Two-dimensional and Periodic Motion Physics Ch.4 Notes Two-dimensional and Periodic Motion

4.1 Displacement in two dimensions The definition of displacement does not change when it is applied to motion in two or more dimensions. Displacement is always the net change in an object’s position and it is always a vector. Both its magnitude (the amount of the displacement) and its direction are required. displacement

4.2 Velocity in two dimensions Like any vector, the velocity vector can be written as the sum of its components, the velocities along the x and y axes Components of velocity vx = v cos θ vy = v sin θ v = speed θ = angle with positive x axis vx = x component of velocity vy = y component of velocity 2D Velocity

4.3 Sample problem: velocity in two dimensions 2D example

4.4 Acceleration in two dimensions Because acceleration is the change in velocity per unit time, it follows that acceleration also can change independently in each dimension. The average acceleration can be calculated using the definition of acceleration, dividing the change in velocity by the elapsed time. The components of the average acceleration can be calculated by dividing the changes in the velocity components by the elapsed time 2D Acceleration      

4.5 Sample problem: acceleration in two dimensions 2D Acceleration example

4.6 Interactive checkpoint: acceleration in two dimensions Practice Problem 2D acceleration

4.7 Projectile motion Projectile Motion- Parabolic path of an object subject to horizontal velocity and vertical acceleration Movement determined by an object’s initial velocity and the constant acceleration of gravity. Projectile Motion

Introduction Projectile Motion: Motion through the air without propulsion Examples:

Part 1. Motion of Objects Projected Horizontally

y v0 x

y x

y x

y x

y x

g = -9.80m/s2 y Motion is accelerated Acceleration is constant, and downward a = g = -9.80m/s2 The horizontal (x) component of velocity is constant The horizontal and vertical motions are independent of each other, but they have a common time g = -9.80m/s2 x

ANALYSIS OF MOTION ASSUMPTIONS: x-direction (horizontal): uniform motion y-direction (vertical): accelerated motion no air resistance QUESTIONS: What is the trajectory? What is the total time of the motion? What is the horizontal range? What is the final velocity?

X Uniform m. Y Accel. m. ax = 0 ay=g=-9.80 m/s2 vx = v0 vy = g t Frame of reference: Equations of motion: x y X Uniform m. Y Accel. m. ACCL ax = 0 ay=g=-9.80 m/s2 VELC vx = v0 vy = g t DSPL x = v0 t y = h + ½ g t2 v0 h g

Trajectory y = h + ½ (g/v02) x2 y = ½ (g/v02) x2 + h x = v0 t y = h + ½ g t2 Parabola, open down Eliminate time, t v02 > v01 v01 t = x/v0 y = h + ½ g (x/v0)2 y = h + ½ (g/v02) x2 y = ½ (g/v02) x2 + h

Total Time, Δt Δt = √ 2h/(-g) Δt = √ 2h/(9.80ms-2) y = h + ½ g t2 Δt = tf - ti y = h + ½ g t2 y x h final y = 0 0 = h + ½ g (Δt)2 ti =0 Solve for Δt: Δt = √ 2h/(-g) Δt = √ 2h/(9.80ms-2) tf =Δt Total time of motion depends only on the initial height, h

Horizontal Range, Δx Δx = v0 Δt Δt = √ 2h/(-g) Δx = v0 √ 2h/(-g) Δx y x h final y = 0, time is the total time Δt Δx = v0 Δt Δt = √ 2h/(-g) Δx = v0 √ 2h/(-g) Δx Horizontal range depends on the initial height, h, and the initial velocity, v0

VELOCITY v = √v02+g2t2 Tan θ = vy/ vx = g t / v0 vx = v0 θ v = √vx2 + vy2 = √v02+g2t2 Tan θ = vy/ vx = g t / v0 vy = g t

FINAL VELOCITY Δt = √ 2h/(-g) v v = √v02+g2(2h /(-g)) vx = v0 Δt = √ 2h/(-g) v tan θ = g Δt / v0 = -(-g)√2h/(-g) / v0 = -√2h(-g) / v0 θ vy = g t v = √vx2 + vy2 v = √v02+g2(2h /(-g)) v = √ v02+ 2h(-g) θ is negative (below the horizontal line)

HORIZONTAL THROW - Summary h – initial height, v0 – initial horizontal velocity, g = -9.80m/s2 Trajectory Half -parabola, open down Total time Δt = √ 2h/(-g) Horizontal Range Δx = v0 √ 2h/(-g) Final Velocity v = √ v02+ 2h(-g) tan θ = -√2h(-g) / v0

4.8 Sample problem: a horizontal cannon

4.9 Interactive problem: the monkey and the professor 4.10 Checkpoint: golfing  

4.11 Projectile motion: juggling Juggling is a form of projectile motion in which the projectiles have initial velocities in both the vertical and horizontal dimensions. This motion takes more work to analyze than when a projectile’s initial vertical velocity is zero, as it was with the horizontally fired cannonball. juggling

Part 2. Motion of objects projected at an angle

x y vi Initial position: x = 0, y = 0 vix viy Initial velocity: vi = vi Velocity components: x- direction : vix = vi cos θ y- direction : viy = vi sin θ θ

y a =g = - 9.80 m/s2 x Motion is accelerated Acceleration is constant, and downward a = g = -9.80 m/s2 The horizontal (x) component of velocity is constant The horizontal and vertical motions are independent of each other, but they have a common time x

ANALYSIS OF MOTION: ASSUMPTIONS x-direction (horizontal): uniform motion y-direction (vertical): accelerated motion no air resistance QUESTIONS What is the trajectory? What is the total time of the motion? What is the horizontal range? What is the maximum height? What is the final velocity?

X Uniform motion Y Accelerated motion Equations of motion: X Uniform motion Y Accelerated motion ACCELERATION ax = 0 ay = g = -9.80 m/s2 VELOCITY vx = vix= vi cos θ vx = vi cos θ vy = viy+ g t vy = vi sin θ + g t DISPLACEMENT x = vix t = vi t cos θ x = vi t cos θ y = h + viy t + ½ g t2 y = vi t sin θ + ½ g t2

X Uniform motion Y Accelerated motion Equations of motion: X Uniform motion Y Accelerated motion ACCELERATION ax = 0 ay = g = -9.80 m/s2 VELOCITY vx = vi cos θ vy = vi sin θ + g t DISPLACEMENT x = vi t cos θ y = vi t sin θ + ½ g t2

4.12 Sample problem: calculating initial velocity in projectile motion 4.13 Interactive problem: the monkey and the professor, part II

Trajectory x = vi t cos  y = vi t sin  + ½ g t2 Parabola, open down Eliminate time, t t = x/(vi cos ) y = bx + ax2 x

Total Time, Δt y = vi t sin  + ½ g t2 Δt = final height y = 0, after time interval Δt 0 = vi Δt sin  + ½ g (Δt)2 Solve for Δt: 0 = vi sin  + ½ g Δt Δt = 2 vi sin  (-g) x t = 0 Δt

4.14 Projectile motion: aiming a cannon 4.15 Sample problem: a cannon’s range 4.16 Interactive checkpoint: clown cannon

4.17 Interactive checkpoint: soccer kick 4.18 Interactive problem: the human cannonball 4.19 Interactive problem: test your juggling!

Horizontal Range, Δx 4.20 The range and elevation equations x = vi t cos  y final y = 0, time is the total time Δt Δx = vi Δt cos  Δt = 2 vi sin  (-g) x sin (2 ) = 2 sin  cos  Δx Δx = 2vi 2 sin  cos  (-g) Δx = vi 2 sin (2 ) (-g)

Horizontal Range, Δx Δx = vi 2 sin (2 ) (deg) Sin (2 ) (-g) 0.00 15 0.50 30 0.87 45 1.00 60 75 90 CONCLUSIONS: Horizontal range is greatest for the throw angle of 450 Horizontal ranges are the same for angles  and (900 – )

Trajectory and horizontal range vi = 25 m/s

Velocity Final speed = initial speed (conservation of energy) Impact angle = - launch angle (symmetry of parabola)

Maximum Height vy = vi sin  + g t y = vi t sin  + ½ g t2 At maximum height vy = 0 0 = vi sin  + g tup hmax = vi t upsin  + ½ g tup2 hmax = vi2 sin2 /(-g) + ½ g(vi2 sin2 )/g2 hmax = vi2 sin2  2(-g) t = vi sin  (-g) tup = Δt/2

Projectile Motion – Final Equations (0,0) – initial position, vi = vi []– initial velocity, g = -9.80m/s2 Trajectory Parabola, open down Total time Δt = Horizontal range Δx = Max height hmax = 2 vi sin  (-g) vi 2 sin (2 ) (-g) vi2 sin2  2(-g)

PROJECTILE MOTION - SUMMARY Projectile motion is motion with a constant horizontal velocity combined with a constant vertical acceleration The projectile moves along a parabola

4.21 Reference frames Reference frame: A coordinate system used to make observations. The choice of a reference frame determines the perception of motion. In the analysis of motion, it is commonly assumed that the observer is standing still. Reference frames are often chosen for the sake of convenience (choosing the Earth’s surface, not the surface of Jupiter, is a logical choice Reference frames

4.23 Sample problem: relative velocity Observers in reference frames moving past one another may measure different velocities for the same object. This concept is called relative velocity. Observer on train 4.23 Sample problem: relative velocity