1
Department of Mechanical, Industrial & Aerospace Engineering
Dynamics (ENGR 243)
LECTURE 10
Dr. Hany Gomaa,
Email:
hany.gomaa@concordia.ca
Office: EV-3.269
Web Site: Access from your “My Concordia” portal - Moodle
Resources Used:
1. Vector Mechanics for Engineers – Dynamics,” by F.P. Beer, E.R. Johnston, Jr., & P.J. Cornwell,
Tenth/ Eleventh Edition in , McGraw-Hill, 2013 / 2016 &
2. Hibbeler Engineering Mechanics: Dynamics, Fourteenth Edition, Pearson, 2016
Dr. Hany Gomaa
Course Information
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Vector Mechanics for Engineers – Dynamics by F.P. Beer, et al.
Topics – Chapters
1. Chapter 11 Introduction to Dynamics (Kinematics of Particles)
Rectilinear Motion of Particles - Curvilinear Motion of Particles
2. Chapter 12 Kinetics of Particles
Newton’s Second Law
3. Chapter 13 Kinetics of Particles
Energy and Momentum Methods
4. Chapter 15 Kinematics of Rigid Bodies
•Translation (Rectilinear – Curvilinear)
•Rotation around a fixed axis
•General Plane Motion
5. Chapter 16 Plane Motion of Rigid Bodies
Forces and Acceleration
6. Chapter 17 Plane Motion of Rigid Bodies
Energy and Momentum Methods
Vector Mechanics for Engineers – Dynamics,”
by F.P. Beer, E.R. Johnston, Jr., & P.J. Cornwell, Tenth Edition in SI Units, McGraw-Hill, 2013
“Vector Mechanics for Engineers – Dynamics,” by F.P. Beer, E.R. Johnston, Jr., & P.J. Cornwell, Tenth Edition in SI Units, McGraw-Hill, 2013
Dr. Hany Gomaa
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3) General Plane Motion
Dr. Hany Gomaa
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Ch. 15: Kinematics of Rigid Bodies
Translation
Rotation
Acceleration analysis
Velocity analysis
Translation
+Rotation
General Plane Motion
Instantaneous
center of rotation
Dr. Hany Gomaa
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APPLICATIONS
In the mechanism for a window, link AC
rotates about a fixed axis through C, and
AB undergoes general plane motion.
Since point A moves along a curved path,
it has two components of acceleration
while point B, sliding in a straight track,
has only one.
The components of acceleration of these
points can be inferred since their motions
are known.
How can we determine the accelerations
of the links in the mechanism?
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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APPLICATIONS
In an automotive engine, the
forces
delivered
to
the
crankshaft, and the angular
acceleration of the crankshaft,
depend on the speed and
acceleration of the piston.
How can we relate the
accelerations of the :
1) piston,
2) connection rod, and
3) crankshaft to each other?
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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APPLICATIONS
In an automotive engine, the
forces
delivered
to
the
crankshaft, and the angular
acceleration of the crankshaft,
depend on the speed and
acceleration of the piston.
How can we relate the
accelerations of the :
1) piston,
2) connection rod, and
3) crankshaft to each other?
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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RELATIVE MOTION ANALYSIS
When a body is subjected to general plane motion, it undergoes a
combination of translation and rotation.
Point A is called the base point in this analysis. It generally has a
known motion. The x’- y’ frame translates with the body, but
does not rotate. The displacement of point B can be written:
Disp. due to translation
drB = drA + drB/A
Disp. due to
translation and rotation
Disp. due to rotation
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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RELATIVE MOTION ANALYSIS: VELOCITY
The velocity at B is given as : (drB/dt) = (drA/dt) + (drB/A/dt) or
vB = vA + vB/A
Since the body is taken as rotating about A,
vB/A = drB/A/dt = w × rB/A
Here w will only have a k component since the axis of rotation is perpendicular to the plane of
translation.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
RELATIVE MOTION ANALYSIS: ACCELERATION
The equation relating the accelerations of two points on the
body is determined by differentiating the velocity equation
vB = vA + vB/A
with respect to time.
dvB
dvA
dv
=
+ B/ A
dt
dt
dt
The result is aB = aA + (aB/A)t + (aB/A)n
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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RELATIVE MOTION ANALYSIS: ACCELERATION
The equation relating the accelerations of two points on the
body is determined by differentiating the velocity equation
vB = vA + vB/A
with respect to time.
dvB
dvA
dv
=
+ B/ A
dt
dt
dt
These are absolute accelerations
of points A and B. They are
measured from a set of fixed x,
y axes.
This term is the acceleration
of B with respect to A and
includes both tangential and
normal components.
The result is aB = aA + (aB/A)t + (aB/A)n
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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RELATIVE MOTION ANALYSIS: ACCELERATION
Graphically:
aB = aA +
(aB/A)t + (aB/A)n
The relative tangential acceleration component (aB/A)t is (α ×rB/A)
and perpendicular to rB/A.
The relative normal acceleration component (aB/A)n is (−ω2 rB/A)
and the direction is always from B towards A.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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RELATIVE MOTION ANALYSIS: ACCELERATION
Since the relative acceleration components can be expressed
as (aB/A)t = α × rB/A and (aB/A)n = - ω2 rB/A, the relative
acceleration equation becomes
aB = aA + α × rB/A − ω2 rB/A
Note that the last term in the relative acceleration equation is
not a cross product. It is the product of a scalar (square of
the magnitude of angular velocity, ω2 and the relative
position vector, rB/A.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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APPLICATION OF THE RELATIVE ACCELERATION
EQUATION
In applying the relative acceleration equation, the two points used in the
analysis (A and B) should generally be selected as points which have a
known motion, such as pin connections with other bodies.
In this mechanism, point B is known to travel along a circular path, so
aB can be expressed in terms of its normal and tangential components.
Note that point B on link BC will have the same acceleration as point B
on link AB.
Point C, connecting link BC and the piston, moves along a straight-line
path. Hence, aC is directed horizontally.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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PROCEDURE FOR ANALYSIS
1. Establish a fixed coordinate system.
2. Draw the kinematic diagram of the body.
1
2
3
4
5
3. Indicate on it aA, aB, ω, α, and rB/A. If the points A and B
move along curved paths, then their accelerations should
be indicated in terms of their tangential and normal
components, i.e., aA = (aA)t + (aA)n and aB = (aB)t + (aB)n.
4. Apply the relative acceleration equation:
aB = aA + α × rB/A − ω2 rB/A
5. If the solution yields a negative answer for an unknown
magnitude, this indicates that the sense of direction of the
vector is opposite to that shown on the diagram.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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Constant Acceleration Eqs.
n-t coordinate system Eqs.
r-θ coordinate system Eqs.
Relative motion velocity
Constant velocity Equ.
Projectile Constant
Velocity in x direction
Projectile Constant
Acceleration in y direction
Rotation about fixed axis equations
GPM Velocity Analysis
(Translation + Rotation)
GPM Velocity Acceleration Analysis
(Translation + Rotation)
Dr. Hany Gomaa
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EXAMPLE I
IC
Given:Point A on rod AB has an
acceleration of 5 m/s2 and a
velocity of 6 m/s at the instant
shown.
Find: The angular acceleration of
the rod and the acceleration at
B at this instant.
Plan: Follow the problem solving
procedure!
Solution: First, we need to find the angular velocity of the rod
at this instant. Locating the instant center (IC) for
rod AB, we can determine ω:
ω = vA/rA/IC = vA / (3) = 2 rad/s
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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EXAMPLE I
Since points A and B both move
along straight-line paths,
aA = -5 j m/s2
aB = aB i m/s2
Applying the relative acceleration equation
aB = aA + α × rB/A – ω2rB/A
aB i = - 5 j + α k × (3 i – 4 j) – 22 (3 i – 4 j)
aB i = - 5 j + 4 α i + 3 α j – (12 i – 16 j)
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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EXAMPLE I (continued)
So with aB i = - 5 j + 4 α i + 3 α j – (12 i – 16 j) , we can solve.
By comparing the i, j components;
aB = 4 α – 12
0 = 11 + 3α
Solving:
aB = - 26.7 m/s2
α = - 3.67 rad/s2
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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BODIES IN CONTACT
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BODIES IN CONTACT
16_005a
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BODIES IN CONTACT
16_005b
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BODIES IN CONTACT
16_005c
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ATTENTION QUIZ
1. Two bodies contact one another
without slipping. If the tangential
component of the acceleration of point
A on gear B is 100 ft/sec2, determine
the tangential component of the
acceleration of point A’ on gear C.
A) 50 ft/sec2
B) 100 ft/sec2
C) 200 ft/sec2
D) None of above.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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ATTENTION QUIZ
1. Two bodies contact one another
without slipping. If the tangential
component of the acceleration of point
A on gear B is 100 ft/sec2, determine
the tangential component of the
acceleration of point A’ on gear C.
A) 50 ft/sec2
B) 100 ft/sec2
C) 200 ft/sec2
D) None of above.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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ATTENTION QUIZ
2. If the tangential component of the acceleration of point A on
gear B is 100 ft/sec2, determine the angular acceleration of
gear B.
A) 50 rad/sec2
B) 100 rad/sec2
C) 200 rad/sec2
D) None of above.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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ATTENTION QUIZ
2. If the tangential component of the acceleration of point A on
gear B is 100 ft/sec2, determine the angular acceleration of
gear B.
A) 50 rad/sec2
B) 100 rad/sec2
C) 200 rad/sec2
D) None of above.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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BODIES IN CONTACT
Consider two bodies in contact with one another without slipping,
where the points in contact move along different paths.
In this case, the tangential components of acceleration will be the
same, i. e.,
(aA)t = (aA’)t (which implies αB rB = αC rC ).
The normal components of acceleration will not be the same.
(aA)n ≠ (aA’)n so aA ≠ aA’
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
ROLLING MOTION
Another common type of problem encountered in dynamics
involves rolling motion without slip; e.g., a ball, cylinder, or disk
rolling without slipping.
This situation can be analyzed using relative velocity and
acceleration equations.
As the cylinder rolls, point G (center) moves along a straight line.
If ω and α are known, the relative velocity and acceleration
equations can be applied to A, at the instant A is in contact with
the ground.
The point A is the instantaneous center of zero velocity, however
it is not a point of zero acceleration.
Dr. Hany Gomaa
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
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ROLLING MOTION
Since no slip occurs, vA = 0 when A is in contact
with ground. From the kinematic diagram:
vG = vA + ω × rG/A
vG i = 0 + (-ωk) × (r j)
vG = ωr or vG = ωr i
Since G moves along a straight-line path, aG is
• Acceleration: horizontal.
Just before A touches ground, its velocity is directed
downward, and just after contact, its velocity is
directed upward.
Thus, point A accelerates upward as it leaves the
ground.
• Velocity:
aG = aA + α × rG/A – ω2rG/A => aG i = aA j + (-α k) × (r j) – ω2(r j)
Evaluating and equating i and j components:
aG = αr and aA = ω2r or aG = αr i and aA = ω2r j
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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EXAMPLE II
Given:The gear rolls on the
fixed rack.
Find: The accelerations of
point A at this instant.
Plan:
Follow the solution procedure!
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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PROCEDURE FOR ANALYSIS
1. Establish a fixed coordinate system.
2. Draw the kinematic diagram of the body.
1
2
3
4
5
3. Indicate on it aA, aB, ω, α, and rB/A. If the points A and B
move along curved paths, then their accelerations should
be indicated in terms of their tangential and normal
components, i.e., aA = (aA)t + (aA)n and aB = (aB)t + (aB)n.
4. Apply the relative acceleration equation:
aB = aA + α × rB/A − ω2 rB/A
5. If the solution yields a negative answer for an unknown
magnitude, this indicates that the sense of direction of the
vector is opposite to that shown on the diagram.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
33
EXAMPLE II
Given:The gear rolls on the
fixed rack.
Find: The accelerations of
point A at this instant.
Plan:
Follow the solution procedure!
Solution: Since the gear rolls on the fixed rack without slip, aO
is directed to the right with a magnitude of:
aO = αr = (6 rad/s2)(0.3 m)=1.8 m/s2
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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EXAMPLE II
So now with aO = 1.8 m/s2, we can apply the relative
acceleration equation between points O and A.
aA = aO + α × rA/O – ω2 rA/O
aA = 1.8i + (-6k)×(0.3j) –122 (0.3j)
= (3.6 i – 43.2j) m/s2
1.8 m/s2
y
x
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
Sample Problem 15.8
35
SOLUTION:
• The angular velocities are determined by
simultaneously solving the component
equations for



vD = vB + vD B
In the position shown, crank AB has a
constant angular velocity ω1 = 20 rad/s
counterclockwise.
• The angular accelerations are determined
by simultaneously solving the component
equations for



aD = aB + aD B
Determine the angular velocities and
angular accelerations of the connecting
rod BD and crank DE.
15 - 35
Dr. Hany Gomaa
Sample Problem 15.8
36
SOLUTION:
• The angular velocities are determined by simultaneously
solving the component equations for



vD = vB + vD B






vD = ω DE × rD = ω DE k × (− 17i + 17 j )


= −17ω DE i − 17ω DE j






vB = ω AB × rB = 20k × (8i + 14 j )


= −280i + 160 j






vD B = ω BD × rD B = ω BD k × (12i + 3 j )


= −3ω BD i + 12ω BD j
x components:
− 17ω DE = −280 − 3ω BD
y components:
− 17ω DE = +160 + 12ω BD

15 - 36
ω BD

= −(29.33 rad s )k

ω DE

= (11.29 rad s )k
Dr. Hany Gomaa
Sample Problem 15.8
37
• The angular accelerations are determined by
simultaneously solving the component equations for



aD = aB + aD B



2 
aD = α DE × rD − ω DE
rD





2
= α DE k × (− 17i + 17 j ) − (11.29 ) (− 17i + 17 j )




= −17α DE i − 17α DE j + 2170i − 2170 j




2 
2 
aB = α AB × rB − ω AB rB = 0 − (20 ) (8i + 14 j )


= −3200i + 5600 j



2 
aD B = α BD × rB D − ω BD
rB D





= α B D k × (12i + 3 j ) − (29.33)2 (12i + 3 j )




= −3α B D i + 12α B D j − 10,320i − 2580 j
x components: − 17α DE + 3α BD = −15,690
15 - 37
y components: − 17α DE − 12α BD = −6010




2
2
α BD = − 645 rad s k
α DE = 809 rad s k
(
)
(
)
Dr. Hany Gomaa
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Kinematics of Rigid Bodies
Translation
Rotation
Velocity analysis
General Plane Motion
Acceleration analysis
Absolute motion Analysis
Translation
+Rotation
Instantaneous
center of rotation
Dr. Hany Gomaa
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APPLICATIONS
1) The position of the piston, x, can be defined as a function
of the angular position of the crank, θ.
2) By differentiating x with respect to time, the velocity of the
piston can be related to the angular velocity, ω, of the crank.
This is necessary when designing an engine:
The stroke of the piston is defined as the total distance moved
by the piston as the crank angle varies from 0 to 180°.
How does the length of crank AB affect the stroke?
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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ABSOLUTE MOTION ANALYSIS (Section 16.4)
The figure below shows the window using a hydraulic cylinder AB.
The absolute motion analysis method relates
the position of a point, B, on a rigid body
undergoing rectilinear motion to the angular
position, θ, of a line contained in the body.
•
Once a relationship in the form of sB = f (θ) is
established,
the velocity and acceleration of point B are
obtained in terms of the angular velocity and
angular acceleration of the rigid body by taking
the first and second time derivatives of the
position function.
Usually the chain rule must be used when taking the derivatives
of the position coordinate equation.
Dr. Hany Gomaa
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
41
Given:Two slider blocks are connected
by a rod of length 2 m.
Also, vA = 8 m/s and aA = 0.
Find: Angular velocity, ω, and
angular acceleration, α, of
the rod when θ = 60°.
Plan: Choose a fixed reference point and define the
position of the slider A in terms of the parameter θ.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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Adv. Of Absolute Motion Analysis
From the geometry,
sA = 2 cos θ
reference
θ
A
sA
From the geometry,
sA = 2 cos θ
By differentiating with respect to time, vA = -2 ω sin θ
Differentiating vA with respect to time, aA = -2α sin θ – 2ω2 cos θ
Using θ = 60° and vA = 8 m/s and solving for ω:
ω = 8/(-2 sin 60°) = - 4.62 rad/s
Differentiating vA and solving for α,
aA = -2α sin θ – 2ω2 cos θ = 0
α = - ω2/tan θ = -12.32 rad/s2
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
Applications for absolute motion analysis
The rolling of a cylinder is an example of general plane motion.
During this motion, the cylinder rotates clockwise while it
translates to the right.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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16_009_EX
04
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
Absolute Motion Analysis
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PROCEDURE FOR ANALYSIS
The absolute motion analysis method (the parametric method) relates
the position of a point, P, on a rigid body undergoing rectilinear
motion to the angular position, θ (parameter), of a line contained in
the body.
(Often this line is a link in a machine.)
Once a relationship in the form of sP = f(θ) is established, the
velocity and acceleration of point P are obtained in terms of the
angular velocity, w, and angular acceleration, a, of the rigid body by
taking the first and second time derivatives of the position function.
Usually the chain rule must be used when taking the derivatives of
the position coordinate equation.
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
Dr. Hany Gomaa
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continued on next slide
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
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continued on next slide
Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
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Engineering Mechanics Dynamics” by Hibbeler.R.C., 12thedition
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NEXT LECTURE
Dr. Hany Gomaa
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Kinematics of Rigid Bodies
Translation
Rotation
Plane Motion
Velocity
Acceleration
Translation
+Rotation
Instantaneous
center of rotation
Absolute
motion
Absolute &
Relative
acceleration
Coriolis
acceleration
Dr. Hany Gomaa