EGM3400  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 HEREand an interesting (imaginary)view of our galaxy and its known properties by going 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.

Go HERE to see a more detailed mathematical discussion of planetary motion.

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 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.

          _{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.

                        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://www.mae.ufl.edu/~uhk/MATHFUNC.htm

    http://www.mae.ufl.edu/~uhk/ANALYSIS.html

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

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

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