Download - ME 208 DYNAMICS
ME 208 DYNAMICS
Dr. Ergin TΓNΓK
Department of Mechanical Engineering
Graduate Program of Biomedical Engineering
http://tonuk.me.metu.edu.tr
5/163 (4th), 5/168 (5th), 5/178 (6th), 5/176 (7th), 5/179 (8th)
For the instant represented, link CB is rotating
counterclockwise at a constant rate N = 4 rad/s, and its pin
A causes a clockwise rotation of the slotted member ODE.
Determine the angular velocity and angular acceleration
of ODE for this instant.
Assume a point P fixed on body ODE instantly
coincident with A.
Τ¦π£π = Τ¦π£π΄ + Τ¦π£π/π΄Τ¦π£π = πππ·πΈ Γ Τ¦ππ/π = πππ·πΈ
π Γ 0.12 ΖΈπ = 0.12πππ·πΈ ΖΈπ
Τ¦π£π΄ = ππ Γ Τ¦ππ΄/πΆ = 4π Γ β0.12 ΖΈπ = 0.48 ΖΈπ π/π
Τ¦π£π/π΄ = π£π/π΄ πππ 45Β° ΖΈπ + π ππ45Β° ΖΈπ
0.12πππ·πΈ ΖΈπ = 0.48 ΖΈπ + π£π/π΄ πππ 45Β° ΖΈπ + π ππ45Β° ΖΈπ
ΖΈπ: 0 = 0.48 + π£π/π΄πππ 45Β°, π£π/π΄ = β0.679 π/π
ΖΈπ: 0.12πππ·πΈ = π£π/π΄π ππ45Β°, πππ·πΈ = β4 πππ/π
Assume a point P fixed on body ODE instantly
coincident with A.Τ¦ππ = Τ¦ππ΄ + Τ¦ππ/π΄Τ¦ππ = Τ¦πππ‘ + Τ¦πππΤ¦ππ = Τ¦πΌππ·πΈ Γ Τ¦ππ/π β πππ·πΈ
2 Τ¦ππ/πΤ¦ππ = πΌππ·πΈ π Γ 0.12 ΖΈπ β 42 β 0.12 ΖΈπΤ¦ππ = β1.920 ΖΈπ + 0.12πΌππ·πΈ ΖΈπ π/π 2
Τ¦ππ΄ = Τ¦ππ΄π‘ + Τ¦ππ΄π = Τ¦πΌπΆπ΄ Γ Τ¦ππ΄/πΆ βπ2 Τ¦ππ΄/πΆ
Τ¦ππ΄ = 0 Γ β0.12 ΖΈπ β 42 β β0.12 ΖΈπ = 1.920 ΖΈπ π/π 2
Τ¦ππ/π΄ = 2πππ·πΈ Γ Τ¦π£π/π΄ + Τ¦ππππΤ¦ππ/π΄ = 2 β β4π Γ β0.679 πππ 45Β° ΖΈπ + π ππ45Β° ΖΈπ + ππππ πππ 45Β° ΖΈπ + π ππ45Β° ΖΈπ
Τ¦ππ/π΄ = 3.84 β ΖΈπ + ΖΈπ + ππππ πππ 45Β° ΖΈπ + π ππ45Β° ΖΈπβ1.92 ΖΈπ + 0.12πΌππ·πΈ ΖΈπ = 1.92 ΖΈπ + 3.84 β ΖΈπ + ΖΈπ + ππππ πππ 45Β° ΖΈπ + π ππ45Β° ΖΈπ
ΖΈπ: β1.92 = β3.84 + πππππππ 45Β°ππππ = 2.72 π/π 2
ΖΈπ: 0.12πΌππ·πΈ = 1.92 + 3.84 + 2.72π ππ45Β°πΌππ·πΈ = 64.0 πππ/π 2
Plane Kinetics of Rigid BodiesChapter 4: Kinetics of Systems of Particles
4/1 IntroductionSince a rigid body is assumed to be a continuous collection of
infinitely many particles of infinitesimal mass (continuum
approximation) with no change in the relative positions of
particles, the kinetic equations are derived using a general
system of particles (rigid or deformable or even flowing) in our
textbook. This approach is very general and when the rigidity
condition is imposed the equations will simplify to what we
will be using in Chapter 6 for kinetics of rigid bodies.
4/2 Generalized Newtonβs Second Law
For a system of particles in a
volume, let Τ¦πΉπ be the
resultant of forces acting on
a particle ππ due to sources
external to the system
boundary whereas Τ¦ππ be the
resultant of forces on ππ due
to sources within the system
boundary. Newtonβs second
law for the particle i is then:
Τ¦πΉπ + Τ¦ππ = ππ Τ¦ππ
Now recall the definition of
mass center (center of mass)
from statics:
πΤ¦ππΊ =
π=1
π
ππ Τ¦ππ = ΰΆ±π
Τ¦π ππ
where
π =
π=1
π
ππ = ΰΆ±π
ππ
Second time derivative of this
equation yields
π α·Τ¦ππΊ =
π=1
π
ππα·Τ¦ππ =
π=1
π
ππ Τ¦ππ
π=1
π
Τ¦πΉπ + Τ¦ππ =
π=1
π
ππ Τ¦ππ = π α·Τ¦ππΊ = πΤ¦ππΊ
However when summed up within the
system boundary, due to Newtonβs third
law, the action-reaction principle,
π=1
π
Τ¦ππ = 0
This leads to the principle of motion of
center of mass:
π=1
π
Τ¦πΉπ = π Τ¦ππΊ
which states that the acceleration of
center of mass of the system of particles
is in the same direction with the
resultant external force and is inversely
proportional with the total mass of the
system of particles. Please note that the
line of action of resultant of external
forces need not pass through the mass
center, G.
π=1
π
Τ¦πΉπ = π Τ¦ππΊ
This vector equation may be
resolved in any convenient
coordinate system:
x-y,
n-t,
r-.
This is sufficient for rigid
bodies.
4/3 Work-Energy
For a particle of mass ππ the work-energy relation:
ππ = βππ + βπππ+ βπππ
The work done by internal forces cancel each other
because for a rigid body the action and reaction
pairs have identical displacements therefore the
work done only by the external forces need to be
considered during summation. However for non-
rigid bodies displacements of action reaction pairs
may be different causing storage of elastic potential
energy or dissipation of mechanical energy which
we will not discuss in this course.
π=1
π
ππ =
π=1
π
βππ +
π=1
π
βπππ+
π=1
π
βπππ
For the system
π1β2 = βπ + βππ + βππ where π1β2 contains only the
work done by external forces.
βπ =
π=1
π1
2πππ£π
2
Τ¦π£π = Τ¦π£πΊ +αΆΤ¦ππ
π£π2 = Τ¦π£π β Τ¦π£π = Τ¦π£πΊ +
αΆΤ¦ππ β Τ¦π£πΊ +αΆΤ¦ππ = π£πΊ
2 + αΆππ2 + 2 Τ¦π£πΊ β
αΆΤ¦ππSubstitution into kinetic energy expression yields
βπ =
π=1
π1
2πππ£πΊ
2 +
π=1
π1
2ππ αΆππ
2 +
π=1
π
ππ Τ¦π£πΊ βαΆΤ¦ππ
βπ =1
2ππ£πΊ
2 +
π=1
π1
2ππ αΆππ
2 + Τ¦π£πΊ β
π=1
π
ππαΆΤ¦ππ
βπ =1
2ππ£πΊ
2 +
π=1
π1
2ππ αΆππ
2 + Τ¦π£πΊ β
π=1
π
ππαΆΤ¦ππ
but
π=1
π
ππ Τ¦ππ = 0
(definition of mass center so its time derivative is
zero too!)
βπ =1
2ππ£πΊ
2 +
π=1
π1
2ππ αΆππ
2
The first term is translational kinetic energy of the
mass center, G, the second term is due to relative
motion of particles with respect to the mass center,
G.
4/4 Impulse Momentum
Linear MomentumΤ¦πΊπ = ππ Τ¦π£π
Τ¦πΊ =
π=1
π
Τ¦πΊπ =
π=1
π
ππ Τ¦π£π =
π=1
π
ππ Τ¦π£πΊ +αΆΤ¦ππ
Τ¦πΊ =
π=1
π
ππ Τ¦π£πΊ +π
ππ‘
π=1
π
ππ Τ¦ππ = Τ¦π£πΊ
π=1
π
ππ + 0 = π Τ¦π£πΊ
αΆΤ¦πΊ =π Τ¦πΊ
ππ‘=
π
ππ‘π Τ¦π£πΊ = π Τ¦ππΊ = Τ¦πΉ
This is valid for constant mass ( αΆπ = 0) systems.
Angular Momentum
β’ Angular Momentum about a Fixed Point O
π»π =
π=1
π
Τ¦ππ Γππ Τ¦π£π
αΆπ»π =
π=1
π
αΆΤ¦ππ Γππ Τ¦π£π +
π=1
π
Τ¦ππ ΓππαΆΤ¦π£π = 0 +
π=1
π
Τ¦ππ Γππ Τ¦ππ
αΆπ»π =
π=1
π
Τ¦ππ Γ Τ¦πΉπ =
π=1
π
ππ
According to Varignonβs principle sum of moments
of forces is equal to the moment of the resultant of
the forces.
Angular Momentum
β’ Angular Momentum about Mass Center G
π»πΊ =
π=1
π
Τ¦ππ Γππ Τ¦π£π , Τ¦π£π = Τ¦π£πΊ +αΆΤ¦ππ
π»πΊ =
π=1
π
Τ¦ππ Γππ Τ¦π£πΊ +αΆΤ¦ππ = β Τ¦π£πΊ Γ
π=1
π
ππ Τ¦ππ +
π=1
π
Τ¦ππ ΓππαΆΤ¦ππ = 0 +
π=1
π
Τ¦ππ ΓππαΆΤ¦ππ
αΆπ»πΊ =
π=1
π
αΆΤ¦ππ ΓππαΆΤ¦ππ +
π=1
π
Τ¦ππ Γππα·Τ¦ππ = 0 +
π=1
π
Τ¦ππ Γππα·Τ¦ππ
Τ¦ππ = Τ¦ππΊ +α·Τ¦ππ ,
α·Τ¦ππ = Τ¦ππ β Τ¦ππΊ
αΆπ»πΊ =
π=1
π
Τ¦ππ Γππ Τ¦ππ β Τ¦ππΊ =
π=1
π
Τ¦ππ Γππ Τ¦ππ + Τ¦ππΊ Γ
π=1
π
Τ¦ππππ =
π=1
π
Τ¦ππ Γ Τ¦πΉπ + 0
According to Varignonβs theorem
αΆπ»πΊ =ππΊ
Angular Momentum
β’ Angular Momentum about an Arbitrary Point P
π»π =
π=1
π
πβ²π Γππ Τ¦π£π
πβ²π = Τ¦ππΊ + Τ¦ππ
π»π =
π=1
π
Τ¦ππΊ + Τ¦ππ Γππ Τ¦π£π = Τ¦ππΊ Γ
π=1
π
ππ Τ¦π£π +
π=1
π
Τ¦ππ Γππ Τ¦π£π = Τ¦ππΊ Γ Τ¦πΊ + π»πΊ
π»π = π»πΊ + Τ¦ππΊ Γ Τ¦πΊOne may write
ππ =ππΊ + Τ¦ππΊ Γ Τ¦πΉ
ππ =αΆ
π»πΊ + Τ¦ππΊ Γπ Τ¦ππΊ
Moment about an arbitrary point is time rate of angular
momentum about mass center plus moment of π Τ¦ππΊ about P.
4/5 Conservation of Energy and Momentum
β’ Conservation of EnergyA conservative system does not lose mechanical energy due
to internal friction or other types of inelastic forces. If no
work is done on a conservative system by external forces
then
π1β2 = 0 = βπ + βππ + βππβ’ Conservation of MomentumIf for an interval of time
Τ¦πΉ = 0,αΆΤ¦πΊ = 0, β Τ¦πΊ = 0
Again for an interval of time if
ππΊ = 0, αΆπ»πΊ = 0, βπ»πΊ = 0
ππ = 0, αΆπ»π = 0, βπ»π = 0
Appendix B: Mass Moment of Inertia
B/1 Mass Moment of Inertia about an Axisππ‘ = ππΌππΉ = ππΌππMoment of this force about O-O
ππ = πππΉ = π2πΌππFor all particles in the rigid body
π = ΰΆ±π
ππ = ΰΆ±π
π2πΌππ = πΌΰΆ±π
π2ππ
Mass moment of inertia of the body about axis O-O is
defined as
πΌπ β‘ ΰΆ±π
π2ππ ππ β π2
Mass is a resistance to linear acceleration, mass
moment of inertia is a resistance to angular
acceleration.
For discrete system of particles
πΌπ =
π=1
π
ππ2ππ
For constant density (homogeneous) rigid bodies
πΌπ = πΰΆ±π
π2ππ
Radius of Gyration
ππ₯ β‘πΌπ₯π, πΌπ₯ = ππ₯
2π
Radius of gyration is a measure of mass
distribution of a rigid body about the axis. A
system with equivalent mass moment of inertia is
a very thin ring of same mass and radius kx.
Transfer of Axis (Parallel Axis Theorem)
πΌπ₯ = πΌπΊ +π ππ 2
ππ₯2 = ππΊ
2 + ππ 2
Composite Bodies
Mass moment of inertia of a composite body about
an axis is the sum of individual mass moments of
each part about the same axis (which may be
calculated utilizing parallel axis theorem if mass
moment of inertia of each part is known about its
mass center).
B/10 (4th), None (5th), B/14 (6th), None (7th), None (8th)
Calculate the mass moment of inertia about the axis O-O
for the steel disk with the hole.
πΌπ = πΌππ ππππ β πΌπβπππFor thin disks
πΌπΊ =1
2ππ2
πΌππ ππππ = πΌπΊ =1
2ππ2 =
1
2ππππ
2π‘ππ2 = 2.96 ππ.π2
πΌπβπππ = πΌπΊ +ππ2 =1
2ππ2 +ππ2 =
1
2πππβ
2π‘πβ2 + πππβ
2π2
= 0.348 ππ.π2
πΌπ = 2.96 β 0.348 = 2.61 ππ.π2
ππ =πΌππ
=2.61
ππππ2π‘ β πππβ
2π‘= 0.230 π
Chapter 6: Plane Kinetics of Rigid Bodies
6/1 IntroductionIn this chapter we will deal with relations among external
forces and moments, and, translational and rotational
motions of rigid bodies in plane motion.
We will write two force and one moment equation (or
equivalent) for the plane motion of rigid bodies.
The equations derived in Chapter 4 will be simplified for a
rigid body and used. Kinematic relations developed in
Chapters 2 and 5 will be utilized.
Drawing correct free body diagrams is essential in
application of Force-Mass-Acceleration method. The other
two methods, similar to kinetics of particles are Work-
Energy and Impulse-Momentum.
A. Direct Application of Newtonβs Second
Law β Force Mass Acceleration Method
for a Rigid Body
6/2 General Equations of Motion
Τ¦πΉ =αΆΤ¦πΊ = π Τ¦ππΊ
ππΊ =αΆ
π»πΊ = πΌπΊ Τ¦πΌ
These are known as Eulerβs first and second laws.
By using statics information one may replace the forces on a
rigid body by a single resultant force passing through mass
center and a couple moment. The equivalent force causes linear
acceleration of the mass center in the direction of force, the
couple moment causes angular acceleration about the axis of the
couple moment.
Plane Motion Equations
Τ¦πΉ = π Τ¦ππΊ
π»πΊ =
π=1
π
Τ¦ππ ΓππαΆΤ¦ππ = ΰΆ±
π
Τ¦π Γ αΆΤ¦πππ
For a rigid body Τ¦ππ = ππππ π‘ thereforeαΆΤ¦π = π Γ Τ¦π
Τ¦π Γ αΆΤ¦π = Τ¦π Γ π Γ Τ¦π = β Τ¦π Γ Τ¦π Γ π = π2π
π»πΊ = ΰΆ±π
π2πππ = πΰΆ±π
π2ππ = ππΌπΊ
For a rigid body IG is constant so αΆ
π»πΊ = πΌπΊαΆπ = πΌπΊ Τ¦πΌ
Τ¦πΉ = π Τ¦ππΊ ,ππΊ = πΌπΊ Τ¦πΌ
Τ¦πΉ = π Τ¦ππΊ
ππΊ = πΌπΊ Τ¦πΌ
Eulerβs laws of motion, generalization of Newtonβs second
law for particles to rigid bodies approximately 50 years
after Newton.
For plane motion the force equation may be resolved in x-y,
n-t or r- coordinates whichever is suitable. For moment
equation it is always normal to the plane of motion
therefore can be expressed in scalar form as CCW or CW.
The moment equation has an alternative derivation yielding
the same result. Please go over it in the textbook.
Alternative Moment Equation
Sometimes it may be more convenient to take moment
about another point rather than the mass center G. In
that case
ππ =αΆ
π»πΊ + Τ¦ππΊ Γπ Τ¦ππΊ , Τ¦ππΊ = ππΊ
This equation can be written as
ππ = πΌπΊπΌ +ππππππ‘ ππ π Τ¦ππΊ ππππ’π‘ π
If point P is a fixed point (like the axis of rotation) then
ππ = πΌπΊπΌ +ππππππ‘ ππ π Τ¦ππΊ ππππ’π‘ π, ππΊπ‘ = ππΊ πΌ,
ππππππ‘ ππ π Τ¦ππΊ ππππ’π‘ π = ππΊ 2πΌ
ππ = πΌπΊπΌ + ππΊ 2ππΌ = πΌπΊ +π ππΊ 2 πΌ = πΌππΌ
In unconstrained motion the two components of
acceleration of the mass center and angular acceleration
are independent of each other as in the case of a rocket.
In constrained motion due to kinematic restrictions there
are relations among two components of the acceleration of
mass center and the angular acceleration of the body.
Therefore these kinematic constraint equations have to be
determined using methods developed in Chapter 5. There
are also reaction forces due to constraints in the direction
of restricted motions which should be included in the free
body diagram.
Analysis Procedure
Kinematics: Determine Τ¦π£πΊ , Τ¦ππΊ , and (or the
kinematic relations among them) if possible.
Diagrams: Draw proper free body and kinetic
diagrams.
Equations of motion: Any force or acceleration in
the direction of positive coordinate is positive.
Count the number of available independent
equations and number of unknowns to be
determined.