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# Lecture 06

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### Lecture 06

1. 1. Linear Kinematics Objectives: • Define the idea of spatial reference frames • Introduce the concepts of position, displacement, distance, velocity, and speed • Learn how to compute displacement, velocity, and speed • Learn the difference between average, instantaneous, and relative velocity Linear KinematicsKinematics• The form, pattern, or sequencing of movement with respect to time• Forces causing the motion are not consideredLinear Motion• All parts of an object or system move the same distance in the same direction at the same timeLinear Kinematics• The kinematics of particles, objects, or systems undergoing linear motion 1
2. 2. Spatial Reference Frames• A spatial reference frame is a set of coordinate axes (1, 2, or 3), oriented perpendicular to each other.• It provides a means of describing and quantifying positions and directions in space1-dimensional (1-D) Reference Frame• Quantifies positions and directions along a line unit of measure origin label x = 2m x (m) -5 -4 -3 -2 -1 0 1 2 3 4 5 – direction + direction 2-D Reference Frame• Quantifies positions and directions in a plane y (m) (x,y) = (1 m, 2 m) 2 +y direction –y direction unit of measure label 90° θ=63° (0,0) x (m) -2 2 origin +x direction –x direction 2
3. 3. 3-D Reference Frame• Quantifies positions and directions in space z (m) 2 (x,y,z) = (1.5m, 2m, 1.8m) (0,0,0) φ=36° y (m) 2 θ=53° x (m) 2 Reference Frames & Motion • The position or motion of a system does not depend on the choice of reference frame • But, the numbers used to describe them do • Always specify the reference frame used! y1 (m) ) (m x2 ) (m y2 1 1 1 x1 (m) 1 3
4. 4. Selecting a Reference Frame• Use only as many dimensions as necessary• Align reference frame with fixed, clearly-defined, and physically meaningful directions. (e.g. compass directions, anatomical planes) Common 2-D conventions: – Sagittal plane: +X = anterior; +Y = upward – Transverse plane: +X = left; +Y = upward – Horizontal plane: +X = anterior; +Y = left• The origin should also have physical meaning. 2-D examples: – X=0 initial position – Y=0 ground height Position • The location of a point, with respect to the origin, within a spatial reference frame • Position is a vector; has magnitude and direction • Or, specify position by the coordinates of the point • Position has units of length (e.g. meters, feet) y (m) (x,y) = (1m, 2m) 2 Point’s position is: • distance of 2.24m at an angle 2.24m of 63° above the +x axis, or • (x,y) position of (1m, 2 m)origin θ=63° x (m) (0,0) 2 4
5. 5. Linear Displacement• Change (directed distance) from a point’s initial position to its final position• Displacement is a vector; has magnitude and direction• Displacement has units of length (e.g. meters, feet) y (m) initial position displacement 1 pinitial final position pfinal x (m) 1 Computing Displacement• Compute displacement (∆p) by vector subtraction ∆p = pfinal – pinitial y (m) initial position 1 pinitial final position pfinal x (m) 1 –pinitial –pinitial ∆p 5
6. 6. Describing Displacement• Can describe displacement by: – Magnitude and direction (e.g. 2.23m at 26.6° below the +x axis) – Components y (m) (change) along each axis 1 pinitial (e.g. 2m in the +x pfinal direction, 1m in 1 the –y direction) θ ∆px x (m) ∆py ∆p -1 Distance• The length of the path traveled between a point’s initial and final position• Distance is a scalar; it has magnitude only• Has units of length (e.g. meters, feet)• Distance ≥ (Magnitude of displacement) Displacement = 64 m West E N Distance = 200m 6
7. 7. Example Problem #1 A box is resting on a table of height 0.3m. A worker lifts the box straight upward to a height of 1m. He carries the box straight backward 0.5m, keeping it at a constant height. He then lowers the box straight downward to the ground. What was the displacement of the box? What distance was the box moved? Linear Velocity• The rate of change of position• Velocity is a vector; has magnitude and direction change in position displacement velocity = = change in time change in time• Shorthand notation: pfinal – pinitial ∆p v = = tfinal – tinitial ∆t• Velocity has units of length/time (e.g. m/s, ft/s) 7
8. 8. Computing Velocity• direction of velocity = direction of displacement• magnitude of velocity = magnitude of displacement change in time• component of velocity = component of displacement change in time y ∆p y V = ∆p / ∆t t /∆ ∆p ∆p ∆py vy = ∆py / ∆t = θ v θ x x ∆px vx = ∆px / ∆t Speed • The distance traveled divided by the time taken to cover it • Equal to the average magnitude of the instantaneous velocity over that time. distance speed = change in time • Speed is a scalar; has magnitude only • Speed has units of length/time (e.g. m/s, ft/s) 8
9. 9. Speed vs. Velocity Displacement = 64 m West E N Distance = 200mAssume a runner takes 25 s to run 200 m: 200 m 64 m West Speed = Velocity = 25 s 25 s = 8 m/s = 2.6 m/s West Example Problem #2During the lifting task of example problem #1, it takes the worker 0.5s to lift the box, 1.3s to carry it backward, and 0.6s to lower it.What were the average velocity and average speed of the box during the first, lifting phase of the task?What were the average velocity and average speed of the box for the task as a whole? 9
10. 10. Velocity as a Slope• Graph x-component of position vs. time• x-component of velocity from t1 to t2 = slope of the line from px at t1 to px at t2 Slope : ∆px / ∆t = vx px (m) ∆px ∆t t1 t2 time (s) Average vs. Instantaneous Velocity• The previous formulas give us the average velocity between an initial time (t1) and a final time (t2)• Instantaneous velocity is the velocity at a single instant in time• Can estimate instantaneous velocity using the central difference method: p (at t1 + ∆t) – p (at t1 – ∆t) v (at t1) = 2 ∆t where ∆t is a very small change in time 10
11. 11. Instantaneous Velocity as a Slope• Graph of x-component of position vs. time slope = instantaneous x-velocity at t1 px (m) slope = average x-velocity from t1 to t2 ∆t t1 t2 time (s) Estimating Velocity from Position Identify points with px (m) zero slope = points with zero velocity Portions of the curve with positive slope time (s) have positive velocity (i.e. velocity in the vx (m/s) + direction) Portions of the curve with negative slope 0 time (s) have negative velocity (i.e. velocity in the – direction) 11
12. 12. Relative Position• Find the position of one point or object relative to another by vector subtraction of their positions p(2 relative to 1) = p2 – p1 p2 = p1 + p(2 relative to 1) y (m) object 2 1 p(2 relative to 1) p2 object 1 p1 x (m) 1 Relative Velocity• Apparent velocity of a second point or object to an observer at a first moving point or object• Compute by vector subtraction of the velocities v(2 relative to 1) = v2 – v1 vy (m/s) v2 = v1 + v(2 relative to 1) object 2 1 v(2 relative to 1) v2 object 1 v1 vx (m/s) 1 12
13. 13. Example Problem #3A runner on a treadmill is running at 3.4 m/s in a direction 10° left of forward, relative to the treadmill belt (resulting in a forward velocity of 3.3 m/s and a leftward velocity of 0.6 m/s, relative to the belt).The treadmill belt is moving backward at 3.6 m/s.What is the runner’s overall velocity? 13