AMME2500
MODULE 1 – DYNAMICS
OF POINT-MASS SYSTEMS
Matthew Cleary
School of Aerospace, Mechanical and Mechatronic
Engineering, University of Sydney
Outline
Module 1 deals with systems which can be treated like “particles” which
have their mass concentrated at a point
Part 1: Kinematics
- position, velocity, acceleration
- various coordinate systems
- relative motion analysis
- connected/constrained particles
Part 2: Kinetics
- Newtons laws
- Work, energy, impulse and momentum
- Impact
- Angular momentum
- Multiple particle systems
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Part 1: Kinematics
Reading: Mariam, Kraige and Bolton Chapter 2
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Kinematics
• Dynamics: the principles governing the state of motion or
rest of bodies under the influence of applied force and
torque
• Kinematics: the “geometry” of motion:
the relationships between position,
velocity, acceleration and rotation and
bodies joined by linkages and other
constraints
• Kinetics: the relationship between motion and it’s cause due to
force, torque, mass and inertia
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Learning Outcomes
This section covers the basic methods for describing
particle motion. The concepts developed later form the
basis for much of dynamics
1. Types of motion
1.
2.
Rectilinear Motion (1D)
Plane curvilinear Motion (2D)
Coordinate Systems
2.
1.
2.
3.
Rectangular/Cartesian coordinates
Normal and tangent coordinates (n-t)
Polar coordinates (r-q)
Absolute motion & Relative motion
4. Application of Vectors and their Time derivatives
5. Constrained Motion
3.
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Learning Outcomes
• Additional practice problems are available in Canvas.
• These problems cover each learning outcome broadly.
• Attempt to solve these problems as you progress through
the lecture slides.
• Use these practice problems to test your understanding.
• If you need help with the practice problems, ask your
tutors for assistance.
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Practice Problem 1
• A single-stage rocket is launched vertically from rest, and
its thrust is programmed to give the rocket a constant
upward acceleration of 6 m/s. If the fuel is exhausted 20 s
after launch, calculate the maximum vm velocity and the
subsequent maximum altitude h reached by the rocket.
Ans: vm = 120m/s, h =1.934 km
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Practice Problem 2
• Ball bearings leave the horizontal trough with a velocity of
magnitude u and fall through the 70-mm diameter hole as
shown. Calculate the permissible range of u which will
enable the balls to enter the hole. Take the dashed
positions to represent the limiting conditions.
Ans: 0.744 < u < 1.135 m/s
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Practice Problem 3
• Race car A follows path a-a
while race car B follows path
b-b on the unbanked track. If
each car has a constant
speed limited to that
corresponding to a lateral
(normal) acceleration of
0.8g, determine the times tA
and tB for both cars to
negotiate the turn as
delimited by the line C-C
Ans: tA= 10.52 s, tB= 10.86 s
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Practice Problem 4
• At a certain instant after jumping from the
airplane A, a skydiver B is in the position
shown and has reached a terminal
(constant) speed vB = 50 m/s. The
airplane has the same constant speed vA
= 50 m/s and after a period of level flight
is just beginning to follow the circular
path shown of radius 𝜌A = 2000 m.
a)
b)
Determine the velocity and acceleration
of the airplane relative to the skydiver.
Determine the time rate of change of the
speed vr of the airplane and the radius
of curvature 𝜌r of its path, both as
observed by the nonrotating skydiver.
Ans: a) 𝑣!/# = 50 𝚤̂ + 50 𝚥̂ m/s, 𝑎!/# = 1.25 𝚥̂ m/s2
b) 𝑣̇$ = 0.884 m/s, 𝜌$ = 5660 𝑚
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Practice Problem 5
• Neglect the diameters of the small pulleys and establish
the relationship between the velocity of A and the velocity
of B for a given value of y.
Ans: 𝑣# = −
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%& (!
) & "*+ "
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Rectilinear Motion: Position
• 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 𝑟⃑ or 𝒓, or the
scalar s. Scalar s can be positive or negative.
• 𝑟⃑ and s have dimensions of length, with SI unit of metre (m).
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Rectilinear Motion: Position
• The displacement of the particle is defined as its change in
position.
• The total distance travelled by the particle, sT, is a positive
scalar that represents the total length of the path over which
the particle travels.
Vector form: D r = r’ - r
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Scalar form: D s = s’ - s
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Rectilinear Motion: Velocity
Velocity: 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 Dt is:
vavg = Dr / Dt
The instantaneous velocity is the
time-derivative of position (limit as ∆𝑡 → 0):
Speed is the magnitude of velocity:
v = ds / dt = | v |
v = dr / dt
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Rectilinear Motion: Velocity
Velocity: 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.
Average speed is the total distance traveled divided by elapsed time:
(vsp)avg = sT / Dt
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Rectilinear Motion: Acceleration
Acceleration is the rate of change in the velocity of a particle. It is a
vector quantity. 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).
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Example Problem 1.1
The position of a particle
is described by:
s(t) = 2t3 – 24t + 6
Find
(a) time required to reach
v = 72 m/s
(b) acceleration at v = 30
m/s
(c) net displacement from
t = 1 to t = 4s
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Example Problem 1.1
The position of a particle
is described by:
s(t) = 2t3 – 24t + 6
Find
(a) time required to reach
v = 72 m/s
(b) acceleration at v = 30
m/s
(c) net displacement from
t = 1 to t = 4s
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(a) v = ds/dt, v = 6t2 – 24 m/s
72 = 6t2 – 24, t = ± 4 = 4 s
(b) 30 = 6t2 – 24, t = 3 s
a = dv/dt, a = 12t m/s2
a(t=3s) = 36 m/s2
(c) Δs = s(t=4) – s(t=1)
= 38 – (-16) = 54 m
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Solving Position and Velocity using Integration
• Integrate acceleration for velocity and position
a = dv/dt
v = ds/dt
Where so and vo represent the initial position and velocity of the particle
at t = 0.
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Solving Position and Velocity using Integration
• Integrate acceleration for velocity and position
a = dv/dt
v = ds/dt
dt = ds/v,
dt = dv/a,
Where so and vo represent the initial position and velocity of the particle
at t = 0.
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Solving Position and Velocity using Integration
• Integrate acceleration for velocity and position
a = dv/dt
v = ds/dt
dt = ds/v,
dt = dv/a,
v dv = a ds
Where so and vo represent the initial position and velocity of the particle
at t = 0.
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Solving Position and Velocity using Integration
• Integrate acceleration for velocity and position
a = dv/dt
v = ds/dt
dt = ds/v,
dt = dv/a,
v dv = a ds
Where so and vo represent the initial position and velocity of the particle
at t = 0.
Be careful, it may be necessary to separate variables, see Ex. 1.1.3
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Example Problem 1.2
The velocity of a particle is described by:
v(t) = 3t2 + 2t
Find the position at t = 3 s.
Assume that the position at t = 0 is s = 0.
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Example Problem 1.3
• A boat is moving at 8 m/s when its engines are stopped.
If it takes 10 minutes for the boat to slow down to 4 m/s,
and the deceleration of the boat is a = -kv2, where k is
constant, determine an equation for the velocity of the
boat as a function of time and the value of k
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Example Problem 1.4
• A metallic particle is subjected to the influence of a
magnetic field between A and B, such that its
acceleration is a = (4s) m/s2. If the particle is
released from rest at point A (s = 0.1 m), determine
the velocity of the particle when it reaches point B:
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Example Problem 1.4
Firstly, which integral relationship would be most useful to solve
this problem?
(A)
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(B)
(C)
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Solving Position and Velocity using Integration
• General approach to solving problems:
Acceleration given as a function of time
2. Acceleration given as a function of position
3. Acceleration given as a function of velocity
1.
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Solving Position and Velocity using Integration
• General approach to solving problems:
Acceleration given as a function of time
2. Acceleration given as a function of position
3. Acceleration given as a function of velocity
1.
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Solving Position and Velocity using Integration
• General approach to solving problems:
Acceleration given as a function of time
2. Acceleration given as a function of position
3. Acceleration given as a function of velocity
1.
Solve using separation of variables
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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
ò dv = ò acdt
yields
v = vo + ac t
ò ds = ò v dt
yields
s = so + v o t + (1/2) ac t 2
ò v dv = ò acds
yields
v2 = (vo) + 2ac(s - so)
vo
o
s
t
so
v
vo
o
s
2
so
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Curvilinear Motion
A particle moving along a curved path undergoes curvilinear motion.
Vectors are 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 Ds along the
curve during time interval Dt, the
displacement is determined by vector
subtraction: D r = r’ - r
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Curvilinear Motion: Velocity
Velocity represents the rate of change in the position of
a particle.
The average velocity of the
particle during the time increment
Dt is vavg = Dr/Dt
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 Ds approaches the magnitude of Dr as Dt→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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Curvilinear Motion: 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 Dt, the average
acceleration during that increment is:
aavg = Dv/Dt = (v’ - v)/Dt
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.
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Rectangular Coordinates
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)0.5
The direction of r is defined by the unit vector: ur = (1/r)r
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Rectangular Coordinates: Velocity
The velocity vector is the time derivative of the position vector:
v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/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
The magnitude of the
velocity vector is
v = [(vx)2 + (vy)2 + (vz)2]0.5
The direction of v is tangent
to the path of motion.
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Rectangular Coordinates: 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
• = •• = dv /dt, a = • = •• = dv /dt,
ax = v
x
x
y
y
vy y
x
az = v• = ••
z = dvz /dt
z
The magnitude of the acceleration vector is: a = [(ax)2 + (ay)2 + (az)2 ]0.5
The direction of a is usually not
tangent to the path of the particle.
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Example 1.6
A cannonball is fired with initial
velocity v at an angle θ from the
horizon. Determine h and s
(maximum height and horizontal
distance).
Assume constant gravitational
acceleration (y-direction), neglecting
air resistance and neglecting
curvature and rotation of the Earth.
Hint: neglection of air resistance
means that motion in each
coordinate is uncoupled
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Curvilinear Motion: Rectangular Coordinate Summary
The position of the particle can
be defined at any instant by the
position vector:
r=xi+yj+zk
The velocity vector is the time derivative of the position vector:
v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt
v = vx i + vy j + vz k
The acceleration vector is the time derivative of the velocity
vector:
a = dv/dt = d2r/dt2 = ax i + ay j + az k
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Normal and Tangential (n-t) Coordinates
• When a particle moves along a curved path, it is sometimes
convenient to describe its motion using coordinates that are
relative to the motion
• When the path of motion is known, normal (n) and tangential
(t) coordinates are often used.
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Normal and Tangent (n-t) Coordinates
• In the n-t coordinate system, the origin is located on the path
• 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 centre of curvature of the curve.
• Unlike rectangular coordinates, these axes rotate as the object
moves
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Normal and Tangent (n-t) Coordinates
The positive n and t directions are defined
by the unit vectors un and ut, respectively.
The centre 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 centre 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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Normal and Tangent (n-t) Coordinates
If 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:
[ 1 + (dy/dx)2 ]3/2
r = ________________
d2y/dx 2
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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 is the unit vector defining the direction of the velocity vector.
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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.
Note:
In rectangular coordinates, the
unit vectors are constant and
their rates of change are zero.
In n-t coordinates, the
coordinate directions vary with
the motion of the object.
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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.
dut/dt = (dθ/dt)(dut/dθ)
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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.
dut/dt = (dθ/dt)(dut/dθ)
(in the limit that dθ -> 0)
dut =undθ
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dut/dθ =un
.
.
ut = θun
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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.
(in the limit)
dut =undθ
ds =r dθ
.
.
ut = θun
dut/dθ =un
.
ds/dt = v = rθ
.
vut = v(v/r)un
= (v2/r) un
.
note that term involving r is
unnecessary as higher order terms
eliminate it
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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.
dut =undθ
ds =r dθ
dut/dθ =un
.
ds/dt = v = rθ
.
.
ut = θun
.
vut = v(v/r)un
= (v2/r) un
.
a = v ut + (v2/r) un = at ut + an un
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Acceleration in the n-t coordinate system
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 centre of curvature of the curve: an = v2/r
• The magnitude of the acceleration vector is
a = [(at)2 + (an)2]1/2
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Special cases of motion in n-t coordinates
1) The particle moves along a straight line.
.
r
¥ => an = v2/r = 0 => 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/r
The normal component represents the time rate of change in the
direction of the velocity.
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Summary: n-t coordinate system
Origin is located on the path, axes move
with the particle, t-axis is tangent to the
path (curve) and n-axis is perpendicular to
the t-axis
Velocity: The magnitude is
determined by taking the time
derivative of the path function,
s(t).
v = v ut
.
where v = s = ds/dt
Acceleration is the time rate
of change of velocity:
.
.
a = dv/dt = d(vut)/dt = vut + vut
.
a = v ut + (v2/r) un = at ut + an un
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n-t coordinates: Example 1.7
• A car accelerates down a ramp which has a
shape dictated by the function: 𝑦 =
,
)-
𝑥)
• If the car accelerates along the slope at a = 2
m/s2, and at an instant when its x-coordinate is
10 m from the origin and it has a speed of 6 m/s,
calculate the total magnitude of acceleration
experienced by the driver
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Polar (Cylindrical) Coordinates
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.
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Polar Coordinates: Velocity
The instantaneous velocity is defined as:
v = dr/dt = d(rur)/dt
.
.
v = rur + rur
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
Proof: text book Section 2/6
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Polar Coordinates: Velocity
.
.
v = rur + rquθ
.
Velocity has a radial component:
vr = r
and a transverse component (θ-component):
vθ = rθ
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.
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Polar Coordinates: Acceleration
The instantaneous acceleration is defined as:
.
.
a = dv/dt = (d/dt)(rur + rquθ)
..
..
..
..
..
= rur + rur+ rquθ + rquθ + rquθ
Using the chain rule again:
duθ/dt = (duθ/dq)(dq/dt)
.
We can prove that duθ/dq = -ur so duθ/dt = -qur
Proof: text book Section 2/6
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Polar Coordinates: Acceleration
After manipulation, the acceleration can be
expressed as:
.
..
..
. .
a = (r – rq 2)ur + (rq + 2rq)uθ
.
..
ar = (r – rq 2) is the radial acceleration
..
..
aq = (rq + 2rq) is the transverse acceleration
The magnitude of acceleration is a =
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..
.
(r – rq
..
..
q + 2rq)2
2)2 + (r
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Polar Coordinates: Example 1.8
• A machinery component is
composed of a radially-slotted
arm with a slider B. The angle θ
of the arm and radius r of the
slider are controlled based on the
functions:
✓ = 0.2t + 0.02t3
r = 0.2 + 0.04t2
Find the magnitude of the velocity and acceleration of the slider at t = 3.0 s
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Circular Motion
• Special case of:
• n-t coordinates when ρ is constant
• Polar coordinates when r is constant
v
at
v = r✓˙
ut uθ
ur
un
r
an
ar= -an
v2
an =
= r✓˙2 = v ✓˙
r
O
θ
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at = v̇ = r✓¨
59
Kinematics of Different Coordinate Systems
• So far we have examined:
• Rectilinear Motion: Motion in a straight line, along a single axis
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Kinematics of Different Coordinate Systems
• So far we have examined:
• Rectilinear Motion: Motion in a straight line, along a single axis
• Curvilinear Motion: Motion along a curve, not necessarily straight
• Rectangular Coordinates: axes remain fixed in space
• Normal-Tangential (n-t) coordinates: axes move with the object,
always aligned to the direction of the particle’s velocity
• Polar (cylindrical) coordinates: axes move with the object, always
aligned along the radial vector of the particle’s position
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Kinematics of Different Coordinate Systems
• So far we have examined:
• Rectilinear Motion: Motion in a straight line, along a single axis
• Curvilinear Motion: Motion along a curve, not necessarily straight
• Rectangular Coordinates: axes remain fixed in space
• Normal-Tangential (n-t) coordinates: axes move with the object,
always aligned to the direction of the particle’s velocity
• Polar (cylindrical) coordinates: axes move with the object, always
aligned along the radial vector of the particle’s position
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Kinematics of Different Coordinate Systems
• So far we have examined:
• Rectilinear Motion: Motion in a straight line, along a single axis
• Curvilinear Motion: Motion along a curve, not necessarily straight
• Rectangular Coordinates: axes remain fixed in space
• Normal-Tangential (n-t) coordinates: axes move with the object,
always aligned to the direction of the particle’s velocity
• Polar (cylindrical) coordinates: axes move with the object, always
aligned along the radial vector of the particle’s position
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Kinematics of Different Coordinate Systems
• Certain coordinate systems are more suited to solving
certain problems depending on how the forces that cause
motion act (with respect to fixed space, radially, or along
the path of the particle)
• But any problem can be solved in any coordinate system
• When we start to discuss relative motion (next), we will
introduce ways of transforming between coordinate
systems
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Relative motion between coordinate systems
and frames of reference
• We have considered up to now coordinate systems with origins
fixed in space, or fixed relative to the moving particle
• Some dynamics problems are simplified by using multiple
coordinate systems that move with respect to each other: are in
relative motion
https://en.wikipedia.org/wiki/
Aerial_refueling
• We will begin by considering only relative translation between
frames, without relative rotation
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Relative motion between coordinate systems
and frames of reference
• Consider two particles A and
B which have separate
motions (in-plane)
y
A
Y
j
B
i
x
X
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Relative motion between coordinate systems
and frames of reference
• Consider two particles A and
B which have separate
motions (in-plane)
• The position of A relative to B
is rAB and we sum its position Y
to rB to determine rA with
respect to the absolute
rB
coordinates at O:
rA = rB + rAB
rAB = xi + yj
AMME2500 Engineering Dynamics |
y
rAB
j
B
A
x
i
rA
X
rA and rB are absolute positions relative
to the fixed coordinates X and Y
67
Relative motion between coordinate systems
and frames of reference
y
• If we differentiate this vector
expression w.r.t time, we
obtain the velocity and
acceleration vectors
Y
rAB
j
B
rB
A
x
i
rA
X
ṙA = ṙB + ṙAB
r̈A = r̈B + r̈AB
vA = vB + vAB
aA = aB + aAB
AMME2500 Engineering Dynamics |
68
Relative motion between coordinate systems
and frames of reference
y
• Since the frames translate
but do not rotate, the di/dt,
dj/dt vector terms are zero
Y
rAB
j
B
rB
A
x
i
rA
X
ṙAB = vAB = ẋi + ẏj
AMME2500 Engineering Dynamics |
r̈AB = aAB = ẍi + ÿj
69
Relative motion between coordinate systems
and frames of reference
y
• We can additionally express
the relative position between
A and B using the difference
of the absolute position
vectors
rAB = rA
rB
vAB = vA
vB
aAB = aA
aB
AMME2500 Engineering Dynamics |
Y
rAB
j
B
rB
A
x
i
rA
X
70
Relative motion between coordinate systems
and frames of reference
y
• We can additionally express
the relative position between
A and B using the difference
of the absolute position
vectors
Y
rBA
j
B
rB
x
i
rA
rAB = rA
rB
vAB = vA
vB
X
aAB = aA
aB
rAB =
rBA
• We can determine the
vAB =
vBA
aAB =
aBA
position of B relative to A
AMME2500 Engineering Dynamics |
A
71
Relative Motion: Example 1.9
• Train A travels with a constant speed
along the tracks of vA = 40 m/s. Car
B is travelling at vB = 20 m/s and
begins to decelerate at 3 m/s2 upon
seeing the train. Determine the
relative velocity and acceleration
vectors of the train relative to the car
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Relative motion between coordinate systems
and frames of reference
• It’s important to remember that the relationships here only
hold for non-rotating frames of reference
• We will re-examine relative motion for rotating reference
frames when we start to consider the kinematics of rigid
bodies
y
Y
A
j
B
i
x
X
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73
Vector Components and Coordinate Transformations
V
• So far we have examined
the kinematics of particles
using various coordinate
systems in 2D and 3D
space (rectangular, n-t,
polar)
• Vectors representing
position, velocity and
acceleration of objects can
be represented in any of
these coordinate systems:
they are still the same
vector, just represented
using different components
AMME2500 Engineering Dynamics |
yA
yB
xB
xA
74
Vector Components and Coordinate Transformations
V
• A vector V can be
represented in the
coordinate system A by
expressing the
components and basis
vectors of A
yA
xA
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75
Vector Components and Coordinate Transformations
• A vector V can be
represented in the
coordinate system A by
expressing the
components and basis
vectors of A
V
vyA
yA
xA
vxA
v = vxA uxA + vyA uyA
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76
Vector Components and Coordinate Transformations
V
• A vector V can be
represented in the
coordinate system A by
expressing the
components and basis
vectors of A
• Alternatively, the same
vector may be represented
in the coordinate system B
by expressing the
components and basis
vectors of B
AMME2500 Engineering Dynamics |
yA
yB
xB
xA
77
Vector Components and Coordinate Transformations
V
• A vector V can be
represented in the
coordinate system A by
expressing the
components and basis
vectors of A
• Alternatively, the same
vector may be represented
in the coordinate system B
by expressing the
components and basis
vectors of B
AMME2500 Engineering Dynamics |
yB
xB
78
Vector Components and Coordinate Transformations
V
• A vector V can be
represented in the
coordinate system A by
expressing the
components and basis
vectors of A
• Alternatively, the same
vector may be represented
in the coordinate system B
by expressing the
components and basis
vectors of B
yB
vyB
xB
vxB
v = vxB uxB + vyB uyB
AMME2500 Engineering Dynamics |
79
Vector Components and Coordinate Transformations
• A vector V can be
represented in the coordinate
system A by expressing the
components and basis
vectors of A
• Alternatively, the same vector
may be represented in the
coordinate system B by
expressing the components
and basis vectors of B
• Here we are only interested in
the components of the vector,
so the origin of either
coordinate system is
irrelevant
AMME2500 Engineering Dynamics |
V
vyA
yA
yB
vyB
xB
vxB
xA
vxA
v = vxA uxA + vyA uyA
v = vxB uxB + vyB uyB
80
Coordinate System Transformations in 2D
• The relative orientation
between coordinate systems
A and B (in 2D) can be
specified by a single angle θ
yA
yB
xB
Θ
AMME2500 Engineering Dynamics |
xA
81
Coordinate System Transformations in 2D
• The relative orientation
between coordinate systems
A and B (in 2D) can be
specified by a single angle θ
• We can represent the
components of any vector in
coordinates A as a function of
the components in coordinate
B and θ
AMME2500 Engineering Dynamics |
yA
yB
xB
Θ
xA
82
Coordinate System Transformations in 2D
• The relative orientation
between coordinate systems
A and B (in 2D) can be
specified by a single angle θ
• We can represent the
components of any vector in
coordinates A as a function of
the components in coordinate
B and θ
v = vxA uxA + vyA uyA
AMME2500 Engineering Dynamics |
vyA
yA
yB
xB
Θ
xA
vxA
83
Coordinate System Transformations in 2D
• The relative orientation
between coordinate systems
A and B (in 2D) can be
specified by a single angle θ
• We can represent the
components of any vector in
coordinates A as a function of
the components in coordinate
B and θ
v = vxA uxA + vyA uyA
v = [vxB cos ✓
AMME2500 Engineering Dynamics |
yA
yB
vyB
Θ
xB
Θ
vyBsinθ
vxB
xA
vxBcosθ
vyB sin ✓]uxA + vyA uyA
84
Coordinate System Transformations in 2D
• The relative orientation
between coordinate systems
A and B (in 2D) can be
specified by a single angle θ
• We can represent the
components of any vector in
coordinates A as a function of
the components in coordinate
B and θ
v = vxA uxA + vyA uyA
yA
yB
vyB
xB
Θ
vyBcosθ
vxB
xA
vxBsinθ
v = [vxB cos ✓
vyB sin ✓]uxA + vyA uyA
v = [vxB cos ✓
vyB sin ✓]uxA + [vxB sin ✓ + vyB cos ✓]uyA
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85
Coordinate System Transformations in 2D
• We can now relate the
components of the vector in
one coordinate system to the
components of the vector in
another coordinate system
yA
yB
xB
Θ
vxA = cos ✓vxB
xA
sin ✓vyB
vyA = sin ✓vxB + cos ✓vyB
AMME2500 Engineering Dynamics |
86
Coordinate System Transformations in 2D
• If we rearrange this into a
matrix expression, where
components are represented
as column vectors, then the
matrix is referred to as a
rotation matrix or
transformation matrix
between the two coordinate
systems
AMME2500 Engineering Dynamics |
yA
yB
xB
Θ
xA
87
Coordinate System Transformations in 2D
• The rotation matrix RBA transforms vectors represented in
coordinate system B components into coordinate system
A components
• The inverse of RBA is RAB and is used to transform from a
vector in A to a vector in B coordinates
• Rotation matrices are orthogonal, so the inverse is equal
to the matrix transpose
A
1
A T
RB
=
(R
)
=
(R
A
B
B)
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88
Coordinate Transforms: Example 1.10
• Consider again our train-car
relative motion problem. If we
consider an n-t coordinate
system attached to the car B,
determine the relative velocity
and acceleration in n-t frame
coordinates
AMME2500 Engineering Dynamics |
ut
nt frame
F frame
un
89
Connected Particles: Constrained Motion
• Some problems in dynamics involve particles that are interconnected
by linkages. Rather than model the elastic forces within linkages, it is
often convenient and sufficient to represent linkages as constraints to
the relative motion of the particles
• Particles connected by a cord or tether, that can be approximated as
inextensible, are a common example of a situation in which kinematic
constraints can be applied
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90
Connected Particles: Constrained Motion
• Motion of block A
produces a corresponding
motion of block B
• We can model the motion
constraint via the length of
the cord:
AMME2500 Engineering Dynamics |
91
Connected Particles: Constrained Motion
• Motion of block A
produces a corresponding
motion of block B
• We can model the motion
constraint via the length of
the cord:
sA + lCD + sB = lT
Note: lengths
are measured
postitive out
from the fixed
pulley centre
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92
Connected Particles: Constrained Motion
• Taking the time derivative
of this equations:
• The displacements,
velocities and
accelerations of blocks A
and B are equal and
opposite
AMME2500 Engineering Dynamics |
sA + lCD + sB = lT
ṡA + 0 + ṡB = 0
vB =
vA
93
Connected Particles: Constrained Motion
• Taking the time derivative
of this equations:
• The displacements,
velocities and
accelerations of blocks A
and B are equal and
opposite
AMME2500 Engineering Dynamics |
sA + lCD + sB = lT
s̈A + 0 + s̈B = 0
aB =
aA
94
Connected Particles: Constrained Motion
• Consider a slightly more
complicated, but still onedegree of freedom, example
• There is still only one chord of
length lT
• The segments of cord shown
in red don’t change length. Let
their combined length be lp
2sB + sA + lp = lT
2ṡB + ṡA + 0 = 0
2s̈B + s̈A + 0 = 0
AMME2500 Engineering Dynamics |
Note: sB and h are coordinates relative
to a fixed datum
95
Connected Particles: Constrained Motion
• Consider a slightly more
complicated, but still onedegree of freedom, example
• There is still only one chord of
length lT
• The segments of cord shown
in red don’t change length. Let
their combined length be lp
2sB + sA + lp = lT
2ṡB + ṡA + 0 = 0
2vB =
vA
2s̈B + s̈A + 0 = 0
2aB =
aA
AMME2500 Engineering Dynamics |
96
Connected Particles: Constrained Motion
• If the velocity of particle A is
vA = 2 m/s to the right, what
is the velocity of particle B?
(A)
vB = 2 m/s down
(D)
vB = 1 m/s up
(B)
vB = 4 m/s down
(E)
vB = 4 m/s up
(C)
vB = 1 m/s down
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97
Connected Particles: Constrained Motion
• If the velocity of particle A is
vA = 2 m/s to the right, what
is the velocity of particle B?
(C)
vB = 1 m/s down
x + 2y + lp = lT
ẋ + 2ẏ + 0 = 0
vA + 2vB + 0 = 0
AMME2500 Engineering Dynamics |
1
1
vB = vA = 2 = 1m/s
2
2
98
Connected Particles: Constrained Motion
• General procedure for analysing connected particles with a
single cord:
• Establish position coordinates from fixed points
• Determine the length of the cord using position coordinates, neglecting
cord segments that do not change length as the particles move
• Evaluate the time derivative of the equation of cord length to relate velocity
and accelerations
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99
Connected Particles: Two Degrees of Freedom
• Problems involving multiple cords
will incorporate additional degrees
of freedom to the relative motion of
the particles
• Consider the following problem:
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100
Connected Particles: Two Degrees of Freedom
• Problems involving multiple cords
will incorporate additional degrees
of freedom to the relative motion of
the particles
• Consider the following problem:
• The constraint applied by each
cord forms a separate equation:
yA + 2yD + lp1 = l1
AMME2500 Engineering Dynamics |
101
Connected Particles: Two Degrees of Freedom
• Problems involving multiple cords
will incorporate additional degrees
of freedom to the relative motion of
the particles
• Consider the following problem:
• The constraint applied by each
cord forms a separate equation:
yB + yC + (yC
yD ) + lp2 = l2
AMME2500 Engineering Dynamics |
102
Connected Particles: Two Degrees of Freedom
• Problems involving multiple cords
will incorporate additional degrees
of freedom to the relative motion of
the particles
• Consider the following problem:
• The constraint applied by each
cord forms a separate equation:
yA + 2yD + lp1 = l1
yB + yC + (yC
yD ) + lp2 = l2
vA + 2vD = 0
vB + 2vC
vD = 0
AMME2500 Engineering Dynamics |
aA + 2aD = 0
aB + 2aC aD = 0
103
Connected Particles: Two Degrees of Freedom
• Problems involving multiple cords
will incorporate additional degrees
of freedom to the relative motion of
the particles
• Consider the following problem:
• The constraint applied by each
cord forms a separate equation
• Eliminating vD and aD :
vA + 2vB + 4vC = 0
aA + 2aB + 4aC = 0
AMME2500 Engineering Dynamics |
104
Connected Particles: Constrained Motion
• General procedure for analysing connected particles with a
multiple cords:
• Establish position coordinates and directions from fixed points as before,
making sure to include coordinates for all components that can move
• Establish N equations for N cords, differentiate length w.r.t time
• Substitute to eliminate components that are not of interest
• The resulting constraints may then be used to evaluate the
velocity and acceleration of certain particles based on the
others
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105
Connected Particles: Example 1.11
• Consider the case that particles A
and B accelerate downwards with
aA = 2 m/s2 and aB = 5 m/s2.
Determine the acceleration of the
mass connected to point C.
AMME2500 Engineering Dynamics |
106
Connected Particles: Example 1.11
• Consider the case that particles A
and B accelerate downwards with
aA = 2 m/s2 and aB = 5 m/s2.
Determine the acceleration of the
mass connected to point C.
vA + 2vB + 4vC = 0
aA + 2aB + 4aC = 0
AMME2500 Engineering Dynamics |
107
Connected Particles: Example 1.11
• Consider the case that particles A
and B accelerate downwards with
aA = 2 m/s2 and aB = 5 m/s2.
Determine the acceleration of the
mass connected to point C.
vA + 2vB + 4vC = 0
aA + 2aB + 4aC = 0
aC =
=
1
(aA + 2aB )
4
1
(2 + 2(5)) = 3m/s2 upwards
4
AMME2500 Engineering Dynamics |
108
Part 2: Kinetics in various coordinate systems
Reading: Mariam, Kraige and Bolton Chapters 3 & 4
AMME2500 Engineering Dynamics |
109
Kinetics of Particles
• Dynamics: the principles governing the state of motion or
rest of bodies under the influence of applied force and
torque
• Kinematics: the “geometry” of motion: the relationships between
position, velocity, acceleration and rotation and bodies joined by
linkages and other constraints
• Kinetics: the relationship between motion
and it’s cause due to force, torque, mass
and inertia
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110
Learning Outcomes
• This section covers basic methods for solving particle kinetics. The
concepts covered lays the foundation for rigid body dynamics
1. Newton’s Laws of Motion
• Drawing Free Body Diagrams
• Inertial and Non-Internal Frames of Reference
• Application of Newton’s laws on different coordinate systems
2.
Work-Energy Principle
• Work done by External Force
• Conservative and Non-Conservative Forces
• Relating Conservative and Non-Conservative work in a system
• Power and Efficiency
3.
Momentum and Impulse
• Understanding the difference between Linear Momentum and Impulse
• Conservation of Linear Momentum
• 1D and 2D Elastic & Inelastic Collision of Particles
4.
Angular Momentum
• Understanding significance of Angular Momentum
• Conservation of Angular Momentum
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Learning Outcomes
• Additional practice problems are available in Canvas.
• These problems cover each learning outcome broadly.
• Attempt to solve these problems as you progress through
the lecture slides.
• Use these practice problems to test your understanding.
• If you need help with the practice problems, ask your
tutors for assistance.
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112
Practice Problem 1
• A man pulls himself up 15∘ the incline by the method
shown. If the combined mass of the man and cart is 100
kg, determine the acceleration of the cart if the man exerts
a pull of 250 N on the rope. Neglect all friction and the
mass of the rope, pulleys, and wheels.
Ans: 𝑎 = 4.96 m/s2 up
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113
Practice Problem 2
• The 7-kg collar A slides with
negligible friction on the fixed
vertical shaft. When the collar is
released from rest at the bottom
position shown, it moves up the
shaft under the action of the
constant force 𝐹 = 200𝑁 applied to
the cable. Calculate the stiffness 𝑘
which the spring must have if its
maximum compression is to be
limited to 75 mm. The position of the
small pulley at B is fixed.
Ans: 𝑘 = 8.79 kN/m
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114
Practice Problem 3
• The spring of modulus 𝑘 = 200 N/m is compressed a
distance of 300 mm and suddenly released with the
system at rest. Determine the absolute velocities of both
masses when the spring is unstretched. Neglect friction
Ans: 𝑣, = 2.05 m/s left, 𝑣) = 0.878 m/s right
AMME2500 Engineering Dynamics |
115
Practice Problem 4
• If the center of the ping-pong ball is to clear the net as
shown, at what height ℎ should the ball be horizontally
served? Also determine ℎ; . The coefficient of restitution
for the impacts between ball and table is 𝑒 = 0.9, and the
radius of the ball is 𝑟 = 0.75 in. (1 in = 25.4 mm)
Ans: ℎ, = 10.94 in, ℎ) = 7.43 in
AMME2500 Engineering Dynamics |
116
Practice Problem 5
• The two spheres are rigidly connected to the rod of
negligible mass and are initially at rest on the smooth
horizontal surface. A force F is suddenly applied to one
sphere in the y-direction and imparts an impulse of 10 N⋅s
during a negligibly short period of time. As the spheres
pass the dashed position, calculate the velocity of each
one.
Ans: 𝑣 = 4.71 m/s both
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117
Newton’s Universal Law of Gravitation
• Any two particles or bodies have a mutually attractive
gravitational force acting between them. Newton postulated the
law governing this gravitational force:
F = G m1m2/r2
where
F = force of attraction between the two bodies,
G = universal constant of gravitation ,
m1, m2 = mass of each body, and
r = distance between centres of the two bodies.
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118
Newton’s Universal Law of Gravitation
• When near the surface of the earth, the only gravitational force
having any sizable magnitude is that between the earth and the
body. This force is called the weight of the body.
• For small ranges of r (relative to REarth), weight force can be
approximated as F = m1g, where g = GmEarth/REarth2 ≈ 9.81 m/s2
AMME2500 Engineering Dynamics |
119
Mass and Weight
• Mass is an absolute property of a body. It is independent of
the gravitational field in which it is measured.
• The mass provides a measure of the resistance of a body
to a change in velocity.
• In SI units mass is measured in kg.
• The weight of a body is not absolute, since it depends on
the gravitational field in which it is measured.
• Weight is a force defined as
W = mg
where g is the acceleration due to gravity measured in m/s2.
In SI units, weight is measured in kg.m/s2, otherwise known
as a Newton (N).
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Newton’s “Laws of Motion”
The motion of a particle is governed by Newton’s three
laws of motion:
First Law: A particle originally at rest, or moving in a
straight line at constant velocity, will remain in this state
if the resultant force acting on the particle is zero.
2. Second Law: If the resultant force on the particle is not
zero, the particle experiences an acceleration in the
same direction as the resultant force. This acceleration
has a magnitude proportional to the resultant force, and
inversely proportional to its mass.
3. Third Law: Mutual forces of action and reaction
between two particles are equal, opposite, and collinear.
1.
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121
Newton’s “Laws of Motion”
• The first and third laws are used in the development of the
concepts of statics. Newton’s second law forms the basis
of the study of dynamics.
• Mathematically, Newton’s second law of motion can be
written:
F = ma
where F is the resultant unbalanced force acting on the
particle, and a is the acceleration of the particle. The
positive scalar m is the mass of the particle.
• Newton’s second law becomes less accurate when the particle’s
speed approaches the speed of light, or if the size of the particle is
extremely small (~ size of an atom), or in the presence of intense
gravitational fields (see special and general relativity)
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Inertial Frame of Reference
• Newton’s equation of motion is only valid if the
acceleration is measured in a Newtonian or inertial frame
of reference.
• An inertial reference frame is both non-rotating and has
zero acceleration (it can be moving !)
• The surface of the Earth is not really an inertial reference frame,
but it is usually sufficient to treat it as one as it has a very low
magnitude of rotation and acceleration, as the Earth moves through
space around the sun
• Problems requiring very high-accuracy often require these effects
to be included when developing equations of motion using
Newton’s second law
• See Meriam, Kraige and Bolton Section 3/2 for an excellent
explanation
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123
Equations of Motion from Newton’s Laws
The motion of a particle is governed by Newton’s second law, relating
the unbalanced forces on a particle to its acceleration. If more than one
force acts on the particle, the equation of motion can be written
åF = FR = ma
where FR is the resultant force, which is a vector summation of all the
forces.
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Example 1.12
• A 75kg person stands on a set of scales in an
elevator. The elevator is released from rest and
the tension in the hoisting cable is 8300N. If the
total mass of the elevator, person and scales is
750kg, find the reading in the scales and the
velocity of the elevator after 3 seconds.
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125
Procedure for Developing Equations of Motion
1) Select a convenient inertial coordinate system. Rectangular,
normal/tangential, or cylindrical coordinates may be used.
2) Draw a free-body diagram showing all external forces applied to the
particle. Resolve forces into their appropriate components.
3) Apply the equations of motion 𝚺F = FR = ma into scalar component
form and solve these equations for the unknowns.
4) It may be necessary to apply the kinematic relations and constraints
to generate additional equations.
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126
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
W = mg
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127
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
T
AMME2500 Engineering Dynamics |
128
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
R
AMME2500 Engineering Dynamics |
129
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
R
AMME2500 Engineering Dynamics |
130
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
R
AMME2500 Engineering Dynamics |
131
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
AMME2500 Engineering Dynamics |
132
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
R
AMME2500 Engineering Dynamics |
133
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
0<μ<1
static: μ = μs
kinetic (moving) μ = μk
AMME2500 Engineering Dynamics |
Direction of
motion or
force
R
<= μR
134
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
AMME2500 Engineering Dynamics |
135
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
Equilibrium
position
AMME2500 Engineering Dynamics |
136
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
Equilibrium
position
AMME2500 Engineering Dynamics |
x = sx
137
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
F = -ksx
Equilibrium
position
AMME2500 Engineering Dynamics |
x = sx
138
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
x = -sx
AMME2500 Engineering Dynamics |
139
Free Body Diagrams and Forces
• Types of force commonly encountered in dynamics problems:
• Weight: Force of gravitational attraction
• Tension: Force owing to tension in an attached cable/tether/cord
• Contact Force: force exerted by a solid surface that maintains a position constraint
between and object and the surface (acts perpendicular to surface at contact point)
• Friction: force exerted by a solid surface on an object, or between objects, that acts
tangential to their surface directions at the contact point
• Friction is often modeled as Coulomb friction for dry surfaces (force proportionate to normal
force between objects)
• Spring/Elastic Force: Force exerted by elastic, spring like objects on one another
based on the compressive displacement between them.
• Spring force is typically modeled using Hooke’s Law, in which the force exerted is
proportional to the compressive displacement w.r.t an equilibrium state
x = -sx
AMME2500 Engineering Dynamics |
F = -ksx
= -k(-sx)
= ksx (to right)
140
Equations of Motion from Newton’s Laws
The motion of a particle is governed by Newton’s second law, relating
the unbalanced forces on a particle to its acceleration. If more than one
force acts on the particle, the equation of motion can be written
åF = FR = ma
where FR is the resultant force, which is a vector summation of all the
forces.
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141
Example 1.13: 1 DOF
• Consider two frictionless pulley
systems. Case 1 has mass with a
10N weight connected via an
inextensible cord to a mass with
a 30N weight. Case two has a
mass with a 10N weight on one
side and a constant tension force
of 30N is applied on the other.
• Which is true? (Hint: which case
has the greatest tension force?)
(A)
For both case 1 and 2, the acceleration of block A is zero
(B)
Block A will accelerate upwards faster for case 1 than case 2
(C)
Block A will accelerate upwards faster for case 2 than case 1
(D)
In both cases, block A will have the same upward acceleration
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142
Equations of Motion: Rectangular Coordinates
Vector Form:
åF = ma
Rectangular coordinates:
åFx i + åFy j + åFz k = m(ax i + ay j + az k)
Equivalent to three scalar equations:
åFx = max, åFy = may, and åFz = maz.
These forms of the rectangular equations of motion are best used when
the problem requires finding forces (especially forces perpendicular to
the path), accelerations, velocities, or mass.
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143
Procedure for Developing Equations of Motion
1) Select a convenient inertial coordinate system. Rectangular,
normal/tangential, or cylindrical coordinates may be used.
2) Draw a free-body diagram showing all external forces applied to the
particle. Resolve forces into their appropriate components.
3) Apply the equations of motion 𝚺F = FR = ma in scalar component
form and solve these equations for the unknowns.
4) It may be necessary to apply the kinematic relations and constraints
to generate additional equations.
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144
Example 1.14: Rectangular Coordinates
• A 200kg cable car runs along a
fixed overhead cable (with slope
5/12) and is pulled along by a
horizontal control cable with
constant tension T = 2.4kN.
Determine (a) the magnitude of
force exerted by the overhead
cable on the car wheels and (b)
the acceleration of the cable car.
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145
Example 1.15: Rectangular Coordinates
• A 200kg log is hauled up a 30o ramp
T
by applying tension T to a rope
attached to one end. If the coefficient
of static friction between the log and
ramp is μs = 0.55 and the coefficient
of kinetic friction is μk = 0.5, determine
the acceleration of the block up the
slope for:
(a) T = 1500N and (b) T = 2500N
AMME2500 Engineering Dynamics |
146
Example 1.16: Rectangular Coordinates
• The same 200kg log is now
connected via a pulley system to a
125kg suspended block and released
from rest as shown. Determine the
velocity of the block when it hits the
ground.
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147
Equations of Motion: n-t Coordinates
Since the equation of motion is a vector
equation, åF = ma,
it may be written in terms of the n & t
coordinates as:
åFtut + åFnun = matut + manun
This vector equation will be satisfied provided the individual
components on each side of the equation are equal, resulting in the two
scalar equations: åFt = mat and åFn = man.
If we also consider forces and motion out of the plane, we can include a
third equation åFz = maz (equivalent to the rectangular z-coordinate)
AMME2500 Engineering Dynamics |
148
Example 1.17: n-t coordinates
• A block is released from rest at the
top of a frictionless slide (A) and
slides onto a conveyor belt at B.
Determine (a) the angular velocity of
the conveyor such that when blocks
arrive at B, they do not slide on the
belt and (b) an expression for the
normal force acting up from the slide
to the block in terms of θ.
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149
Work and Energy
• Two of the problems we examined today involved the integration of
forces as a function of displacement:
• Spring forces vary as a function of compression
• Component of force acting along a slide depends on slide shape, which varies as a
function of displacement
• Integration with respect to displacement leads to the equations of
work and energy
• Using principles of work and energy:
• simplifies the analysis of many problems
• allows us just to examine the start and end points of motion, while not needing to
evaluate accelerations resulting from unbalanced forces as a function of time
• Integration with respect to time leads to the equations of impulse and
momentum (see next week’s lecture)
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150
Work of a Force
•
•
Consider a force F acting on a particle as it travels along a path
between points 1 and 2.
A force does work on a particle when the particle undergoes a
displacement along the line of action of the force.
• The total work done by F is:
r2
U1-2 =
∫
F • dr
r1
•
The magnitude of the dot product
expression F • dr is equal to the
component of the force projected
onto the path at each path
segment:
AMME2500 Engineering Dynamics |
U1-2 =
s2
s2
∫
Fcosθds = ∫
s1
s1
Ftds
151
Work of a Force
•
Work is a scalar and has units N.m, or Joules (J) (the work done by
a force of 1N acting through 1m)
•
Work is positive if the force is acting in the direction of movement
•
Forces in the direction of motion do work. They are called active
forces
•
Forces acting perpendicular to the direction of motion produce no
work. Such constraints are called reactive forces
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152
Work by a Constant Force
•
If both F and q are constant (F = Fc), work further simplifies to
U1-2 = Fc cos q (s2 - s1)
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153
Work by Friction
• Friction can be modeled as the work done by a constant
force, if the friction is constant (e.g. constant normal force)
• However, in general the friction force is not constant, e.g.
the circular slide example we have been looking at in this
lecture
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154
Work by a Weight
The work done by the gravitational force acting on a particle (or
weight of an object) can be calculated by using
U1 2 =
U1 2 =
U1 2 =
Z 2
1
F · dr
1
( mgj) · (dxi + dyj)
Z 2
mg
Z y2
dy =
mg(y2
y1 )
y1
The work of a weight is the product of the magnitude of the
particle’s weight and its vertical displacement.
If motion is upward, the work is negative since the weight force
always acts downward.
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155
Work by an Elastic Force
When stretched, a linear elastic
spring develops a force of
magnitude Fs = ks, via Hooke’s law
x1
x2
The work of the spring force moving from position s1 to position s2 is:
U1 2 =
U1 2 =
U1 2 =
Z 2
1
F · dr
1
( kxi) · dxi
Z 2
Z x2
1
U1 2 = k(x21
2
x22 )
kxdx
x1
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156
Work and Kinetic Energy
• If we consider F to be the resultant force vector of all forces acting on
a particle, the total work done by F = ma is:
U1 2 =
Z 2
1
U1 2 =
Z 2
F · dr
1
Z v2
ma · dr
a · dr = at ds = vdv
1
U1 2 =
mvdv = m(v22
2
v1
1
U1 2 = mv22
2
v12 )
1
mv12 = T2
2
T1
Where T2 and T1 are the kinetic energies (units = Joule) of the particle
at path positions 1 and 2
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157
Work and Kinetic Energy
• The relationship between the work done on a particle and
its change in kinetic energy can be used to determine the
velocity of a particle when constrained to a path,
independent of the path itself:
T2 =
X
U 1 2 + T1
• Since constraining forces always act perpendicular to the
path, they do no work (no component of force in the
direction of motion):
• Problems may be greatly simplified when examining particles under
certain types of force
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158
Example 1.18: Work and Kinetic Energy
• A block is released from rest at the
top of a frictionless slide (A) and
slides onto a conveyor belt at B.
Determine the velocity of the block at
the bottom of the slide at B.
AMME2500 Engineering Dynamics |
159
Example 1.19: Work and Kinetic Energy
• Calculate the velocity of the 50kg
crate when it reaches the bottom of
the flat slide at B if the initial velocity
is 4 m/s down the slide at A and the
coefficient of kinetic friction is μk = 0.3
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160
Conservative Forces
•
A force F is said to be conservative if the work done is independent
of the path followed as a particle moves from A to B.
•
This also means that the work done by the force F in a closed path
(i.e., from A to B and then back to A) is zero.
•
Work is conserved when:
I
•
z
F · dr = 0
The work done by a conservative force
depends only on the positions of the
particle, and is independent of its
velocity or acceleration.
AMME2500 Engineering Dynamics |
F
B
A
y
x
161
Conservative Forces
• Generally, forces whose magnitude and direction are only a
function of position, and are independent of other variables that
change along the path are conservative
• Forces due to gravity (even when accounting for changes in
the magnitude of gravitational acceleration) and elastic
elements (such as springs) are conservative forces
• In general friction and aerodynamic/hydrodynamic drag are not
conservative forces, as their work is dependent on the path
taken
• Drag is a function of particle velocity, hence non-conservative
due to force magnitude changes and direction dependent on
velocity, not position
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162
Potential Energy
Potential energy (V) is a measure of the amount of work a
conservative force will do when a body changes position.
Gravitational Potential Energy: is the weight force (W = mg)
multiplied by its elevation from a datum. The datum can be defined at
any convenient location.
Vg = mg y
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163
Potential Energy
Elastic Potential Energy: is the force provided by a spring multiplied
by its distance from the uncompressed position.
V
AMME2500 Engineering Dynamics |
=
e
1 k s2
2
164
Conservation of Energy
• When a particle is acted upon by a system of conservative forces,
the work done by these forces is conserved and the sum of kinetic
energy and potential energy remains constant.
• As the particle moves, kinetic energy is converted to potential
energy and vice versa. This principle is called the principle of
conservation of energy and is expressed as
T +V
1
1
= T 2 + V 2 = constant
• T1 and T2 are the kinetic energies at states 1 and 2
• V1 and V2 are the potential energies at states 1 and 2
AMME2500 Engineering Dynamics |
165
Conservation of Energy
• If the system is additionally subjected to other forces that are
not conservative, then the change in the sum of potential and
kinetic energies from state 1 to state 2 is equal to the work
done by these non-conservative forces:
T +V
1
1
+ + U′1 −2 = T 2 + V 2
• + U′1 −2 is the work done by the non-conservative forces.
AMME2500 Engineering Dynamics |
166
Example 1.20: Potential Energy
• A block is released from rest at the
top of a frictionless slide (A) and
slides onto a conveyor belt at B.
Determine the velocity of the block at
the bottom of the slide at B:
AMME2500 Engineering Dynamics |
167
Work and Kinetic Energy
• The relationship between the work done on a particle and
its change in kinetic energy can be used to determine the
velocity of a particle when constrained to a path,
independent of the path itself:
1
T = mv 2
X
2
T2 =
U 1 2 + T1
T: kinetic energy, measured in the
same units as work, Joules (J)
• Since constraining forces always act perpendicular to the
path, they do no work (no component of force in the
direction of motion):
• Problems may be greatly simplified when examining particles under
certain types of force
AMME2500 Engineering Dynamics |
168
Conservation of Energy
• When a particle is acted upon by a system of conservative
forces, the work done by these forces is conserved and the
sum of kinetic energy and potential energy remains constant.
• As the particle moves, kinetic energy is converted to potential
energy and vice versa. This principle is called the principle of
conservation of energy and is expressed as
T +V
1
1
+ U’1-2 = T 2 + V 2
Where T1 and T2 are the kinetic energies at state 1 and 2
respectively and V1 and V2 are the potential energy for state
1 and 2, U’1-2 is the work done by non-conservative external
forces.
AMME2500 Engineering Dynamics |
169
Power and Efficiency
•
Power is defined as the amount of work performed per unit of
time.
•
If a machine or engine performs a certain amount of work, dU,
within a given time interval, dt, the power generated can be
calculated as
P = dU/dt
•
Since the work can be expressed as dU = F • dr, the power can
be written
P = dU/dt = (F • dr)/dt = F • (dr/dt) = F • v
•
Thus, power is a scalar defined as the product of the force and
velocity components acting in the same direction.
AMME2500 Engineering Dynamics |
170
Power
•
Using scalar notation, power can be written:
P = F • v = F v cos q
where q is the angle between the force and velocity vectors.
•
So if the velocity of a body acted on by a force F is known, the
power can be determined by calculating the dot product or by
multiplying force and velocity components.
•
The unit of power in the SI system is the Watt (W) where
1 W = 1 J/s = 1 (N · m)/s.
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171
Efficiency
•
The mechanical efficiency of a machine is the ratio of the useful
power produced (output power) to the power supplied to the machine
(input power) or
e = (power output) / (power input)
•
If energy input and removal occur at the same time, efficiency may
also be expressed in terms of the ratio of output energy to input
energy or:
e = (energy output) / (energy input)
•
Machines will always have frictional forces. Since frictional forces
dissipate energy, additional power will be required to overcome these
forces. Consequently, the efficiency of a machine is always less than
1.
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172
Power: Example 1.21
• A powered winch A hoists a 360kg log
up the 30o incline with a constant speed
of 1.2 m/s. If the maximum power
provided by the winch is 4 kW, compute
the maximum coefficient of kinetic
friction between the log and ground,
such that the log may be lifted.
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173
Constrained Motion: Example 1.22
• If the weight of block B is 100 N, how
A
much power is required to lift the block
at a constant speed of 2 m/s if hauled
by pulling the cable at A to the left?
Neglect friction.
(A)
200 W
(D)
50 W
(B)
100 W
(E)
0W
(C)
400 W
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174
Impulse and Momentum
• Previously we examined the principles of work and
energy:
• Allowed us to examine the start and end points of motion, thus
simplifying the analysis greatly for problems where forces and
accelerations are integrated w.r.t position
• Now will examine the concepts of impulse and
momentum:
• These analysis techniques can facilitate problem solving for
systems for which force and acceleration are integrated w.r.t time
• The methods developed here are particularly useful for problems in
which applied forces act over very small intervals of time, such as
in problems involving the impact of two particles
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175
Linear Impulse and Momentum
• The linear momentum G of a particle is defined as:
G = mv
• m is the particle’s mass and v is the velocity vector.
• Linear momentum is a vector property: its magnitude is G =
mv, where v is the magnitude of velocity, and its direction is
coincident to the direction of the velocity vector v
• Units for linear momentum are kg.m/s or N.s
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176
Linear Impulse and Momentum
• Computing the time rate of change of momentum for
constant mass:
d(mv)
dv X
=m
=
F
dt
dt
Ġ =
X
F
• ΣF is the resultant force vector acting on the particle, and
(for now) we assume the mass of the particle does not
change.
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177
Linear Impulse and Momentum
• If we now integrate our expression w.r.t time, we develop
a relationship between the change in momentum between
two times, based on the force acting across the time
period:
X
Fdt = dG
Z t2 X
t1
Z t2 X
Z G2
dG
Fdt = G2
G1
Fdt =
G1
t1
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178
Linear Impulse and Momentum
• If we now integrate our expression w.r.t time, we develop
a relationship between the change in momentum between
two times, based on the force acting across the time
period:
X
Fdt = dG
Z t2 X
t1
Z t2 X
Fdt =
Z G2
G2 =
Z t2 X
t1
dG
Where:
G1
Fdt = G2
t1
AMME2500 Engineering Dynamics |
Fdt + G1
G1
I=
Z t2 X
Fdt
t1
is the linear impulse I acting on the
particle.
179
Linear Impulse and Momentum
• The linear impulse-momentum principle states that the change
in momentum of a particle is equal to the impulse applied:
G2 =
Z t2 X
Fdt + G1
t1
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180
Linear Impulse and Momentum
• The linear impulse-momentum principle states that the change
in momentum of a particle is equal to the impulse applied:
G2 =
Z t2 X
Fdt + G1
t1
• Impulse is a vector and the
magnitude of a component of I is
equal to the area under the curve
of a force vs. time plot for the
same component of F
• If F is constant, I is:
I = F(t2
AMME2500 Engineering Dynamics |
t1 )
181
Linear Impulse and Momentum
• The linear impulse-momentum principle states that the change
in momentum of a particle is equal to the impulse applied:
G2 =
Z t2 X
Fdt + G1
t1
• The expression can be
broken up into three separate
equations describing the
different x-y-z components of
G and I:
m(v1 )x +
Z t2 X
Fx dt = m(v2 )x
t1
m(v1 )y +
Z t2 X
Fy dt = m(v2 )y
t1
m(v1 )z +
Z t2 X
Fz dt = m(v2 )z
t1
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182
Example 1.23: Linear Impulse and Momentum
• A tennis player hits a tennis ball
with their racket when the ball is at
the uppermost point in its trajectory
after bouncing. The horizontal
speed just before impact is v1 = 15
m/s, and after impact is v2 = 21 m/s,
directed 15o up from the horizontal.
• If the 60g ball is in contact with the
racket for 0.02s, determine the
average force exerted by the racket
on the ball.
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183
Conservation of Linear Momentum for two particles
• Consider the principle of linear momentum for two particles A
and B, at times 1 and 2
• The linear impulse-momentum principle applied to each gives:
GA2 =
Z t2 X
FA dt + GA1
GB2 =
Z t2 X
FB dt + GB1
t1
t1
A
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B
184
Conservation of Linear Momentum for two particles
• We can determine the sum these to provide an expression for
the linear momentum-impulse principle for the system of
particles:
GA2 + GB2 =
Z t2 X
t1
A
AMME2500 Engineering Dynamics |
FA dt +
Z t2 X
FB dt + GA1 + GB1
t1
B
185
Conservation of Linear Momentum for two particles
• If particle A applies a force to particle B, we know via Newton’s 3rd law
“When one body exerts a force on a second body, the second
body simultaneously exerts a force equal in magnitude and
opposite in direction on the first body.”
F
A
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B
186
Conservation of Linear Momentum for two particles
• If particle A applies a force to particle B, we know via Newton’s 3rd law
“When one body exerts a force on a second body, the second
body simultaneously exerts a force equal in magnitude and
opposite in direction on the first body.”
F
A
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B
187
Conservation of Linear Momentum for two particles
• If particle A applies a force to particle B, we know via Newton’s 3rd law
“When one body exerts a force on a second body, the second
body simultaneously exerts a force equal in magnitude and
opposite in direction on the first body.”
• Equivalently, if B applies a force on A, that same force must be
applied back to B, in equal magnitude and opposite direction
F
A
AMME2500 Engineering Dynamics |
F
B
188
Conservation of Linear Momentum for two particles
• If particle A applies a force to particle B, we know via Newton’s 3rd law
“When one body exerts a force on a second body, the second body
simultaneously exerts a force equal in magnitude and opposite in
direction on the first body.”
• Equivalently, if B applies a force on A, that same force must be applied
back to B, in equal magnitude and opposite direction
• Hence ΣFB = -ΣFA when no external force act on the two particles:
GA2 + GB2 =
Z t2 X
t1
A
AMME2500 Engineering Dynamics |
FA dt +
Z t2 X
0
FB dt + GA1 + GB1
t1
B
189
Conservation of Linear Momentum for two particles
• When the sum of all external impulses acting on two particles is
zero, the sum of linear momentum of both particles is
conserved. This is known as the principle of conservation of
momentum.
GA2 + GB2 = GA1 + GB1
A
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B
190
Example 1.24: Conservation of Linear Momentum for two
particles
• A 100 kg astronaut throws a 20 kg box. If
the astronaut observes the relative velocity
of the box (after throwing it) to be 4 m/s,
determine the change in the astronaut’s
velocity induced by throwing the box
(a)
(b)
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191
Linear Impulse and Momentum
• The linear impulse-momentum principle states that the change
in momentum of a particle is equal to the impulse applied:
G2 =
Z t2 X
Fdt + G1
t1
• In rectangular coordinates it
can be written as three
separate scalar equations
describing the different x-y-z
components of G and I:
m(v1 )x +
Z t2 X
Fx dt = m(v2 )x
t1
m(v1 )y +
Z t2 X
Fy dt = m(v2 )y
t1
m(v1 )z +
Z t2 X
Fz dt = m(v2 )z
t1
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192
Conservation of Linear Momentum for two particles
• When the sum of all external impulses acting on two particles is
zero, the sum of linear momentum of both particles is
conserved. This is known as the principle of conservation of
linear momentum.
GA2 + GB2 = GA1 + GB1
A
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B
193
Impact
• Impact refers to the collision between two bodies
• It is typically characterised by the generation of relatively large
contact forces which act over a very short time interval
• Problems involving impact are therefore well suited to analysis using
principles of impulse and the conservation of momentum
• The fine details of impact are very complex, involving material
deformation and recovery and the generation of heat and sound
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Impact
• During impact, the line of impact is a line through the mass centres of
the colliding particles.
• In general, there are two types of impact:
Central impact occurs when the directions
of motion of the two colliding particles are
along the line of impact.
Oblique impact occurs when the direction of
motion of one or both of the particles is at an
angle to the line of impact.
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195
Central Impact
Axis for positive direction of motion
• Before impact: vA1 > vB1 in
order for a collision to occur
• During impact: forces are
acting inwards (a) period
from initial contact until
maximum deformation, when
vA = vB = vmd (b) after
maximum deformation, there
is a period of restoration until
the contact area between the
objects becomes zero
• After impact: vA2 < vB2,
Line of impact
A
B
VA1
VB1
Vmd
FAB
-FAB
A B
A
VA2
momentum is always
conserved
mA vA1 + mB vB1 = mA vA2 + mB vB2
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B
VB2
non-impact forces
assumed to be negligible
196
Coefficient of Restitution
• The coefficient of restitution, e, of two bodies during impact represents the
capacities of the two bodies to recover from deformation after impact
• Highly elastic objects will recover their shape during the period of restoration,
whereas highly plastic bodies may become permanently deformed
• The coefficient of restitution is defined as the ratio of the magnitude of
impulse between the bodies during restoration, to the magnitude of impulse
during deformation:
Rt P
Fr dt
t0
e = R t0 P
Fd dt
0
where Fr is the forces during restitution, Fd is the forces during deformation,
initial contact occurs at t = 0, maximum deformation at t = t0 and separation
at t = t
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Coefficient of Restitution
• The coefficient of restitution e is between 0 and 1
• e = 1 corresponds to a perfectly elastic collision, with no associated
energy loss
• e = 0 is when the objects stick together, and there is a maximal loss of
kinetic energy (lost to noise, work associated with permanent
deformation, etc)
• Note that e is always associated with a pair of bodies
• For real bodies, e usually depends on the actual impact (typically
lower e for higher speed) as the power of the impact may dictate
whether the resulting deformation is within the elastic region of the
material or not.
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Coefficient of Restitution
• The coefficient of restitution is then related to the initial and
final velocities of each particle via:
Impulse acting on particle A:
Rt P
Fr dt
t0
e = R t0 P
Fd dt
0
mA [ vA2
e=
mA [ vmd
vmd
e=
vA1
( vmd )]
( vA1 )]
vA2
vmd
Where vmd is the velocity of both particles
at maximum deformation
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199
Coefficient of Restitution
• The coefficient of restitution is then related to the initial and
final velocities of each particle via:
Impulse acting on particle B:
Rt P
Fr dt
t0
e = R t0 P
Fd dt
0
mB [vB2
e=
mB [vmd
vB2
e=
vmd
vmd ]
vB1 ]
vmd
vB1
Where vmd is the velocity of both particles
at maximum deformation
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200
Coefficient of Restitution
• The coefficient of restitution e is then related to the initial and
final velocities of each particle via:
Combining equations to cancel for vmd:
Rt P
Fr dt
t0
e = R t0 P
Fd dt
0
vB2
e=
vA1
vA2
vB1
• Hence, the coefficient of restitution is the ratio of the magnitude
of the relative velocity of separation to the magnitude of the
relative velocity of approach
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201
Coefficient of Restitution
• Combining the equations of impact restitution and conservation of
linear momentum, provides a way to solve the two unknowns arising
from problems in which only the initial speeds of each object are
known
mA vA1 + mB vB1 = mA vA2 + mB vB2
vB2
e=
vA1
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vA2
vB1
202
Oblique Impact
• Oblique impact occurs when the direction of motion of one or both of the
particles is at an angle to the line of impact
• Define a normal coordinate parallel to the line of impact and a tangent
coordinate parallel to the plane of contact
• We can isolate the direction of the change in momentum associated with the
collision:
t
n
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203
Oblique Impact
• The impact forces acting in the tangent direction are zero,
hence the momentum components of both particles in this
direction are conserved:
mA (vA1 )t = mA (vA2 )t
mB (vB1 )t = mB (vB2 )t
t
n
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204
Oblique Impact
• The impact forces acting in the normal direction are maximal,
and, as in the case of central impact, the momentum of the
system in conserved:
mA (vA1 )n + mB (vB1 )n = mA (vA2 )n + mB (vB2 )n
(vB2 )n
e=
(vA1 )n
t
(vA2 )n
(vB1 )n
n
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205
Oblique Impact
• Problems involving oblique impact can
therefore be treated as a problem of
central impact along the line of impact,
while ensuring the momentum of each
colliding object is conserved in the
direction tangent to the line of impact
mA (vA1 )n + mB (vB1 )n = mA (vA2 )n + mB (vB2 )n
(vB2 )n
e=
(vA1 )n
(vA2 )n
(vB1 )n
AMME2500 Engineering Dynamics |
mA (vA1 )t = mA (vA2 )t
mB (vB1 )t = mB (vB2 )t
206
Example 1.25: Impact
• Particle A (a sphere) has a velocity of 6 m/s in
the direction shown and collides with a spherical
particle B of equal mass and diameter, initially at
rest. Determine the resulting motion of both
particles given the coefficient of restitution of the
collision is e = 0.6
B
VA1 = 6 m/s
A
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207
Example 1.26: Impact with the Earth
• A ball is dropped from a height of
1m. If the coefficient of restitution
e = 0.5 for the impact with the
ground, how high does the ball
bounce on its first bounce?
(A)
(B)
(C)
(D)
(E)
0.5 m
0.75 m
0.25 m
0m
No Idea!
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208
Angular Momentum
• Consider a particle with linear
momentum G = mv
• Its position vector is r with
respect to some origin O
• The moment of the linear
momentum about the origin is
defined as the angular
momentum:
Ho = r ⇥ mv
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209
Angular Momentum
• The vectors r and v form a
plane A
• Ho is normal to this plane,
with direction dictated by the
right-hand rule for cross
products
• Angular momentum (like
linear momentum) is a vector
• Units for angular momentum
are kg.m2/s or N.m.s
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Angular Momentum
• The scalar components of Ho are
obtained from expansion of the
cross product:
Ho = r ⇥ mv
Ho = m(vz y
vz x)j + m(vy x
vy z)i + m(vx z
Ho,x = m(vz y
vy z)
Ho,y = m(vx z
vz x)
Ho,z = m(vy x
vx y)
AMME2500 Engineering Dynamics |
vx y)k
Where r = xi + yj + zk and
v = vxi + vyj + vzk are the
position and velocity vectors,
respectively
211
Angular Momentum
• If we consider the angular momentum
within the plane made by the vectors r and
v, then the magnitude of the angular
momentum is the product of the magnitude
of linear momentum mv with the moment
arm r.sinθ:
Ho = |r ⇥ mv|
Ho = mvr sin ✓
• The direction of the vector Ho points out of the page, with positive
angular momentum corresponding to the anti-clockwise direction
(right hand rule)
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Angular Momentum
• The time of change of angular momentum Ho is:
d(r ⇥ mv)
= ṙ ⇥ mv + r ⇥ mv̇
dt
= v ⇥ mv + r ⇥ mv̇
= r ⇥ mv̇
=r⇥
Ḣo =
AMME2500 Engineering Dynamics |
X
X
F=
X
Mo
Mo
213
Angular Momentum
• The time rate of change of angular momentum Ho is:
Ḣo =
X
Mo
where Mo is the moment of all forces acting on the particle, about
the origin O (each force multiplied by its lever-arm from O)
• This can be stated as:
the time rate of change of angular momentum of a particle about an origin
O is equal to the moment of all forces acting on the particle, measured
about the same origin O
• Note that the angular momentum (and hence force moment)
depends on the point O about which it is measured
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Example 1.27: Angular Impulse and Momentum
• Consider our box moving down a
circular slide problem (neglect
friction). Use angular momentum and
its rate of change to develop an
expression for the acceleration of the
block down the slide.
AMME2500 Engineering Dynamics |
O
v
215
Angular Impulse-Momentum Principle
• Re-arrange and integrate the equation for the rate of
change of angular momentum:
X
Z t2 X
Mo dt = dHo
Mo dt = Ho2
Ho1
t1
Ho2 =
Z t2 X
Mo dt + Ho1
t1
• The integral of Mo w.r.t time is referred to the angular
impulse acting on the particle (about O) and is equal to
the change in angular momentum between times 1 and 2
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216
Angular Impulse-Momentum Principle
• If r, v and ΣF all act in the same plane between times 1
and 2, then motion occurs within the plane and we can
consider the scalar components of angular impulse and
momentum acting in the plane:
Ho2 =
Z t2 X
F r sin ✓dt + Ho1
t1
mv2 d2 =
Z t2 X
F r sin ✓dt + mv1 d1
t1
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217
Conservation of Angular Momentum
• If the resultant moment about a fixed point 0 is zero
during a time interval then the angular moment is said to
be conserved:
Ho2 =
Z t2 X
0
Mo dt + Ho1
t1
Ho2 = Ho1
• By extension, if during the collision of two particles the
only unbalanced forces acting the particles are the equal
and opposite collision forces then the total angular
momentum of the two-particle system is conserved.
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Example 1.28: Conservation of Angular Momentum
The assembly of the light
rod and two end masses is
at rest when it is struck by
the falling wad of putty
traveling with speed v1 as
shown. The putty adheres to
and travels with the righthand end mass. Determine
the angular velocity of the
assembly just after impact.
The pivot at O is frictionless,
and all three masses may be
assumed to be particles.
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