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MEE 214 (Dynamics)
Tuesday 8.30-11.20
• Dr. Soratos Tantideeravit (สรทศ ตนตธรวทย) [email protected]
• Lecture Notes, Course updates, Extra problems, etc
•• No Homework
• Final Exam (Date & Time – TBD)
12/03/58 MEE214 – Dynamics 1
Course Overview
• Kinematics of a Particle
– Rectilinear and Curvilinear Motion
• Kinetics of a Particle
– Force and Acceleration– Force and Acceleration
– Work and Energy
– Impulse and Momentum
• Kinetics of a System of Particles
12/03/58 MEE214 – Dynamics 2
Introduction to Dynamics
EngineeringMechanics
Statics
12/03/58 MEE214 – Dynamics 3
Mechanics
Dynamics
Kinematics Kinetics
Kinematics of a particle
Objectives
• Concepts of position, displacement, velocity, and
acceleration.
• Particle motion along a straight line
• Particle motion along a curved path using • Particle motion along a curved path using
different coordinate systems.
• Analysis of dependent motion of two particles.
• Principles of relative motion of two
particles using translating axes.
12/03/58 MEE214 – Dynamics 5
Rectilinear Kinematics
• Origin
– Define a fixed point in space
• Position
12/03/58 MEE214 – Dynamics 6
• Position
– Defined by a position vector r
or an algebraic scalar s
Rectilinear Kinematics
• Displacement
– Change in position
• Velocity
s∆
12/03/58 MEE214 – Dynamics 7
• Velocity
t
svavg
∆
∆=
dt
dsv =
Rectilinear Kinematics
• Acceleration
t
vaavg
∆
∆=
12/03/58 MEE214 – Dynamics 8
2
2
dt
sd
dt
dva ==
dvvdsa =
Constant Acceleration
caa =tavv c+= 0
21tatvss ++=
12/03/58 MEE214 – Dynamics 9
2
002
1tatvss c++=
)(2 0
2
0
2 ssavv c −+=
Problem 12-31
The acceleration of a particle along a straight
line is defined by a=(2t-9) m/s2, where t is in
seconds. At t=0, s=1 m and v=10 m/s. When
t=9 s, determine (a) the particle’s position, (b)
12/03/58 MEE214 – Dynamics 10
t=9 s, determine (a) the particle’s position, (b) the total distance traveled and (c) the velocity.
General Curvilinear Motion
Curvilinear motion occurs when the particle moves
along a curved path
Position. The position of the particle, measured from a
fixed point O, is designated by the position vector
12/03/58 MEE214 – Dynamics 11
fixed point O, is designated by the position vectorr = r(t).
General Curvilinear Motion
Displacement. Suppose during a small time interval
Δt the particle moves a distance Δs along the curve to a
new position P`, defined by r` = r + Δr. The displacementΔr represents the change in the particle’s position.
12/03/58 MEE214 – Dynamics 12
Δr represents the change in the particle’s position.
srt ∆=∆→∆ ,0
General Curvilinear Motion
Velocity
t
rvavg
∆
∆=
vv
12/03/58 MEE214 – Dynamics 13
dt
ds
dt
rdvins ==
vv
General Curvilinear Motion
Acceleration.
t
vaavg
∆
∆=
vv
12/03/58 MEE214 – Dynamics 14
2
2
2
2
dt
sd
dt
rd
dt
vda ===
vvv
Curvilinear Motion:
Rectangular Components
kzjyixr ˆˆˆ ++=v
Position. Position vector is defined by
The magnitude of is always positive and defined asrv
12/03/58 MEE214 – Dynamics 15
222 zyxr ++=
The direction of r is specified by
the components of the unit
vector rrur /ˆv
=
Curvilinear Motion:
Rectangular Components
Velocity.
zvyvxv
kvjvivdt
rdv
zyx
zyx
&&&
vv
===
++== ˆˆˆ
12/03/58 MEE214 – Dynamics 16
vvuv /ˆv
=
The velocity has a magnitude defined as the positive value of
222
zyx vvvv ++=
Curvilinear Motion:
Rectangular Components
Acceleration.
yva
xva
kajaiadt
vda
xx
zyx
&&&
&&&
vv
==
==
++== ˆˆˆ
12/03/58 MEE214 – Dynamics 17
zva
yva
zz
yy
&&&
&&&
==
==
The acceleration has a magnitude defined as the
positive value of
222
zyx aaaa ++=
Curvilinear Motion:
Rectangular Components
• The acceleration has a direction specified by
the components of the unit vector .
• Since a represents the time rate of change in
aaua /ˆv
=
12/03/58 MEE214 – Dynamics 18
• Since a represents the time rate of change in velocity, a will not be tangent to the path.
Motion of Projectile
• Constant downward acceleration, no air
resistance
• Mathema<cal expressions, ↑ [=] +, → [=] +
−= x 0=&&
12/03/58 MEE214 – Dynamics 19
2
00
0
2
1gttvyy
gtvy
gy
y
y
−+=
−=
−=
&
&&
tvxx
vx
x
x
x
00
0
0
+=
=
=
&
&&
Example 12.12
The chipping machine is designed to eject wood chips
at vO = 25 ft/s. If the tube is oriented at 30° from the
horizontal, determine how high, h, the chips strike the
pile if they land on the pile 20 ft from the tube.
12/03/58 MEE214 – Dynamics 20
Curvilinear Motion:
Normal and Tangential Components
• When the path of motion of a particle is known,
describe the path using n and t coordinates which act
normal and tangent to the path
•Consider origin located at the particle
12/03/58 MEE214 – Dynamics 21
tu
nu
Tangential direction
Normal direction
Curvilinear Motion:
Normal and Tangential Components
Velocity.• Since the particle is moving, s is a function of time
• Particle’s velocity v has direction that is always tangent to the path and a magnitude that is determined
by taking the time derivative of the path function s = s(t)
12/03/58 MEE214 – Dynamics 22
by taking the time derivative of the path function s = s(t)
sv
uvv t
&
v
=
= ˆ
Curvilinear Motion:
Normal and Tangential Components
Acceleration• Acceleration of the particle is the time rate of change
of velocity
tt uvuvdt
uvdva &&&vv
ˆˆ)ˆ(
+===
12/03/58 MEE214 – Dynamics 23
tt uvuvdt
va & ˆˆ +===
2
2
dt
sd
dt
dvv ==&
Curvilinear Motion:
Normal and Tangential Components
Acceleration• Find tu
&'ˆˆˆttt uudu =+
nuntt uduud ˆˆ =
θθ dddut == )1(
12/03/58 MEE214 – Dynamics 24
'ˆtu
tu tud ˆθd
t
nt udud ˆˆ θ=
nnnt uv
us
uu ˆˆˆˆρρ
θ ===&
&&
Curvilinear Motion:
Normal and Tangential Components
vdvdsava
uauaa
tt
nntt
==
+=
&
vˆˆ
2v
12/03/58 MEE214 – Dynamics 25
ρ
2van =
22
2/32
/
])/(1[
dxyd
dxdy+=ρ
Problem 12-120
The automobile is originally at rest at s=0. If it
then starts to increase its speed at
ft/s2 where t is in seconds, determine the
magnitudes of its velocity and acceleration at
)05.0( 2tv =&
12/03/58 MEE214 – Dynamics 26
magnitudes of its velocity and acceleration at s = 550 ft.
Problem 12-131
At a given instant the train engine at E has a
speed of 20 m/s and an acceleration of 14
m/s2 acting in the direction shown. Determine
the rate of increase in the train’s speed and
12/03/58 MEE214 – Dynamics 27
the rate of increase in the train’s speed and the radius of curvature of the path.
Problem 12-152
If the speed of the box at
point on the track is 30ft/s
which is increasing at the rate
of ft/s2 , determine the 5=v&
12/03/58 MEE214 – Dynamics 28
of ft/s2 , determine the
magnitude of the acceleration of the box at this instant.
5=v&
Curvilinear Motion:
Cylindrical Components
• Fixed origin
u Radial direction
12/03/58 MEE214 – Dynamics 29
ru
θu
Radial direction
Transverse direction
Curvilinear Motion:
Cylindrical Components
Position
urr ˆ=v
12/03/58 MEE214 – Dynamics 30
rurr ˆ=v
Curvilinear Motion:
Cylindrical Components or Polar
Velocity
rr ururrv &&&vv
ˆˆ +==
θθuu ˆˆ && =
12/03/58 MEE214 – Dynamics 31
θθururv rˆˆ &&
v+=
θθuur ˆˆ && =
θθ uvuvv rrˆˆ +=
v
Curvilinear Motion:
Cylindrical Components
Acceleration
θθθ θθθ urururururva rr&&&&&&&&&&&
vvˆˆˆˆˆ ++++==
ruu ˆˆ θθ&& −=
12/03/58 MEE214 – Dynamics 32
ruu ˆˆ θθ −=
θθ uauaa rrˆˆ +=
v
2θ&&& rrar −=
θθθ&&&& rra 2+=
Example 12-19
The searchlight casts a spot of light along the face of a wall that is
located 100m from the searchlight. Determine the magnitudes of
the velocity and acceleration at which the spot appears to travel
across the wall at the instant θ = 45°. The searchlight is rotating at
a constant rate of 4 rad/s
12/03/58 MEE214 – Dynamics 33
Problem 12-184
The slotted arm AB drives pin C through the spiral groove
described by the equation r = (1.5Ө) ft, where Ө is in radians. If
the arm starts from rest when Ө = 60˚ and is driven at an angular
velocity of Ө) = (4t) rad/s, where t is in seconds, determine the
radial and transverse components of velocity and acceleration of
the pin C when t=1 s.
12/03/58 MEE214 – Dynamics 34
the pin C when t=1 s.
Absolute Dependent Motion
• Dependent motions of two particles are
normally associated with systems of
connected masses via inextensible cords and pulleys.
12/03/58 MEE214 – Dynamics 35
pulleys.
Absolute Dependent Motion
BA ssl 3+=
vv 30 +=
12/03/58 MEE214 – Dynamics 36
BA vv 30 +=
BA aa 30 +=
Problem 12-206
If the hydraulic cylinder at H draws in rod BC
by 200 mm at 2ft/s, determine how far the slider A moves and the speed of the slider.
12/03/58 MEE214 – Dynamics 37
Example 3
A man at A is hoisting a safe Sby walking to the right with a
constant velocity vA = 0.5m/s.
Determine the velocity and
12/03/58 MEE214 – Dynamics 38
Determine the velocity and
acceleration of the safe when it
reaches the elevation at E. The
rope is 30m long and passes
over a small pulley at D.
Problem 12-208
If block A is moving downward with a speed of
6 ft/s while C is moving down at 18 m/s, determine the speed of block B.
12/03/58 MEE214 – Dynamics 39
Relative Motion Analysis
The relative position of B with respect to A is given by
The relative velocity and acceleration of B with respect
ABAB rrrvvv
−=/
12/03/58 MEE214 – Dynamics 40
The relative velocity and acceleration of B with respect
to A are given by
ABAB vvvvvv
−=/
ABAB aaavvv
−=/
Example 4
A train, traveling at a constant speed of 60 mi/h, crosses
over a road. If automobile A is traveling t 45 mi/h along
the road, determine the magnitude and direction of
relative velocity of the train with respect to the
automobile.
12/03/58 MEE214 – Dynamics 41
automobile.
Problem 12-149
The two particles A and B start at the origin O and travel in opposite directions along the circular path at constant speeds vA=0.7 m/s and vB=1.5 m/s respectively.
12/03/58 MEE214 – Dynamics 42
Determine at t= 2s, (a) the displacement along the path of each particle, (b) the position vector to each particle, and (c) the magnitude of the acceleration of particle B.