1
ESD6233
APPLIED DYNAMICS
Chapter 1
Kinematics of Particles
Dynamics, Fourteenth Edition in SI Units
Instructor: Low KO
R.C. Hibbeler
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Trimester 2510
2
Chapter Outline
▪ Introduction
▪ Rectilinear Kinematics : Continuous Motion
▪ Rectilinear Kinematics : Erratic Motion
▪ General Curvilinear Motion
▪ Curvilinear Motion : Rectangular Components
▪ Motion of a Projectile
▪ Curvilinear Motion : Normal and Tangential Components
▪ Curvilinear Motion : Cylindrical Components
▪ Absolute Dependent Motion Analysis of Two Particles
▪ Relative Motion Analysis of Two Particles Using
Translating Axes
Dynamics, Fourteenth Edition in SI Units
R.C. Hibbeler
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INTRODUCTION &
RECTILINEAR KINEMATICS: CONTINUOUS MOTION
Today’s Objectives:
Students will be able to:
1. Find the kinematic quantities
(position, displacement, velocity,
and acceleration) of a particle
traveling along a straight path.
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4
APPLICATIONS (continued)
A sports car travels along a straight road.
If the car accelerates at a constant rate, how can we
determine its position and velocity at some instant?
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5
An Overview of Mechanics
Mechanics: The study of how bodies
react to the forces acting on them.
Statics: The study of
bodies in equilibrium.
Dynamics, Fourteenth Edition in SI Units
R.C. Hibbeler
Dynamics:
1. Kinematics – concerned with
the geometric aspects of motion
2. Kinetics - concerned with
the forces causing the motion
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RECTILINEAR KINEMATICS:
CONTINIOUS MOTION (Section 12.2)
A particle travels along a straight-line path
defined by the coordinate axis s.
The position of the particle at any instant,
relative to the origin, O, is defined by the
position vector r, or the scalar s. Scalar s
can be positive or negative. Typical units
for r and s are meters (m).
The displacement of the particle is
defined as its change in position.
Vector form:  r = r’ - r
Scalar form:  s = s’ - s
The total distance traveled by the particle, sT, is a positive scalar
that represents the total length of the path over which the particle
travels.
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7
VELOCITY
Velocity is a measure of the rate of change in the position of a particle.
It is a vector quantity (it has both magnitude and direction). The
magnitude of the velocity is called speed, with units of m/s.
The average velocity of a particle during a
time interval t is
vavg = r / t
The instantaneous velocity is the time-derivative of position.
v = dr / dt
Speed is the magnitude of velocity: v = ds / dt
Average speed is the total distance traveled divided by elapsed time:
(vsp)avg = sT / t
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8
ACCELERATION
Acceleration is the rate of change in the velocity of a particle. It is a
vector quantity. Typical units are m/s2.
The instantaneous acceleration is the time
derivative of velocity.
Vector form: a = dv / dt
Scalar form: a = dv / dt = d2s / dt2
Acceleration can be positive (speed
increasing) or negative (speed decreasing).
As the text shows, the derivative equations for velocity and
acceleration can be manipulated to get a ds = v dv
Dynamics, Fourteenth Edition in SI Units
R.C. Hibbeler
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SUMMARY OF KINEMATIC RELATIONS:
RECTILINEAR MOTION
• Differentiate position to get velocity and acceleration.
v = ds/dt ;
a = dv/dt or a = v dv/ds
• Integrate acceleration for velocity and position.
Position:
Velocity:
v
t
v
s
 dv =  a dt or  v dv =  a ds
s
t
 ds =  v dt
vo
o
vo
so
so
o
• Note that so and vo represent the initial position and
velocity of the particle at t = 0.
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R.C. Hibbeler
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10
CONSTANT ACCELERATION
The three kinematic equations can be integrated for the special case
when acceleration is constant (a = ac) to obtain very useful equations.
A common example of constant acceleration is gravity; i.e., a body
freely falling toward earth. In this case, ac = g = 9.81 m/s2 downward.
Velocity as function of time
v
t
 dv =  a dt
c
vo
o
o
Assume initially v = v0 when s = s0
s
 v dv =  ac ds
vo
s = s o + v ot + (1/2) a c t 2
yields
Velocity as function of position
v
Assume initially s = s0 when t = 0
t
 ds =  v dt
so
v = vo + act
yields
Position as function of time
s
Assume initially v = v0 when t = 0
yields
v 2 = (vo )2 + 2ac(s - so)
so
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(a = ac)
11
EXAMPLE
Given: A particle travels along a straight line to the right
with a velocity of v = ( 4 t – 3 t2 ) m/s where t is
in seconds. Also, s = 0 when t = 0.
Find: The position and acceleration of the particle
when t = 4 s.
Plan: Establish the positive coordinate, s, in the direction the
particle is traveling. Since the velocity is given as a
function of time, take a derivative of it to calculate the
acceleration. Conversely, integrate the velocity
function to calculate the position.
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EXAMPLE (continued)
Solution:
v = ( 4 t – 3 t2 ) , s = 0 when t = 0.
1) Take a derivative of the velocity to determine the acceleration.
a = dv / dt = d(4 t – 3 t2) / dt = 4 – 6 t
 a = – 20 m/s2 (or in the  direction) when t = 4 s
2) Calculate the distance traveled in 4s by integrating the
velocity using so = 0:
s
t
v = ds / dt  ds = v dt   ds =  (4 t – 3 t2) dt
so
o
 s – so = 2 t 2 – t 3
 s – 0 = 2(4)2 – (4)3  s = – 32 m (or )
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13
GROUP PROBLEM SOLVING
Given: A sandbag is dropped from a balloon
ascending vertically at a constant speed
of 6 m/s. The bag is released with the
same upward velocity of 6 m/s at t = 0 s
and hits the ground when t = 8 s.
Find: The speed of the bag as it hits the ground and the altitude
of the balloon at this instant.
Plan: The sandbag is experiencing a constant downward
acceleration of 9.81 m/s2 due to gravity. Apply the
formulas for constant acceleration, with ac = - 9.81 m/s2.
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GROUP PROBLEM SOLVING (continued)
Solution:
The bag is released when t = 0 s and hits the
ground when t = 8 s.
Calculate the distance using a position equation.
+ sbag = (sbag )o + (vbag)o t + (1/2) ac t2
sbag = 0 + (-6) (8) + 0.5 (9.81) (8)2 = 265.9 m
During t = 8 s, the balloon rises
+ sballoon = (vballoon) t = 6 (8) = 48 m
Therefore, altitude is of the balloon is (sbag + sballoon).
Altitude = 265.9 + 48 = 313.9 = 314 m.
Dynamics, Fourteenth Edition in SI Units
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GROUP PROBLEM SOLVING (continued)
Calculate the velocity when t = 8 s, by applying a velocity
equation.
+ vbag = (vbag )o + ac t
vbag = -6 + (9.81) 8 = 72.5 m/s 
Dynamics, Fourteenth Edition in SI Units
R.C. Hibbeler
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16
RECTILINEAR KINEMATICS: ERRATIC MOTION
Today’s Objectives:
Students will be able to:
1. Determine position, velocity, and acceleration of a particle using
graphs, i.e non-linear or discontinuity
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R.C. Hibbeler
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17
APPLICATIONS
In many experiments, a
velocity versus position (v-s)
profile is obtained.
If we have a v-s graph for the
tank truck, how can we
determine its acceleration at
position s = 1500 m?
Dynamics, Fourteenth Edition in SI Units
R.C. Hibbeler
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18
APPLICATIONS (continued)
The velocity of a car is
recorded from a experiment.
The car starts from rest and
travels along a straight track.
If we know the v-t plot, how
can we determine the distance
the car traveled during the
time interval 0 < t < 30 s or
15 < t < 25 s?
Dynamics, Fourteenth Edition in SI Units
R.C. Hibbeler
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19
ERRATIC MOTION (Section 12.3)
Graphing provides a good way to
handle complex motions that
would be difficult to describe
with formulas.
Graphs also provide a visual
description of motion and
reinforce the calculus concepts of
differentiation and integration as
used in dynamics.
The approach builds on the facts that slope and differentiation
are linked and that integration can be thought of as finding the
area under a curve.
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R.C. Hibbeler
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20
S-T GRAPH
Plots of position versus time can
be used to find velocity versus
time curves. Finding the slope of
the line tangent to the motion curve
at any point is the velocity at that
point (or v = ds/dt).
Therefore, the v-t graph can be
constructed by finding the slope at
various points along the s-t graph.
Given s-t graph, construct v-t graph
Slope of s-t graph = velocity
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21
V-T GRAPH
Plots of velocity versus time can be used
to find acceleration versus time curves.
Finding the slope of the line tangent to
the velocity curve at any point is the
acceleration at that point (or a = dv/dt).
Therefore, the acceleration versus time
(or a-t) graph can be constructed by
finding the slope at various points along
the v-t graph.
Also, the displacement of the particle is the
area under the v-t graph during time t.
Given v-t graph, construct a-t graph
Slope of v-t graph = acceleration
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22
A-T GRAPH
Given the acceleration versus
time or a-t curve, the change in
velocity (v) during a time
period is the area under the a-t
curve.
So we can construct a v-t graph
from an a-t graph if we know the
initial velocity of the particle.
Given a-t graph, construct v-t graph
Change in velocity = area under a-t graph
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23
A-S GRAPH
A more complex case is presented by the
acceleration versus position or a-s graph.
The area under the a-s curve represents the
change in velocity.
s2
½ (v1² – vo²) =  a ds = area under the
s1
a-s graph
This equation can be solved for v1, allowing
you to solve for the velocity at a point. By
doing this repeatedly, you can create a plot of
velocity versus distance.
Given a-s graph, construct v-s graph
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24
V-S GRAPH
Another complex case is presented by the velocity
versus distance or v-s graph. By reading the
velocity v at a point on the curve and multiplying it
by the slope of the curve (dv/ds) at this same point,
we can obtain the acceleration at that point. Recall
the formula
a = v (dv/ds)
Acceleration = velocity x slope of v-s graph
Thus, we can obtain an a-s plot from the v-s curve.
Given v-s graph, construct a-s graph
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25
EXAMPLE
Given: The v-t graph for a dragster moving along a straight road.
Find: The a-t graph and s-t graph over the time interval shown.
What is your plan of attack for the problem?
Dynamics, Fourteenth Edition in SI Units
R.C. Hibbeler
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26
EXAMPLE (continued)
Solution: The a-t graph can be constructed by finding the slope
of the v-t graph at key points. What are those?
when 0 < t < 5 s; v0-5 = ds/dt = d(30t)/dt = 30 m/s2
when 5 < t < 15 s; v5-15 = ds/dt = d(-15t+225)/dt = -15 m/s2
a(m/s2)
a-t graph
30
5
-15
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15 t(s)
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EXAMPLE (continued)
Now integrate the v - t graph to build the s – t graph.
when 0 < t < 5 s; s =  v dt = [15 t2 ] t = 15 t2 m
0
when 5 < t < 15 s; s − 15 (52) =  v dt = [(-15) (1/2) t 2 + 225 t]
t
5
s = - 7.5 t 2 + 225 t − 562.5 m
s(m)
s-t graph
1125
-7.5 t2 + 225 t − 562.5
375
t(s)
15t2
5
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GROUP PROBLEM SOLVING I
Given: The v-t graph shown.
v = 20 -0.25 t
Find: The a-t graph, average
speed, and distance
traveled for the 0 - 80 s
interval.
Plan: Find slopes of the v-t curve and draw the a-t graph.
Find the area under the curve to get the distance traveled.
Calculate average speed (using basic definitions).
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GROUP PROBLEM SOLVING I (continued)
Solution:
Find the a–t graph.
For 0 ≤ t ≤ 40
a = dv/dt = 0 m/s²
For 40 ≤ t ≤ 80
a = dv/dt = -10 / 40 = -0.25 m/s²
a(m/s²)
0
a-t graph
40
80
-0.25
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t(s)
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GROUP PROBLEM SOLVING I (continued)
Now find the distance traveled:
s0-40 =  v dt =  10 dt = 10 (40) = 400 m
Area 1
s40-80 =  v dt
=  (20 − 0.25 t) dt
80
2
= [ 20 t -0.25 (1/2) t ]40 = 200 m
Area 2
s0-90 = 400 + 200 = 600 m
vavg(0-90) = total distance / time
= 600 / 80
v = 10
v = 20 -0.25 t
Area 1
Area 2
= 7.5 m/s
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GROUP PROBLEM SOLVING II
Given: The v-t graph shown.
Find: The a-t graph and
distance traveled for
the 0 - 15 s interval.
Plan: Find slopes of the v-t curve and draw the a-t graph.
Find the area under the curve to get the distance traveled.
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32
GROUP PROBLEM SOLVING II (continued)
Solution:
Find the a–t graph:
For 0 ≤ t ≤ 4
a = dv/dt = 1.25 m/s²
For 4 ≤ t ≤ 10
a = dv/dt = 0 m/s²
For 10 ≤ t ≤ 15
a = dv/dt = -1 m/s²
a(m/s²)
a-t graph
1.25
4
10
15 t(s)
-1
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GROUP PROBLEM SOLVING II (continued)
Now find the distance traveled:
s0-4 =  v dt
4
2
= [ (1.25) (1/2) t ]0 =
s4-10 =  v dt = [ 5 t ] 4 = 30 m
10
s10-15 =  v dt
10 m
A1
A2
15
2
= [ - (1/2) t + 15 t]10 =
12.5 m
A3
s0-15= 10 + 30 + 12.5 = 52.5 m
A2
A1
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A3
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CURVILINEAR MOTION:
GENERAL & RECTANGULAR COMPONENTS
Today’s Objectives:
Students will be able to:
1. Describe the motion of a particle traveling along a curved path
(curvilinear motion).
2. Relate kinematic quantities in terms of the rectangular components
(x, y, z) of the vectors.
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35
APPLICATIONS (continued)
A roller coaster car travels down
a fixed, helical path at a constant
speed.
How can we determine its
position or acceleration at any
instant?
If you are designing the track, why is it important to be
able to predict the acceleration of the car?
Dynamics, Fourteenth Edition in SI Units
R.C. Hibbeler
Copyright ©2017 by Pearson Education, Ltd.
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GENERAL CURVILINEAR MOTION
(Section 12.4)
A particle moving along a curved path undergoes curvilinear motion.
Since the motion is often three-dimensional, vectors are usually used
to describe the motion.
A particle moves along a curve
defined by the path function, s.
The position of the particle at any instant is designated by the vector
r = r(t). Both the magnitude and direction of r may vary with time.
If the particle moves a distance s along the
curve during time interval t, the
displacement is determined by vector
subtraction: r = r’ - r
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37
VELOCITY
Velocity represents the rate of change in the position of a
particle.
The average velocity of the particle
during the time increment t is
vavg = r/t .
The instantaneous velocity is the
time-derivative of position
v = dr/dt .
The velocity vector, v, is always
tangent to the path of motion.
The magnitude of v is called the speed. Since the arc length s
approaches the magnitude of r as t→0, the speed can be
obtained by differentiating the path function (v = ds/dt). Note
that this is not a vector!
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38
ACCELERATION
Acceleration represents the rate of change in the
velocity of a particle.
If a particle’s velocity changes from v to v’ over a
time increment t, the average acceleration during
that increment is:
aavg = v/t = (v’ - v)/t
The instantaneous acceleration is the timederivative of velocity:
a = dv/dt = d2r/dt2
A plot of the locus of points defined by the arrowhead
of the velocity vector is called a hodograph. The
acceleration vector is tangent to the hodograph, but
not, in general, tangent to the path function.
Dynamics, Fourteenth Edition in SI Units
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CURVILINEAR MOTION:
RECTANGULAR COMPONENTS (Section 12.5)
It is often convenient to describe the motion of a particle in
terms of its x, y, z or rectangular components, relative to a fixed
frame of reference.
The position of the particle can be
defined at any instant by the
position vector
r=xi+yj+zk .
The x, y, z-components may all be
functions of time, i.e.,
x = x(t), y = y(t), and z = z(t) .
The magnitude of the position vector is: r = √(x2 + y2 + z2)
The direction of r is defined by the unit vector: ur = r / r
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40
RECTANGULAR COMPONENTS: VELOCITY
The velocity vector is the time derivative of the position vector:
v = dr/dt = d(x i)/dt + d(y j)/dt + d(z k)/dt
Since the unit vectors i, j, k are constant in magnitude and
direction, this equation reduces to v = vx i + vy j + vz k
•
•
•
where vx = x = dx/dt, vy = y = dy/dt, vz = z = dz/dt
Dot (●) represents 1st time derivative of x = x (t), y = y(t), z = z(t)
The magnitude of the velocity vector is
v = √[(vx)2 + (vy)2 + (vz)2]
The direction of v is always tangent to
the path of motion.
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41
RECTANGULAR COMPONENTS: ACCELERATION
The acceleration vector is the time derivative of the velocity
vector (second derivative of the position vector).
a = dv/dt = d2r/dt2 = ax i + ay j + az k
where
The magnitude of the acceleration vector is
a = (ax )2 +(ay )2 +(az )2
The direction of a is usually not
tangent to the path of the particle.
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42
EXAMPLE
Given: The box slides down a slope described by the equation
y = (0.05x2) m, where x is in meters.
vx = -3 m/s, ax = -1.5 m/s2 at x = 5 m.
Find: The y components of the velocity and the acceleration
of the box at at x = 5 m.
Plan: Note that the particle’s velocity can be found by taking
the first time derivative of the path’s equation. And the
acceleration can be found by taking the second time
derivative of the path’s equation.
Take a derivative of the position to find the component
of the velocity and the acceleration.
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43
EXAMPLE (continued)
Solution:
Find the y-component of velocity by taking a time
derivative of the position y = (0.05x2)

 y = 2 (0.05) x x = 0.1 x x
Find the acceleration component by taking a time
derivative of the velocity y
 = 0.1 x x + 0.1 x x
y

Substitute the x-component of the acceleration, velocity at

x=5 into y and y.
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44
EXAMPLE (continued)


Since x = vx = -3 m/s, x = ax = -1.5 m/s2 at x = 5 m
 = 0.1 x x = 0.1 (5) (-3) = -1.5 m/s
y

 y = 0.1 x x + 0.1 x x
= 0.1 (-3)2 + 0.1 (5) (-1.5)
= 0.9 – 0.75
= 0.15 m/s2
At x = 5 m
vy = – 1.5 m/s = 1.5 m/s 
OR express y = f(t), then derive
for v = f(t) and a = f(t),
ay = 0.15 m/s2 
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45
GROUP PROBLEM SOLVING
Given: The particle travels along the path y = 0.5 x2.
When t = 0, x = y = z = 0.
Find: The particle’s distance and the magnitude of its
acceleration when t = 1 s, if vx = (5 t) m/s, where t is in
seconds.
Plan:
1) Determine x and ax by integrating and differentiating
vx, respectively, using the initial conditions.
2) Find the y-component of velocity & acceleration by
taking a time derivative of the path.
3) Determine the magnitude of the acceleration &
position.
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46
GROUP PROBLEM SOLVING (continued)
Solution:
1) x-components:
Velocity known as:
Position:

•
vx = x = (5 t ) m/s  5 m/s at t=1s
t

vxdt = (5t) dt  x = 2.5 t2  2.5 m at t=1s
0
••
Acceleration: ax = x = d/dt (5 t)  5 m/s2 at t=1s
2) y-components:
 3.125 m at t=1s
Position known as : y = 0.5 x2
•
•
•
Velocity: y = 0.5 (2) x x = x x
••
• •
••
Acceleration: ay = y = x x + x x
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 12.5 m/s at t=1s
 37.5 m/s2 at t=1s
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GROUP PROBLEM SOLVING (continued)
3) The position vector and the acceleration vector are
Position vector: r = [ x i + y j ] m
where x= 2.5 m, y= 3.125 m
Magnitude: r = 2.52 + 3.1252 = 4.00 m
Acceleration vector: a = [ ax i + ay j] m/s2
where ax = 5 m/s2, ay = 37.5 m/s2
Magnitude: a = 52 + 37.52 = 37.8 m/s2
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MOTION OF A PROJECTILE
Today’s Objectives:
Students will be able to:
1. Analyze the free-flight motion of a projectile.
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APPLICATIONS
A good kicker instinctively knows at what angle, , 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?
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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?
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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).
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MOTION OF A PROJECTILE (Section 12.6)
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.
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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
Slide 11
Why is ax equal to zero (what assumption must
be made if the movement is through the air)?
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KINEMATIC EQUATIONS: VERTICAL MOTION
Since the positive y-axis is directed upward, ay = – g.
Application of the constant acceleration equations yields:
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EXAMPLE I
Given: vA and θ
Find: Horizontal distance it travels and vC.
Plan: Apply the kinematic relations in x- and y-directions.
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EXAMPLE I
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EXAMPLE I (continued)
Velocity components at C are;
vCx = 10 cos 30
= 8.66 m/s →
vCy = 10 sin 30 – (9.81) (1.019)
= -5 m/s = 5 m/s 
vC =
8.662 + (−5)2 =10 m/s
Since y = 0 at C
0 = (10 sin 30) t – ½ (9.81) t2  t = 0, 1.019 s
Only the time of 1.019 s makes sense!
Horizontal distance the ball travels is;
x = (10 cos 30) t
x = (10 cos 30) 1.019 = 8.83 m
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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: How will you proceed?
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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.
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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 + 0.5 (– 9.81) tAB2
Solving for tAB first, tAB = 19.89 s.
Then, R = 120 tAB = 120 (19.89) = 2387 m
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GROUP PROBLEM SOLVING I
Given: A skier leaves the ski
jump ramp at A = 25o
and hits the slope at B.
Find: The skier’s initial speed vA.
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.
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GROUP PROBLEM SOLVING I (continued)
Solution:
Motion in x-direction:
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GROUP PROBLEM SOLVING I (continued)
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GROUP PROBLEM SOLVING II
Given: The golf ball is struck
with a velocity of 24
m/s as shown.
y
x
Find: Distance d to where it
will land.
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.
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GROUP PROBLEM SOLVING II (continued)
Solution:
Motion in x-direction:
Using xB = xA + vox(tAB)
 d cos10 = 0 + 24 (cos 55) tAB
y
x
tAB = 0.07154 d
Motion in y-direction:
Using yB = yA + voy(tAB) – ½ g(tAB)2
 d sin10 = 0 + 24(sin 55)(0.07154 d) – ½ (9.81) (0.07154 d)2
 0 = 1.2328 d – 0.025104 d2
d = 0, 49.1 m Only the non-zero answer is meaningful.
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CURVILINEAR MOTION:
NORMAL AND TANGENTIAL COMPONENTS
Today’s Objectives:
Students will be able to:
1. Determine the normal and
tangential components of
velocity and acceleration of
a particle traveling along a
curved path.
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67
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?
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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
(thus the origin and coordinate
system move 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.
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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, , 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.
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70
VELOCITY IN THE n-t COORDINATE SYSTEM
The velocity vector is always
tangent to the path of motion
(t-direction).
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ACCELERATION IN THE n-t COORDINATE SYSTEM
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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/
• The magnitude of the acceleration vector is
a = (an )2 +(at )2
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73
SPECIAL CASES OF MOTION
There are some special cases of motion to consider.
1) The particle moves along a straight line.
.
2

 => an = v / =  = 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 = 0 => a = an = v2/
The normal component represents the time rate of change in the
direction of the velocity.
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SPECIAL CASES OF MOTION (continued)
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EXAMPLE I
Given: A car travels along the road
with a speed of v = (2s) m/s,
where s is in meters.
 = 50 m
Find: The magnitudes of the car’s
acceleration at s = 10 m.
Plan:
1) Calculate the velocity when s = 10 m using v(s).
2) Calculate the tangential and normal components of
acceleration and then the magnitude of the
acceleration vector.
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EXAMPLE I (continued)
Solution:
1) The velocity vector is v = v ut , where the magnitude is
given by v = (2s) m/s.
When s = 10 m: v = 20 m/s
2) The acceleration vector is
Tangential component:
.
Since at = v = dv/dt = (dv/ds) (ds/dt) = (dv/ds) v
where v = 2s  at = d(2s)/ds (v)= 2 v
At s = 10 m: at = 40 m/s2
Normal component: an = v2/
When s = 10 m: an = (20)2 / (50) = 8 m/s2
The magnitude of the acceleration is
a = (an )2 +(at )2 = 402 + 82 = 40.8 m/s2
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EXAMPLE II
Given: A boat travels around a
circular path,  = 40 m, at a
speed that increases with
time, v = (0.0625 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.
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EXAMPLE II (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 = 0.0625 t2 = 0.0625 (10)2 = 6.25 m/s
2) The acceleration vector is
.
Tangential component: at = v = d(0.0625 t2 )/dt = 0.125 t m/s2
At t = 10s: at = 0.125t = 0.125(10) = 1.25 m/s2
Normal component: an = v2/ m/s2
At t = 10s: an = (6.25)2 / (40) = 0.9766 m/s2
The magnitude of the acceleration is
a = (an )2 +(at )2 = 1.252 + 0.97662 = 1.59 m/s2
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GROUP PROBLEM SOLVING I
Given: The train engine at E has a
speed of 20 m/s and an
acceleration of 14 m/s2 acting
in the direction shown.
Find: The rate of increase in the
train’s speed and the radius of
curvature  of the path.
Plan:
1. Determine the tangential and normal components of the
acceleration.
2. Calculate vሶ from the tangential component of the acceleration.
3. Calculate  from the normal component of the acceleration.
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GROUP PROBLEM SOLVING I (continued)
Solution:
1) Acceleration
Tangential component :
at =14 cos(75) = 3.623 m/s2
Normal component :
an = 14 sin(75) = 13.52 m/s2
2) The tangential component of
acceleration is the rate of increase
of the train’s speed, so
at = vሶ = 3.62 m/s2.
3) The normal component of acceleration is
an = v2/  13.52 = 202 / 
 =  m
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GROUP PROBLEM SOLVING II
Given: Starting from rest, a bicyclist travels around a
horizontal circular path,  = 10 m, at a speed of
v = (0.09 t2 + 0.1 t) m/s.
Find: The magnitudes of her velocity and acceleration when
she has traveled 3 m.
Plan:
The bicyclist starts from rest (v = 0 when t = 0).
1) Integrate v(t) to find the position s(t).
2) Calculate the time when s = 3 m using s(t).
3) Calculate the tangential and normal components of
acceleration and then the magnitude of the
acceleration vector.
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GROUP PROBLEM SOLVING II (continued)
Solution:
1) The velocity vector is v = (0.09 t2 + 0.1 t) m/s, where t is in
seconds. Integrate the velocity and find the position s(t).
Position:

v dt =

(0.09 t2 + 0.1 t) dt
s (t) = 0.03 t3 + 0.05 t2
2) Calculate the time, t when s = 3 m.
3 = 0.03 t3 + 0.05 t2
Solving for t, t = 4.147 s
The velocity at t = 4.147 s is,
v = 0.09 (4.147 ) 2 + 0.1 (4.147 ) = 1.96 m/s
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GROUP PROBLEM SOLVING II (continued)
3) The acceleration vector is
Tangential component:
.
at = v = d(0.09 t2 + 0.1 t) / dt = (0.18 t + 0.1) m/s2
At t = 4.147 s : at = 0.18 (4.147) + 0.1 = 0.8465 m/s2
Normal component:
an = v2/ m/s2
At t = 4.147 s : an = (1.96)2 / (10) = 0.3852 m/s2
The magnitude of the acceleration is
a = (an )2 +(at )2 = 0.84652 + 0.38522 = 0.930 m/s2
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CURVILINEAR MOTION: CYLINDRICAL
COMPONENTS
Today’s Objectives:
Students will be able to:
1. Determine velocity and
acceleration components
using cylindrical
coordinates.
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85
APPLICATIONS
A 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
check to see if the package
will fly off the ramp?
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CYLINDRICAL COMPONENTS
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,  is measured counterclockwise (CCW) from the horizontal.
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VELOCITY in POLAR COORDINATES
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88
ACCELERATION (POLAR COORDINATES)
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CYLINDRICAL COORDINATES
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EXAMPLE
Given: The platform is rotating such
that, at any instant, its angular
position is  = (4t3/2) rad, where
t is in seconds.
A ball rolls outward so that its
position is r = (0.1t3) m.
Find: The magnitude of velocity and acceleration of the
ball when t = 1.5 s.
Plan: Use a polar coordinate system and related
kinematic equations
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EXAMPLE (continued)
Solution:
𝑟 = 0.1𝑡 3 , rሶ = 0.3 t 2 , rሷ = 0.6 t
𝜃 = 4 t3/2, 𝜃ሶ = 6 t1/2, 𝜃ሷ = 3 t−1/2
At t=1.5 s,
r = 0.3375 m, rሶ = 0.675 m/s, rሷ = 0.9 m/s2
𝜃 = 7.348 rad, 𝜃ሶ = 7.348 rad/s, 𝜃ሷ = 2.449 rad/s2
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EXAMPLE (continued)
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GROUP PROBLEM SOLVING
Given: The arm of the robot is
extending at a constant rate
𝑟ሶ = 1.5 m/s when r = 3 m,
z = (4t2) m, and  = (0.5 t) rad,
where t is in seconds.
Find: The velocity and acceleration
of the grip A when t = 3 s.
Plan: Use cylindrical coordinates.
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GROUP PROBLEM SOLVING (continued)
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GROUP PROBLEM SOLVING (continued)
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ABSOLUTE DEPENDENT MOTION ANALYSIS
OF TWO PARTICLES
Today’s Objectives:
Students will be able to:
1. Relate positions, velocities, and
accelerations of particles
undergoing dependent motion.
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97
APPLICATIONS (continued)
Rope and pulley arrangements
are often used to assist in lifting
heavy objects. The total lifting
force required from the truck
depends on both the weight and
the acceleration of the cabinet.
How can we determine the
acceleration and velocity of
the cabinet if the acceleration
of the truck is known?
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DEPENDENT MOTION (Section 12.9)
In many kinematics problems, the motion of one object will
depend on the motion of another object.
The blocks in this figure are
connected by an inextensible cord
wrapped around a pulley.
If block A moves downward along
the inclined plane, block B will
move up the other incline.
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DEPENDENT MOTION (continued)
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DEPENDENT MOTION (continued)
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101
DEPENDENT MOTION EXAMPLE
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DEPENDENT MOTION EXAMPLE (continued)
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103
DEPENDENT MOTION EXAMPLE (continued)
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DEPENDENT MOTION: PROCEDURES
These procedures can be used to relate the dependent motion of
particles moving along rectilinear paths (only the magnitudes of
velocity and acceleration change, not their line of direction).
1. Define position coordinates from fixed datum lines,
along the path of each particle. Different datum lines can
be used for each particle.
2. Relate the position coordinates to the cord length.
Segments of cord that do not change in length during the
motion may be left out.
3. If a system contains more than one cord, relate the
position of a point on one cord to a point on another
cord. Separate equations are written for each cord.
4. Differentiate the position coordinate equation(s) to relate
velocities and accelerations. Keep track of signs!
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EXAMPLE
Given: In the figure on the left, the cord
at A is pulled down with a speed
of 3 m/s.
Find: The speed of block D.
Plan: There is only one cord involved in
the motion, so only one
position/length equation is required.
Define position coordinates for block
D and cable lengths that change,
write the position relation and then
differentiate it to find the relationship
between the two velocities.
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EXAMPLE (continued)
Solution:
1) A datum line can be drawn through the upper, fixed pulleys.
Two coordinates must be defined: one for block D (sD) and
one for the changing cable length (sA).
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EXAMPLE (continued)
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GROUP PROBLEM SOLVING I
Given: In the figure on the left, the
cord at A is pulled down with
a speed of 2 m/s.
Find: The speed of block B.
Plan: There are two cords involved
in the motion in this example.
There will be two position
equations (one for each cord).
Write these two equations,
combine them, and then
differentiate them.
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GROUP PROBLEM SOLVING I (continued)
Solution:
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GROUP PROBLEM SOLVING I (continued)
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GROUP PROBLEM SOLVING II
Given: In this pulley system, block A is
moving downward with a speed
of 6 m/s while block C is
moving down at 18 m/s.
Find: The speed of block B.
Plan:
All blocks are connected to a single
cable, so only one position/length
equation will be required. Define position
coordinates for each block, write out the
position relation, and then differentiate it
to relate the velocities.
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112
GROUP PROBLEM SOLVING II (continued)
Solution:
1) A datum line can be drawn through the upper, fixed, pulleys
and position coordinates defined from this line to each block
(or the pulley above the block).
The velocity of block B is 21 m/s up.
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RELATIVE-MOTION ANALYSIS OF TWO PARTICLES
USING TRANSLATING AXES
Today’s Objectives:
Students will be able to:
1. Understand translating frames of reference.
2. Use translating frames of reference to analyze relative motion.
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114
APPLICATIONS
When you try to hit a
moving object, the position,
velocity, and acceleration of
the object all have to be
accounted for by your mind.
You are smarter than you
thought!
Here, the boy on the ground is at d = 10 m when the girl in
the window throws the ball to him.
If the boy on the ground is running at a constant speed of 4 m/s,
how fast should the ball be thrown?
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115
RELATIVE POSITION (Section 12.10)
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116
RELATIVE VELOCITY
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117
RELATIVE ACCELERATION
The time derivative of the relative
velocity equation yields a similar
vector relationship between the
absolute and relative accelerations
of particles A and B.
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118
SOLVING PROBLEMS
Since the relative motion equations are vector equations,
problems involving them may be solved in one of two ways.
For instance, the velocity vectors in vB = vA + vB/A could be
written as two dimensional (2-D) Cartesian vectors and the
resulting 2-D scalar component equations solved for up to
two unknowns.
Alternatively, vector problems can be solved “graphically” by
use of trigonometry. This approach usually makes use of the
law of sines or the law of cosines.
Could a CAD system be used to solve these types of problems?
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119
LAWS OF SINES AND COSINES
Since vector addition or subtraction forms
a triangle, sine and cosine laws can be
applied to solve for relative or absolute
velocities and accelerations. As a review,
their formulations are provided below.
C
b
a
B
A
c
Law of Sines:
a
sin A
Law of Cosines:
b
=
sin B
c
=
sin C
a 2 = b 2 + c 2 − 2 bc cos A
b = a + c
2
2
− 2 ac cos B
2
c = a + b − 2 ab cos C
2
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2
2
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120
EXAMPLE
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121
EXAMPLE (continued)
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122
EXAMPLE (continued)
o
VB
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123
GROUP PROBLEM SOLVING
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124
GROUP PROBLEM SOLVING (continued)
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125
GROUP PROBLEM SOLVING (continued)
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126
Reference
Slides are adapted from:
R.C. Hibbeler, "Engineering Mechanics - Dynamics", 14th S.I.
Edition, Prentice Hall, 2017.
Dynamics, Fourteenth Edition in SI Units
Instructor: Low KO
R.C. Hibbeler
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Trimester 2, 2020/2021