PC11 Fundamentals of Physics I Lectures 7 and 8 Motion in Two Dimensions Dr Tay Sen Chuan 1 Ground Rules Switch off your handphone and paer Switch off your laptop computer and keep it No talkin while lecture is oin on No ossipin while the lecture is oin on Raise your hand if you have question to ask Be on time for lecture Be on time to come back from the recess break to continue the lecture Brin your lecturenotes to lecture Position and Displacement The position of an object is described by its position vector, r The displacement of the object is defined as the chane in its position Δr r f - r i In two- or threedimensional kinematics, everythin is the same as in one-dimensional motion except that we must now Averae Velocity The averae velocity is the ratio of the displacement to the time interval for the displacement Δ v r Δt The direction of the averae velocity is the direction of the displacement vector, Δr The averae velocity between points is independent of the path taken It is dependent on the use full vector notation 3 displacement Plane View 4
Instantaneous Velocity The instantaneous velocity is the limit of the averae velocity as Δt approaches zero lim Δr dr v Δt dt Δ t 0 The direction of small displacement (chane in positions) tells the direction that the particle is headin at the moment. 5 Instantaneous Velocity, cont The direction of the instantaneous velocity vector at any point in a particle s path is alon a line tanent to the path at that point and in the direction of motion The manitude of the instantaneous velocity vector is the speed The speed is a scalar quantity 6 Averae Acceleration Averae Acceleration, cont The averae acceleration of a particle as it moves is defined as the chane in the instantaneous velocity vector divided by the time interval durin which that chane occurs. a v v Δv t t Δt f i f i As a particle moves, Δv can be found in different ways The averae acceleration is a vector quantity directed alon Δv Plane View 7 8
Instantaneous Acceleration The instantaneous acceleration is the limit of the averae acceleration as Δt approaches zero Δv a lim Δt Δ t 0 dv dt The direction of small chane in velocities tells the direction of acceleration. x 9 Producin An Acceleration Various chanes in a particle s motion may produce an acceleration, such as: The manitude of the velocity vector may chane The direction of the velocity vector may chane Even if the manitude remains constant Both may also chane simultaneously 10 Kinematic Equations for Two- Dimensional Motion When the two-dimensional motion has a constant acceleration, a series of equations can be developed that describe the motion These equations will be similar to those of one-dimensional kinematics Kinematic Equations, Position vector Velocity r xˆi+ yˆj dr v v ˆ ˆ xi+ vyj dt Since acceleration is constant, we can also find an expression for the velocity as a function of time: v f v i + at 11 1
Kinematic Equations, 3 The velocity vector can be represented by its components v f is enerally not alon the direction of either v i or at Kinematic Equations, 4 The position vector can also be expressed as a function of time: r f r i + v i t + ½ at This indicates that the position vector is the sum of three other vectors: The initial position vector r i The displacement resultin from v i t The displacement resultin from ½ at v f v i + at 13 14 Kinematic Equations, 5 Kinematic Equations, Components The vector representation of the position vector r f is enerally not in the same direction as v i or as a i r f and v f are enerally not in the same direction The equations for final velocity and final position are vector equations, therefore they may also be written in component form This shows that two-dimensional motion at constant acceleration is equivalent to two independent motions One motion in the x-direction and r f r i + v i t + ½ at the other in the y-direction 15 16
Kinematic Equations, Component Equations v f v i + at becomes v xf v xi + a x t and v yf v yi + a y t r f r i + v i t + ½ at becomes x f x i + v xi t + ½ a x t and y f y i + v yi t + ½ a y t Projectile Motion An object may move in both x and y directions simultaneously The form of twodimensional motion we will deal with is called projectile motion 17 18 Assumptions of Projectile Motion The free-fall acceleration is constant over the rane of motion is directed downward The effect of air friction is neliible With these assumptions, an object in projectile motion will follow a parabolic path This path is called the trajectory Verifyin the Parabolic Trajectory Reference frame chosen y is vertical with upward positive Acceleration components a y - and a x 0 Initial velocity components v xi v i cos θ and v yi v i sin θ v i Θ 19 0
Verifyin the Parabolic Trajectory, cont Rane and Maximum Heiht of a Projectile Displacements x f v xi t (v i cos θ) t y f v yi t + ½a y t (v i sin θ)t -½t Combinin the equations and removin tives: y ( tanθ ) x x i vi cos θi This is in the form of y ax bx which is the standard form of a parabola 1 When analyzin projectile motion, two characteristics are of special interest The rane, R, is the horizontal distance of the projectile The maximum heiht the projectile reaches is h Heiht of a Projectile, equation The maximum heiht of the projectile can be found in terms of the initial velocity vector: h v i sin i θ This equation is valid only for symmetric motion Rane of a Projectile, equation The rane of a projectile can be expressed in terms of the initial velocity vector: v i sin θ R i This is valid only for symmetric trajectory 3 We will derive them. 4
Projectile Formulation Let the initial velocity be u. Let the anle subscripted by u and x-axis be θ. Alon the x-direction, x t ( u cos θ ) t x u cos θ Alon the Y-direction, 1 ( u sin ) t t y θ 1 Substitute (1) into (): y y y ( u sinθ ) ( tanθ ) u θ x u cosθ x x u cos 1 θ u x x cos θ Maximum occurs when the y-component of the velocity is equal to 0, i.e. the object is at the moment of comin down. v u sinθ t 0 u sinθ t 0 u sinθ t Substitute (3) into (): u sinθ y h u sinθ u sin θ u sin θ h u h sin θ v or i 1 3 u sinθ sin θi h To derive the maximum rane (R) alon the x-direction, we let y 0 in equation. 1 y ( u sinθ ) t t R The distance travelled in x-direction Why? 1 ( u sin θ ) t t t u sin θ t u sin θ u sin θ t t 0 0 4 for y t x x x u sin θ seconds ( u cos θ ) ( u cos θ ) u u R t u sin θ ( sin θ cos θ ) sin θ is : What is the maximum value of R, and the correspondin? Since sin θ 1, sin θ 1, i.e., u R u R u R R θ When θ 45 is 90 o ( sin( 45 ) o ( sin 90 ) θ o, maximum o θ 45 when o
Rane of a Projectile, final The maximum rane occurs at θ i 45 o Complementary anles will produce the same rane But the maximum heiht will be different for the two anles The times of the fliht will be different for the two anles Projectile Motion Problem Solvin Hints Select a coordinate system Resolve the initial velocity into x and y components Analyze the horizontal motion usin constant velocity techniques Analyze the vertical motion usin constant acceleration techniques Remember that both directions share the same time 9 30 Non-Symmetric Projectile Motion Follow the eneral rules for projectile motion Break the y-direction into parts up and down or symmetrical back to initial heiht and then the rest of the heiht May be non-symmetric in other ways Example. A firefihter, a distance d from a burnin buildin, directs a stream of water from a fire hose at anle Ө i above the horizontal as in Fiure. If the initial speed of the stream is v i, at what heiht h does the water strike the buildin? 31 3
Answer: Uniform Circular Motion Uniform circular motion occurs when an object moves in a circular path with a constant speed An acceleration exists since the direction of the motion is chanin This chane in velocity is related to an acceleration The velocity vector is always tanent to the path of the object 33 34 Chanin Velocity in Uniform Circular Motion The chane in the velocity vector is due to the chane in direction The vector diaram shows Δv v f - v i Centripetal Acceleration The acceleration is always perpendicular to the path of the motion The acceleration always points toward the center of the circle of motion This acceleration is called the centripetal acceleration 35 36
Centripetal Acceleration, cont Period The manitude of the centripetal acceleration vector is iven by a C v r By the use of same ratio, this can be derived as follows: Δv Δr v r Δv v Δr r Δv v Δr x Δt r Δt a C v r v 37 The period, T, is the time required for one complete revolution (one complete cycle) As one complete cycle is the distance of the circumference. The speed of the particle would be the circumference of the circle of motion divided by the period. Speed distance/time, so time distance/speed π r Therefore, the period is T v 38 so 8.1 m/s Example. The astronaut orbitin the Earth in Fiure is preparin to dock with a Westar VI satellite. The satellite is in a circular orbit 600 km above the Earth's surface, where the free-fall acceleration is 8.1 m/s. Take the radius of the Earth as 6400 km. Determine the speed of the satellite and the time interval required to Speed of satellite in order to maintain the orbit Time to complete 1 cycle of orbit complete one orbit around the Earth. 39 40 and
Tanential Acceleration The manitude of the velocity could also be chanin In this case, there would be a tanential acceleration Observe what happens at the center (my hand) when I make the ball to rotate faster Circular motion on Total Acceleration The tanential acceleration causes the chane in the speed of the particle The radial acceleration comes from a chane in the direction of the velocity vector. (In the textbook by Serway, the direction or radial acceleration is away from center not important.) a a t + a r my hand 41 4 a a r a t Total Acceleration, equations The tanential acceleration: a t The radial acceleration: a r a c The total acceleration: Manitude a a + a r t Why the strin does not slack and the ball does not fly towards the center? dv dt v r 43 Example A train slows down as it rounds a sharp horizontal turn, slowin from 90.0 km/h to 50.0 km/h in the 15.0 s that it takes to start to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50.0 km/h. Assume it continues to slow down at this time at the same rate. 44
We assume the train is still slowin down at the instant in question. v [50 x (1000/3600)] ac 1.9 m s 1.9 m/s r 150 3 1 h Δ v ( 40.0 km h)( 10 m km )( ) at 0.741 m s Δ t 15.0 s 1.48 m/s 90 km/h ( 1.9 m s ) ( 0.741 m s ) c t a a + a + 1 at 1 0.741 at an anle of tan tan a 1.9 c 9.9 50 km/h 3600 s 90 km/h to 50 km/h in 15 seconds 45 Outcome of Derailments This dent is symmetrical (the beautiful v shape). This situation seldom occurs because the total acceleration (which is the direction of the force) is usually not toward the center due to the existence of tanential acceleration. a The dent is caused by the force acted by the train on the rail. This dent is in the direction of the total acceleration (a). This situation usually occurs. 46 Relative Velocity Relative Velocity, eneralized Two observers movin relative to each other enerally do not aree on the outcome of an experiment For example, observers A and B below see different paths for the ball Reference frame S is stationary Reference frame S is movin at v 0 This also means that S moves at v 0 relative to S Define time t 0 as that time when the oriins coincide 47 48
Relative Velocity, equations The positions as seen from the two reference frames (r is the position seen from stationary frame, r seen from movin frame) are related throuh the velocity r r v 0 t The derivative of the position equation will ive the velocity equation v v v 0 These are called the Galilean transformation equations (d1-9) 49 Acceleration in Different Frames of Reference The derivative of the velocity equation will ive the acceleration equation The acceleration of the particle measured by an observer in one frame of reference is the same as that measured by any other observer movin at a constant velocity relative to the first frame. Why? r r v 0 t v v v v 0 is a constant, so v 0 0 has no effect on the chane in velocity. 50 Example. How lon does it take an automobile travelin in the left lane at 60.0 km/h to pull alonside a car travelin in the riht lane at 40.0 km/h if the cars' front bumpers are initially 100 m apart? (Disk ) Examples. d-03, d-04 Answer: The bumpers are initially 100 m apart (0.1 km). After time t the bumper of the leadin car travels 40.0 thrs, while the bumper of the chasin car travels 60.0t. Since the cars are side by side at time t, we have 60 t 40 km/h 40 t Constant speed (A) Horizontal Track 0.1 + 40 t 60 t t 0.1/0 0.005 hr 0.005 x 3600 sec 18 s 60 km/h 0.1 km 51 (B) Tilted Track 5
Examples. d-03, d-04 This time the car will be pulled alon the horizontal track by a strin and a weiht hanin over a pulley, and will be acceleratin constantly. The ball is fired at point X. Where will the ball land this time? (The horizontal track is lon enouh for the car not to fall off when the ball lands.) Point X (C) The car is pulled alon by a strin and a weiht hanin over a pulley 53