Page first set up-Sept 1, 1996                                             Latest Update-July 28, 2016
 

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 . 


Click HEREfor the COURSE OUTLINE , TEST AND GRADE DETERMINATION METHOD.

Lecture 1- Introductory Remarks. Velocity and Acceleration. Curvilinear Motion. Simple Harmonic Motion. Motion of a Projectile. Go HERE to see a schematic of the Projectile Problem and it's Solution.

Lecture 2- Kinematics in Cylindrical Coordinates including discussion of the three Base Vectors used. Radial and Angular Acceleration. Motion along a Spiral. Normal and Tangential Coordinates. Relative position and velocity.

A PROBLEM IN RELATIVE MOTION: In class today we discussed, among other things, the kinematics of particles and their absolute and relative motions as described by position vectors. Click HERE to see a graph for the solution of the two vehicle problem based on relative velocities.

MOTION ALONG A SPIRAL DESCRIBED IN POLAR COORDINATES: In addition to describing the motion of a particle in cartesian coordinates, it is often to advantage to express things in polar or cylindrical coordinates.
In polar coordinates one has the position vector r=r*e[r] with velocity v=(dr/dt)*e[r]+r*d(e[r])/dt. Here e[r] is the unit base vector in the radial direction and one also has the unit base vector in the theta direction of e[theta]. The two base vectors are orthogonal to each other so that the dot product between them vanishes. Simple geometry also shows that d(e[r])/dt=e[q]*wand d(e[q])/dt=-e[r]*w, where w=d(q)/dt. Using these last identities one finds that v=dr/dt*e[r]+r*w*e[q] and the acceleration in polar coordinates becomes A={d^2r/dt^2-r*(w)^2}*e[r]+{2*dr/dt*w+r*d(w)/dt}*e[q]. Click HERE to see a specific calculation for particle motion in polar coordinates leading to the famous spiral of Bernoulli.(Bernoulli was so proud of this spiral that he had it engraved on his tombstone. Click HERE to see me pointing to it during a recent visit to Basel, Switzerland). Extention to 3D problems will involve cylindrical coordinates which have the extra base vector e[z] pointing in the z direction.

TANGENTIAL AND NORMAL COORDINATES: An alternative method for expressing velocities and accelerations of a particle moving along a curve is by the use of unit tangential e[t] and unit normal e[n] base vectors. For this type of description the velocity is simply V=v(t)e[t] and the acceleration is A=dV/dt=(dv/dt)e[t]+(de[t]/dt)v . But de[t]/dt=d(icos(q)+jsin(q)/dt which by use of the chain rule reduces to A=(dv/dt)e[t]+(v^2/r)e[n]. Here r is the radius of curvature of the curve which generally changes with position and in 2D is given from calculus by r=(1+y'^2)^(3/2)/y". Click HERE to see a calculation for the acceleration of a particle moving with constant speed along a cubic curve.

Lecture 3-Kinetics of Particles.(Chapter 13 of Hibbeler) Newton's Laws of Motion. The role of Weight and Thrust on the Acceleration of a Mass according to Newton's Second Law. Distance a hockey puck will slide. Pendulum jumping off of the Tampa Bay Bridge. Atwood machine. Geosynchronous Satellite.

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.


SOME EARTH DIMENSIONS: I noticed in yesterdays discussion on earth dimensions that some of you were not very familiar with what is the precise value of the earth's radius. Lets quickly go over this and some other earth properties. The earth's equatorial circumference is 24,902 miles=40,075 km, so that its equatorial radius is R=3963.5 miles=6,378 kilometers. Because it spins at an omega of 2p/(24 x 3600)=7.2722x10-5 rad/sec, any point on the equator moves to the east(relative to the earth center) at 24,902/24=1037.6 miles/hr. At a latitude of theta this number drops by cos(q), so that in Gainesville, which is at about 30 degrees latitude , we are moving to the east at 898.6 mph. This eastward rotational speed is used to advantage when launching an object into orbit from the cape here in Florida. A nautical mile corresponds to one minute of longitude along the equator and thus equals 24,902/(360x60)=1.15 miles=6076 ft. Note that , due to the spin of the earth , the near spherical shape of the earth is slightly flattened at the poles so that the polar radius is actually some 13.32 miles less than its equatorial value. If the earth were spinning at w=sqrt(g/R)=sqrt[32.2/(3963.5 x 5280)] x 3600=4.47 rad/hr or 17.06 revolutions per present earth day , a mass sitting on the equator would become weightless because the downward gravitational force would then just be cancelled by the outward centrifugal force , just like for a satellite. We also have that GMm/a^2=mg, where a is the earth radius , G=[6.6754 plus or minus 0.0005]x10^(-11) is the universal gravitational constant in SI units, M is the earth's mass and m a test mass sitting at the earth's surface. From this last equality one sees that the earth's mass equals
M=ga^2/G=9.81x(6.378x10^6)^2 /(6.6754x10^-11)=5.97x10^24  kg as first found over 200 years ago via the famous torsion fiber experiment for measuring G by Henry Cavendish(1798).

Lecture 5-Newton's Law of Gravitation and Orbital Mechanics. Kepler's Laws of Planetary Motion. Abhelion, Perihelion and Eccentricity. Other sample problems involving particle kinetics including the acceleration of a dragster and the motion of a mass spring system down an incline with friction.

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.


DETERMINING THE MASS OF THE SUN: Have you ever wondered how astronomers are able to know the mass of various astronomical objects? The determination is really quite simple and just based on the assumed validity of Newton's universal Law of Gravitation F=-[GMm/r^2] everywhere in the universe. Take for example the sun's mass. It is found by balancing the attractive force between the sun and earth with the earth centrifugal force in its essentially circular orbit. I show you  HERE the calculation which yields a mass of M=1.98x10^30 kg. Note, since one usually does not recall the precise value of the universal gravitational constant G, it becomes convenient to use the identity Gm=ga^2, where m is the earth mass , g the acceleration of gravity, and 'a' the earth radius.

Lecture 6-Work-Energy Principle. Conservation of Energy. Energy stored in a Spring. Gravitational Potential.

WORK-ENERGY PRINCIPLE APPLIED TO A SLIDING MASS-SPRING SYSTEM: The Work-Energy Principle for a particle states that the sum of the work done on the particle equals the change in its kinetic energy. To demonstrate this principle consider the problem of releasing a mass attached to a linear spring on a rough incline. Here the spring work is Ws=-(1/2)kx^2, the work against gravity is +mgxsin(q), where theta is the angle of the incline, and -(m)mgcos(q)x is the friction work with m the coefficient of friction and x the distance down the plane the mass has slid. Details of the problem are found HERE
 Note that the mass first speeds up, reaches a maximum speed and then slows down and comes to rest before reversing its motion. It is important to remember that the friction work  always acts to reduce the kinetic energy and this explains why the present mass will never return to its starting point.

THE SIMPLE PENDULUM ANALYZED BY THE ENERGY METHOD: For non-dissipative dynamical systems one has the well known fact that the sum of the kinetic energy(T)and potential (V)energy is a constant. Lets apply this conservation law to a simple pendulum of mass m held by a weightless rod of length L. Measuring the potential energy from the bottom of the swing we have V=mgL[1-cos(q)] as its value at angle q with respect to the vertical. The corresponding kinetic energy is T=mv2/2. Neglecting all friction, we then have , from the conservation of energy law, that Const=[gl(1-cos(q)+v2/2]. Thus if we start the pendulum at rest from its inverted position at q=p , its speed at the bottom(q=0)of the swing will reach the value of v=2sqrt(gl). At the same time if one started the pendulum at the bottom with a speed greater than this last value, it would go over the top and continue to rotate about the pivot point instead of oscillating about it. This behavior can be well described by looking at the pendulum's phase plane trajectory with q lying along the horizontal axis and v/sqrt(2gL) along the vertical axis. Click HEREto see the phase plane diagram for the simple pendulum.

G FORCES ON ROLLERCOASTER RIDES: Another interesting  dynamics problem treated by energy methods is that of determining the g forces experienced at the bottom of the first drop in some of the new rollercoaster rides. These accelerations,  can become considerable and approach values comparible to those experienced by fighter pilots and astronauts. Lets quickly  show the analysis and take as our model the newly opened Millenium Force steel rollercoaster at Cedar Point in Sandusky, Ohio. It has a 300 ft drop and extends for some 6600 ft. We represent the track height in ft as y=300*cos(0.004*x))^2*exp(-0.002*x) and look what happens in the first 2000 ft of horizontal distance. We know from the conservation of total energy that the speed is v(x)=sqrt[2*(300-y(x))] , since (1/2)mv^2=mg[300-y(x)]. Also the centripetal  acceleration will be v^2/R, with R=[1+y'^2]^(3/2)/y" from calculus. Thus the centripetal acceleration(and hence a corresponding centrifugal force per mass) measured in units of g will be 2(300-y(x))*y"(x)/[1+y'(x)^2]^1.5. A plot of both the assumed track and the normal acceleration experienced is shown HERE. Note that in our model the maximum centripetal acceleration is 3g at the bottom of the first drop and  becomes -1g at the top of the next rise. You have to add a downward acceleration of 1g onto these results due to the earth's attraction on any body at the earth's surface. The speed at the bottom is about 140ft./sec. The Millenium coaster people  claim their coaster reaches 92mph=135ft/sec. This slightly lower value must be due to drag forces not included in our conservation of energy calculations. You probably heard about a lady recently being killed on one of these rides in California due to a burst blood vessel in her head. Healthy individuals can withstand accelerations as high as 7g before blacking out and rollercoasters are designed to not exceed 3g. One should also recognize that in addition to the centripetal acceleration considered here there will also be an acceleration along the track given by v*dv/ds plus a sideways acceleration if the  rollercoaster makes turns.

IMPACT VELOCITY OF AN ASTEROID OR OTHER CELESTIAL BODY HITTING THE EARTH: In discussing the conservation of energy law for conservative systems, we asked the question "With what velocity will a rock of mass m released from rest at distance H above the earth's surface hit the earth?". Neglecting all frictional losses, one simply has that the sum of
the potential energy V plus kinetic energy T at the beginning and end of the trip are the same. We thus have that [0-GMm/(R+H)]=[mv2/2-GMm/R]. This evaluates to v2=2gR[1-1/(1+H/R)] , when using the identity GM=gR2. If we now start the rock from rest at H=R above the earth's surface, it will impact with v=sqrt[gR]=4.91 miles/sec since the earth radius is R=3960 miles and the acceleration constant has the value of g=32.2 ft/sec. You can see from this example the tremendous
amount of kinetic energy which would be carried by a large asteroid impacting the earth and why it is not implausible that such a collision might have led to the demise of the dinosauers. Note that a rock starting from rest at infinity would impact the earth with the escape velocity of v=sqrt(2gR)=6.95 miles/sec=11.18 km/sec. Meteor impact speeds (before being slowed by the
atmosphere) can actually be much higher than this last value and range up to 72 km/sec. This last value would result from a head on collision and is derived by adding the sun's escape velocity at earth distance plus the earth's orbital speed around the sun of about 30 km/sec [i.e. 30 +30 sqrt(2) = 72]. A major problem the space agencies around the world are running into is space junk. There presently are thousands of pieces of man made junk from earlier orbit missions floating around the earth at orbital speeds of about 11.18/sqrt(2)=7.9km/sec. The larger pieces(greater than 1meter) can be tracked by radar and so can be avoided, but pieces between a few mm to several cm in dimension  pose a real hazard to astronauts in near earth orbit(see the June 2002 issue of Science). A cheap way to install a shield against ICBMs would be not try to shoot down an incoming missile with another one but rather to deploy( from orbiting satellites) millions of retro-orbiting small particles into the path of the incoming ICBMs. The result would be a very effective space flak. However, a major drawback of this approach would be that one has no kown economic way to remove these particles (and also other space junk)once the missle threat is over.

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 9-FIRST HOUR EXAM.

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. 


Lecture 11-More on Impulse-Momentum Principle. Ballistic Pendulum. Oblique Collisions for specified values of e. Angular Impulse and Conservation of Angular Momentum. Systems of Particles. Force of a Water Jet and functioning of a Rocket.  Go HERE to see an interesting animation of oblique billiard ball collisions. You can vary the initial conditions and the mass ratio. Go HERE to view a pdf file in which I show you the mathematical details of an oblique collision process.

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 12-Completion of discussion on Impulse Momentum Principle, Systems of Particles. Forces of a Water Jet on a Turbine Blade. Angular Impulse and Conservation of Angular Momentum.

FORCE ON A TURBINE BLADE: Another interesting application of the impulse momentum principle is the turbine blade problem shown in the accompanying figure( Click HERE). We know from the impulse momentum principle that the force exerted on a body equals the net momentum flux out of a system bounded by a control volume. For the turbine blade we have an incoming fluid jet which breaks up into two parts at the turbine blade and shoots two streams out at angle theta with respect to the axis of the incoming jet. The mass flow rate  in this case is dq/dt=rAoV and the reaction force Fx=(dq/dt)*V*[1+cos(theta)]. Note such forces can become quite large and can be used to advantage in hydroelectric generating plants. Cavitation damage remains a major problem for water driven turbine blades.

FAMOUS PEOPLE IN THE HISTORY OF MECHANICS: As a small diversion from our dynamics discussions, I thought you might be interested in seing some of the most famous individuals historically associated with mechanics. Click HERE to see a thumbnail gallery. Contributions of these and other scientist and mathematicians can be found at
http://www-groups.dcs.st-and.ac.uk/~history/BiogIndex.html

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.


CRANK-PISTON MECHANISM: One of the more interesting problems treated by the kinematic formulas of a rigid body is the crank-piston mechanism in your automobile. We have there a drive shaft rotating at essentially constant angular velocity wo connected by a pin link B to a connecting rod which in turn links to a piston at C. We show you HERE the mathematical  development using the basic velocity formula Va=Vb+wk x ra/b  for the resultant piston velocity. Note that your piston velocity becomes nearly a simple harmonic motion when the connecting rod length L is large compared to the distance l from the drive shaft to the connecting pin B. To see an applet showing the functioning of a simple crank-piston mechanism go HERE.

Lecture 14- Kinematics of Plane Motion. Instantaneous Center of Rotation. Also a discussion on Rotating Frame of References. Coriolis Acceleration. Review for Exam #2.

THE SCOTCH YOKE: A device which can produces pure sinusoidal back and forth motion from a constant rotational motion is the Scotch Yoke. We can explain its functioning via the kinematics of rigid bodies as expained in the previous two lectures. The velocity of the peg sitting on the periphery of a  rotating wheel of radius R and constant angular velocity w is V= wR[icos(wt)+jsin(wt)]. Now the slot in which the peg slides is part of larger rigid body confined to move strictly in the x direction by the shown constraints. The axial position the slot takes is then simply the x component of V integrated over time, namely, x=Rsin(wt). This is a pure sinusoidal motion and finds application in areas such as  internal combustion engines, electric jig-saw drives, and some of our own work on oscillatory heat transfer. Go to-

                            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.

VELOCITY AND ACCELERATION IN A ROTATING REFERENCE FRAME: Two of the most important , and at the same time most difficult to grasp, equations encountered in dynamics are the kinematic relations for velocity and acceleration expressed in a rotating reference frame. Relative to an inertial reference frame located at point O, these functions have the form given HERE. Study them carefully.

EPITROCHOID PATH OF A POINT P: In our class discussion we derived the velocity and acceleration of a point P on a rigid rotating body when measured relative to an inertial frame of reference. We use this procedure here to determine the rather complicated path a point P located at the periphery of a rotating disc of radius R and angular velocity w2 when the disc axis located at moving point A is attached to a bar of length L at one end while the other is fixed at O. The position vector of P relative to O is then rP/O=rA/O+rP/A= {I*Lcos(w1*t)+Lsin(w1*t)}+ {i*Rcos(w2*t)+j*Rsin(w2*t)}, where w1 is the angular velocity of the bar L. Now from the geometry we have that i=I*cos(w1*t) +J*sin(w1*t) and j=-I*sin(w1*t)+J*cos(w1*t). Substituting this relation between the base vectors i,j and I,J , we find the X and Y components of the position vector rP/O to be-

          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.


BASIC LAWS OF MOTION FOR A RIGID BODY: We have now reached the point in our dynamics course at which you are capable of calculating the behaviour of a translating and rotating rigid body subjected to a collection of forces and moments. You need only use the two basic equations of kinetics, namely, F=mA and M=dH/dt, where A is the acceleration of the center of mass of the body , F is the sum of the externally acting forces, M represents the sum of the moments acting about the center of mass( or a zero velocity point) of the body, and H is the angular momentum about the same point. In 2D these two vector expressions reduce to a total of four algebraic equations, which when used in conjunction with our kinematic relations for a rigid body, make most dynamics problems soluble. Note that one recovers the basic laws of statics when A and H are zero.

Lecture 17-More problems worked for Rigid Body Plane Motion. The falling rod hinged at one end. Acceleration of an automobile. Center of Percussion. The Compound Pendulum. Falling Rod on Smooth Surface.

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.


Lecture 18-Work Energy Principle for Rigid Bodies. Kinetic Energy for a Rotating Body. Solving the "cylinder rolling down an incline" problem by the work-energy method. Rolling cylinder produced by spring force. Some  Conservation of Energy Problems. Introductionto Impulse-Momentum Principle for Rigid Bodies.

Lecture 19-(Last Lecture on New Material)- More on the  Impulse-Momentum Principle. Conservation of Angular Momentum. Angular Impact. Center of Percussion. Oscillation Frequency of a Rigid Body. Brief discussion of Vibrations.

A BILLIARD BALL PROBLEM: A problem often posed in dynamics is -At what height h above a table should one hit a stationary billiard ball of radius R in order to have it roll without slipping? This is a problem involving angular impulse plus the kinematic condition that when rolling we have V(the speed of the mass center)=omega*R. Applying the angular impulse formula about the contact point C between the ball and the table, we have that h*Fdt=IC*omega, with IC=2/5 m r^2+mr^2=7/5 m r^2 by the parallel axis theorem. Also, applying the linear impulse principle to the ball's mass center, we have Fdt=mV, if we assume the contact friction is neglibgible. So eliminating the impulse term Fdt, one finds that omega*R*h=(7/5)R^2*omega. On cancelling the R and omega terms, this says that the billiard ball will roll without slipping if it is hit by the cue stick at height h=(7/5)R above the table surface. Try it the next time you are playing billiards.

Lecture 20-Review for Third Hour Exam. Emphasis will be on Planar Kinetics, Work-Energy for Rigid Bodies, Impulse-Momentum for Rigid Bodies, and Vibrations. Same format as first two exams with 3 out of 4 questions to answer from material covered in the book and lectures. Closed book, but one 3"x5" card allowed. Filling out of teacher evaluation forms.

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. To remind you, course grades are determined as follows:

                        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.


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    https://web.mae.ufl.edu/uhk/ANALYSIS.html

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    https://web.mae.ufl.edu/uhk/HOMEPAGE.html

    https://web.mae.ufl.edu/uhk/STATICS.html

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