Mechanics: the study of how bodies react to forces acting on them
1) Particles, Rigid Bodies
Statics: the study
of bodies in
equilibrium
2) Deformable Bodies
Dynamics:
1. Kinematics – concerned
with the geometric aspects of
motion
2. Kinetics - concerned with
the forces causing the motion
Mechanics: the study of how bodies react to forces acting on them
1) Particles, Rigid Bodies
2) Deformable Bodies
Continuum Mechanics:
-Elasticity
-Fluid Mechanics
1
Mechanics: the study of how bodies react to forces acting on them
1) Particles, Rigid Bodies
Gives rise to Ordinary
Differential Equations
2) Deformable Bodies
Gives rise to Partial
Differential Equations
Solution of a mechanics problem usually
requires solving a differential equation.
This course is about class 1: particles and rigid bodies
1- INTRODUCTION TO DYNAMICS
APPLICATIONS OF DYNAMICS: Particles
The motion of large objects,
such as rockets, airplanes, or
cars, can often be analyzed
as if they were particles.
Why?
Typical question:
Can we predict the altitude,
velocity and acceleration of
this rocket as a function of
time?
2
APPLICATIONS (continued)
A train travels along a straight length of track.
Can we treat the train as a particle?
Typical Question:
If the train accelerates at a constant rate, how can we
determine its position and velocity at some instant?
APPLICATIONS (continued)
What about a baseball?
What is the correct model: particle, rigid body or
deformable body?
Depends:
• A particle model will describe the approximate
displacement when thrown or hit.
• A rigid body model will describe the effect of spin.
• A deformable body model will describe the transfer
of energy when hit by a bat.
3
Goals of This Course:
• Develop modeling skills for realistic problems
• Apply a logical, mathematical framework to realistic
problems
• Understand the role of force in motion
2- KINEMATICS OF PARTICLES
2/1 - 2/2- INTRODUCTION & RECTILINEAR MOTION
Objectives:
Introduce the kinematic quantities
(position, displacement, velocity, and acceleration)
of a particle traveling along a straight path.
4
POSITION AND DISPLACEMENT
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) or feet (ft).
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.
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 or ft/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
5
ACCELERATION
Acceleration is the rate of change in the velocity of a particle. It is a
vector quantity. Typical units are m/s2 or ft/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).
From the chain rule, the derivative equations for velocity and
acceleration can be manipulated to get
a ds = v dv
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
∫ ds = ∫ v dt
t
vo
o
vo
so
so
o
• Note that so and vo represent the initial position and
velocity of the particle at time t = 0.
6
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. These equations are:
v
t
∫ dv = ∫ a dt
vo
o
s
t
∫ ds = ∫ v dt
so
v
v = vo + a t
yields
s = s o + v ot + (1/2) a t 2
yields
v2 = (vo )2 + 2a (s - so)
o
s
∫ v dv = ∫ a
vo
yields
ds
so
EXAMPLE
Given: A motorcyclist travels along a straight road at a speed
of 27 m/s. When the brakes are applied, the
motorcycle decelerates at a rate of -6t m/s2.
Find: The distance the motorcycle travels before it stops.
Plan: Establish the positive coordinate s in the direction the
motorcycle is traveling. Since the acceleration is given
as a function of time, integrate it once to calculate the
velocity and again to calculate the position. Use the
initial conditions to get final answer.
7
EXAMPLE (continued)
Solution:
1) Integrate acceleration to determine the vvelocity.
t
=
a = dv / dt => dv = a dt => ∫ dv ∫ (−6t )dt
vo
o
=> v – vo = -3t2 => v = -3t2 + vo
2) We can now determine the amount of time required for
the motorcycle to stop (v = 0). Use vo = 27 m/s (why?).
0 = -3t2 + 27 => t = 3 s
3) Now calculate the distance traveled in 3s by integrating the
velocity using so = 0:
s
t
=
v = ds / dt => ds = v dt => ∫ ds ∫ ( −3t 2 + vo) dt
so
o
=> s – so = -t3 + vot
=> s – 0 = -(3)3 + (27)(3) => s = 54 m
Assignment : Introductory Problem 2/4
or (2/3 in 6th edition)
P.S.:The problem numbers, page numbers, etc., refer to the 7th
edition (SI units) of Meriam and Kraige (M&K), unless
indicated otherwise.
8
Last Class:
1. Reviewed vectors.
This week:
1. Discuss kinematics of
curvilinear motion (3-D) of
particles.
2. Discussed kinematics of
rectilinear motion of
particles (position,
velocity, acceleration).
2. See that there are different, but
equivalent, ways of describing
the kinematics.
3. Introduced special case of
constant acceleration.
3. Look at examples of the
different ways.
Week # 2: Lecture hour # 4-6
2/3- PLANE CURVILINEAR MOTION
2/4- RECTANGULAR COORDINATES & MOTION OF A PROJECTILE
2/5- NORMAL AND TANGENTIAL COMPONENTS (n-t)
Multi-dimensional kinematics
3 ways to keep track of a particle motion:
•
Fixed axes (Cartesian coordinates). Measure position from
a fixed origin.
•
Moving axes (path variables). Axes move with the particle.
•
Combination of the above.
The choice of method depends on the particular problem.
1
CURVILINEAR MOTION: RECTANGULAR COMPONENTS
(Sections 2/3 )
We first study a special case of fixed axes:
Cartesian coordinates
Later we will see moving axes.
APPLICATIONS of FIXED AXES
The path of motion of each airplane in
this formation can be tracked with
radar and their x, y, and z coordinates
(relative to a point on earth) recorded
as a function of time.
Typical Problem:
How can we determine the velocity
or acceleration of each plane at any
instant?
And should they be the same for each
aircraft?
2
APPLICATIONS (continued)
A roller coaster car travels down
a fixed, helical path at a constant
speed.
Typical Problem:
How can we determine its
position or acceleration at any
instant?
And if you are designing the track, why is it important to
be able to predict the acceleration of the car?
REVIEW: POSITION AND DISPLACEMENT
A particle moving along a curved path undergoes curvilinear motion.
Since the motion is often three-dimensional, vectors are commonly
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
3
REVIEW: 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. Why?
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!
REVIEW: 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
Note that the acceleration vector is not, in general,
tangent to the path function.
4
2/4- RECTANGULAR COMPONENTS: POSITION
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 position vector is: r = (x2 + y2 + z2)1/2
The direction of r is defined by the unit vector: er = (1/r)r
RECTANGULAR COMPONENTS: VELOCITY
The velocity vector is the time derivative of the position vector:
v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt
The unit vectors i, j, k are constant in magnitude and direction,
so this equation reduces to v = vxi + vyj + vzk
•
•
•
where vx = x = dx/dt, vy = y = dy/dt, vz = z = dz/dt
The magnitude of the velocity
vector is
v = [(vx)2 + (vy)2 +(vz)2]1/2
The direction of v is tangent to
the path of motion.
5
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 = axi + ay j + azk
where
•
•
•
••
••
ax = vx = x = dvx /dt, ay = vy = y = dvy /dt,
••
az = vz = z = dvz /dt
The magnitude of the acceleration vector is
a = [(ax)2 + (ay)2 + (az)2 ]1/2
The direction of a is usually
not tangent to the path of the
particle.
EXAMPLE
Given:The motion of two particles (A and B) is described by
the position vectors
rA = [3t i + 9t(2 – t) j] m
rB = [3(t2 –2t +2) i + 3(t – 2) j] m
Find: The point at which the particles collide and their
speeds just before the collision.
Plan: 1) The particles will collide when their position
vectors are equal, or rA = rB .
2) Their speeds can be determined by differentiating
the position vectors.
6
EXAMPLE (continued)
Solution:
1) The point of collision requires that rA = rB, so
xA = xB and yA = yB .
x-components: 3t = 3(t2 – 2t + 2)
Simplifying: t2 – 3t + 2 = 0
Solving:
t = {3 ± [32 – 4(1)(2)]1/2}/2(1)
=> t = 2 or 1 s
Which one?
y-components: 9t(2 – t) = 3(t – 2)
Simplifying:
3t2 – 5t – 2 = 0
Solving: t = {5 ± [52 – 4(3)(–2)]1/2}/2(3)
=> t = 2 or – 1/3 s
So, the particles collide when t = 2 s. Substituting this
value into rA and rB yields
xA = xB = 6 m
and yA = yB = 0,
so rA=6i and rB=?
EXAMPLE (continued)
2) Differentiate rA and rB to get the velocity vectors.
.
.
vA = drA/dt = x. A i + yA j = [3i + (18 – 18t)j] m/s
At t = 2 s: vA = [3i – 18j] m/s
•
•
vB = drB/dt = xBi + yBj = [(6t – 6)i + 3j] m/s
At t = 2 s: vB = [6i + 3j] m/s
(why not equal?)
Speed is the magnitude of the velocity vector.
vA = (32 + 182) 1/2 = 18.2 m/s
vB = (62 + 32) 1/2 = 6.71 m/s
7
MOTION OF A PROJECTILE
An application of fixed Cartesian Coordinates
APPLICATION OF CARTESIAN COORDINATES:
PROJECTILE MOTION
Projectile motion can be treated as two rectilinear motions,
• one in the horizontal direction with zero acceleration
• the other in the vertical direction with constant acceleration
(i.e., gravity).
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. Both balls are subjected to
the same downward acceleration since they
remain at the same elevation at any instant.
The horizontal distance between successive
photos of the yellow ball is constant since the
velocity in the horizontal direction is constant.
8
KINEMATIC EQUATIONS: HORIZONTAL MOTION
Since ax = 0, the velocity in the horizontal direction remains
constant (vx = vox) so the position in the x direction is
determined by:
x = xo + (vox)(t)
Why is ax equal to zero (assuming movement through the air)?
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 gt2
• vy2 = voy2 – 2g(y – yo)
For any given problem, only two of these three
equations can be used. Why?
9
Example 1
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
vy = vo sin θ
and
We can write: x = (vo cos θ)t or
t =
x
vo cos θ
y = (vo sin θ)t – ½ (g)t2
By substituting for t:
y = (vo sin θ)
(
x
vo cos θ
) ( )(
–
g
2
2
x
vo cos θ
)
Example 1 (continued):
Simplifying the last equation, we get:
y = (x tanθ) –
(
g x2
2vo2
)
(1 + tan2θ)
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.
Why?
10
GROUP PROBLEM SOLVING
Given: Skier leaves the ramp
at θA = 25o and hits
the slope at B.
Find: The skier’s initial speed
vA.
Plan: ???
GROUP PROBLEM SOLVING
Given: Skier leaves the ramp
at θA = 25o and hits
the slope at B.
Find: The skier’s initial speed
vA.
Plan: Establish a fixed x,y coordinate system (in the
solution here, the origin of the coordinate system is
placed at A). Apply the kinematic relations in x and
y-directions.
11
GROUP PROBLEM SOLVING (continued)
Solution:
Motion in x-direction:
xB = xA + (vox)(tAB)
Using
tAB=
(4/5)100
vA (cos 25)
=
88.27
vA
Motion in y-direction:
Using
yB = yA + (voy)(tAB) – ½ (g)(tAB)2
-64 = 0 + vA(sin 25)
80
– ½ (9.81)
vA (cos 25)
88.27
vA
2
vA = 19.42 m/s
CURVILINEAR MOTION:
NORMAL AND TANGENTIAL COMPONENTS (n-t)
(Section 2/5)
Moving axes (path variables).
12
NORMAL AND TANGENTIAL COMPONENTS
When a particle moves along a curved path, it is sometimes
convenient to use 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).
NORMAL AND TANGENTIAL COMPONENTS (continued)
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.
The positive n and t directions are defined by the unit vectors en
and et, respectively.
The radius of curvature, ρ, is 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.
13
NORMAL AND TANGENTIAL
COMPONENTS (continued)
The velocity vector is always
tangent to the path of motion
(t-direction).
VELOCITY IN THE n-t COORDINATE SYSTEM
The magnitude is determined by taking the time derivative of
the path function, s(t).
.
v = vet where v = s = ds/dt
Here v defines the magnitude of the velocity (speed) and
et defines the direction of the velocity vector.
Acceleration is the time rate of change of velocity:
.
.
a = dv/dt = d(vet)/dt = vet + vet
14
ACCELERATION IN THE n-t COORDINATE SYSTEM
.
Here v represents the change in
.
the magnitude of velocity and et
represents the rate of change in
the direction of et.
After mathematical manipulation,
the acceleration vector can be
expressed as:
.
2
a = vet + (v /ρ)en = atet + anen
ACCELERATION IN THE n-t COORDINATE SYSTEM
(continued)
There are two components to the
acceleration vector:
a = at et + an en
• The tangential component is tangent to the curve and in the
direction of increasing or decreasing velocity.
.
at = v or at ds = v dv (using chain rule!)
• The normal or centripetal component is always directed
toward the center of curvature of the curve. an = v2/ρ
• The magnitude of the acceleration vector is
a = [(at)2 + (an)2]1/2
15
Circular Motion
EXAMPLE PROBLEM
Given: Starting from rest, a motorboat
travels around a circular path of
ρ = 50 m at a speed that
increases with time,
v = (0.2 t2) m/s.
Find: The magnitudes of the boat’s
velocity and acceleration at
the instant t = 3 s.
Plan: The boat starts from rest (v = 0 when t = 0).
1) Calculate the velocity at t = 3s using v(t).
2) Calculate the tangential and normal components of
acceleration and then the magnitude of the
acceleration vector.
16
EXAMPLE (continued)
Solution:
1) The velocity vector is v = v et , where the magnitude is given
by v = (0.2t2) m/s. At t = 3s:
v = 0.2t2 = 0.2(3)2 = 1.8 m/s
.
2) The acceleration vector is a = atet + anen = vet+(v2/ρ)en.
.
Tangential component: at = v = d(.2t2)/dt = 0.4t m/s2
At t = 3s: at = 0.4t = 0.4(3) = 1.2 m/s2
Normal component: an = v2/ρ = (0.2t2)2/(ρ) m/s2
At t = 3s: an = [(0.2)(32)]2/(50) = 0.0648 m/s2
The magnitude of the acceleration is
a = [(at)2 + (an)2]1/2 = [(1.2)2 + (0.0648)2]1/2
= 1.20 m/s2
Assignment : Introductory Problem
(in 7th edition)
or
(in 6th edition)
2/60
(2/61)
2/77
(2/83)
2/104
(2/129)
17
Week # 3: Lecture hour # 7-8
2. Kinematics of Particles
Curvilinear Motion:
2/6- Polar Components (r - θ)
2/7 Space Curvilinear Motion
2/8 Relative Motion (Translating Axes)
CURVILINEAR MOTION:
2/6- POLAR COMPONENTS (r - θ)
APPLICATIONS
Cylindrical coordinates are
commonly used in cases
where the particle moves
along a helical curve.
z const: Polar coordinates
Typical Problem:
The boy slides down the
slide at a constant speed of 2
m/s. How fast is his
elevation from the ground
.
changing (i.e., what is z )?
1
POSITION (POLAR COORDINATES)
The location of A in polar coordinates is r
= rer.
The radial direction, r, extends outward from the fixed origin, O,
and the transverse coordinate, θ, is measured counter-clockwise
from the horizontal.
VELOCITY (POLAR COORDINATES)
The instantaneous velocity is defined as:
v = dr/dt = d(rer)/dt
de
.
v = re + r r
r
dt
Using the chain rule:
der/dt = (der/dθ)(dθ/dt)
der/dθ = eθ so der/dt =. (dθ/dt) eθ
.
Therefore: v = rer + rθeθ
.
Thus, the velocity vector has two components:
r,
.
called the radial component, and rθ, 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 θ )2 + ( r )2
2
ACCELERATION (POLAR COORDINATES)
The instantaneous acceleration is defined as:
.
.
a = dv/dt = (d/dt)(rer + r θeθ)
But: deθ/dt= - dθ/dt ur.
So the acceleration can be expressed as
..
..
.. .
a = (r – rθ2)er + (r θ + 2rθ)eθ
.. .
The term (r – rθ2) is the radial acceleration
or ar.
..
..
The term (rθ + 2rθ) is the transverse
acceleration or aθ
..
..
.. .
The magnitude of acceleration is a = (r – rθ2)2 + (rθ + 2rθ)2
2/7- SPACE CURVILINEAR MOTION
CYLINDRICAL COORDINATES (r- θ−z)
In the figure: ur=er , uθ=eθ , uz = k
If the particle P moves along a space
curve, its position can be written as
rP = R = rer + zez
Velocity:
Taking time derivatives and using
the chain rule (ez constant):
.
.
.
vP = rer + rθeθ + z ez
..
..
.. .
..
Acceleration: aP = (r – rθ2)er + (rθ + 2rθ)eθ + z ez
3
EXAMPLE
Given: r. = 5 cos(2θ) [m]
θ = 3t2 [rad/s]
θo = 0
Find: Velocity and acceleration at θ = 30°.
.
..
Plan: Apply chain rule to determine r and r
and evaluate at θ = 30°.
t
t
.
θ = ∫ θ dt = ∫ 3t2 dt = t3
Solution:
to= 0
At θ = 30°,
0
π
θ = = t3. Therefore: t = 0.806 s.
6
.
θ = 3t2 = 3(0.806)2 = 1.95 rad/s
EXAMPLE (continued)
..
θ = 6t = 6(0.806) = 4.836 rad/s2
r = 5 cos(2θ) = 5 cos(60) = 2.5m
.
.
r = -10 sin(2θ)θ = -10 sin(60)(1.95) = -16.88 m/s
.
..
..
r = -20 cos(2θ)θ2 – 10 sin(2θ)θ
= -20 cos(60)(1.95)2 – 10 sin(60)(4.836) = -80 m/s2
Substitute in the equation
for velocity
.
.
v = rer + rθeθ
v = -16.88er + 2.5(1.95)eθ
v =
(16.88)2 + (4.87)2 = 17.57 m/s
4
EXAMPLE (continued)
Substitute in the equation for acceleration:
..
..
.. .
a = (r – rθ2)er + (rθ + 2rθ)eθ
a = [-80 – 2.5(1.95)2]er + [2.5(4.836) + 2(-16.88)(1.95)]eθ
a = -89.5er – 53.7eθ m/s2
a =
(89.5)2 + (53.7)2 = 104.4 m/s2
2/8- RELATIVE MOTION (TRANSLATING AXES)
rA = rB + rA/B
A/B means “A relative to B” or
“A with respect to B”
Differentiation position vector with respect to time gives
OR
vA = vB + vA/B
Acceleration
aA = aB + aA/B
5
Sample Problem 2/14:
Car A is accelerating in the direction of its motion at the rate of
1.2 m/s2. Car B is rounding a curve of 150m radius at a constant
speed of 54 km/h. Determine the velocity and acceleration which
car B appears to have to an observer in car A if car A has reached
a speed of 72 km/h for the positions represented.
Solution: choose non-rotating
reference axes attached car to A
since the motion of B with respect
to A is considered.
54 km/h = 15 m/s
72 km/h = 20 m/s
Velocity Using law of sine or cosine or using
vector algebra
VB/A = 18.03 and Ɵ= 46.1
Acceleration a A is given (1.2 m/s2)
aB is constant an = v2 /p
= 152 /150 = 1.5 m/s2
(a B/A)x = 1.5cos30 – 1.2 = 0.099 m/s2
(a B/A)y = 1.5sin30 = 0.750 m/s2
a B/A = (0.0992 + 0.752) 1/2 = 0.757 m/s2
6
Using sine rule :
β = 97.5
Assignment : Introductory Problem
2/131
or
(2/137 in 6th Edition)
2/183
or
(2/193 in 6th Edition)
7