Ground Rules. PC1221 Fundamentals of Physics I. Position and Displacement. Average Velocity. Lectures 7 and 8 Motion in Two Dimensions

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PC11 Fndamentals of Physics I Lectres 7 and 8 Motion in Two Dimensions A/Prof Tay Sen Chan 1 Grond Rles Switch off yor handphone and paer Switch off yor laptop compter and keep it No talkin while

1 PC11 Fndamentals of Physics I Lectres 7 and 8 Motion in Two Dimensions A/Prof Tay Sen Chan 1 Grond Rles Switch off yor handphone and paer Switch off yor laptop compter and keep it No talkin while lectre is oin on No ossipin while the lectre is oin on Raise yor hand if yo have qestion to ask Be on time for lectre Be on time to come back from the recess break to contine the lectre Brin yor lectrenotes to lectre Position and Displacement Averae Velocity 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 mst now se fll vector notation Plane View 3 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 displacement Plane View 4

2 Instantaneos Velocity The instantaneos 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 Instantaneos Velocity, cont The direction of the instantaneos 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 manitde of the instantaneos velocity vector is the speed The speed is a scalar qantity 6 Averae Acceleration Averae Acceleration, cont The averae acceleration of a particle as it moves is defined as the chane in the instantaneos velocity vector divided by the time interval drin which that chane occrs. a v v Δv t t Δt f i f i As a particle moves, Δv can be fond in different ways The averae acceleration is a vector qantity directed alon Δv Plane View 7 8

Instantaneos Acceleration The instantaneos 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

3 Instantaneos Acceleration The instantaneos 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 Prodcin An Acceleration Varios chanes in a particle s motion may prodce an acceleration, sch as: The manitde of the velocity vector may chane The direction of the velocity vector may chane Even if the manitde remains constant Both may also chane simltaneosly 10 Kinematic Eqations for Two- Dimensional Motion When the two-dimensional motion has a constant acceleration, a series of eqations can be developed that describe the motion These eqations will be similar to those of one-dimensional kinematics Kinematic Eqations, 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 fnction of time: v f v i + at 11 1

Kinematic Eqations, 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 Eqations, 4 The position vector can also be

The displacement resltin from ½ at v f v i + at 13 14 Kinematic Eqations, 5 Kinematic Eqations, Components The vector representation of the position vector r f is enerally not in the same direction

4 Kinematic Eqations, 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 Eqations, 4 The position vector can also be expressed as a fnction of time: r f r i + v i t + ½ at This indicates that the position vector is the sm of three other vectors: The initial position vector r i The displacement resltin from v i t The displacement resltin from ½ at v f v i + at Kinematic Eqations, 5 Kinematic Eqations, 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 eqations for final velocity and final position are vector eqations, therefore they may also be written in component form This shows that two-dimensional motion at constant acceleration is eqivalent 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

5 Kinematic Eqations, Component Eqations Projectile Motion 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 17 An object may move in both x and y directions simltaneosly The form of twodimensional motion we will deal with is called projectile motion Front View 18 Assmptions 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 assmptions, 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 pward 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 Maximm 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 eqations and removin t ives: y (

6 Verifyin the Parabolic Trajectory, cont Rane and Maximm 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 eqations and removin t ives: 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 maximm heiht the projectile reaches is h Heiht of a Projectile, eqation The maximm heiht of the projectile can be fond in terms of the initial velocity vector: h v i sin i θ This eqation is valid only for symmetric motion Rane of a Projectile, eqation 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

7 Projectile Formlation Let the initial velocity be. Let the anle sbscripted by and x-axis be θ. Alon the x-direction, x t ( cos θ ) t x cos θ Alon the Y-direction, 1 ( sin ) t t y θ 1 Sbstitte (1) into (): y θ x Maximm occrs when the y-component of the velocity is eqal to 0, i.e. the object is at the moment of comin down. v sinθ t 0 sinθ t 0 sinθ t Sbstitte (3) into (): sinθ y h sinθ sin θ sin θ h h sin θ v or i 1 3 sinθ sin θi h To derive the maximm rane (R) alon the x-direction, we let y 0 in eqation. 1 y ( sinθ ) t t R The distance travelled in x-direction Why? 1 ( sin θ ) t t t sin θ t sin θ sin θ t t 0 0 sin (A+B) sin A cos B + cos A sin B 4 for y t x x x sin θ seconds ( cos θ ) ( cos θ ) R t sin θ ( sin θ cos θ ) sin θ is : What is the maximm vale of R, and the correspondin? Since sin θ 1, sin θ 1, i.e., θ 90 When θ 45 R R R R is a maximm when o ( sin( 45 ) o ( sin 90 ) θ o, o θ 45 o

8 Rane of a Projectile, final The maximm rane occrs at θ i 45 o Complementary anles will prodce the same rane Bt the maximm 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 sin constant velocity techniqes Analyze the vertical motion sin constant acceleration techniqes Remember that both directions share the same time 9 30 Non-Symmetric Projectile Motion Follow the eneral rles for projectile motion Break the y-direction into parts p and down or symmetrical back to initial heiht and then the rest of the heiht May be non-symmetric in other ways Example. One stratey in a snowball fiht is to throw a snowball at a hih anle over level rond. Then, while yor opponent is watchin that snowball, yo throw a second one at a low anle timed to arrive before or at the same time as the first one. Assme that both snowballs are thrown with a speed of 5.0 m/s. The first is thrown at an anle of 70.0 with respect to the horizontal. The release point of snowball and the hit point of the opponent are at the same level. (a) At what anle shold the second snowball be thrown to arrive at the same point as the first? (b) How many seconds later shold the second snowball be thrown after the first in order for both to arrive at the same time? 31 3

9 We have derived (b) R sin θ Since R,, are constant, sin θ will also be a constant. Let the soltion of θ be α, where α 90. We have θ α, or θ 180 α. R So, θ α, or θ 90 α Answer: (a) The second snowball shold be thrown at Uniform Circlar Motion Chanin Velocity in Uniform Circlar Motion Uniform circlar motion occrs when an object moves in a circlar 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 Plane View The chane in the velocity vector is de to the chane in direction The vector diaram shows Δv v f - v i 35 36

Centripetal Acceleration Centripetal Acceleration, cont The acceleration is always perpendiclar to the path of the motion The acceleration always points toward the center of the circle of motion This

10 Centripetal Acceleration Centripetal Acceleration, cont The acceleration is always perpendiclar 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 37 The manitde of the centripetal acceleration vector is iven by a C v r By the se 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 38 Period The period, T, is the time reqired for one complete revoltion (one complete cycle) As one complete cycle is the distance of the circmference. The speed of the particle wold be the circmference of the circle of motion divided by the period. Speed distance/time, so time distance/speed π r Therefore, the period is T v 39 Example. The astronat orbitin the Earth in Fire is preparin to dock with a Westar VI satellite. The satellite is in a circlar orbit 600 km above the Earth's srface, where the free-fall acceleration is 8.1 m/s. Take the radis of the Earth as 6400 km. Determine the speed of the satellite and the time interval reqired to complete one orbit arond the Earth. 40

11 so 8.1 m/s Tanential Acceleration and Speed of satellite in order to maintain the orbit Time to complete 1 cycle of orbit 41 The manitde of the velocity cold also be chanin In this case, there wold be a tanential acceleration Observe what happens at the center (my hand) when I make the ball to rotate faster Circlar motion on my hand 4 Total Acceleration The tanential acceleration cases 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 a a t a r 43 Total Acceleration, eqations dv The tanential acceleration: a t dt The radial acceleration: a r a c The total acceleration: Manitde a a + a r t Why the strin does not slack and the ball does not fly towards the center? v r 44

90 km/h 50 km/h Example A train slows down as it ronds a sharp horizontal trn, slowin from 90.0 km/h to 50.

Compte the acceleration at the moment the train speed reaches 50.

Assme it contines to slow down at this time at the same We assme the train is still slowin down at the

741 m s Δ t 15.0 s 1.48 m/s ( 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 3600 s 90 km/h to 50 km/h in 15 seconds Relative Velocity Two observers movin relative to each other

12 90 km/h 50 km/h Example A train slows down as it ronds a sharp horizontal trn, slowin from 90.0 km/h to 50.0 km/h in the 15.0 s that it takes to start to rond the bend. The radis of the crve is 150 m. Compte the acceleration at the moment the train speed reaches 50.0 km/h when it starts to trn as indicated by the dot in the fire. Assme it contines to slow down at this time at the same We assme the train is still slowin down at the instant in qestion. v [50 x (1000/3600)] ac 1.9 m s 1.9 m/s r h Δ v ( 40.0 km h)( 10 m km )( ) at m s Δ t 15.0 s 1.48 m/s ( 1.9 m s ) ( m s ) c t a a + a + 1 at at an anle of tan tan a rate. c s 90 km/h to 50 km/h in 15 seconds Relative Velocity Two observers movin relative to each other enerally do not aree on the otcome of an experiment For example, observers A and B below see different paths for the ball Relative Velocity, eneralized 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

13 Relative Velocity, eqations Acceleration in Different Frames of Reference The positions as seen from the two reference frames (r is the position seen from stationary frame, r seen from movin frame) are related throh the velocity r r v 0 t The derivative of the position eqation will ive the velocity eqation v v v 0 These are called the Galilean transformation eqations 49 The derivative of the velocity eqation will ive the acceleration eqation The acceleration of the particle measred by an observer in one frame of reference is the same as that measred by any other observer movin at a constant velocity relative to the first frame. Why? r r v 0 t v 0 is a constant, so v 0 has no effect on the v v v 0 chane in velocity, Ie, a a. 50 Example. How lon does it take an atomobile travelin in the left lane at 60.0 km/h to pll alonside a car travelin in the riht lane at 40.0 km/h if the cars' front bmpers are initially 100 m apart? Examples. Answer: The bmpers are initially 100 m apart (0.1 km). After time t the bmper of the leadin car travels 40.0t km, while the bmper of the chasin car travels 60.0t km. Since the cars are side by side at time t we have 60 t 40 km/h 40 t Constant speed (A) Horizontal Track t 60t t 0.1/ hr x 3600 sec 18 s 60 km/h 0.1 km 51 (B) Tilted Track 5

Examples. This time the car will be plled alon the horizontal track by a strin and a weiht hanin over a plley, and will be acceleratin constantly. The ball is fired at point X.

14 Examples. This time the car will be plled alon the horizontal track by a strin and a weiht hanin over a plley, and will be acceleratin constantly. The ball is fired at point X. Where will the ball land this time? (The horizontal track is lon enoh for the car not to fall off when the ball lands.) Point X (C) The car is plled alon by a strin and a weiht hanin over a plley. 53

Ground Rules. PC1221 Fundamentals of Physics I. Position and Displacement. Average Velocity. Lectures 7 and 8 Motion in Two Dimensions

Ground Rules. PC1221 Fundamentals of Physics I. Position and Displacement. Average Velocity. Lectures 7 and 8 Motion in Two Dimensions 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