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PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 109 L12. KINEMATICS I MOTION IN ONE and THREE DIMENSIONS READING ASSIGNMENT: Textbook: Chapters 2 and 4 KINEMATICS I Definitions for 1-D and 3-D motions Kinematic Equations for 1-D motion Free Fall Motion Diagrams 2 DEMONSTRATIONS Special Presentation: Bio-Bullet 1: Velocity and Acceleration in Animal Kingdom
PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 110 Kinematics describes motion while ignoring the agents that caused the motion. Concept of Motion: Motion is a change of position of a body (or body elements) with respect to some selected frame of reference. Independence of Motion Principle: Any complex 3D motion may be described by considering motion along each of the spatial coordinates independently. Types of Motion Translational Motion: all elements of a body undergo the same change of position Rotational Motion : the body changes orientation in space. Vibrational Motion : the body oscillates about the equilibrium position. More complex types of motion may be analyzed as a combination of two or more simple motions:
PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 111 Concept of a Particle (Material Point) We can effectively describe the TRANSLATIONAL MOTION of a body using the idealization of a particle (dimensionless point of the same mass as a body (material point)) By doing this, we are neglecting various effects, which are related to the body’s physical dimensions (such as air-resistance, buoyancy, and other possible body rotational effects). For many of the problems discussed in our introductory course, this is an excellent approximation and we can describe many aspects of motion using this simplified model. Fundamental definitions for 1-D motion: Position Displacement Fundamental definitions for 3-D motion (vectors!): Position: Displacement: Average velocity Instantaneous velocity definition Average acceleration definition Instantaneous acceleration definition M Particle (material point) of mass M 2 2 0 0 lim lim ) ( dt r d dt dr dt d dt dv t v a t a t avg t = ÷ ÷ ø ö ç ç è æ = = D D = ® D ® D = ® D ® D = = D D = = r dt dr t r v t v t avg t lim lim 0 0 ) ( t x erval int time nt displaceme v v avg D D = = ñ á = 1 2 1 2 t t v v t v erval int time velocity in change a a avg - - = D D = = ñ á = ^ ^ ^ ^ ^ ^ ^ ^ ^ z y x z y x i z i y i x k z j y i x k r j r i r r + + = + + = + + = ) , , ( ) , , ( z y x r r r r z y x = = ^ ^ ^ ^ 1 2 ^ 1 2 ^ 1 2 ^ 1 ^ 1 ^ 1 ^ 2 ^ 2 ^ 2 1 2 ) ) ( ) ( ) ( ) ( ) ( k z j y i x k z z j y y i x x r k z j y i x k z j y i x r r r r D + D + D = - + - + - = D + + - + + = D - = D 2 2 0 0 lim lim ) ( dt x d dt dx dt d dt dv t v a t a t avg t = ÷ ø ö ç è æ = = D D = ® D ® D = ® D ® D = = D D = = x dt dx t x v t v t avg t lim lim 0 0 ) ( t x erval int time nt displaceme v v avg D D = = ñ á = t s D = = interval time distance total speed average 1 2 1 2 t t v v t v erval int time velocity in change a a avg - - = D D = = ñ á = i f x x x - = D
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PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 112 KINEMATIC EQUATIONS: Kinematic Equations – valid only for particles moving under constant acceleration! Given our general definitions of instantaneous velocity and acceleration, it is possible to obtain a set of kinematic equations when acceleration is constant. ࠵? " = ࠵?࠵? + ࠵? ࠵? " = ) * ࠵?࠵? * + ࠵? ࠵? + ࠵? ࠵? " * −࠵? * = 2࠵?(࠵? " − ࠵? ) ࠵? " = ) * /࠵? + ࠵? " 0࠵? + ࠵? These kinematic equations may be used to solve any problem involving one-dimensional motion with a constant acceleration! One may need to use two or more equations to solve a problem. Typically, there is more than one way to solve a problem. When we deal with 3-D motion with constant acceleration, the same equations describe each coordinate separately! An Example of 1-D motion with constant acceleration: Free Fall A free falling object is any object moving under the influence of gravity alone. ࠵? 1 = −࠵? ; ࠵? 1’ , the initial velocity may be positive, negative or 0 Dropped – released from rest ࠵? 1’ = 0 * Thrown downward ࠵? 1’ < 0 * Thrown upward ࠵? 1’ > 0 * * NOTE: these signs are only valid if we pick the direction of the y axis to be “up”! Exercise: Attributing signs to the kinematic equations: general approach
PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 113 FREE FALL TEXTBOOK ANALYSIS: The acceleration of an object in free fall is directed downward, regardless of the initial motion The magnitude of free fall acceleration is g = 9.80 m/s2 g decreases with increasing altitude g varies with latitude 9.80 m/s2 is the average at the Earth’s surface We will neglect air resistance Free fall motion is constantly accelerated motion in one dimension Let upward be positive Use the kinematic equations with ay = - g = -9.80 m/s2 Example from the textbook Initial velocity at A is upward (+) and acceleration is g (-9.8 m/s2) At B, the velocity is 0 and the acceleration is g (-9.8 m/s2) At C, the velocity has the same magnitude as at A, but is in the opposite direction The displacement is –50.0 m (it ends up 50.0 m below its starting point) Exercise (DIY) READING OF THE x(t), v(t), AND a(t) MOTION DIAGRAMS
PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 114 Suggested Problems 1. Particle moves from initial position r i to final position r f in 2.0s a) Find average velocity vector during this move, b) Find the magnitude of the average velocity vector 2. Time dependent vector r(t) describes the position of the Particle A at any instant of time: a) Find x ,y, z components of instantaneous velocity v(t) vector. b) Find magnitude of the velocity vector at time t=2s 3. An ion’s position vector is initially r = 5.0 i - 6.0 j +2.0 k , and 10s later it is r = -2.0 i + 8.0 j -2.0 k , all in meters. What is its average velocity during this time interval ? 4. The position of an electron is given by r = (3.00t) i – (4.00t) j +(2.00) k, / t in second and r in meters / a) What is the electron’s velocity v ? b) What is electron’s velocity at t=2s (in unit-vector notation)? c) What is a magnitude and angle relative to the positive direction of the x axis? 5 A radar station detects an airplane approaching directly from the east. At first observation the range to the plane is 360m at 40 o above the horizon. The airplane is tracked for another 123o in the vertical east-west plane, the range at final contact being 790 m. Find the displacement of the airplane during the period of observation. 6 An object A moves along the x-axis according to the equation: x(t) = 3.0t 2 +3.3t+10.4, where x is in meters, and t is in seconds. The other object (object B) moves along the x-axis with constant speed +9 m/s. When does A have the same instantaneous velocity as B ? State your answer to nearest 0.01s. 7 Particle A moves along the line y=30 m with a constant velocity v of magnitude 3.0m/s and directed parallel to the positive x axis. Particle B starts at origin with zero speed and constant acceleration a (of magnitude0.40m/s2) at the same instant that particle A passes the y axis. What angle θ between a and the positive axis would result in a collision between these two particles? 8 The ball is shot vertically with the initial velocity of 7 m/s from the 13m level. What is the maximum height it will reach? Neglect the air resistance, and take g=9.8 m/s 2 . Give the answer to the nearest tenth of a meter. 9 The ball is shot vertically with the initial velocity of 3 m/s from the 10m level. How long it will take for the ball to reach the ground? Neglect the air resistance, and take g=9.8 m/s 2 . Give the answer to the nearest hundredth of a second. ^ ^ ^ 2 2 6 2 ) ( k t j t i t t r i + - = ^ ^ ^ ^ ^ ^ 7 4 ; 6 2 k j i r k j i r f i - + = + - =
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PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 115 10 A window cleaner uses an external elevator while moving between the floors of a skyscraper in downtown Manhattan. At the moment when the elevator is at 5 m above the ground, moving up with the speed of 15 m/s and acceleration of 1 m/s 2 , a small brush is dropped from the elevator. How long before it will hit the ground? Round your answer to the nearest tenth of a second. 11 Motion of the rocket is given by the following equation: x=2+1t+7t 2 +3t 3 where t is in seconds, x is in meters. At 17s a small object detaches itself from the body of the rocket. What is the initial height of this object? Round your answer to the nearest tenth of a meter. 12 The maximum height from which a person can safely jump is 2.45 m. What is the maximum allowable landing speed for a parachutist 13 A body covers 64% of the total distance fallen in the last second. From what height did it fall? 14 Cheetah can reach 105km/h in 2 s and maintain this speed for 15 s. After this time it must rest. An antelope can reach 90km/h in 2s and sustain it for a long time. Suppose they are initially separated by 100m and the antelope reacts in 0.5s. a) can cheetah get the antelope? b) if not, how close does it get? Assume both start from the rest. 15 A climber can estimate the height of a cliff by dropping a stone and noting the time at which he hears the impact on the ground. Suppose this time is 2.5s. Find the height of the cliff under the following conditions: a) by assuming the speed of sound is large enough to be ignored b) by taking the speed of the sound to be 330m/s.
PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 116 L13. KINEMATICS II EXAMPLES OF MOTION IN TWO DIMENSIONS READING ASSIGNMENT: Textbook: Chapters 2 and 4 KINEMATICS II Projectile Motion Uniform Circular Motion Non-uniform Circular Motion Relative Velocity 4 DEMONSTRATIONS Projectile Motion Projectile Motion with varying g-value Uniform Circular Motion Falling Monkey Demo 14 Suggested Problems Bio-Bullet 1: Velocity and Acceleration
PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 117 Two-dimensional motion: projectile motion In projectile motion, the horizontal motion and the vertical motion are independent of each other: They do not affect each other and may be considered separately THE TRAJECTORY: The path of a projectile that is launched at t=0 (x 0 =0, y 0 =0) with initial velocity v 0 (v 0x ,v 0y ) Initial velocity and velocities at various points along its path are shown. R (Range) is the horizontal distance the projectile has traveled when it returns to its launch height
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PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 118 Projectile Motion Analyzed Horizontal motion: There is no acceleration in the horizontal direction. At any time t, the projectile’s horizontal displacement x-x 0 from the initial position x 0 is given by: Vertical Motion: This is the motion of a body shot up vertically with an initial speed of v 0y and experiencing constant gravitational acceleration directed downward Equation for Projectile Trajectory To find the equation of the projectile’s path (trajectory ) of the form y= y(x) we will eliminate t from x = x(t) and y = y(t) equations which leads to the following expression for y as a function of x THIS IS THE EQUATION OF A PARABOLA!! ) cos ( ; ) cos ( 0 0 0 0 0 q q v v because t v t v x x x x = = = - ) sin ( ; 2 1 ) sin ( 2 1 0 0 2 0 2 0 0 q q v v because gt t v gt t v y y y y = - = - = - 0 0 ) cos ( x t v x + = q 0 2 0 2 1 ) sin ( y gt t v y + - = q 0 0 0 0 = = y x q cos 0 v x t = 2 2 0 2 0 0 0 ) cos ( 2 ) (tan ) cos ( 2 1 cos ) sin ( x v g x v x g v x v y q q q q q - = - = 2 2 0 ) cos ( 2 ) (tan x v g x y q q - =
PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 119 1 Find the algebraic expression for the horizontal range The horizontal range R of the projectile is the horizontal distance the projectile has traveled when it returns to its initial (launch) height. To find R let’s put (x-x 0 ) = R, and y-y 0 =0 in our equations for projectile motion ࠵? = ࠵? − ࠵? 7 and so ࠵? = ࠵? − ࠵? 7 = (࠵? 7 ࠵?࠵?࠵?࠵?)࠵? ࠵? ࠵? 7 ࠵?࠵?࠵?࠵? = ࠵? ࠵? − ࠵? 7 = ࠵? 7 ࠵?࠵?࠵?࠵? ࠵? ࠵? 7 ࠵?࠵?࠵?࠵? ࠵? 2 ? ࠵? ࠵? 7 ࠵?࠵?࠵?࠵? @ * 0 = ࠵? ࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵? ࠵?࠵? * 2࠵? 7 * ? 1 ࠵?࠵?࠵?࠵? @ * 2 Show that the vertical velocity component v y obeys the equation for the free fall: 2 0 0 2 1 ) sin ( 0 gt t v y y - = - = q ÷ ø ö ç è æ ÷ = q q q q cos ) (cos 2 cos sin 2 2 0 2 R v g R R ÷ ÷ ø ö ç ç è æ ÷ = q q q cos 2 cos 2 sin 2 0 2 0 v g v g R g v g v R q q q 2 sin cos sin 2 2 0 2 0 = = gt v gt v t v y y - = - = ) sin ( ) ( 0 0 q 2 0 2 0 2 0 2 ) ( ) sin ( 2 ) sin ( ) sin ( )] ( [ gt gt v v gt v t v y + - = - = q q q 2 0 2 0 0 ) ( ) ( 2 ] 2 1 ) [( 2 ] ) sin [( 2 t g g y y g gt y y g t v + - = + - = q g y y v t v gt t g g y y v t v gt t g g y y v t v y y y ) ( 2 ) sin ( )] ( [ ) ( ) ( ) ( 2 ( ) sin ( )] ( [ ) ( ] ) ( ) ( 2 [( ) sin ( )] ( [ 0 2 0 2 2 2 0 2 0 2 2 2 0 2 0 2 - - = + - - - = + + - - = q q q
PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 120 Effects of the Air: Real Projectiles VACUUM AIR RANGE 177 m 99 m MAX H 77 m 53 m T.O.F. 7.9 s 6.6 s In reality, projectile motion is more complex than we assumed. Depending on the shape of the projected object, there is a certain amount of air resistance proportional to the square of the speed or to the speed) Real ballistic trajectories are not parabolic! V=600 m/s theta=60 Uniform Circular Motion A Particle is in a uniform circular motion if it travels around a circle in a circular arc at a constant (uniform) speed Even though the speed is constant, the particle is accelerating. Magnitude of centripetal acceleration Period of rotation (T) r v a a 2 = = v r T p 2 =
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PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 121 PROOF THAT ACCELARATION IS CENTRIPETAL 1)Derive equations for uniform circular motion. 2)Show that the acceleration direction is truly towards the center This means that acceleration vector a is directed along the radius r , toward the circle’s center! ^ ^ ^ ^ ) cos ( ) sin ( j v i v j v i v v y x Q + Q - = + = r y and r x p p = Q = Q sin cos ^ ^ ) ( ) ( j r x v i r y v v p p + - = ^ ^ ^ ^ ) sin ( ) cos ( ) ( ) ( j v r v i v r v j v r v i v r v a x y Q - + Q - = + - = dt dx dt dy and v v v v v v x y x y = = - = Q = Q ; sin ; cos r v r v r v r v a a a a a a y x y x 2 2 2 2 2 2 2 2 2 2 2 2 ) (cos ) (sin ) sin ( ) cos ( = Q + Q = = Q - + Q - = + = + = = q j q q q j = = - - = = tan ) cos ( ) sin ( tan 2 2 r v r v a a x y
PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 122 Non-uniform Circular Motion The radial acceleration associated with changes in the direction of velocity is When the speed varies, there is also an acceleration along the tangent to the path. This is tangential acceleration So that total acceleration is: In the special case of motion in a circle, it is convenient to use polar unit vectors Θ is a unit vector tangent to the circle at each instant, it is positive if pointing CCW r is a unit vector directed radially outward the center to the of motion Magnitudes of these vectors are constant (=1) but their directions change constantly in time a r is always negative a t may be positive or negative Relative Motion Two observers moving relative to each other generally do not agree on the outcome of an experiment For example, observers A and B below see different paths for the ball The motion of any body has to be described relative to some frame of reference such as ground, centre of the Earth etc. In our work up to this point we have always assumed that this frame of reference by itself is not moving. This is often not realistic assumption ( everything is in motion!), and it is often necessary to describe motion of one body with respect to another body moving with respect to “resting “frame of reference. Two observers A and B watch the same car P passing by. Observer B moves with a constant speed v BA with respect to A. r v a r 2 = dt dv a t = t r a a t a + = ) ( ^ ^ ) ( q dt dv r a t a t r + =
PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 123 At some instant both A and B measure the P position and velocity. How will their observation relate to each other? Taking derivative of both sides of this equation we get: Acceleration in the Relative Motion What about the acceleration? To find out let’s take derivative of the equation describing velocity relationship: CONCLUSION: Observers in two different reference frames, moving at constant velocity relative to each other, will measure the same acceleration for moving particle. Relative motion in 3-D: We may easily generalize these 1-D results to 3-D PB BA PA x x x + = dt dx dt dx dt dx BA PB PA + = BA PB PA v v v + = PB PA PB PA const v BA PB PA a a dt dv dt dv dt dv dt dv dt dv BA = Û = Û + = = BA PB PA v v v + = PB BA PA r r r + = BA PB PA v v v + = PB PA PB PA const v BA PB PA a a dt v d dt v d dt v d dt v d dt v d BA = Û = Û + = =
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PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 124 Suggested Problems 1 What angle of launch results in a maximum horizontal range for projectile? 2 Mountain rescue officers trigger an avalanche by firing a gun at a distant mountain slope. The muzzle velocity of the gun is 150 m/s. The gun muzzle is at sea level (y=0) The horizontal and vertical distances from the gun to the slope are 1000 m and 400 m, respectively. Find the lowest angle between the barrel of the gun and the horizontal plane. Round your answer to the nearest tenth of degree. 3 The artillery gun with a muzzle velocity of v 0 =150m/s is fired at the slope of the mountain at horizontal and vertical distance of 840 m and 180m. The gun was shot at the smallest angle possible to achieve this goal. At the point where it hits the mountain the projectile trajectory makes, angle α with the horizontal plane. Round yoru answer to the nearest thousandth. 4. Pirate ship 560m from a fort defending the harbor entrance of an island. A defense canon, located at sea level, fires grenade at initial speed of v 0 = 82m/s. a) At what angle from the horizontal must grenade be fired to hit the ship? b) How far should the pirate ship be from the cannon if it is to be beyond the maximum range of the cannonballs? 5. Figure below illustrates the flight of a human cannonball over three Ferris wheels located as shown and each 18 meters high. The acrobat is launched with speed v 0 =26.5m/s at the angle of θ 0 =53 o up from the horizontal and with an initial height of 3.0 meters above the ground. The net on which he lands is at the same height. a) Find the distance between the Ferris Wheel and the acrobat when he is exactly above the first wheel b) How far from the cannon should the net be placed? c) How tall should the tent be? DEMONSTRATION RELATED PROBLEMS 6 The falling monkey is shot at by a projectile launched at 40 degrees above horizontal at horizontal distance of 4 meters from the tree. Knowing that the monkey is hit while being 30cm below the branch from which it was released, find the speed of the bullet. 7 The projectile is launched on the inclined smooth surface towards the target. The launcher barrel is at (0,0) the target is at (50, 25), the speed of the projectile is 10cm/s. The projectile is aimed at 70 deg above the horizontal find the angle of the incline for which the projectile makes it into the target. 8 An Earth satellite moves in a circular orbit 640 km above Earth’s surface with a period of 98 .0 min. What are: a) Speed b) The magnitude of centripetal acceleration of the satellite 9 When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about 20km. The neutron star rotates once every second. a) What is the speed of a particle on the star’s equator b) What is the magnitude of the particle’s centripetal acceleration c) Repeat (a) and (b) for the particle at surface of the star, 45 o above the equator.

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