Lecture 8 Linear Kinematics 2024 PDF

Lecture 8 Linear Kinematics 2024 Thursdays 8:00 – 9:50 am Linear Kinematics At the end of the lecture, you should be able to: Define Linear Kinematics Understand the concepts of...

Lecture 8 Linear Kinematics 2024 Thursdays 8:00 – 9:50 am Linear Kinematics At the end of the lecture, you should be able to: Define Linear Kinematics Understand the concepts of Linear & Angular Distance vs Displacement Kinematics Speed vs Velocity Acceleration  What is Kinematics? Kinematics is the study of motion in isolation from the forces that cause the motion. Linear kinematics describes movement: i. displacement ii. velocity iii. acceleration  Calculus - Derivatives There are three basic kinematic variables 1. Displacement, position (where to and how far?) the position of an object is simply its location in space changes in position can be described by distance or displacement 2. Velocity (how fast?) Displacement/time the velocity of an object is how fast it is changing its position 3. Acceleration (changes in velocity?) Velocity/time the acceleration of an object is how fast the velocity is changing  Types of motion patterns in kinematics : position of an object All movements can be classified as either translational, rotational or a combination of both. Translational: all points in a body follow parallel path. Can be either rectilinear or curvilinear Rotational: Particles within the body have rotated with respect to their original frame of reference Both: Motion that happens in a single plane and is some combination of rotation and translation (most human movements)  Linear Kinematics Used to describe rectilinear translational motions In linear kinematics we are concerned with the following variables: Distance vs Displacement Speed vs Velocity Acceleration  Distance vs Displacement DISTANCE DISPLACEMENT Scalar Vector Measure of path length Measure of straight line Path dependent distance in a specific direction Path independent Velocity Velocity: the rate of change of displacement over time. Therefore velocity is a vector quantity, as is displacement. Velocity always has an associated direction. The average velocity of an object may be calculated by dividing the change in displacement over the duration of time Where sf and si are the final and initial displacements in meters of the object and ∆t is the time elapsed in seconds. v = ∆s /∆ t [m/s] v = sf - si /∆ t [m/s]  Instantaneous Velocity Instantaneous velocity (v) is very important Specifies how fast and in what direction one is moving at one particular point in time Achieved by reducing the change in time drastically (e.g. 1000fps) i.e.: time between pictures is 0.001 seconds Average velocity vs instantaneous velocity NOTE: For planar motion, use vx and vy vx = sfx - six /∆ t [m/s] vy = sfy - siy /∆ t [m/s] So what is the difference between Speed and Velocity? SPEED VELOCITY Scalar quantity Vector quantity Rate of motion Rate of motion with a specific direction Average or instantaneous s = d / Δt Average or instantaneous –Where: s = speed v = Δd/ Δt = ( df - di ) / ( tf - ti ) d = distance –Where: v = velocity Δt = time taken Δd = displacement Δt = time taken Units are m/s Units are m/s [direction]  Distance vs Displacement example problem Susan cycles 20 km north, then 13km east, then 16 km south. What distance has Susan travelled? Using the A: 20km + 13km + 16km = 49 km Pythagorean theory: 42 + 132 = R2 What is Susan’s displacement? 185 = R2 13km R = 13.6 km NE 20km 16km 13km 4km ?km  Speed vs Velocity – when does it matter? Average speed: Typically has a greater magnitude than average velocity Straight line & rectilinear motion → speed and velocity identical e.g.: 100m dash Motion changing direction – speed and velocity not the same magnitude e.g.: 100 m swim – swimmer changes direction and average velocity = 0 because displacement = 0 In a race average speed is often more important than average velocity When start and finish = same → average velocity = 0m/s Average Speed in a Marathon Marathon example t = 2:12:36 t= 2 hrs (3600s/1 hr) + 12 min (60 s/ 1min) + 36 s = 7,956 s t = 2:26:05 = 8,765 s average speed = distance/time speed = 42,195m/7956 s = 5.3 m/s Average velocity? speed = 42,195/8765 s = 4.8 m/s Example Problem Problem: During an ice hockey game, Phil had two shots on goal – one shot from 5m away at 10m/s (snapshot) and one shot from 10m away at 40m/s (slap shot). Which shot did the hockey goalie have a better chance of blocking? Hint: consider how much time the goalie has to block the shots Calculate time using Answer: Closer shot is easier v = d/t to block b/c goalie has more t = d/v time Snap shot t1 = 5m/10m/s = 0.5s Slap shot t2 = 10m/40m/s = 0.25s Click View then Header and Footer to change this footer ACCELERATION  Acceleration ▪ When a body changes its velocity, it is said to undergo acceleration Rate at which the velocity is changing (differentiation) ▪ Important to set-up coordinate system and record the movement because you can’t “see” acceleration ▪ If the signs of displacement and acceleration are the same speed INCREASES ▪ If the signs of displacement and acceleration are different speed DECREASES v v f − v i m / s m v n +1 − v n −1 a= = = = 2 an = t t f − t i s s t n +1 − t n −1 Direction of motion does not indicate direction of acceleration - 0 + (Displacement, acceleration) ▪Speeding up in positive direction (++) = pos acc ▪Slowing down in positive direction (-+) = neg acc ▪Speeding up in a negative direction (+-) = neg acc ▪Slowing down in a negative direction (--) = pos acc  Average Acceleration v0.0 = 0 m/s v2.5 = 5 m/s v5.0 = 0 m/s 5 4.5 4 velocity (m/s) 3.5 3 2.5 2 1.5 1 0.5 0 0 1 2 3 4 5 time (s)  Average Acceleration 5 4.5 4 velocity (m/s) 3.5 3 2.5 2 1.5 1 0.5 0 0 1 2 3 4 5 time (s) 1st interval v −v m 5m −0 m a0.0 → 2.5 = = =+ 2.0 2 2.5 0.0 s s 2.5−0 2.5s s Note: velocity is positive and acceleration is positive.  Average Acceleration 5 4.5 4 velocity (m/s) 3.5 3 2.5 2 1.5 1 0.5 0 0 1 2 3 4 5 time (s) 2nd interval v −v 0 m −5 m s =− 2.0 m a2.5 → 5.0 = 5.0 2.5 = s 5.0− 2.5 2.5 m s s 2 Note: velocity is positive but acceleration is negative.  5 4.5 4 velocity (m/s) 3.5 3 2.5 2 1.5 1 0.5 0 whole interval 0 1 2 3 4 5 time (s) m m a0.0→5.0 = v 5.0 - v 0.0 = 0 -0 s s = 0 m2 m m s t 5.0 - t 0.0 5 -0 s s CONSTANT LINEAR VELOCITY Where do we encounter constant velocity in sport? running at top velocity?? jogging (marathon) horizontal velocity of a projectile.  Constant linear velocity ▪Occurs when the resultant force acting on an object is 0 ▪Object will continue moving at constant velocity along the same straight line ▪If we know the object’s velocity and line of motion, we can determine the position of the object at any point in time ▪We can determine the final displacement of an object (sf) by knowing the object’s initial displacement (si) and the amount of time elapsed (t) Sf = Si + vi ∆t  Constant Linear Velocity ▪Situation: Resultant force acting on body is zero Body continues to move at constant velocity in a straight line until something changes (Force in Newtons) ▪Solution: Determine final displacement (sf) if we know: Initial displacement (si) Initial velocity (vi) Time elapsed (∆t) ▪Equation: sf = si + vi∆t ▪Or use variations on the equation with known quantities.  Constant linear velocity – example 5.1.1 Sf = Si + vi∆t If a person is on a train that is moving in a straight line at the constant velocity of 100.0 km/h, after 2.00 hrs what is the position of the person (assuming an initial displacement of zero kilometers)? Sf = si + vi∆t Sf = 0 + 100 km/h x 2 h Sf = 200 km CONSTANT LINEAR ACCELERATION Where in sport do we encounter constant acceleration? Constant linear acceleration ▪ Occurs whenever the resultant force acting on the object is a constant. Examples: gravity, friction. ▪ If we know the initial position and initial velocity of the object, we can predict the position and velocity of the object at any instant in time. Constant Linear Acceleration ▪In constant linear acceleration: Resultant force acting on system is constant Rate of acceleration is known ▪Solution Determine position and velocity or time when given: Constant rate of acceleration Initial conditions ▪Equations I. vf = vi + a∆t II. sf = si + vi∆t + ½a∆t2 III. vf2 = vi2 + 2a∆d IV. sf = si + ½(vi + vf) ∆t Constant linear acceleration equations ▪ Let’s take a closer look at the equations: Rearrangement of eqn for constant acceleration Prediction of final position of an object moving at constant lin. velocity plus displ. from constant acceleration kinematic description of motion – duration of motion is unknown or irrelevant Derived from above equations  Example – Constant Acceleration A runner starts from rest, 1) uniformly accelerates at 3 m/s 2 for 3 seconds, 2) then runs at a constant velocity for 5 seconds, 3) then decelerates at -2 m/s2 for 2 seconds. How far does the runner travel during this 10-second period? Three parts to this question: 1. Acceleration 2. Constant velocity 3. Deceleration 1) Time 3 sec, V initial = 0, Acceleration average = 3 m/s2 V f = a x time ? V f = 3 m/s2 x 3 sec, V f = 9m/s Vi - V f x t = disc 3 sf = si + vi∆t + ½a∆t2 3m/s – 2 m/s x 2 = disc 3 sf = 0 + 0 ∆t + ½ 3m/s2 x 32 = 13.5 m Decelerates -2m/s2 2) Time 5 s Time total = 10 sec Constant velocity 5 s 5 s x V f = disc 2 5 s x 9 m/s =45 m Total displacement at 8 seconds is 58.3 m 3) Time 2 seconds sf = si + vi∆t + ½a∆t2 sf = 58.5 + 9 m/s x 2 + ½ -2m/s2 x 22 = 58.5m + 18m – 4m = 72.5m in 10 seconds Things to review lecture 8 ▪ Define linear and rotational kinematics. ▪ Understand the concepts of: Distance vs Displacement (units of measurement) Speed vs Velocity (units of measurement) Acceleration (units of measurement) ▪ Define translational, rotational and combined motion. More things to review lecture 8 ▪ Define the difference between instantaneous velocity and average velocity and describe what they represent. ▪ How can understanding the relationship between displacement, velocity and time be applied to coaching, teaching and therapy. ▪ Which principles of human movement apply kinematics to improve human movement? More things to review lecture 8 ▪ Be able to interpret a velocity graph in what they tell us about displacement and acceleration. ▪ Give an example of where you would find constant linear velocity in sport. ▪ Give two examples of where you would find constant acceleration in sport and therapy? ▪ Know these constant acceleration equations: vf = vi + a∆t sf = si + vi∆t + ½a∆t2 vf2 = vi2 + 2a∆d sf = si + ½(vi + vf) ∆t