ENGINEERING MECHANICS-DYNAMICS Instructor: U.H.Kurzweg |

**INTRODUCTION:-The material presented below is an an
extended outline for a 2 credit dynamics course( EGM 3400)
which I have taught here at the University of Florida on and
off for nearly three decades. This course and my other
classes in mechanics and applied mathematics have won
numerous teaching awards including five from the College of
Engineering and three University wide awards. We meet twice
a week for a total of 21 contact hours and the book we have
been using most often is that by
R.C.Hibbeler's,"Engineering Mechanics-Dynamics". You can contact me anytime at kurzweg@ufl.edu . **

**PENDULUM JUMP:(Click Here)
BANKING
OF A RACETRACK:(Click Here)**

** ****KINETICS OF A DRAGSTER:
As another example of a particle dynamics problem consider the
time it takes for a dragster to accelerate from 0 to
60mph. Click HEREto see the mathematical
development of this problem based on F=ma. Note that the time
to reach a speed v for a car of mass M and power P is at
least t=Mv^2/(2P). If I plug the numbers of P=500HP and
W=3600lb (with driver)applicable for the 10 cylinder Dodge
Viper, the minimum time to reach 60mph will be 1.57 sec. The
latest issue of Car and Driver gives the actual value for the
Viper to be 4 seconds even. The factor of two difference
clearly has to do with presence of air resistance at higher
speeds and transmission losses among other things. Nevetheless
the formula clearly shows that one needs small mass and high
engine power to achieve large accelerations. For my 1967
Camaro rated at 275HP the formula gives a time of 2.62sec
which is again shorter by an approximate factor of two to
the six seconds it actually takes me to reach 60
mph. Can you explain, in view of the above formula , why a
bicyclist will beat an automobile everytime for the first few
feet after a standing start?
**

**Lecture 4-More
applications of F=ma . Mass-Spring System. Geosynchronous
Satellite. Determination of Escape Velocity. Sorry about
the poor acoustics of todays lecture in NEB100, have contacted
instructional resources to get the PA system fixed.**

**PERIOD OF A SATELLITE IN
CIRCULAR ORBIT: A simple
calculation involving a polar coordinate description is the
determination of the orbit of a satellite of mass m moving
in a circular orbit about a heavy mass M. The attractive
force on such a satellite is GMm/r^2, where r is the
distance between the mass centers according to Newton's Law
of Universal Gravitation. This force is balanced by
the satellite mass times its radial acceleration, which for
a constant radius circular orbit, is just mv^2/r. One thus
has that the orbital velocity must be v=sqrt(GM/r).
But GMm/a^2=mg, where 'a' is the radius of the large mass
and , for the earth, g=9.81m/s^2=32.2ft/s^2. So
v=a*sqrt(g/r) and the satellite period becomes T=2*p*r/v=(2*p*r^1.5)/(a*sqrt(g)).
For
a near earth satellite one has that r is approximately 'a',
so that there the orbital period becomes 2*p*sqrt(a/g), which turns out to be
about 1hr 24min for the case of a near earth satellite where
a=6378km=3960 miles. What should be the approximate
orbit time of the moon located an average distance of some
239,000 miles from the earth center? If you click HERE you
can see the calculation which gives the height above the
earth's surface a geosynchronous satellite must be placed in
order for it to have a period of one day and hence appear
stationary.**

**CALCULATING THE ORBIT OF
A PLANET: One of the most
famous problems in all of mechanics was Newton's analysis of
the orbit of planets based on his Universal Law of
Gravitation. Kepler, using observational data of Tycho Brahe
, had already determined several years earlier that the
orbits of planets about the sun were in the form of ellipses
of very small eccentricity with the sun at a focus(Kepler's
First Law), that the trajectories swept out equal area per
time(Kepler's Second Law), and that the square of the
orbital period is proportional to the cube of the semi-major
axis(Kepler's Third law). You can see a summary of Newton's
analysis for the earth-sun system by going HERE You can also see a beautiful confirmation
of Kepler's Third Law by clicking HERE. An
interesting (imaginary)view of our galaxy and its known
properties is shown HERE. We can in addition use these laws to
calculate the distance a geosynchronus satellite must be
placed above the earth's equator. Knowing that the time for
a near-earth satellite to go once around is 1.4 hrs, a
geosynchronus satellite, which has a period T= 24 hrs and
obeys Kepler's third law, must be at 'a'=(24/1.4)^(2/3)=6.6
earth radii from the earth's center or (a-1)*3960=22,200
miles above the earth's surface.
**

**Lecture 7-More on
the Work-Energy Principle. Conservation of Energy
T+U=Const in the absence of friction. Pendulum and Satellite
Motion. Phase plane concepts. Click HERE to see the Loop the Loop
problem.**

**PARTICLE FALLING THROUGH A SHAFT
DRILLED THROUGH THE EARTH: Another very good application of the conservation
of energy for a mass m is to determine the time, speed
and acceleration of this mass as it is dropped through a
straight-line shaft drilled through the earth between
two points A and C on its surface. Go HERE for
a discussion of this problem. We show that the period of
the resulant SHM motion will always be the same
regardless of the off-set position d of the shaft
relative to the earth's rotation axis.
**

Lecture 8-Review for First Hour Exam-There will be 3 out of 4 questions to answer. Exam will be during the class period. It will be closed book but you can bring one 3" x 5" card. You are responsible for the material in the Hibbeler book covered in our first seven lectures plus all material covered in class , shown on this WEB page, and encountered while doing your homework.

**Lecture 10-
Impulse-Momentum Principle. Conservation of Linear Momentum in
Collisions. The Coefficient of
Restitution. Bouncing Balls, Gun Recoil, and Car Wreck.**

**DEMONSTRATION OF THE
CONSERVATION OF MOMENTUM LAW: We
have shown you in class that the total momentum is conserved
during the collision of bodies. Lets now apply this law to
two masses m and M moving along the x axis with speeds v and
V, respectively. If v>V and m follows M, the two bodies
will eventually collide . We want to find their speeds after
this collision. Applying the conservation of linear momentum
we have mv+MV=mv'+MV', with the primes indicating the speeds
after collision. To make this problem soluable we need a
second condition, namely, that the coefficient of
restitution equals e=-(V'-v')/(V-v).
Solving for V' from this last equality and plugging into the
momentum conservation result, then yields v'=[v(m-eM)+MV(1+e)]/[m+M].
This is an interesting result. It shows, for example, that
if we have an elastic collision(e=1)
and the two masses are equal, that v'=V and V'=v. So if V=0
then little m stops completely and M=m travels forward with
the original speed v of m. The billard players among you
will recognize this fact. Note that the energy loss in a
elastic(e=1) collision is zero,
but that there will be losses whenever e<1
and
that this loss becomes large during plastic collisions where
e=0. Click HERE to
see a pictoral development of the collision formula. **

**THE BALLISTIC PENDULUM:
An interesting application of the
conservation of momentum law coupled with the conservation
of energy concerns the ballistic pendulum. The
ballistic pedulum is a device used to measure the speed of a
bullet by noting how high a woodden block attached to a
swing arm will rise from a position of rest after the
bullet becomes embeded. The analysis consistes of a two part
consideration . First during initial impact the linear
momentum is conserved. Second after the bullet is embeded in
the block one has a conservation of total energy where the
kinetic energy at the bottom is entirely converted to
potential energy at the top of the subsequent swing. Click HEREto
see the details of the analysis.**

**ROCKET PROPULSION: An interesting application of the Impulse
Momentum Principle is the determination of the speed of
a rocket as a function of time.HERE
is the analysis. The very small payload boosted
to orbital speed compared to the initial launch
mass(even with staging) shows you, for example, why you
won't be taking vacations in earth orbit until someone
comes up with a much cheaper method than the use of
rocket propulsion using chemical fuel.(I take back
this comment I made several years ago, since in May of 2001
Dennis Tito, a wealthy eccentric businessman from
California, did go up into earth orbit at the cost to him of
some twenty million dollars, or about $100,000 per pound.
The russians took advantage of him since NASA claims they
can launch things at ten times less per pound. Costs with
chemical fuel launches are not expected to ever drop much
below about $1000/lb for orbiting a body, although the
actual kinetic energy equivalent which must be expended for
a mass m in orbit is only T=(1/2)mv^2=mgR/2, since the near
earth orbital speed is v=sqrt(gR), with R equal to the earth
radius. Thus each kilogram in a near earth circular
orbit has about 31 megajoules of kinetic energy which is
close to the 44 megajoules chemical energy contained
in a kilogram of gasoline. It is the need to boost the
heavy peripheral equipment, including rocket motors , fuel
tanks, etc , to high speeds which makes the chemical launch
process expensive. For example, the space shuttle
requires some 4 million pounds of solid and
liquid fuel for a ground launch from the cape . There ought
to a much less expensive way to put something into orbit,
but so far no one has come up with a good alternative.)**

**Lecture 13- Kinematics of Rigid Bodies. Angular
Velocity and Angular Acceleration. Velocity and Acceleration
at any Point on a Body in Translation and Rotation.
Crank-Piston Mechanism.**

**VELOCITY AND
ACCELERATION AT ANY POINT OF A ROTATING AND TRANSLATING
RIGID BODY: Consider a rigid
body containing two points A and B. We can relate the
velocity and acceleration at these two points to each other
by a simple vector addition involving the angular velocity
omega and angular acceleration alpha of the rigid body and
the position vector between points A and B. HERE
are the formulas and an application for a rolling wheel.**

http://www.brockeng.com/mechanism/ScotchYoke.htm

to see an animation of a Scotch Yoke. You will need to download the latest Java SE6 plug-in to view things.

_{
X=L*cos[w1*t]+R*cos[(w1+w2)*t]
and
Y=L*sin[w1*t]+R*sin[(w1+w2)*t].}

**This is the parametric
representaion of the famous epitrochoid. We show you HEREthe
result
when L=1, R=0.5, w1=1 and w2=6. The ability to visualize
such paths in two and three dimensions is a valuable skill
possesed by many design engineers such as Felix Wankel , the
inventor of the Wankel rotary engine. Go HERE
to see the epitrochoidal shape of a Wankel engine housing
generated by the above formulas when L=1, R=0.2, w1=1, and
w2=3.**

**Lecture 15- SECOND HOUR
EXAM . Will follow the same format as the first exam
with 3 out of 4 questions to answer. You are responsible for
all materials covered in class since the first exam, however,
the emphasis will be on Impulse and Momentum andthe
Kinematics of Rigid Bodies, and the material covered in the
homeworks and the lectures. The exam is closed book, except
you can bring one 3"x5" card.**

**Lecture 16-Introduction
to
Kinetics of Rigid Bodies. Basic Laws of Plane Motion. Spin-up
of a FlyWheel . Disc rolling down an incline. Review of Mass
Moments of Inertia and the Identity for Plane Motion that H=I*w or dH/dt=I*a. Moments
of Inertia for the Disc, Rod, Plate and Sphere. Parallel Axis
Theorem.**

**THE FALLING ROD PROBLEM:
In class today we discussed one
of the more interesting problems encoutered in dynamics,
namely, the behaviour of a uniform rod initially standing
vertically on a smooth floor. I summarize the governing
kinetic and kinematic conditions governing the rod HERE. Note that
the rod's angle relative to the vertical as a function of
time is determined by a solution of a highly non-linear
second order differential equation for theta as a function
of time which can only be solved numerically. Alternatively,
you can plot the square of the angular velocity versus angle
directly as done HERE. This last result is also possible to obtain
directly by use of energy methods. A numerical
integration(using Runge-Kutta) for q versus
time
shows that it takes about 0.88 sec for the rod to hit the
floor when L=1meter and the rod is started from rest
at**

**q(0)=0.01 rad. This
time increases with increasing rod length L and decreasing
acceleration of gravity g.**

**OSCILLATION FREQUENCY OF
A MASS VIA THE RAYLEIGH PRINCIPLE: Consider a mass m free to rotate about a pin
at A located at distance d from its mass center C. Under
resting conditions point C will lie on the same vertical
line as A. Next put a small clockwise displacement on C
corresponding to a very small displacement angle theta about
point A of line A-C. Letting go of the mass at this new
angle, where the potential energy is V=mgd[1-cos(theta)] or
approximately V=(1/2)*mgd*theta^2, will result in the
oscillation of the mass. The maximum kinetic energy will be
present at the bottom of the subsequent swings and equals
T=1/2*(mk^2+md^2)*[d(theta)/dt]^2. Representing the
resultant oscillatory motion by theta=(max
theta)*sin[omega*t] and realizing from the conservation of
energy and Rayleigh's Principle that Vmax=Tmax, we find that
(omega)^2=gd/(k^2+d^2), where k is the radius of gyration
about C. Thus a yard stick will oscillate about its end with
an omega of sqrt[3g/2L]=4.01r/s or a period of
tau=2*pi/omega=1.56 seconds.**

**PERIOD OF A COMPOUND
PENDULUM: A classic problem in
the area of oscillations is that of the period of a compound
pendulum. The question which is asked is what is the period
of oscillation when one pivots an arbitrary shaped mass
about a point other than its center of gravity and releases
the mass with its cg slighly away from the vertical line
passing through the pivot point. Clearly this sets the mass
into oscillation and for small maximum swing angle leads to
the result that w=sqrt(mgL/I) as
shown by clicking HERE.
For several years now I have had students from Mechanical
Engineering come by and ask me how one might determine the
moment of inertia of a connecting rod experimentally. The
usual answer one gives is to use an Atwood machine. But a
really much simpler way is to treat the connecting rod as a
compound pendulum and measure its oscillation period
experimentally.**

Lecture 21-THIRD HOUR EXAM. Can pick up graded exams in front of my office in three days. There will be no final exam for the course.

**
G=[(Sum
of Three Class Exams)/90]x90+[(Sum of Homeworks)/21]x10
**

**This means 90% for the three tests and 10%
for the 21 homework problems you worked on during the
semester.
**

** http://www2.mae.ufl.edu/~uhk/MATHFUNC.htm**

** http://www2.mae.ufl.edu/~uhk/ANALYSIS.html**

http://www2.mae.ufl.edu/~uhk/HOMEPAGE.html

http://www2.mae.ufl.edu/~uhk/STATICS.html

http://www2.mae.ufl.edu/~uhk/STRENGTH.html