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

2. 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

3. 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. 4.3 Sample problem: velocity in two dimensions
2D example

5. 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
     

6. 4.5 Sample problem: acceleration in two dimensions
2D Acceleration example

7. 4.6 Interactive checkpoint: acceleration in two dimensions
Practice Problem 2D acceleration

8. 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

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

10. Part 1. Motion of Objects Projected Horizontally

11. y v0 x

12. y x

13. y x

14. y x

15. y x

16. 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

17. 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?

18. 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

19. 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

20. 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

21. Horizontal Range, Δx Δx = v0 Δt Δt = √ 2h/(-g) Δx = v0 √ 2h/(-g) Δxy 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

22. 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

23. 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)

24. 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

25. 4.8 Sample problem: a horizontal cannon

26. 4.9 Interactive problem: the monkey and the professor
4.10 Checkpoint: golfing  

27. 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

28. Part 2. Motion of objects projected at an angle

29. 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

30. y a =g = - 9.80 m/s2 x Motion is accelerated
Acceleration is constant, and downward
a = g = 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

31. 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?

32. X Uniform motion Y Accelerated motion
Equations of motion: X
Uniform motion Y
Accelerated motion
ACCELERATION
ax = 0
ay = g = 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

33. X Uniform motion Y Accelerated motion
Equations of motion: X
Uniform motion Y
Accelerated motion
ACCELERATION
ax = 0
ay = g = 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

34. 4.12 Sample problem: calculating initial velocity in projectile motion
4.13 Interactive problem: the monkey and the professor, part II

35. 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

36. 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 = Δt

37. 4.14 Projectile motion: aiming a cannon
4.15 Sample problem: a cannon’s range
4.16 Interactive checkpoint: clown cannon

38. 4.17 Interactive checkpoint: soccer kick
4.18 Interactive problem: the human cannonball
4.19 Interactive problem: test your juggling!

39. 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)

40. 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 – )

41. Trajectory and horizontal range
vi = 25 m/s

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

43. 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

44. 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)

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

46. 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

47. 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