00 ) m / s . Find 9 7 ı –  –  (a) the accelera%on of the par%cle and (b) its coordinates at any %me t. Projec%le Mo%on Object moving in two dimensions under the influence of Earth's gravity ax=0, ay=- g •  An object may move in both the x and y direc%ons simultaneously –  It moves in two dimensions •  The form of two dimensional mo%on we will deal with is called projec>le mo>on Rules of Projec%le Mo%on •  •  The x- and y- direc%ons of mo%on are completely independent of each other The x- direc%on is uniform mo>on •  ax = 0 •  The y- direc%on is free fall •  ay = - g •  The ini%al velocity can be broken down into its x- and y- components vOx = vO cos θ O vOy = vO sin θ O vx=v0x=v0cosθ0; vy=v0y-gt=v0sinθ0-gt; x=(v0cosθ0) t y=(v0sinθ0) t - gt2/2 g y = (tanθ 0 )x − 2 x2 2v0 cos2 θ 0 Velocity, maximum height and maximum range •  The velocity of the projec%le at any point of its mo%on is the vector sum of its x and y components at that point •  Maximum height (when vy=0) v 2 sin 2 θ 0 h= =0 2g 2g 2 v0 y •  Horizontal range (when y=0) v 2 sin(2θ 0 ) R= 0 g The maximum range occurs at a projection angle of 45o Problem- Solving Strategies 1.  Select a coordinate system and sketch the path of the projec%le 1.  Include ini%al and final posi%ons, veloci%es, and accelera%ons 2.  List known and unknown quan%%es in x- and y- column. –  If the ini%al velocity is known, Resolve the ini%al velocity into x- and y- components first 3.  Treat the horizontal and ver%cal mo%ons independently, write down x(t) and y(t) equa%ons. 4.  See which direc%on can be solved first and then use the result of t to solve the other direc%on. Example A football is kicked at an angle θ0 = 37.0° with a velocity of 20.0 m/s, as shown. Calculate (a) the maximum height, (b) the time of travel before the football hits the ground, (c) how far away it hits the ground, (d) the velocity vector at the maximum height, and (e) the acceleration vector at maximum height. Assume the ball leaves the foot at ground level, and ignore air resistance and rotation of the ball. Example A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), (a) how far in advance of the recipients (horizontal distance) must the package be dropped? (b) Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position? (c) With what speed does the package land in the latter case? Example •  A catapult launches a rocket at an angle of 53.0° above the horizontal with an ini%al speed of 100 m/s. The rocket engine immediately starts a burn, and for 3.00 s the rocket moves along its ini%al line of mo%on with an accelera%on of 30.0 m/s2. Then its engine fails, and the rocket proceeds to move in free- fall. Find (a) the maximum al%tude reached by the rocket, (b) its total %me of flight, and (c) its horizontal range. Rela%ve Velocity •  Two observers moving rela%ve to each other generally do not agree on the outcome of an experiment •  For example, the two cars are not moving against each other; observers A and B below see different paths for the ball Rela%ve Velocity, generalized •  Reference frame S is sta%onary •  Reference frame S’ is moving at vo –  This also means that S moves at –vo rela%ve to S’ •  Define %me t = 0 as that %me when the origins coincide Rela%ve Velocity, equa%ons •  The posi%ons as seen from the two reference frames are related through the velocity r’ = r – vo t •  The deriva%ve of the posi%on equa%on will give the velocity equa%on v’ = v – vo •  These are called the Galilean transforma>on equa>ons Example A boat’s speed in still water is vBW = 2.00 m/s. If the boat is to travel directly across a river whose current has speed vWS = 1.20 m/s, at what upstream angle must the boat head? Example •  A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 10.0 m/s. The student throws a ball into the air along a path that he judges to make an ini%al angle of 60.0° with the horizontal and to be in line with the track. The student's professor, who is standing on the ground nearby, observes the ball to rise ver%cally. How high does she see the ball rise? ...
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This note was uploaded on 12/12/2011 for the course PHYS 3A taught by Professor Casper during the Fall '07 term at UC Irvine.

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