me 208 dynamics
TRANSCRIPT
ME 208 DYNAMICS
Dr. Ergin TÃNÃK
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.
ÔŠð£ð = ÔŠð£ðŽ + ÔŠð£ð/ðŽÔŠð£ð = ððð·ðž à Ԋðð/ð = ððð·ðž
ð Ã 0.12 Æžð = 0.12ððð·ðž Æžð
ÔŠð£ðŽ = ðð à ԊððŽ/ð¶ = 4ð à â0.12 Æžð = 0.48 Æžð ð/ð
ÔŠð£ð/ðŽ = ð£ð/ðŽ ððð 45° Æžð + ð ðð45° Æžð
0.12ððð·ðž Æžð = 0.48 Æžð + ð£ð/ðŽ ððð 45° Æžð + ð ðð45° Æžð
Æžð: 0 = 0.48 + ð£ð/ðŽððð 45°, ð£ð/ðŽ = â0.679 ð/ð
Æžð: 0.12ððð·ðž = ð£ð/ðŽð ðð45°, ððð·ðž = â4 ððð/ð
Assume a point P fixed on body ODE instantly
coincident with A.ÔŠðð = ÔŠððŽ + ÔŠðð/ðŽÔŠðð = ÔŠððð¡ + ÔŠðððÔŠðð = ÔŠðŒðð·ðž à Ԋðð/ð â ððð·ðž
2 ÔŠðð/ðÔŠðð = ðŒðð·ðž ð à 0.12 Æžð â 42 â 0.12 ÆžðÔŠðð = â1.920 Æžð + 0.12ðŒðð·ðž Æžð ð/ð 2
ÔŠððŽ = ÔŠððŽð¡ + ÔŠððŽð = ÔŠðŒð¶ðŽ à ԊððŽ/ð¶ âð2 ÔŠððŽ/ð¶
ÔŠððŽ = 0 à â0.12 Æžð â 42 â â0.12 Æžð = 1.920 Æžð ð/ð 2
ÔŠðð/ðŽ = 2ððð·ðž à Ԋð£ð/ðŽ + ÔŠððððÔŠðð/ðŽ = 2 â â4ð à â0.679 ððð 45° Æžð + ð ðð45° Æžð + ðððð ððð 45° Æžð + ð ðð45° Æžð
ÔŠðð/ðŽ = 3.84 â Æžð + Æžð + ðððð ððð 45° Æžð + ð ðð45° Æžðâ1.92 Æžð + 0.12ðŒðð·ðž Æžð = 1.92 Æžð + 3.84 â Æžð + Æžð + ðððð ððð 45° Æžð + ð ðð45° Æžð
Æžð: â1.92 = â3.84 + ððððððð 45°ðððð = 2.72 ð/ð 2
Æžð: 0.12ðŒðð·ðž = 1.92 + 3.84 + 2.72ð ðð45°ðŒðð·ðž = 64.0 ððð/ð 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 Newtonâs Second Law
For a system of particles in a
volume, let ÔŠð¹ð be the
resultant of forces acting on
a particle ðð due to sources
external to the system
boundary whereas ÔŠðð be the
resultant of forces on ðð due
to sources within the system
boundary. Newtonâs second
law for the particle i is then:
ÔŠð¹ð + ÔŠðð = ðð ÔŠðð
Now recall the definition of
mass center (center of mass)
from statics:
ðÔŠððº =
ð=1
ð
ðð ÔŠðð = නð
ÔŠð ðð
where
ð =
ð=1
ð
ðð = නð
ðð
Second time derivative of this
equation yields
ð á·ÔŠððº =
ð=1
ð
ððá·ÔŠðð =
ð=1
ð
ðð ÔŠðð
ð=1
ð
ÔŠð¹ð + ÔŠðð =
ð=1
ð
ðð ÔŠðð = ð á·ÔŠððº = ðÔŠððº
However when summed up within the
system boundary, due to Newtonâs third
law, the action-reaction principle,
ð=1
ð
ÔŠðð = 0
This leads to the principle of motion of
center of mass:
ð=1
ð
ÔŠð¹ð = ð ÔŠððº
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.
ð=1
ð
ÔŠð¹ð = ð ÔŠððº
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 ðð the work-energy relation:
ðð = âðð + âððð+ âððð
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.
ð=1
ð
ðð =
ð=1
ð
âðð +
ð=1
ð
âððð+
ð=1
ð
âððð
For the system
ð1â2 = âð + âðð + âðð where ð1â2 contains only the
work done by external forces.
âð =
ð=1
ð1
2ððð£ð
2
ÔŠð£ð = ÔŠð£ðº +á¶ÔŠðð
ð£ð2 = ÔŠð£ð â ÔŠð£ð = ÔŠð£ðº +
á¶ÔŠðð â ÔŠð£ðº +á¶ÔŠðð = ð£ðº
2 + á¶ðð2 + 2 ÔŠð£ðº â
á¶ÔŠððSubstitution into kinetic energy expression yields
âð =
ð=1
ð1
2ððð£ðº
2 +
ð=1
ð1
2ðð á¶ðð
2 +
ð=1
ð
ðð ÔŠð£ðº âá¶ÔŠðð
âð =1
2ðð£ðº
2 +
ð=1
ð1
2ðð á¶ðð
2 + ÔŠð£ðº â
ð=1
ð
ððá¶ÔŠðð
âð =1
2ðð£ðº
2 +
ð=1
ð1
2ðð á¶ðð
2 + ÔŠð£ðº â
ð=1
ð
ððá¶ÔŠðð
but
ð=1
ð
ðð ÔŠðð = 0
(definition of mass center so its time derivative is
zero too!)
âð =1
2ðð£ðº
2 +
ð=1
ð1
2ðð á¶ðð
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 MomentumÔŠðºð = ðð ÔŠð£ð
ÔŠðº =
ð=1
ð
ÔŠðºð =
ð=1
ð
ðð ÔŠð£ð =
ð=1
ð
ðð ÔŠð£ðº +á¶ÔŠðð
ÔŠðº =
ð=1
ð
ðð ÔŠð£ðº +ð
ðð¡
ð=1
ð
ðð ÔŠðð = ÔŠð£ðº
ð=1
ð
ðð + 0 = ð ÔŠð£ðº
á¶ÔŠðº =ð ÔŠðº
ðð¡=
ð
ðð¡ð ÔŠð£ðº = ð ÔŠððº = ÔŠð¹
This is valid for constant mass ( á¶ð = 0) systems.
Angular Momentum
⢠Angular Momentum about a Fixed Point O
ð»ð =
ð=1
ð
ÔŠðð Ãðð ÔŠð£ð
á¶ð»ð =
ð=1
ð
á¶ÔŠðð Ãðð ÔŠð£ð +
ð=1
ð
ÔŠðð Ãððá¶ÔŠð£ð = 0 +
ð=1
ð
ÔŠðð Ãðð ÔŠðð
á¶ð»ð =
ð=1
ð
ÔŠðð à Ԋð¹ð =
ð=1
ð
ðð
According to Varignonâs principle sum of moments
of forces is equal to the moment of the resultant of
the forces.
Angular Momentum
⢠Angular Momentum about Mass Center G
ð»ðº =
ð=1
ð
ÔŠðð Ãðð ÔŠð£ð , ÔŠð£ð = ÔŠð£ðº +á¶ÔŠðð
ð»ðº =
ð=1
ð
ÔŠðð Ãðð ÔŠð£ðº +á¶ÔŠðð = â ÔŠð£ðº Ã
ð=1
ð
ðð ÔŠðð +
ð=1
ð
ÔŠðð Ãððá¶ÔŠðð = 0 +
ð=1
ð
ÔŠðð Ãððá¶ÔŠðð
á¶ð»ðº =
ð=1
ð
á¶ÔŠðð Ãððá¶ÔŠðð +
ð=1
ð
ÔŠðð Ãððá·ÔŠðð = 0 +
ð=1
ð
ÔŠðð Ãððá·ÔŠðð
ÔŠðð = ÔŠððº +á·ÔŠðð ,
á·ÔŠðð = ÔŠðð â ÔŠððº
á¶ð»ðº =
ð=1
ð
ÔŠðð Ãðð ÔŠðð â ÔŠððº =
ð=1
ð
ÔŠðð Ãðð ÔŠðð + ÔŠððº Ã
ð=1
ð
ÔŠðððð =
ð=1
ð
ÔŠðð à Ԋð¹ð + 0
According to Varignonâs theorem
á¶ð»ðº =ððº
Angular Momentum
⢠Angular Momentum about an Arbitrary Point P
ð»ð =
ð=1
ð
ðâ²ð Ãðð ÔŠð£ð
ðâ²ð = ÔŠððº + ÔŠðð
ð»ð =
ð=1
ð
ÔŠððº + ÔŠðð Ãðð ÔŠð£ð = ÔŠððº Ã
ð=1
ð
ðð ÔŠð£ð +
ð=1
ð
ÔŠðð Ãðð ÔŠð£ð = ÔŠððº à Ԋðº + ð»ðº
ð»ð = ð»ðº + ÔŠððº à ԊðºOne may write
ðð =ððº + ÔŠððº à Ԋð¹
ðð =á¶
ð»ðº + ÔŠððº Ãð ÔŠððº
Moment about an arbitrary point is time rate of angular
momentum about mass center plus moment of ð ÔŠððº about P.
4/5 Conservation of Energy and Momentum
⢠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
ð1â2 = 0 = âð + âðð + âðð⢠Conservation of MomentumIf for an interval of time
ÔŠð¹ = 0,á¶ÔŠðº = 0, â ÔŠðº = 0
Again for an interval of time if
ððº = 0, á¶ð»ðº = 0, âð»ðº = 0
ðð = 0, á¶ð»ð = 0, âð»ð = 0
Appendix B: Mass Moment of Inertia
B/1 Mass Moment of Inertia about an Axisðð¡ = ððŒðð¹ = ððŒððMoment of this force about O-O
ðð = ððð¹ = ð2ðŒððFor all particles in the rigid body
ð = නð
ðð = නð
ð2ðŒðð = ðŒà¶±ð
ð2ðð
Mass moment of inertia of the body about axis O-O is
defined as
ðŒð ⡠නð
ð2ðð ðð â ð2
Mass is a resistance to linear acceleration, mass
moment of inertia is a resistance to angular
acceleration.
For discrete system of particles
ðŒð =
ð=1
ð
ðð2ðð
For constant density (homogeneous) rigid bodies
ðŒð = ðනð
ð2ðð
Radius of Gyration
ðð¥ â¡ðŒð¥ð, ðŒð¥ = ðð¥
2ð
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)
ðŒð¥ = ðŒðº +ð ðð 2
ðð¥2 = ððº
2 + ðð 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.
ðŒð = ðŒðð ðððð â ðŒðâðððFor thin disks
ðŒðº =1
2ðð2
ðŒðð ðððð = ðŒðº =1
2ðð2 =
1
2ðððð
2ð¡ðð2 = 2.96 ðð.ð2
ðŒðâððð = ðŒðº +ðð2 =1
2ðð2 +ðð2 =
1
2ðððâ
2ð¡ðâ2 + ðððâ
2ð2
= 0.348 ðð.ð2
ðŒð = 2.96 â 0.348 = 2.61 ðð.ð2
ðð =ðŒðð
=2.61
ðððð2ð¡ â ðððâ
2ð¡= 0.230 ð
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 Newtonâs Second
Law â Force Mass Acceleration Method
for a Rigid Body
6/2 General Equations of Motion
ÔŠð¹ =á¶ÔŠðº = ð ÔŠððº
ððº =á¶
ð»ðº = ðŒðº ÔŠðŒ
These are known as Eulerâs 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
ÔŠð¹ = ð ÔŠððº
ð»ðº =
ð=1
ð
ÔŠðð Ãððá¶ÔŠðð = න
ð
ÔŠð à á¶ÔŠððð
For a rigid body ÔŠðð = ðððð ð¡ thereforeá¶ÔŠð = ð à Ԋð
ÔŠð à á¶ÔŠð = ÔŠð à ð à Ԋð = â ÔŠð à Ԋð à ð = ð2ð
ð»ðº = නð
ð2ððð = ðනð
ð2ðð = ððŒðº
For a rigid body IG is constant so á¶
ð»ðº = ðŒðºá¶ð = ðŒðº ÔŠðŒ
ÔŠð¹ = ð ÔŠððº ,ððº = ðŒðº ÔŠðŒ
ÔŠð¹ = ð ÔŠððº
ððº = ðŒðº ÔŠðŒ
Eulerâs laws of motion, generalization of Newtonâs 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
ðð =á¶
ð»ðº + ÔŠððº Ãð ÔŠððº , ÔŠððº = ððº
This equation can be written as
ðð = ðŒðºðŒ +ðððððð¡ ðð ð ÔŠððº ðððð¢ð¡ ð
If point P is a fixed point (like the axis of rotation) then
ðð = ðŒðºðŒ +ðððððð¡ ðð ð ÔŠððº ðððð¢ð¡ ð, ððºð¡ = ððº ðŒ,
ðððððð¡ ðð ð ÔŠððº ðððð¢ð¡ ð = ððº 2ðŒ
ðð = ðŒðºðŒ + ððº 2ððŒ = ðŒðº +ð ððº 2 ðŒ = ðŒððŒ
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 ÔŠð£ðº , ÔŠððº , 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.