1. MOTION OF A PROJECTILE
Today’s Objectives:
Students will be able to:
Analyze the free-flight motion of a projectile.
In-Class Activities:
• Check Homework
• Reading Quiz
• Applications
• Kinematic Equations for Projectile Motion
• Concept Quiz
• Group Problem Solving
• Attention Quiz

2. The downward acceleration of an object in free-flight motion is
READING QUIZ
The downward acceleration of an object in free-flight motion is
A) zero. B) increasing with time.
C) 9.81 m/s2. D) 9.81 ft/s2.
2. The horizontal component of velocity remains _________ during a free-flight motion.
A) zero B) constant
C) at 9.81 m/s2 D) at 32.2 ft/s2
Answers: C B 2

3. APPLICATIONS
A good kicker instinctively knows at what angle, q, and initial velocity, vA, he must kick the ball to make a field goal.
For a given kick “strength”, at what angle should the ball be kicked to get the maximum distance?

4. APPLICATIONS (continued)
A basketball is shot at a certain angle. What parameters should the shooter consider in order for the basketball to pass through the basket?
Distance, speed, the basket location, … anything else ?

5. APPLICATIONS (continued)
A firefighter needs to know the maximum height on the wall she can project water from the hose. What parameters would you program into a wrist computer to find the angle, q, that she should use to hold the hose?

6. MOTION OF A PROJECTILE (Section 12.6)
Projectile motion can be treated as two rectilinear motions, one in the horizontal direction experiencing zero acceleration and the other in the vertical direction experiencing constant acceleration (i.e., from gravity).
For illustration, consider the two balls on the left. The red ball falls from rest, whereas the yellow ball is given a horizontal velocity. Each picture in this sequence is taken after the same time interval. Notice both balls are subjected to the same downward acceleration since they remain at the same elevation at any instant. Also, note that the horizontal distance between successive photos of the yellow ball is constant since the velocity in the horizontal direction is constant.

7. KINEMATIC EQUATIONS: HORIZONTAL MOTION
Since ax = 0, the velocity in the horizontal direction remains constant (vx = vox) and the position in the x direction can be determined by:
x = xo + (vox) t
No air resistance is assumed!
Why is ax equal to zero (assuming movement through the air)? 7

8. KINEMATIC EQUATIONS: VERTICAL MOTION
Since the positive y-axis is directed upward, ay = – g. Application of the constant acceleration equations yields:
vy = voy – g t
y = yo + (voy) t – ½ g t2
vy2 = voy2 – 2 g (y – yo)
For any given problem, only two of these three equations can be used. Why?

9. Find: The equation that defines y as a function of x.
EXAMPLE I
Given: vo and θ
Find: The equation that defines y as a function of x.
Plan: Eliminate time from the kinematic equations.
Solution: Using vx = vo cos θ and vy = vo sin θ
We can write: x = (vo cos θ)t or
y = (vo sin θ) t – ½ g (t)2
t = x
vo cos θ
y = (vo sin θ) { } { }2
x g x
vo cos θ 2 vo cos θ –
By substituting for t: 9

10. EXAMPLE I (continued)
Simplifying the last equation, we get:
y = (x tanq) –
g x2
2vo2
(1 + tan2q)
The above equation is called the “path equation” which describes the path of a particle in projectile motion.
The equation shows that the path is parabolic.

11. EXAMPLE II
Given: Projectile is fired with vA=150 m/s at point A.
Find: The horizontal distance it travels (R) and the time in the air.
Plan:
Establish a fixed x, y coordinate system (in this solution, the origin of the coordinate system is placed at A). Apply the kinematic relations in x- and y-directions.

12. EXAMPLE II (continued)
Solution:
1) Place the coordinate system at point A.
Then, write the equation for horizontal motion.
+ xB = xA + vAx tAB
where xB = R, xA = 0, vAx = 150 (4/5) m/s
Range, R will be R = 120 tAB
2) Now write a vertical motion equation. Use the distance equation.
+ yB = yA + vAy tAB – 0.5 g tAB2
where yB = – 150, yA = 0, and vAy = 150(3/5) m/s
We get the following equation: –150 = 90 tAB (– 9.81) tAB2
Solving for tAB first, tAB = s.
Then, R = 120 tAB = 120 (19.89) = 2387 m

13. A) (vo sin q)/g B) (2vo sin q)/g C) (vo cos q)/g D) (2vo cos q)/g
CONCEPT QUIZ
1. In a projectile motion problem, what is the maximum number of unknowns that can be solved?
A) B) 2
C) D) 4
2. The time of flight of a projectile, fired over level ground, with initial velocity Vo at angle θ, is equal to?
A) (vo sin q)/g B) (2vo sin q)/g
C) (vo cos q)/g D) (2vo cos q)/g
Answers: C B 13

14. GROUP PROBLEM SOLVING
Establish a fixed x,y coordinate system (in this solution, the origin of the coordinate system is placed at A). Apply the kinematic relations in x and y-directions. x y
Given: A skier leaves the ski jump ramp at qA = 25o and hits the slope at B.
Find: The skier’s initial speed vA.
Plan:

15. GROUP PROBLEM SOLVING (continued)
Solution:
Motion in x-direction:
Using xB = xA + vox(tAB) => (4/5)100 = 0 + vA (cos 25) tAB =
tAB= 80
vA (cos 25)
88.27 vA
– 64 = 0 + vA(sin 25) { }
88.27 vA
– ½ (9.81) { }2
Motion in y-direction:
Using yB = yA + voy(tAB) – ½ g(tAB)2
vA = m/s

16. ATTENTION QUIZ 1. A projectile is given an initial velocity
vo at an angle f above the horizontal.
The velocity of the projectile when it
hits the slope is ____________ the
initial velocity vo.
A) less than B) equal to
C) greater than D) None of the above.
2. A particle has an initial velocity vo at angle q with respect to the horizontal. The maximum height it can reach is when
A) q = 30° B) q = 45°
C) q = 60° D) q = 90°
Answers:
1. A
2. D 16

17. End of the Lecture
Let Learning Continue

18. Section 2.6 – Polar Coordinates (r,θ)
Particle is located by the radial distance r from a fixed point O, and
An angular measurement θ from a fixed line passing through O.
Note: by convention the fixed line is chosen to be the x-axis.
Note: by convention θ is taken as positive in the counter-clockwise direction.
Note: conventions for unit vectors.

19. Polar Coordinates
Polar coordinates are most useful for problems where:
the motion is Constrained through the control of a radial distance or an angular position, or
Unconstrained motion that is observed by measuring a radial distance and an angular position.
The position vector is given by:
We wish to find the velocity vector:
We must find the derivative of the unit vectors!

20. Derivative of the Unit Vectors

21. Velocity
The r-component of the velocity vr is due to the rate at which stretches or contracts, and
The θ-component of the velocity vθ is due to the rate at which rotates.

22. Acceleration

23. Circular Motion

24. Example Problem

25. Normal & Tangential Coordinates
Uses path variables. i.e. measurements made along the tangent and normal to the path.
Recall that the velocity vector is always in the direction tangent to the path of the motion.
Provides a natural description for Curvilinear Motion
Coordinates move along the path with the particle.

26. Normal & Tangential Coordinates
The positive direction of the normal direction ên is towards the centre of curvature of the path.
The positive direction for the tangential direction êt is in the direction of the motion.

27. Velocity
The velocity in n-t coordinates has only one component:

28. Acceleration Lets look at the velocity vectors at t and at t
Radius of Curvature

29. Example
450

30. Practice Problem
Try this problem at home. We will go over the solution during the next lecture. Use normal-tangential coordinates to solve it!

31. Space Curvilinear Coordinates
Describes the 3-D motion of a particle.
We will consider three coordinate systems:
Rectangular (x, y, z)
Cylindrical (r, θ, z)
Spherical (R, θ, )

32. Rectangular Coordinates (x, y, z)
Add the z-coordinate
Use R instead of r as the position vector (keep r for the position in the x-y plane).
Very simple since the unit vectors are fixed in space.

33. Cylindrical Coordinates (r, θ, z)
An extension of polar coordinates
Simply add z-coordinate and its time derivatives.

34. Spherical Coordinates (R, θ, )
where

35. Linear algebraic transformations exist between these three coordinate systems.
Spherical coordinates will bog us down here, but you will study them in detail in other courses.
We will try to avoid spherical coordinates except where the simplification they introduce exceeds their added complexity.

36. We could also use n-t coordinates here; however
While this would be natural for many plane motion problems it is awkward in 3-D (space curvilinear) motion.
We will rarely use n-t coordinates in this course.
You will use them in fluid mechanics.

37. Example Problem (Prob. 2-177)
An aircraft P takes off at A with a velocity v0 of 250 km/h and climbs in the vertical yz -plane at the constant 15 angle with an acceleration along its flight path of 0.8 m/s2. Flight progress is monitored by radar at point O. Resolve the velocity of the aircraft P into spherical coordinate components 60 s after takeoff and find: , and for that instant.
Problem repeat for cylindrical coordinates.

38. Problems & 2-177

39. CURVILINEAR MOTION: CYLINDRICAL COMPONENTS
Today’s Objectives:
Students will be able to:
Determine velocity and acceleration components using cylindrical coordinates.
In-Class Activities:
Check Homework
Reading Quiz
Applications
Velocity Components
Acceleration Components
Concept Quiz
Group Problem Solving
Attention Quiz 39

40. A) transverse velocity. B) radial velocity.
READING QUIZ
1. In a polar coordinate system, the velocity vector can be written as v = vrur + vθuθ = rur + rquq. The term q is called
A) transverse velocity. B) radial velocity.
C) angular velocity. D) angular acceleration. .
2. The speed of a particle in a cylindrical coordinate system is
A) r B) rq
C) (rq)2 + (r)2 D) (rq)2 + (r)2 + (z)2 .
Answers:
1. C
2. D 40

41. APPLICATIONS
The cylindrical coordinate system is used in cases where the particle moves along a 3-D curve.
In the figure shown, the box slides down the helical ramp. How would you find the box’s velocity components to know if the package will fly off the ramp?

42. CYLINDRICAL COMPONENTS
(Section 12.8)
We can express the location of P in polar coordinates as r = r ur. Note that the radial direction, r, extends outward from the fixed origin, O, and the transverse coordinate, q, is measured counter-clockwise (CCW) from the horizontal. 42

43. VELOCITY in POLAR COORDINATES)
The instantaneous velocity is defined as:
v = dr/dt = d(rur)/dt
v = rur + r
dur dt .
Using the chain rule:
dur/dt = (dur/dq)(dq/dt)
We can prove that dur/dq = uθ so dur/dt = quθ
Therefore: v = rur + rquθ . .
Thus, the velocity vector has two components: r, called the radial component, and rq called the transverse component. The speed of the particle at any given instant is the sum of the squares of both components or
v = (r q )2 + ( r )2 43

44. ACCELERATION (POLAR COORDINATES)
The instantaneous acceleration is defined as:
a = dv/dt = (d/dt)(rur + rquθ) .
After manipulation, the acceleration can be expressed as
a = (r – rq 2)ur + (rq + 2rq)uθ .. . .
The magnitude of acceleration is a = (r – rq 2)2 + (rq + 2rq) 2
The term (r – rq 2) is the radial acceleration or ar .
The term (rq + 2rq) is the transverse acceleration or aq . .. 44

45. CYLINDRICAL COORDINATES
If the particle P moves along a space curve, its position can be written as
rP = rur + zuz
Taking time derivatives and using the chain rule:
Velocity: vP = rur + rquθ + zuz
Acceleration: aP = (r – rq2)ur + (rq + 2rq)uθ + zuz .. . 45

46. Given: A car travels along acircular path.
EXAMPLE
Given: A car travels along acircular path.
r = 300 ft, q = 0.4 (rad/s),
q = 0.2 (rad/s2)
Find: Velocity and acceleration
Plan: Use the polar coordinate system. . ..
Solution:
r = 300 ft, r = r = 0, and q = 0.4 (rad/s), q = 0.2 (rad/s2) .. .
Substitute in the equation for velocity
v = r ur + rq uθ = 0 ur (0.4) uθ
v = (0)2 + (120)2 = 120 ft/s . 46

47. Substitute in the equation for acceleration:
EXAMPLE (continued)
Substitute in the equation for acceleration:
a = (r – rq 2)ur + (rq + 2rq)uθ
a = [0 – 300(0.4)2] ur + [300(0.2) + 2(0)(0.4)] uθ
a = – 48 ur + 60 uθ ft/s2
a = (– 48)2 + (60)2 = ft/s2 .. . 47

48. CONCEPT QUIZ .. . 1. If r is zero for a particle, the particle is
A) not moving. B) moving in a circular path.
C) moving on a straight line. D) moving with constant velocity. .
2. If a particle moves in a circular path with constant velocity, its radial acceleration is
A) zero B) r .
C) − rq D) 2rq . .. .
Answers:
1. B
2. C 48

49. GROUP PROBLEM SOLVING
Given: The car’s speed is constant at 1.5 m/s.
Find: The car’s acceleration (as a vector).
Plan:
f = tan-1( ) = °
Also, what is the relationship between f and q? 12
2p(10)
Hint: The tangent to the ramp at any point is at an angle
Plan: Use cylindrical coordinates. Since r is constant, all derivatives of r will be zero.

50. GROUP PROBLEM SOLVING (continued)
Solution: Since r is constant, the velocity only has 2 components:
vq = rq = v cos f and vz = z = v sin f .
Therefore: q = ( ) = rad/s
q = 0
v cosf r . ..
vz = z = v sinf = m/s
z = 0
r = r = 0 . .. .. .
a = (r – rq 2)ur + (rq + 2rq)uθ + zuz
a = (-rq 2)ur = -10(0.147)2ur = ur m/s2

51. C) greater than its transverse component.
ATTENTION QUIZ
1. The radial component of velocity of a particle moving in a circular path is always
A) zero.
B) constant.
C) greater than its transverse component.
D) less than its transverse component.
2. The radial component of acceleration of a particle moving in a circular path is always
A) negative.
B) directed toward the center of the path.
C) perpendicular to the transverse component of acceleration.
D) All of the above.
Answers:
1. A
2. D 51

52. End of the Lecture
Let Learning Continue

53. NORMAL AND TANGENTIAL COMPONENTS
CURVILINEAR MOTION:
NORMAL AND TANGENTIAL COMPONENTS
Today’s Objectives:
Students will be able to:
Determine the normal and tangential components of velocity and acceleration of a particle traveling along a curved path.
In-Class Activities:
• Check Homework
• Reading Quiz
• Applications
• Normal and Tangential Components of Velocity and Acceleration
• Special Cases of Motion
• Concept Quiz
• Group Problem Solving
• Attention Quiz 53

54. A) positive. B) negative. C) zero. D) constant.
READING QUIZ
1. If a particle moves along a curve with a constant speed, then its tangential component of acceleration is
A) positive. B) negative.
C) zero. D) constant.
2. The normal component of acceleration represents
A) the time rate of change in the magnitude of the velocity.
B) the time rate of change in the direction of the velocity.
C) magnitude of the velocity.
D) direction of the total acceleration.
Answers:
1. C
2. B 54

55. APPLICATIONS
Cars traveling along a clover-leaf interchange experience an acceleration due to a change in velocity as well as due to a change in direction of the velocity.
If the car’s speed is increasing at a known rate as it travels along a curve, how can we determine the magnitude and direction of its total acceleration?
Why would you care about the total acceleration of the car?

56. APPLICATIONS (continued)
A roller coaster travels down a hill for which the path can be approximated by a function
y = f(x).
The roller coaster starts from rest and increases its speed at a constant rate.
How can we determine its velocity and acceleration at the bottom?
Why would we want to know these values?

57. NORMAL AND TANGENTIAL COMPONENTS (Section 12.7)
When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian. When the path of motion is known, normal (n) and tangential (t) coordinates are often used.
In the n-t coordinate system, the origin is located on the particle (the origin moves with the particle).
The t-axis is tangent to the path (curve) at the instant considered, positive in the direction of the particle’s motion.
The n-axis is perpendicular to the t-axis with the positive direction toward the center of curvature of the curve.

58. NORMAL AND TANGENTIAL COMPONENTS (continued)
The positive n and t directions are defined by the unit vectors un and ut, respectively.
The center of curvature, O’, always lies on the concave side of the curve.
The radius of curvature, r, is defined as the perpendicular distance from the curve to the center of curvature at that point.
The position of the particle at any instant is defined by the distance, s, along the curve from a fixed reference point.

59. VELOCITY IN THE n-t COORDINATE SYSTEM
The velocity vector is always tangent to the path of motion (t-direction).
The magnitude is determined by taking the time derivative of the path function, s(t).
v = v ut where v = s = ds/dt .
Here v defines the magnitude of the velocity (speed) and
ut defines the direction of the velocity vector.

60. ACCELERATION IN THE n-t COORDINATE SYSTEM
Acceleration is the time rate of change of velocity:
a = dv/dt = d(vut)/dt = vut + vut .
Here v represents the change in the magnitude of velocity and ut represents the rate of change in the direction of ut. . .
a = v ut + (v2/r) un = at ut + an un.
After mathematical manipulation, the acceleration vector can be expressed as:

61. ACCELERATION IN THE n-t COORDINATE SYSTEM (continued)
So, there are two components to the acceleration vector:
a = at ut + an un
• The tangential component is tangent to the curve and in the direction of increasing or decreasing velocity.
at = v or at ds = v dv .
• The normal or centripetal component is always directed toward the center of curvature of the curve. an = v2/r
• The magnitude of the acceleration vector is
a = [(at)2 + (an)2]0.5

62. SPECIAL CASES OF MOTION
There are some special cases of motion to consider.
1) The particle moves along a straight line.
r  => an = v2/r = => a = at = v
The tangential component represents the time rate of change in the magnitude of the velocity. .
2) The particle moves along a curve at constant speed.
at = v = => a = an = v2/r .
The normal component represents the time rate of change in the direction of the velocity.

63. SPECIAL CASES OF MOTION (continued)
3) The tangential component of acceleration is constant, at = (at)c.
In this case,
s = so + vo t + (1/2) (at)c t2
v = vo + (at)c t
v2 = (vo)2 + 2 (at)c (s – so)
As before, so and vo are the initial position and velocity of the particle at t = 0. How are these equations related to projectile motion equations? Why?
4) The particle moves along a path expressed as y = f(x).
The radius of curvature, r, at any point on the path can be calculated from
r = ________________
]3/2
(dy/dx)2 1 [ + 2
d2y/dx

64. THREE-DIMENSIONAL MOTION
If a particle moves along a space curve, the n and t axes are defined as before. At any point, the t-axis is tangent to the path and the n-axis points toward the center of curvature. The plane containing the n and t axes is called the osculating plane.
A third axis can be defined, called the binomial axis, b. The binomial unit vector, ub, is directed perpendicular to the osculating plane, and its sense is defined by the cross product ub = ut × un.
There is no motion, thus no velocity or acceleration, in the binomial direction.

65. EXAMPLE
Given: A boat travels around a circular path, r = 40 m, at a speed that increases with time, v = ( t2) m/s.
Find: The magnitudes of the boat’s velocity and acceleration at the instant t = 10 s.
Plan:
The boat starts from rest (v = 0 when t = 0).
1) Calculate the velocity at t = 10 s using v(t).
2) Calculate the tangential and normal components of acceleration and then the magnitude of the acceleration vector.

66. EXAMPLE (continued)
Solution:
1) The velocity vector is v = v ut , where the magnitude is given by v = (0.0625t2) m/s. At t = 10s:
v = t2 = (10)2 = m/s
2) The acceleration vector is a = atut + anun = vut + (v2/r)un. .
Tangential component: at = v = d(.0625 t2 )/dt = t m/s2
At t = 10s: at = 0.125t = 0.125(10) = 1.25 m/s2 .
Normal component: an = v2/r m/s2
At t = 10s: an = (6.25)2 / (40) = m/s2
The magnitude of the acceleration is
a = [(at)2 + (an)2]0.5 = [(1.25)2 + (0.9766)2]0.5 = 1.59 m/s2

67. CHECK YOUR UNDERSTANDING QUIZ
1. A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing at a constant rate of 4 m/s2. What is the magnitude of its total acceleration at this instant?
A) 3 m/s2 B) 4 m/s2
C) 5 m/s2 D) -5 m/s2
2. If a particle moving in a circular path of radius 5 m has a velocity function v = 4t2 m/s, what is the magnitude of its total acceleration at t = 1 s?
A) 8 m/s B) 8.6 m/s
C) 3.2 m/s D) 11.2 m/s
Answers:
1. C
2. B 67

68. 2. Calculate the radius of curvature of the path at B.
GROUP PROBLEM SOLVING
Given: A roller coaster travels along a vertical parabolic path defined by the equation y = 0.01x2. At point B, it has a speed of 25 m/s, which is increasing at the rate of 3 m/s2.
Find: The magnitude of the roller coaster’s acceleration when it is at point B.
Plan:
1. The change in the speed of the car (3 m/s2) is the tangential component of the total acceleration.
2. Calculate the radius of curvature of the path at B.
3. Calculate the normal component of acceleration.
4. Determine the magnitude of the acceleration vector. 68

69. GROUP PROBLEM SOLVING (continued)
Solution:
1) The tangential component of acceleration is the rate of increase of the roller coaster’s speed, so at = v = 3 m/s2. .
2) Determine the radius of curvature at point B (x = 30 m):
dy/dx = d(0.01x2)/dx = 0.02x, d2y/dx2 = d (0.02x)/dx = 0.02
At x =30 m, dy/dx = 0.02(30) = 0.6, d2y/dx2 = 0.02
=> r = = [1 + (0.6)2]3/2/(0.02) = 79.3 m
[1+(dy/dx)2]3/2
d2y/dx2
3) The normal component of acceleration is
an = v2/r = (25)2/(79.3) = m/s2
4) The magnitude of the acceleration vector is
a = [(at)2 + (an)2]0.5 = [(3)2 + (7.881)2]0.5 = 8.43 m/s2 69

70. 1. The magnitude of the normal acceleration is
ATTENTION QUIZ
1. The magnitude of the normal acceleration is
A) proportional to radius of curvature.
B) inversely proportional to radius of curvature.
C) sometimes negative.
D) zero when velocity is constant.
2. The directions of the tangential acceleration and velocity are always
A) perpendicular to each other. B) collinear.
C) in the same direction. D) in opposite directions.
Answers:
1. B
2. B 70

71. End of the Lecture
Let Learning Continue 71