Chapter 3. Analyzing motion of systems of particles

Transcription

1 Chapter 3 Analyzng moton of systems of partcles In ths chapter, we shall dscuss 1. The concept of a partcle. Poston/velocty/acceleraton relatons for a partcle 3. Newton s laws of moton for a partcle 4. How to use Newton s laws to calculate the forces needed to mae a partcle move n a partcular way 5. How to use Newton s laws to derve `equatons of moton for a system of partcles 6. How to solve equatons of moton for partcles by hand or usng a computer. The focus of ths chapter s on settng up and solvng equatons of moton we wll not dscuss n detal the behavor of the varous examples that are solved. 3.1 Equatons of moton for a partcle We start wth some basc defntons and physcal laws Defnton of a partcle A `Partcle s a pont mass at some poston n space. It can move about, but has no characterstc orentaton or rotatonal nerta. It s characterzed by ts mass. Examples of applcatons where you mght choose to dealze part of a system as a partcle nclude: 1. Calculatng the orbt of a satellte for ths applcaton, you don t need to now the orentaton of the satellte, and you now that the satellte s very small compared wth the dmensons of ts orbt.. A molecular dynamc smulaton, where you wsh to calculate the moton of ndvdual atoms n a materal. Most of the mass of an atom s usually concentrated n a very small regon (the nucleus) n comparson to nter-atomc spacng. It has neglgble rotatonal nerta. Ths approach s also sometmes used to model entre molecules, but rotatonal nerta can be mportant n ths case. Obvously, f you choose to dealze an obect as a partcle, you wll only be able to calculate ts poston. Its orentaton or rotaton cannot be computed Poston, velocty, acceleraton relatons for a partcle (Cartesan coordnates) In most practcal applcatons we are nterested n the poston or the velocty (or speed) of the partcle as a functon of tme. But Newton s laws wll only tell us ts acceleraton. We therefore need equatons that relate the poston, velocty and acceleraton. Poston vector: In most of the problems we solve n ths course, we wll specfy the poston of a partcle usng the Cartesan components of ts poston vector wth respect to a convenent orgn. Ths means x O y r P z

2 1. We choose three, mutually perpendcular, fxed drectons n space. The three drectons are descrbed by unt vectors,,. We choose a convenent pont to use as orgn. 3. The poston vector (relatve to the orgn) s then specfed by the three dstances (x,y,z) shown n the fgure. r x( t) y( t) z( t) In dynamcs problems, all three components can be functons of tme. Velocty vector: By defnton, the velocty s the dervatve of the poston vector wth respect to tme (followng the usual machnery of calculus) r( t t) r( t) v lm t0 t Velocty s a vector, and can therefore be expressed n terms of ts Cartesan components v v v v x y z You can vsualze a velocty vector as follows The drecton of the vector s parallel to the drecton of moton x y z The magntude of the vector v v v v v s the speed of the partcle (n meters/sec, for example). When both poston and velocty vectors are expressed n terms Cartesan components, t s smple to,, are constant calculate the velocty from the poston vector. For ths case, the bass vectors (ndependent of tme) and so d dx dy dz vx vy vz x( t) y( t) z( t) Ths s really three equatons one for each velocty component,.e. vx dx v dy dz y vz Acceleraton vector: The acceleraton s the dervatve of the velocty vector wth respect to tme; or, equvalently, the second dervatve of the poston vector wth respect to tme. v( t t) v( t) a lm t0 t The acceleraton s a vector, wth Cartesan representaton a ax ay az. Le velocty, acceleraton has magntude and drecton. Sometmes t may be possble to vsualze an acceleraton vector for example, f you now your partcle s movng n a straght lne, the acceleraton vector must be parallel to the drecton of moton; or f the partcle moves around a crcle at constant speed, ts acceleraton s towards the center of the crcle. But sometmes you can t trust your ntuton regardng the magntude and drecton of acceleraton, and t can be best to smply wor through the math. The relatons between Cartesan components of poston, velocty and acceleraton are

3 dvx d x dvy d y dvz d z x y z a a a Examples usng poston-velocty-acceleraton relatons It s mportant for you to be comfortable wth calculatng velocty and acceleraton from the poston vector of a partcle. You wll need to do ths n nearly every problem we solve. In ths secton we provde a few examples. Each example gves a set of formulas that wll be useful n practcal applcatons. Example 1: Constant acceleraton along a straght lne. There are many examples where an obect moves along a straght lne, wth constant acceleraton. Examples nclude free fall near the surface of a planet (wthout ar resstance), the ntal stages of the acceleraton of a car, or and arcraft durng taeoff roll, or a spacecraft durng blastoff. Suppose that The partcle moves parallel to a unt vector The partcle has constant acceleraton, wth magntude a At tme t t0 the partcle has speed v 0 At tme t t0 the partcle has poston vector r x 0 The poston, velocty acceleraton vectors are then 1 r x0 v0 ( t t0) a( t t0) v v at a a 0 Verfy for yourself that the poston, velocty and acceleraton () have the correct values at t=0 and () are related by the correct expressons (.e. dfferentate the poston and show that you get the correct expresson for the velocty, and dfferentate the velocty to show that you get the correct expresson for the acceleraton). HEALTH WARNING: These results can only be used f the acceleraton s constant. In many problems acceleraton s a functon of tme, or poston n ths case these formulas cannot be used. People who have taen hgh school physcs classes have used these formulas to solve so many problems that they automatcally apply them to everythng ths wors for hgh school problems but not always n real lfe! Example : Smple Harmonc Moton: The vbraton of a very smple sprngmass system s an example of smple harmonc moton. In smple harmonc moton () the partcle moves along a straght lne; and () the poston, velocty and acceleraton are all trgonometrc functons of tme. x(t),l 0 m For example, the poston vector of the mass mght be gven by

4 r x( t) X0 X sn( t / T) X X Here X 0 s the average length of the sprng, 0 s the maxmum length of the sprng, and T s the tme for the mass to complete one complete cycle of oscllaton (ths s called the `perod of oscllaton). Harmonc vbratons are also often characterzed by the frequency of vbraton: The frequency n cycles per second (or Hertz) s related to the perod by f=1/t The angular frequency s related to the perod by /T The moton s plotted n the fgure on the rght. The velocty and acceleraton can be calculated by dfferentatng the poston, as follows dx( t) X v cos( t / T) T d x( t) 4 X a sn( t / T) T Note that: The velocty and acceleraton are also harmonc, and have the same perod and frequency as the dsplacement. If you now the frequency, and ampltude and of ether the dsplacement, velocty, or acceleraton, you can mmedately calculate the ampltudes of the other two. For example, f X, V, A denote the ampltudes of the dsplacement, velocty and acceleraton, we have that V X A X V T T T Example 3: Moton at constant speed around a crcular path Crcular moton s also very common examples nclude any rotatng machnery, vehcles travelng around a crcular path, and so on. The smplest way to mae an obect move at constant speed along a crcular path s to attach t to the end of a shaft (see the fgure), and then rotate the shaft at a constant angular rate. Then, notce that The angle ncreases at constant rate. We can wrte t, where s the (constant) angular speed of the shaft, n radans/seconds. The speed of the partcle s related to by V R. To see ths, notce that the crcumferental dstance traveled by the partcle s s V ds / Rd / R. R Rcos sn t cos n t Rsn R. Therefore, For ths example the poston vector s r Rcos Rsn

5 The velocty can be calculated by dfferentatng the poston vector. dr d d v R sn R cos R( sn cos ) Here, we have used the chan rule of dfferentaton, and noted that d /. The acceleraton vector follows as dv d d a R( cos sn ) R (cos sn ) Note that () The magntude of the velocty s V R, and ts drecton s (obvously!) tangent to the path (to see ths, vsualze (usng trg) the drecton of the unt vector t ( sn cos ) () The magntude of the acceleraton s R and ts drecton s towards the center of the crcle. To see ths, vsualze (usng trg) the drecton of the unt vector n (cos sn ) We can wrte these mathematcally as dr dv V v Rt Vt a R n n R Example 4: More general moton around a crcular path We next loo at more general crcular moton, where the partcle stll moves around a crcular path, but does not move at constant speed. The angle s now a general functon of tme. We can wrte down some useful scalar relatons: d Angular rate: d d Angular acceleraton d Speed V R R Rate of change of speed dv d d R R R sn t cos R n Rsn Rcos We can now calculate vector veloctes and acceleratons r Rcos Rsn The velocty can be calculated by dfferentatng the poston vector. dr d d v R sn R cos R( sn cos ) The acceleraton vector follows as dv d d d a R ( sn cos ) R( cos sn ) R ( sn cos ) R (cos sn )

6 It s often more convenent to re-wrte these n terms of the unt vectors n and t normal and tangent to the crcular path, notng that t ( sn cos ), n (cos sn ). Then dv V v Rt Vt a R t R n t n R These are the famous crcular moton formulas that you mght have seen n physcs class. Usng MAPLE to dfferentate poston-velocty-acceleraton relatons If you fnd that your calculus s a bt rusty you can use MAPLE to do the tedous wor for you. You already now how to dfferentate and ntegrate n MAPLE the only thng you may not now s how to tell MAPLE that a varable s a functon of tme. Here s how ths wors. To dfferentate the vector r x( t) y( t) z( t) you would type It s essental to type n the (t) after x,y,and z f you don t do ths, Mupad assumes that these varables are constants, and taes ther dervatve to be zero. You must enter (t) after _any_ varable that changes wth tme. Here s how you would do the crcular moton calculaton f you only now that the angle s some arbtrary functon of tme, but don t now what the functon s

7 As you ve already seen n EN3, Matlab can mae very long and complcated calculatons farly panless. It s a godsend to engneers, who generally fnd that every real-world problem they need to solve s long and complcated. But of course t s mportant to now what the program s dong so eep tang those math classes Velocty and acceleraton n normal-tangental and cylndrcal polar coordnates. In some cases t s helpful to use specal bass vectors to wrte down velocty and acceleraton vectors, nstead of a fxed {,,} bass. If you see that ths approach can be used to qucly solve a problem go ahead and use t. If not, ust use Cartesan coordnates ths wll always wor, and wth MAPLE s not very hard. The only beneft of usng the specal coordnate systems s to save a couple of lnes of rather tedous trgonometrc algebra whch can be extremely helpful when solvng an exam queston, but s generally nsgnfcant when solvng a real problem. Normal-tangental coordnates for partcles movng along a prescrbed planar path In some problems, you mght now the partcle speed, and the x,y coordnates of the path (a car travelng along a road s a good example). In ths case t s often easest to use normal-tangental coordnates to descrbe forces and moton. For ths purpose we Introduce two unt vectors n and t, wth t pontng tangent to the path and n pontng normal to the path, towards the center of curvature Introduce the radus of curvature of the path R. If you happen to now the parametrc equaton of the path (.e. the x,y coordnates are nown n terms of some varable ), then dx dy dy dx 1 d r d d d d r x( ) y( ) t n dr d dx dy dx dy d d d d d The sgn of n should be selected so that t t R n

8 The radus of curvature can be computed from d x d y 0 d d n dx d y dy d x 1 d d d d R dx dy d d The radus of curvature s always postve. The drecton of the velocty vector of a partcle s tangent to ts path. The magntude of the velocty vector s equal to the speed. The acceleraton vector can be constructed by addng two components: the component of acceleraton tangent to the partcle s path s equal dv / The component of acceleraton perpendcular to the path (towards the center of curvature) s equal to V / R. Mathematcally dv V v Vt a t n R Example: Desgn speed lmt for a curvy road: As a consultng frm specalzng n hghway desgn, we have been ased to develop a desgn formula that can be used to calculate the speed lmt for cars that travel along a curvy road. The followng procedure wll be used: The curvy road wll be approxmated as a sne wave y Asn( x / L) as shown n the fgure for a gven road, engneers wll measure values of A and L that ft the path. Vehcles wll be assumed to travel at constant speed V around L the path your msson s to calculate the value of V For safety, the magntude of the acceleraton of the car at any pont along the path must be less than 0.g, where g s the gravtatonal acceleraton. (Agan, note that constant speed does not mean constant acceleraton, because the car s drecton s changng wth tme). Our goal, then, s to calculate a formula for the magntude of the acceleraton n terms of V, A and L. The result can be used to deduce a formula for the speed lmt. Calcluaton: We can solve ths problem qucly usng normal-tangental coordnates. Snce the speed s constant, the acceleraton vector s a V n R The poston vector s r x Asn( x / L), so we can calculate the radus of curvature from the formula 3/ V

9 dx d y dy d x 1 d d d d R 3/ dx dy d d Note that x acts as the parameter for ths problem, and y Asn( x / L), so and the acceleraton s d y 1 dx A( / L) sn( x / L) dy 1 dx R 3/ 3/ 1 ( A / L) cos( x / L) a A V L x L ( / ) sn( / ) 1 ( A / L)cos( x / L) We are nterested n the magntude of the acceleraton A( V / L) sn( x / L) a 1 ( A / L)cos( x / L) We see from ths that the car has the bggest acceleraton when x L/. The maxmum acceleraton follows as amax A( V / L) The formula for the speed lmt s therefore V ( L / ) 0. g / A Now we send n a bll for a bg consultng fee 3/ 3/ n Polar coordnates for partcles movng n a plane When solvng problems nvolvng central forces (forces that attract partcles towards a fxed pont) t s often convenent to descrbe moton usng polar coordnates. e e r Polar coordnates are related to x,y coordnates through 1 r x y tan ( y / x) Suppose that the poston of a partcle s specfed by ts polar coordnates ( r, ) relatve to a fxed orgn, as shown n the fgure. Let e r be a unt vector pontng n the radal drecton, and let e be a unt vector pontng n the tangental drecton,.e er cos sn e sn cos The velocty and acceleraton of the partcle can then be expressed as r

10 dr d v er r e d r d d dr d a r er r e Dervng these results taes some tedous algebra, but t s conceptually smple here s what we do: 1. Wrte down the poston vector n terms of ( r, ) n a fxed (,,) coordnate system. Tae the tme dervatves to fnd acceleraton and velocty n the (,,) coordnate system 3. Convert the results to the { er, e, e z} coordnate system. To do ths remember that the component of v parallel to e r can be found usng a dot product: vr ve r. Smlarly v ve Here are the detals wth Mupad tang care of the tedous algebra. These are the answers stated. Example The robotc manpulator shown n the fgure rotates wth constant angular speed about the axs. Fnd a formula for the maxmum allowable (constant) rate of extenson dl / f the acceleraton of the grpper may not exceed g. We can smply wrte down the acceleraton vector, usng polar coordnates. We dentfy d / and r=l, so that a dl 4 dl dl 1 L e r L 4 g g / L e a 4 L O O e L e r

11 Other examples usng polar coordnates can be found n sectons below Measurng poston, velocty and acceleraton If you are desgnng a control system, you wll need some way to detect the moton of the system you are tryng to control. A vast array of dfferent sensors s avalable for you to choose from: see for example the lst at A very short lst of common sensors s gven below 1. GPS determnes poston on the earth s surface by measurng the tme for electromagnetc waves to travel from satelltes n nown postons n space to the sensor. Can be accurate down to cm dstances, but the sensor needs to be left n poston for a long tme for ths nd of accuracy. A few m s more common.. Optcal or rado frequency poston sensng measure poston by (a) montorng deflecton of laser beams off a target; or measurng the tme for sgnals to travel from a set of rado emtters wth nown postons to the sensor. Precson can vary from cm accuracy down to lght wavelengths. 3. Capactatve dsplacement sensng determne poston by measurng the capactance between two parallel plates. The devce needs to be physcally connected to the obect you are tracng and a reference pont. Can only measure dstances of mm or less, but precson can be down to mcron accuracy. 4. Electromagnetc dsplacement sensng measures poston by detectng electromagnetc felds between conductng cols, or col/magnet combnatons wthn the sensor. Needs to be physcally connected to the obect you are tracng and a reference pont. Measures dsplacements of order cm down to mcrons. 5. Radar velocty sensng measures velocty by detectng the change n frequency of electromagnetc waves reflected off the travelng obect. 6. Inertal accelerometers: measure acceleratons by detectng the deflecton of a sprng actng on a mass. Accelerometers are also often used to construct an nertal platform, whch uses gyroscopes to mantan a fxed orentaton n space, and has three accelerometers that can detect moton n three mutually perpendcular drectons. These acceleratons can then be ntegrated to determne the poston. They are used n arcraft, marne applcatons, and space vehcles where GPS cannot be used Newton s laws of moton for a partcle Newton s laws for a partcle are very smple. Let 1. m denote the mass of the partcle. F denote the resultant force actng on the partcle (as a vector) 3. a denote the acceleraton of the partcle (agan, as a vector). Then F ma

12 Occasonally, we use a partcle dealzaton to model systems whch, strctly speang, are not partcles. These are: 1. A large mass, whch moves wthout rotaton (e.g. a car movng along a straght lne). A sngle partcle whch s attached to a rgd frame wth neglgble mass (e.g. a person on a bcycle) In these cases t may be necessary to consder the moments actng on the mass (or frame) n order to calculate unnown reacton forces. 1. For a large mass whch moves wthout rotaton, the resultant moment of external forces about the center of mass must vansh.. For a partcle attached to a massless frame, the resultant moment of external forces actng on the frame about the partcle must vansh. M 0 C It s very mportant to tae moments about the correct pont n dynamcs problems! Forgettng ths s the most common reason to screw up a dynamcs problem If you need to solve a problem where more than one partcle s attached to a massless frame, you have to draw a separate free body dagram for each partcle, and for the frame. The partcles must obey Newton s laws F ma. The forces actng on the frame must obey F 0 and M C 0, (because the frame has no mass). The Newtonan Inertal Frame. Newton s laws are very famlar, and t s easy to wrte them down wthout much thought. They do have a flaw, however. When we use Newton s laws, we assume that we can dentfy a convenent orgn somewhere that we regard as `fxed. In adon, to wrte down an acceleraton vector, we need to be able to choose a set of fxed drectons n space. For engneerng calculatons, ths usually poses no dffculty. If we are solvng problems nvolvng terrestral moton over short dstances compared wth the earth s radus, we smply tae a pont on the earth s surface as fxed, and tae three drectons relatve to the earth s surface to be fxed. If we are solvng problems nvolvng moton n space near the earth, or modelng weather, we tae the center of the earth as a fxed pont, (or for more complex calculatons the center of the sun); and choose axes to have a fxed drecton relatve to nearby stars. But n realty, an unambguous nertal frame does not exst. We can only descrbe the relatve moton of the mass n the unverse, not ts absolute moton. The general theory of relatvty addresses ths problem and n dong so explans many small but notceable dscrepances between the predctons of Newton s laws and experment. It would be fun to cover the general theory of relatvty n ths course but regrettably the mathematcs needed to solve any realstc problem s horrendous. As engneers, we always have to solve realstc problems, and we usually can t afford to spend a long tme dong complcated calculatons, so we use the smplest theory that wll allow us to mae the correct desgn decsons. Newton s laws are fne for us

13 3. Calculatng forces requred to cause prescrbed moton of a partcle Newton s laws of moton can be used to calculate the forces requred to mae a partcle move n a partcular way. We use the followng general procedure to solve problems le ths (1) Decde how to dealze the system (what are the partcles?) () Draw a free body dagram showng the forces actng on each partcle (3) Consder the nematcs of the problem. The goal s to calculate the acceleraton of each partcle n the system you may be able to start by wrtng down the poston vector and dfferentatng t, or you may be able to relate the acceleratons of two partcles (eg f two partcles move together, ther acceleratons must be equal). (4) Wrte down F=ma for each partcle. (5) If you are solvng a problem nvolvng a massless frames (see, e.g. Example 3, nvolvng a bcycle wth neglgble mass) you also need to wrte down M C 0 about the partcle. (5) Solve the resultng equatons for any unnown components of force or acceleraton (ths s ust le a statcs problem, except the rght hand sde s not zero). It s best to show how ths s done by means of examples. Example 1: Estmate the mnmum thrust that must be produced by the engnes of an arcraft n order to tae off from the dec of an arcraft carrer (the fgure s from We wll estmate the acceleraton requred to reach taeoff speed, assumng the arcraft accelerates from zero speed to taeoff speed along the dec of the carrer, and then use Newton s laws to deduce the force. Data/ Assumptons: 1. The flght dec of a Nmtz class arcraft carrer s about 300m long ( but only a fracton of ths s used for taeoff (the angled runway s used for landng). We wll tae the length of the runway to be d=00m. We wll assume that the acceleraton durng taeoff roll s constant. 3. We wll assume that the arcraft carrer s not movng (ths s wrong actually the arcraft carrer always moves at hgh speed durng taeoff. We neglect moton to mae the calculaton smpler) 4. The FA18 Super Hornet s a typcal arcraft used on a carrer t has max catapult weght of m=15000g 5. The manufacturers are somewhat retcent about performance specfcatons for the Hornet but vt 150 nots (77 m/s) s a reasonable guess for a mnmum controllable arspeed for ths arcraft. Calculatons: 1. Idealzaton: We wll dealze the arcraft as a partcle. We can do ths because the arcraft s not rotatng durng taeoff.

14 . FBD: The fgure shows a free body dagram. F T represents the (unnown) force exerted on the arcraft due to ts engnes. 3. Knematcs: We must calculate the acceleraton requred to reach taeoff speed. We are gven () the dstance to taeoff d, () the taeoff speed v t and () the arcraft s at rest at the start of the taeoff roll. We can therefore wrte down the poston vector r and velocty v of the arcraft at taeoff, and use the straght lne moton formulas for r and v to calculate the tme t to reach taeoff speed and the acceleraton a. Tang the orgn at the ntal poston of the arcraft, we have that, at the nstant of taeoff 1 r d at v vt at Ths gves two scalar equatons whch can be solved for a and t 1 vt d d at vt at a t d vt 4. EOM: The vector equaton of moton for ths problem s v F t T ma m d 5. Soluton: The component of the equaton of moton gves an equaton for the unnown force n terms of nown quanttes v F t T m d Substtutng numbers gves the magntude of the force as F= N. Ths s very close, but slghtly greater than, the 00N (44000lb) thrust quoted on the spec sheet for the Hornet. Usng a catapult to accelerate the arcraft, speedng up the arcraft carrer, and ncreasng thrust usng an afterburner buys a margn of safety. F T NA mg N B Example : Mechancs of Magc! You have no doubt seen the smple `tablecloth trc n whch a tablecloth s whpped out from underneath a fully set table (f not, you can watch t at Tablecloth Table a In ths problem we shall estmate the crtcal acceleraton that must be mposed on the tablecloth to pull t from underneath the obects placed upon t. We wsh to determne conons for the tablecloth to slp out from under the glass. We can do ths by calculatng the reacton forces actng between the glass and the tablecloth, and see whether or not slp wll occur. It s best to calculate the forces requred to mae the glass move wth the tablecloth (.e. to prevent slp), and see f these forces are bg enough to cause slp.

15 1. Idealzaton: We wll assume that the glass behaves le a partcle (agan, we can do ths because the glass does not rotate). FBD. The fgure shows a free body dagram for the glass. The forces nclude () the weght; and () the normal and tangental components of reacton at the contact between the tablecloth and the glass. The normal and tangental forces must act somewhere nsde the contact area, but ther poston s unnown. For a more detaled dscusson of contact forces see Sects.4 and Knematcs We are assumng that the glass has the same acceleraton as the tablecloth. The table cloth s movng n the drecton, and has magntude a. The acceleraton vector s therefore a a. 4. EOM. Newton s laws of moton yeld F ma T ( N mg) ma 5. Soluton: The and components of the vector equaton must each be satsfed (ust as when you solve a statcs problem), so that T ma N mg 0 N mg Fnally, we must use the frcton law to decde whether or not the tablecloth wll slp from under the glass. Recall that, for no slp, the frcton force must satsfy T N where s the frcton coeffcent. Substtutng for T and N from (5) shows that for no slp a g To do the trc, therefore, the acceleraton must exceed g. For a frcton coeffcent of order 0.1, ths gves an acceleraton of order 1 m/ s. There s a specal trc to pullng the tablecloth wth a large acceleraton but that s a secret. mg T N Example 3: Bcycle Safety. If a be rder braes too hard on the front wheel, hs or her be wll tp over (the fgure s from In ths example we nvestgate the conons that wll lead the be to capsze, and dentfy desgn varables that can nfluence these conons. If the be tps over, the rear wheel leaves the ground. If ths happens, the reacton force actng on the wheel must be zero so we can detect the pont where the be s ust on the verge of tppng over by calculatng the reacton forces, and fndng the conons where the reacton force on the rear wheel s zero. 1. Idealzaton: a. We wll dealze the rder as a partcle (apologes to be racers but that s how we thn of you ). The partcle s located at the center of mass of the rder. The fgure shows the most mportant desgn parameters- these are the heght of the h A d B L

16 rder s COM, the wheelbase L and the dstance of the COM from the rear wheel. b. We assume that the be s a massless frame. The wheels are also assumed to have no mass. Ths means that the forces actng on the wheels must satsfy F 0 and M 0 - and can be analyzed usng methods of statcs. If you ve forgotten how to thn about statcs of wheels, you should re-read the notes on ths topc n partcular, mae sure you understand the nature of the forces actng on a freely rotatng wheel (Secton.4.6 of the reference notes). c. We assume that the rder braes so hard that the front wheel s prevented from rotatng. It must therefore sd over the ground. Frcton wll resst ths sldng. We denote the frcton coeffcent at the contact pont B by. d. The rear wheel s assumed to rotate freely. e. We neglect ar resstance.. FBD. The fgure shows a free body dagram for the rder and for the be together. Note that a. A normal and tangental force acts at the contact pont on the front wheel (n general, both normal and tangental forces always act at contact ponts, unless the contact happens to be frctonless). Because the contact s slppng t s essental to draw the frcton force n the correct drecton the force must resst the moton of the be; b. Only a normal force acts at the contact pont on the rear wheel because t s freely rotatng, and behaves le a -force member. N A W T B N B 3. Knematcs The be s movng n the drecton. As a vector, ts acceleraton s therefore a a, where a s unnown. 4. EOM: Because ths problem ncludes a massless frame, we must use two equatons of moton ( F ma and M C 0). It s essental to tae moments about the partcle (.e. the rder s COM). F ma gves T ( N N W) ma B A B It s very smple to do the moment calculaton by hand, but for those of you who fnd such calculatons unbearable here s a Mupad scrpt to do t. The scrpt smply wrtes out the poston vectors of ponts A and B relatve to the center of mass as 3D vectors, wrtes down the reactons at A and B as 3D vectors, and calculates the resultant moment (we don t bother ncludng the weght, because t acts at the orgn and so exerts zero moment) The two nonzero components of scalar equatons F ma and the one nonzero component of M C 0 gve us three

17 T B ( N N W ) 0 A ma B N ( L d) N d T h 0 B A B We have four unnowns the reacton components NA, NB, T B and the acceleraton a so we need another equaton. The mssng equaton s the frcton law T N 5. Soluton: Here s the soluton wth Mupad. It s easy to get the same answer by hand as well. B B We are nterested n fndng what maes the reacton force at A go to zero (that s when the be s about to tp). So h ( L d) N A W 0 ( L d) / h h L Ths tells us that the be wll tp f the frcton coeffcent exceeds a crtcal magntude, whch depends on the geometry of the be. The smplest way to desgn a tp-resstant be s to mae the heght of the center of mass h small, and the dstance (L-d) between the front wheel and the COM as large as possble. A `recumbent be s one way to acheve ths the fgure (from shows an example. The recumbent desgn offers many other sgnfcant advantages over the classc bcycle besdes tppng resstance.

18 Example 4: A stupd problem that you mght fnd n the FE professonal engneerng exam. The purpose of ths problem s to show what you need to do to solve problems nvolvng more than one partcle. Two weghts of mass m A and m B are connected by a cable passng over two freely rotatng pulleys as shown. They are released, and the system begns to move. Fnd an expresson for the tenson n the cable connectng the two weghts. m A m B 1. Idealzaton The masses wll be dealzed as partcles; the cable s nextensble and the mass of the pulleys s neglected. Ths means the nternal forces n the cable, and the forces actng between cables/pulleys must satsfy F 0 and M 0, and we can treat them as though they were n statc equlbrum. T T. FBD we have to draw a separate FBD for each partcle. Snce the pulleys and cable are massless, the tenson T n the cable s constant. m A m B 3. Knematcs We now that both masses must move n the drecton. We also now that the masses always move at the same speed but n opposte drectons. Therefore, ther acceleratons must be equal and opposte. We can express ths mathematcally as a a A B m A g m B g 4. EOM: We must wrte down two equatons of moton, as there are two masses ( T mag) maaa ( T mbg) mbab We now have three equatons for three unnowns (the unnowns are a, a and T). 5. Soluton: As pad up members of ALE (the Academy of Lazy Engneers) we use Mupad to solve the equatons A B So the tenson n the cable s We pass! T m m A m B A m B g

19 Example 5: Another stupd FE exam problem: The fgure shows a small bloc on a rotatng bar. The contact between the bloc and the bar has frcton coeffcent. The bar rotates at constant angular speed. Fnd the crtcal angular velocty that wll ust mae the bloc start to slp when 0. Whch way does the bloc slde? The general approach to ths problem s the same as for the Magc trc example we wll calculate the reacton force exerted by the bar on the bloc, and see when the forces are large enough to cause slp at the contact. We analyze the moton assumng the slp does not occur, and then fnd out the conons where ths can no longer be the case. r 1. Idealzaton We wll dealze the bloc as a partcle. Ths s dangerous, because the bloc s clearly rotatng. We hope that because t rotates at constant rate, the rotaton wll not have a sgnfcant effect but we can only chec ths once we now how to deal wth rotatonal moton.. FBD: The fgure shows a free body dagram for the bloc. The bloc s T subected to a vertcal gravtatonal force, and reacton forces at the contact wth the bar. Snce we have assumed that the contact s not N slppng, we can choose the drecton of the tangental component of the mg reacton force arbtrarly. The resultant force on the bloc s F T cos N sn ( N cos T sn mg) 3. Knematcs We can use the crcular moton formula to wrte down the acceleraton of tbe bloc (see secton 3.1.3) a r (cos sn ) 4. EOM: The equaton of moton s T cos N sn ( N cos mg) mr (cos sn ) 5. Soluton: The and components of the equaton of moton can be solved for N and T Mupad maes ths panless To fnd the pont where the bloc ust starts to slp, we use the frcton law. Recall that, at the pont of slp T N For the bloc to slp wth 0 r g so the crtcal angular velocty s g/ r. Snce the tangental tracton T s negatve, and the frcton force must oppose sldng, the bloc must slde outwards,.e. r s ncreasng durng slp.

20 Alternatve method of soluton usng normal-tangental coordnates We wll solve ths problem agan, but ths tme we ll use the short-cuts descrbed n Secton to wrte down the acceleraton vector, and we ll wrte down the vectors n Newton s laws of moton n terms of the unt vectors n and t normal and tangent to the obect s path. () Acceleraton vector If the bloc does not slp, t moves wth speed V raround a crcular arc wth radus r. Its acceleraton vector has magntude V / r and drecton parallel to the unt vector n. t () The force vector can be resolved nto components parallel to n and t. Smple trg on the free body dagram shows that () Newton s laws then gve N mg cos mg sn T F t n cos sn F ma N mg t mg T n m rn The components of ths vector equaton parallel to t and n yeld two equatons, wth soluton N mg cos T mg sn m r Ths s the same soluton as before. The short-cut maes the calculaton slghtly more straghtforward. Ths s the man purpose of usng normal-tangental components. r n t n T N mg Example 6: Wndow blnds. Have you ever wondered how wndow shades wor? You gve the shade a lttle downward er, let t go, and t wnds tself up. If you pull the shade down slowly, t stays down. The fgure shows the mechansm (whch probably only costs a few cents to manufacture) that acheves ths remarable feat of engneerng. It s called an `nertal latch the same prncple s used n the nerta reels on the seatbelts n your car. The pcture shows an enlarged end vew of the wndow shade. The hub, shown n brown, s fxed to the bracet supportng the shade and cannot rotate. The drum, shown n peach, rotates as the shade s pulled up or down. The drum s attached to a torsonal sprng, whch tends to cause the drum to rotate counterclocwse, so wndng up the shade. The rotaton s prevented by the small dogs, shown n red, whch engage wth the teeth on the hub. You can pull the shade downwards freely, snce the dogs allow the drum to rotate counterclocwse. To rase the shade, you need to gve the end of the shade a er downwards, and then release t. When the drum rotates suffcently qucly (we wll calculate how qucly shortly) the Statonary Hub Rotatng Drum Statonary Hub Wndow shade

21 dogs open up, as shown on the rght. They reman open untl the drum slows down, at whch pont the topmost dog drops and engages wth the teeth on the hub, thereby locng up the shade once more. We wll estmate the crtcal rotaton rate requred to free the rotatng drum. 1. Idealzaton We wll dealze the topmost dog as a partcle on the end of a massless, nextensble rod, as shown n the fgure. a. We wll assume that the drum rotates at constant angular rate. Our goal s to calculate the crtcal speed where the dog s ust on the pont of droppng down to engage wth the hub. b. When the drum spns fast, the partcle s contacts the outer rm of the drum a normal force acts at the contact. When the dog s on the pont of droppng ths contact force goes to zero. So our goal s to calculate the contact force, and then to fnd the crtcal rotaton rate where the force wll drop to zero. c. We neglect frcton. R. FBD. The fgure shows a free body dagram for the partcle. The partcle s subected to: () a reacton force N where t contacts the rm; () a tenson T n the ln, and () gravty. The resultant force s F T cos( ) N cos ( N sn T sn( ) mg) 3. Knematcs We can use the crcular moton formula to wrte down the acceleraton of the partcle(see secton 3.1.3) a R (cos sn ) 4. EOM: The equaton of moton s T cos( ) N cos ( N sn T sn( ) mg) a R (cos sn ) 5. Soluton: The and components of the equaton of moton can be solved for N and T Mupad maes ths panless T mg N normal reacton force s therefore N mg cos( ) / sn mr We are loong for the pont where ths can frst become zero or negatve. Note that max{cos( )} 1 at the pont where =0. The smallest value of N therefore occurs at ths pont, and has magntude The crtcal speed where N=0 follows as Nmn mg / sn mr g/ ( Rsn )

22 Changng the angle and the radus R gves a convenent way to control the crtcal speed n desgnng an nertal latch. Alternatve soluton usng polar coordnates We ll wor through the same problem agan, but ths tme handle the vectors usng polar coordnates. 1. FBD. The fgure shows a free body dagram for the partcle. The partcle s subected to: () a reacton force N where t contacts the rm; () a tenson T n the ln, and () gravty. The resultant force s F ( N T cos mg sn ) er ( T sn mg cos ) e. Knematcs The acceleraton vector s now ar e r 3. EOM: The equaton of moton s ( N T cos mg sn ) e ( T sn mg cos ) e R e r 4. Soluton: The er, e components of the equaton of moton can be solved for N and T agan, we can use Mupad for ths T e r mg N e r The normal reacton force s therefore N mg cos( ) / sn mr We are loong for the pont where ths can frst become zero or negatve. Note that max{cos( )} 1 at the pont where =0. The smallest value of N therefore occurs at ths pont, and has magntude The crtcal speed where N=0 follows as Nmn mg / sn mr g/ ( Rsn ) Changng the angle and the radus R gves a convenent way to control the crtcal speed n desgnng an nertal latch.

23 Example 7: Arcraft Dynamcs Arcraft performng certan nstrument approach procedures (such as holdng patterns or procedure turns) are requred to mae all turns at a standard rate, so that a complete 360 degree turn taes mnutes. All turns must be made at constant alttude and constant speed, V. People who desgn nstrument approach procedures need to now the radus of the resultng turn, to mae sure the arcraft won t ht anythng. Engneers desgnng the arcraft are nterested n the forces needed to complete the turn specfcally, the load factor, whch s the rato of the lft force on the arcraft to ts weght. In ths problem we wll calculate the radus of the turn R and the ban angle requred, as well as the load factor caused by the maneuver, as a functon of the arcraft speed V. Before startng the calculaton, t s helpful to understand what maes an arcraft travel n a crcular path. Recall that 1. If an obect travels at constant speed around a crcle, ts acceleraton vector has constant magntude, and has drecton towards the center of the crcle. A force must act on the arcraft to produce ths acceleraton.e. the resultant force on the arcraft must act towards the center of the crcle. The necessary force comes from the horzontal component of the lft force the plot bans the wngs, so that the lft acts at an angle to the vertcal. Wth ths nsght, we expect to be able to use the equatons of moton to calculate the forces. 1. Idealzaton The arcraft s dealzed as a partcle t s not obvous that ths s accurate, because the arcraft clearly rotates as t travels around the curve. However, the forces we wsh to calculate turn out to be fully determned by F=ma and are not nfluenced by the rotatonal moton.. FBD. The fgure shows a free body dagram for the arcraft. It s subected to () a gravtatonal force (mg); () a thrust from the engnes F T, () a drag force F D, actng perpendcular to the drecton of moton, and (v) a lft force F L, actng perpendcular to the plane of the wngs. R The resultant force s ( F F ) cos F sn sn ( F F )sn F sn cos F cos mg T D L D T L L R (you may fnd the components of the lft force dffcult to vsualze to see where these come from, note that the lft force can be proected onto components along OR and the drecton as FL FL snro FL cos. Then note that RO sn cos.) O F T F L mg F D 3. Knematcs a. The arcraft moves at constant speed around a crcle, so the angle t, where s the (constant) angular speed of the lne OP. Snce the arcraft completes a turn n two mnutes, we now that / (60) / 60 rad/sec b. The poston vector of the plane s r Rsnt Rcost

24 We can dfferentate ths expresson wth respect to tme to fnd the velocty v R(cos t sn t) c. The magntude of the velocty s V R, so f the arcraft fles at speed V, the radus of the turn must be R V / d. Dfferentatng the velocty gves the acceleraton a R (sn t cos t ) 4. EOM: The equaton of moton s ( F F )cos F sn sn ( F F )sn F sn cos F cos mg T D L D T L L mr (sn cos ) mv(sn cos ) 5. Soluton: The and components of the equaton of moton gve three equatons that can be solved for F T, F L and. We assume that the drag force s nown, snce ths s a functon of the arcraft s speed. Ths soluton s correct, but you need a PhD to understand what t means (other symbolc manpulaton programs le Maple and Mathematca gve a comprehensble soluton). Fortunately, I happen to have a PhD The soluton can be smplfed to 1 tan ( V / g) FL mg 1 V / g FT FD As we wll see below, f you choose to solve ths problem n normal-tangental coordnates, you don t need a PhD to We can calculate values of, R V / and the load factor F / mg for a few arcraft 0 a. Cessna 150 V=70nots (36 m/s) : 11 R=690m, F / mg 1.0 L L

25 0 b. Boeng 747: V=00 nots (10 m/s) 8 R=1950m, F / mg c. F111 V=300 nots (154 m/s) 39 R=950m, F / mg 1.3 L L Alternatve soluton usng normal-tangental coordnates Ths problem can also be solved rather more qucly usng normal and tangental bass vectors. () Acceleraton vector. The arcraft travels around a crcular path at constant speed, so ts acceleraton s V a n Vn R where n s a unt vector pontng towards the center of the crcle. () Force vector. The force vector can be wrtten n terms of the unt vectors n,t, as F ( F F ) t F sn n ( F cos mg) T D L L O R n F L F T R mg t F D () Newton s law F ( FT FD ) t FL sn n ( FL cos mg) mvn The n, t and components of ths equaton gve three equatons that can be solved for F T, Ths tme t s easy to solve the equatons by hand 1 tan ( V / g) FL mg 1 V / g FT FD F L and. Ths example agan shows why normal-tangental coordnates are useful descrbng forces, and solvng the resultng force-acceleraton relatons are much smpler than worng wth a fxed coordnate system. 3.3 Dervng and solvng equatons of moton for systems of partcles We next turn to the more dffcult problem of predctng the moton of a system that s subected to a set of forces General procedure for dervng and solvng equatons of moton for systems of partcles It s very straghtforward to analyze the moton of systems of partcles. You should always use the followng procedure 1. Introduce a set of varables that can descrbe the moton of the system. Don t worry f ths sounds vague t wll be clear what ths means when we solve specfc examples.. Wrte down the poston vector of each partcle n the system n terms of these varables 3. Dfferentate the poston vector(s), to calculate the velocty and acceleraton of each partcle n terms of your varables; 4. Draw a free body dagram showng the forces actng on each partcle. You may need to ntroduce varables to descrbe reacton forces. Wrte down the resultant force vector. 5. Wrte down Newton s law F ma for each partcle. Ths wll generate up to 3 equatons of moton (one for each vector component) for each partcle.

26 6. If you wsh, you can elmnate any unnown reacton forces from Newton s laws. If you are tryng to solve the equatons by hand, you should always do ths; of you are usng MATLAB, t s not usually necessary you can have MATLAB calculate the reactons for you. The result wll be a set of dfferental equatons for the varables defned n step (1) 7. If you fnd you have fewer equatons than unnown varables, you should loo for any constrants that restrct the moton of the partcles. The constrants must be expressed n terms of the unnown acceleratons. 8. Identfy the ntal conons for the varables defned n (1). These are usually the values of the unnown varables, ther tme dervatves, at tme t=0. If you happen to now the values of the varables at some other nstant n tme, you can use that too. If you don t now ther values at all, you should ust ntroduce new (unnown) varables to denote the ntal conons. 9. Solve the dfferental equatons, subect to the ntal conons. Steps (3) (6) and (8) can usually be done on the computer, so you don t actually have to do much calculus or math. Sometmes, you can avod solvng the equatons of moton completely, by usng conservaton laws conservaton of energy, or conservaton of momentum to calculate quanttes of nterest. These shortcuts wll be dscussed n the next chapter. 3.. Smple examples of equatons of moton and ther solutons The general process descrbed n the precedng secton can be llustrated usng smple examples. In ths secton, we derve equatons of moton for a number of smple systems, and fnd ther solutons. The purpose of these examples s to llustrate the straghtforward, step-by-step procedure for analyzng moton n a system. Although we solve several problems of practcal nterest, we wll smply set up and solve the equatons of moton wth some arbtrary values for system parameter, and won t attempt to explore ther behavor n detal. More detaled dscussons of the behavor of dynamcal systems wll follow n later chapters. Example 1: Traectory of a partcle near the earth s surface (no ar resstance) At tme t=0, a proectle wth mass m s launched from a poston X0 X Y Z wth ntal velocty vector V. Calculate ts traectory as a functon 0 Vx Vy Vz of tme. X0 V 0 1. Introduce varables to descrbe the moton: We can smply use the Cartesan coordnates of the partcle ( x( t), y( t), z( t )). Wrte down the poston vector n terms of these varables: r x( t) y( t) z( t) 3. Dfferentate the poston vector wth respect to tme to fnd the acceleraton. For ths example, ths s trval

27 dx dy dz d x d y d z v a 4. Draw a free body dagram. The only force actng on the partcle s gravty the free body dagram s shown n the fgure. The force vector follows as Fmg. 5. Wrte down Newton s laws of moton. Ths s easy d x d y d z F ma mg m The vector equaton actually represents three separate dfferental equatons of moton d x d y d z 0 0 g mg 6. Elmnate reactons ths s not needed n ths example. 7. Identfy ntal conons. The ntal conons were gven n ths problem we have that dx dy dz x X 0 V x y Y0 V y z Z0 V z 8. Solve the equatons of moton. In general we wll use MAPLE or matlab to do the rather tedous algebra necessary to solve the equatons of moton. Here, however, we wll ntegrate the equatons by hand, ust to show that there s no magc n MAPLE. The equatons of moton are d x d y d z 0 0 g It s a bt easer to see how to solve these f we defne dx v dy dz x vy vz The equaton of moton can be re-wrtten n terms of ( vx, vy, v z ) as dv dv x y dvz 0 0 g We can separate varables and ntegrate, usng the ntal conons as lmts of ntegraton vx t vy t vz t dv 0 dv 0 dv g x x z Vx 0 Vy 0 Vz 0 v V v V v V gt x x y y z z Now we can re-wrte the velocty components n terms of (x,y,z) as dx dy dz Vx Vy Vz gt dy Agan, we can separate varables and ntegrate

28 y x t t z t dx V dy V dz V gt x y z X0 0 Y0 0 Z0 0 1 x X 0 Vxt y Y0 Vyt z Z0 Vzt gt so the poston and velocty vectors are 1 r X Vxt Y Vyt Z Vzt gt v V V V gt x y z Here s how to ntegrate the equatons of moton usng Mupad Applcatons of traectory problems: It s traonal n elementary physcs and dynamcs courses to solve vast numbers of problems nvolvng partcle traectores. These nvarably nvolve beng gven some nformaton about the traectory, whch you must then use to wor out somethng else. These problems are all somewhat tedous, but we wll show a couple of examples to uphold the fne traons of a 19 th century educaton. Estmate how far you could throw a stone from the top of the Kremln palace. Note that 1. The horzontal and vertcal components of velocty at tme t=0 follow as V v cos V 0 V v sn x 0 y z 0. The components of the poston of the partcle at tme t=0 are X 0, Y 0, Z H 3. The traectory of the partcle follows as 1 r v0cost H v0snt gt D 4. When the partcle hts the ground, ts poston vector s r D. Ths must be on the traectory, so 1 v0costi H v0snti gti D where t I s the tme of mpact. 5. The two components of ths vector equaton gves us two equatons for the two unnowns { t, D }, whch can be solved I H v 0

29 The RootOf n MAPLE s scary any tme that MAPLE gves you a scary result, loo n the help and see f there s a `convert functon that mght mae t less scary. For a rough estmate of the dstance we can use the followng numbers 1. Heght of Kremln palace 71m. Throwng velocty maybe 5mph? (pretty pathetc, I now - you can probably do better, especally f you are on the baseball team). 3. Throwng angle 45 degrees. Substtutng numbers gves 36m (118ft). If you want to go wld, you can maxmze D wth respect to, but ths won t mprove your estmate much. Slcon nanopartcles wth radus 0nm are n thermal moton near a flat surface. At the surface, they have an average velocty T / m, where m s ther mass, T s the temperature and = s the Boltzmann constant. Estmate the maxmum heght above the surface that a typcal partcle can reach durng ts thermal moton, assumng that the only force actng on the partcles s gravty 1. The partcle wll reach ts maxmum heght f t happens to be travelng vertcally, and does not collde wth any other partcles.. At tme t=0 such a partcle has poston X 0, Y 0, Z 0 and velocty Vx 0 Vy 0 Vz T / m 3. For tme t>0 the poston vector of the partcle follows as T / m t 1 r gt Its velocty s v T / m gt 4. When the partcle reaches ts maxmum heght, ts velocty must be equal to zero (f you don t see ths by vsualzng the moton of the partcle, you can use the mathematcal statement that f r y s a maxmum, then dr / v ( dy / ) 0 ). Therefore, at the nstant of maxmum heght v T / m gtmax 0 5. Ths shows that the nstant of max heght occurs at tme tmax T / m / g 6. Substtutng ths tme bac nto the poston vector shows that the poston vector at max heght s

30 T 1 T T r mg mg mg 7. S has a densty of about 330 g/m^3. At room temperature (93K) we fnd that the dstance s surprsngly large: 10mm or so. Gravty s a very wea force at the nano-scale surface forces actng between the partcles, and the partcles and the surface, are much larger. Example : Free vbraton of a suspenson system. A vehcle suspenson can be dealzed as a mass m supported by a sprng. The sprng has stffness and un-stretched length L 0. To test the suspenson, the vehcle s constraned to move vertcally, as shown n the fgure. It s set n moton by stretchng the sprng to a length L and then releasng t (from rest). Fnd an expresson for the moton of the vehcle after t s released. m b,l 0 O x(t) As an asde, t s worth notng that a partcle dealzaton s usually too crude to model a vehcle a rgd body approxmaton s much better. In ths case, however, we assume that the vehcle does not rotate. Under these conons the equatons of moton for a rgd body reduce to F ma and M 0, and we shall fnd that we can analyze the system as f t were a partcle. 1. Introduce varables to descrbe the moton: The length of the sprng xt () s a convenent way to descrbe moton.. Wrte down the poston vector n terms of these varables: We can tae the orgn at O as shown n the fgure. The poston vector of the center of mass of the bloc s then b r xt () 3. Dfferentate the poston vector wth respect to tme to fnd the acceleraton. For ths example, ths s trval dx d x v a 4. Draw a free body dagram. The free body dagram s shown n the fgure: R Ax the mass s subected to the followng forces Gravty, actng at the center of mass of the vehcle The force due to the sprng R Bx Reacton forces at each of the rollers that force the vehcle to move F s vertcally. Recall the sprng force law, whch says that the forces exerted by a sprng act parallel to ts length, tend to shorten the sprng, and are proportonal to the dfference between the length of the sprng and ts unstretched length. mg

31 5. Wrte down Newton s laws of moton. Ths s easy d x F ma ( RAx RBx ) mg ( x L0 ) m The and components gve two scalar equatons of moton ( R R ) 0 Ax Bx d x g ( x L 0) m 6. Elmnate reactons ths s not needed n ths example. 7. Identfy ntal conons. The ntal conons were gven n ths problem at tme t=0, we now that x L and dx / 0 8. Solve the equatons of moton. Agan, we wll frst ntegrate the equatons of moton by hand, and then repeat the calculaton wth MAPLE. The equaton of moton s d x g ( x L 0) m We can re-wrte ths n terms of dx vx Ths gves dvx dvx dx dvx vx g ( x L0 ) dx dx m We can separate varables and ntegrate vx x vxdvx g ( x L0 ) dx m 0 L 1 vx g( x L) x L L0 ( x L) m m mg mg vx L L0 x L0 m Don t worry f the last lne loos mysterous wrtng the soluton n ths form ust maes the algebra a bt smpler. We can now ntegrate the velocty to fnd x dx mg mg vx L L0 x L0 m x dx L mg mg L L0 x L0 m The ntegral on the left can be evaluated usng the substtuton t 0

32 so that x L0 mg / L L0 mg / cos 0 t snd 1 x L0 mg / 0 cos L L mg / 1 cos m Here s the MAPLE soluton 0 t m x L0 mg / cos t L L0 mg / m x L L0 mg / cos t L0 mg / m Note that t s mportant to nclude the assume() statements, otherwse Mupad gves the soluton n the form of an exponental of a complex number. The soluton n ths other form s also correct, but s dffcult to vsualze. The soluton s plotted n the fgure. The behavor of vbratng systems wll be dscussed n more detal later n ths course, but t s worth notng some features of the soluton: 1. The average poston of the mass s x L0 mg /. Here, mg/ s the statc deflecton of the sprng.e. the deflecton of the sprng due to the weght of the vehcle (wthout moton). Ths means that the car vbrates symmetrcally about ts statc deflecton.. The ampltude of vbraton s L L0 mg /. Ths corresponds to the dstance of the mass above ts average poston at tme t=0. 3. The perod of oscllaton (the tme taen for one complete cycle of vbraton) s T m / 1 4. The frequency of oscllaton (the number of cycles per second) s f m (note f=1/t). Frequency s also sometmes quoted as angular frequency, whch s related to f by f / m. Angular frequency s n radans per second.

33 An nterestng feature of these results s that the statc deflecton s related to the frequency of oscllaton - so f you measure the statc deflecton mg /, you can calculate the (angular) vbraton frequency as g / Example 3: Slly FE exam problem Ths example shows how polar coordnates can be used to analyze moton. The rod shown n the pcture rotates at constant angular speed n the horzontal plane. The nterface between bloc and rod has frcton coeffcent. The rod pushes a bloc of mass m, whch starts at r=0 wth radal speed V. Fnd an expresson for r(t). L r e e r 1. Introduce varables to descrbe the moton the polar coordnates r, wor for ths problem. Wrte down the poston vector and dfferentate to fnd acceleraton we don t need to do ths we can ust wrte down the standard result for polar coordnates d r dr a r e r e 3. Draw a free body dagram shown n the fgure note that t s mportant to draw the frcton force n the correct drecton. The bloc wll slde radally outwards, and frcton opposes the slp. 4. Wrte down Newton s law 5. Elmnate reactons d r dr Ter Ne m r e r e F=ma gves two equatons for N and T. A thrd one comes from the frcton law T e e r T N N The thrd soluton can be rearranged nto an equaton of moton for r

34 d r dr r 0 6. Identfy ntal conons Here, r=0 dr/=v at tme t=0. 7. Solve the equaton: If you ve taen AM33 you wll now how to solve ths equaton But f not, or you are lazy, you can use MAPLE to solve t for you. Ths can be smplfed slghtly by hand: V t r( t) e snh( 1 t) 1 Example 4: Moton of a pendulum (R-rated verson) A pendulum s a ubqutous engneerng system. You are, of course, famlar wth how a pendulum can be used to measure tme. But t s used for a varety of other scentfc applcatons. For example, Professor Crsco s lab uses pendulum to measure propertes of human onts, see Crsco Joseph J; Futa Lndsey; Spencner Davd B, The dynamc flexon/extenson propertes of the lumbar spne n vtro usng a novel pendulum system. Journal of bomechancs 007;40(1): l In ths example, we wll wor through the basc problem of dervng and solvng the equatons of moton for a pendulum, neglectng ar resstance. 1. Introduce varables to descrbe moton: The angle () t shown n the fgure s a convenent varable.. Wrte down the poston vector as a functon of the varables We ntroduce a Cartesan coordnate system wth orgn at O, as shown n the pcture. Elementary geometry shows that r lsn lcos

35 Note that we have assumed that the cable remans straght ths wll be true as long as the nternal force n the cable s tensle. If calculatons predct that the nternal force s compressve, ths assumpton s wrong. But there s no way to chec the assumpton at ths pont so we smply proceed, and chec the answer at the end 3. Dfferentate the poston vector to fnd the acceleraton: The computer maes ths panless. 4. The free body dagram s shown n the fgure. The force exerted by the cable on the partcle s ntroduced as an unnown reacton force. The force vector s F Rsn Rcos mg 5. Newton s laws of moton can be expressed as F ma d d d d Rsn Rcos mg ml sn l cos m l cos l sn R mg Equatng the and components gves two equatons for the two unnowns d d Rsn m l sn l cos d d Rcos mg m l cos l sn 6. Elmnate the reacton forces. In ths problem, t s helpful to elmnate the unnown reacton force R. You can do ths on the computer f you le, but n ths case t s smpler to do ths by hand. You can smply multply the frst equaton by cos and the second equaton by sn and then add them. Ths yelds d mg sn msn cos d g sn 0 l 7. Identfy ntal conons. Some calculatons are necessary to determne the ntal conons n ths problem. We are gven that 0 at tme t=0, and the horzontal velocty s V 0 at tme t=0, but to solve

36 the equaton of moton, we need the value of d / 0. We can fnd the relatonshp we need by dfferentatng the poston vector to fnd the velocty d v lcos lsn Settng vv0 and 0 at t=0 shows that d V0 l so d / V0 / l 8. Solve the equatons of moton Ths equaton of moton s too dffcult for MAPLE but actually the soluton does exst and s very well nown ths s a classc problem n mathematcal physcs. Wth ntal conons 0, d / V0 / l t 0 the soluton s 1 V sn 0 g V sn, 0 V t 0 1 gl l gl gl () t V am 0 gl V t, 0 1 l V 0 gl The frst soluton descrbes swngng moton of the pendulum, whle the second soluton descrbes the moton that occurs f you push the pendulum so hard that t whrls around on the pvot. The equatons may loo scary, but you can smply use MAPLE to calculate and plot them. 1. In the frst equaton, sn( x, ) s a specal functon called the `sn ampltude. You can thn of t as a sort of trg functon for adults n fact for =0, sn( x,0) sn( x) and we recover the PG verson. You can compute t n Mupad usng acobsn(x,). Smlarly, am( x, ) s a functon called the `Ampltude. You can calculate t n Mupad usng acobam(x,). In Mupad, the am functon has range - <am( x, ), so the soluton predcts that as the pendulum whrls around the pvot, the angle ncreases from 0 to, then umps to, ncreases to agan, and so on. You mght have solved the pendulum problem already n an elementary physcs course, and mght remember a dfferent soluton. Ths s because you probably only derved an approxmate soluton, by assumng that the angle remans small. Ths occurs when the ntal velocty satsfes V / gl 1, n whch case the soluton can be approxmated by V g ( t) sn t gl l Numercal solutons to equatons of moton usng MATLAB In the precedng secton, we were able to solve all our equatons of moton exactly, and hence to fnd formulas that descrbe the moton of the system. Ths should gve you a warm and fuzzy feelng t appears that wth very lttle wor, you can predct everythng about the moton of the system. You may even have vsons of runnng a consultng busness from your yacht n the Carbbean, wth nothng more than your chef, your masseur (or masseuse) and a laptop wth a copy of MAPLE.

37 Unfortunately real lfe s not so smple. Equatons of moton for most engneerng systems cannot be solved exactly. Even very smple problems, such as calculatng the effects of ar resstance on the traectory of a partcle, cannot be solved exactly. For nearly all practcal problems, the equatons of moton need to be solved numercally, by usng a computer to calculate values for the poston, velocty and acceleraton of the system as functons of tme. Vast numbers of computer programs have been wrtten for ths purpose some focus on very specalzed applcatons, such as calculatng orbts for spacecraft (STK); calculatng moton of atoms n a materal (LAMMPS); solvng flud flow problems (e.g. fluent, CFDRC); or analyzng deformaton n solds (e.g. ABAQUS, ANSYS, NASTRAN, DYNA); others are more general purpose equaton solvng programs. In ths course we wll use MATLAB, whch s wdely used n all engneerng applcatons. You should complete the MATLAB tutoral before proceedng any further. In the remander of ths secton, we provde a number of examples that llustrate how MATLAB can be used to solve dynamcs problems. Each example llustrates one or more mportant technque for settng up or solvng equatons of moton. Example 1: Traectory of a partcle near the earth s surface (wth ar resstance) As a smple example we set up MATLAB to solve the partcle traectory problem dscussed n the precedng secton. We wll extend the calculaton to account for the effects of ar resstance, however. We wll assume that our proectle s sphercal, wth dameter D, and we wll assume that there s no wnd. You may fnd t helpful to revew the dscusson of aerodynamc drag forces n Secton.1.7 before proceedng wth ths example. 1. Introduce varables to descrbe the moton: We can smply use the Cartesan coordnates of the partcle ( x( t), y( t), z( t )) X0 V 0 F D mg. Wrte down the poston vector n terms of these varables: r x( t) y( t) z( t) 3. Dfferentate the poston vector wth respect to tme to fnd the acceleraton. Smple calculus gves dx dy dz d x d y d z v a 4. Draw a free body dagram. The partcle s now subected to two forces, as shown n the pcture. Gravty as always we have Fg mg. Ar resstance. The magntude of the ar drag force s gven by F D 1 CDDV, where

38 s the ar densty, C D s the drag coeffcent, D s the proectle s dameter, and V s the magntude of the proectle s velocty relatve to the ar. Snce we assumed the ar s statonary, V s smply the magntude of the partcle s velocty,.e. dx dy dz V The Drecton of the ar drag force s always opposte to the drecton of moton of the proectle through the ar. In ths case the ar s statonary, so the drag force s smply opposte to the drecton of the partcle s velocty. Note that v /V s a unt vector parallel to the partcle s velocty. The drag force vector s therefore The total force vector s therefore 1 v 1 D D D D D dx dy dz F C V C V 4 V 4 1 D dx dy dz F mg CD V 4 5. Wrte down Newton s laws of moton. F 1 D dx dy dz d x d y d z m a mg CD V m 4 It s helpful to smplfy the equaton by defnng a specfc drag coeffcent c D C D, so that 8m dx dy dz d x d y d z g cv The vector equaton actually represents three separate dfferental equatons of moton d x dx d y dy d z dz cv cv g cv 6. Elmnate reactons ths s not needed n ths example. 7. Identfy ntal conons. The ntal conons were gven n ths problem - we have that dx dy dz x X 0 V x y Y0 V y z Z0 V z 8. Solve the equatons of moton. We can t use the magc dsolve command n MAPLE to solve ths equaton t has no nown exact soluton. So nstead, we use MATLAB to generate a numercal soluton.

39 Ths taes two steps. Frst, we must turn the equatons of moton nto a form that MATLAB can use. Ths means we must convert the equatons nto frst-order vector valued dfferental equaton of the dy general form f( t, ) y. Then, we must wrte a MATLAB scrpt to ntegrate the equatons of moton. Convertng the equatons of moton: We can t solve drectly for (x,y,z), because these varables get dfferentated more than once wth respect to tme. To fx ths, we ntroduce the tme dervatves of (x,y,z) as new unnown varables. In other words, we wll solve for ( x, y, z, vx, vy, v z ), where vx dx v dx dx y vz These defntons are three new equatons of moton relatng our unnown varables. In adon, we can re-wrte our orgnal equatons of moton as dv dv x y dv cvv z x cvvy g cvvz So, expressed as a vector valued dfferental equaton, our equatons of moton are x vx y v y d z v z vx cvvx v y cvv y v z g cvv z MATLAB scrpt. The procedure for solvng these equatons s dscussed n the MATLAB tutoral. A basc MATLAB scrpt s lsted below. functon traectory_3d % Functon to plot traectory of a proectle % launched from poston X0 wth velocty V0 % wth specfc ar drag coeffcent c % stop_tme specfes the end of the calculaton g = 9.81; % gravtatonal accel c=0.5; % The constant c X0=0; Y0=0; Z0=0; % The ntal poston VX0=10; VY0=10; VZ0=0; stop_tme = 5; ntal_w = [X0,Y0,Z0,VX0,VY0,VZ0]; % The soluton at t=0 [tmes,sols] = ode45(@eom,[0,stop_tme],ntal_w); plot3(sols(:,1),sols(:,),sols(:,3)) % Plot the traectory functon dw = eom(t,w) % Varables stored as follows w = [x,y,z,vx,vy,vz] %.e. x = w(1), y=w(), z=w(3), etc x = w(1); y=w(); z=w(3); vx = w(4); vy = w(5); vz = w(6); vmag = sqrt(vx^+vy^+vz^);

40 dx = vx; dy = vy; dz = vz; dvx = -c*vmag*vx; dvy = -c*vmag*vy; dvz = -c*vmag*vz-g; dw = [dx;dy;dz;dvx;dvy;dvz]; end end Ths produces a plot that loos le ths (the plot s been eed to add the grd,etc) Example : Smple satellte orbt calculaton The fgure shows satellte wth mass m orbtng a planet wth mass M. At tme t=0 the satellte has poston vector r R and velocty vector vv. The planet s moton may be neglected (ths s accurate as long as M>>m). Calculate and plot the orbt of the satellte. M R V m 1. Introduce varables to descrbe the moton: We wll use the (x,y) coordnates of the satellte.. Wrte down the poston vector n terms of these varables: r x y 3. Dfferentate the poston vector wth respect to tme to fnd the acceleraton. dx dy d x d y v a F g

41 4. Draw a free body dagram. The satellte s subected to a gravtatonal force. GMm The magntude of the force s Fg, where r G s the gravtatonal constant, and r x y s the dstance between the planet and the satellte The drecton of the force s always towards the orgn: r / r s therefore a unt vector parallel to the drecton of the force. The total force actng on the satellte s therefore GMm r F GMm x y r 3 r r 5. Wrte down Newton s laws of moton. GMm d x d y F ma 3 x y m r The vector equaton represents two separate dfferental equatons of moton d x GM d y GM x y 3 3 r r 6. Elmnate reactons ths s not needed n ths example. 7. Identfy ntal conons. The ntal conons were gven n ths problem - we have that dx dy x R 0 y 0 V 8. Solve the equatons of moton. We follow the usual procedure: () convert the equatons nto MATLAB form; and () code a MATLAB scrpt to solve them. Convertng the equatons of moton: We ntroduce the tme dervatves of (x,y) as new unnown varables. In other words, we wll solve for ( x, y, v, v ), where x y dx dx vx vy These defntons are new equatons of moton relatng our unnown varables. In adon, we can rewrte our orgnal equatons of moton as dvx GM dvy GM x y 3 3 r r So, expressed as a vector valued dfferental equaton, our equatons of moton are x vx y v d y v 3 x GMx / r vy 3 GMy / r Matlab scrpt: Here s a smple scrpt to solve these equatons.

42 functon satellte_orbt % Functon to plot orbt of a satellte % launched from poston (R,0) wth velocty (0,V) GM=1; R=1; V=1; Tme=100; w0 = [R,0,0,V]; % Intal conons [t_values,w_values] = ode45(@odefunc,[0,tme],w0); plot(w_values(:,1),w_values(:,)) functon dw = odefunc(t,w) x=w(1); y=w(); vx=w(3); vy=w(4); r = sqrt(x^+y^); dw = [vx;vy;-gm*x/r^3;-gm*y/r^3]; end end Runnng the scrpt produces the result shown (the plot was annotated by hand) Do we beleve ths result? It s a bt surprsng the satellte seems to be spralng n towards the planet. Most satelltes don t do ths so the result s a bt suspcous. The Frst Law of Scentfc Computng states that ` f a computer smulaton predcts a result that surprses you, t s probably wrong. So how can we test our computaton? There are two good tests: 1. Loo for any features n the smulaton that you can predct wthout computaton, and compare your predctons wth those of the computer.. Try to fnd a specal choce of system parameters for whch you can derve an exact soluton to your problem, and compare your result wth the computer We can use both these checs here. 1. Conserved quanttes For ths partcular problem, we now that () the total energy of the system should be constant; and () the angular momentum of the system about the planet should be constant (these conservaton laws wll be dscussed n the next chapter for now, ust tae ths as gven). The total energy of the system conssts of the potental energy and netc energy of the satellte, and can be calculated from the formula GMm 1 GMm 1 E mv mvx vy r r E GM 1 vx vy m r The total angular momentum of the satellte (about the orgn) can be calculated from the formula

43 ( x y ) y x H r mv x y m v v m xv yv H xvy yvx m (If you don t now these formulas, don t panc we wll dscuss energy and angular momentum n the next part of the course) We can have MATLAB plot E/m and H / m, and see f these are really conserved. The energy and momentum can be calculated by addng these lnes to the MATLAB scrpt for =1:length(t) r = sqrt(w_values(,1)^ + w_values(,)^) vmag = sqrt(w_values(,1)^ + w_values(,)^) energy() = -GM/r + vmag^/; angularm() = w_values(,1)*w_values(,4)-w_values(,)*w_values(,3); end You can then plot the results (e.g. plot(t_values,energy)). The results are shown below. These results loo really bad nether energy, nor angular momentum, are conserved n the smulaton. Somethng s clearly very badly wrong. Comparson to exact soluton: It s not always possble to fnd a convenent exact soluton, but n ths case, we mght guess that some m specal ntal conons could set the satellte movng on a crcular path. R A crcular path mght be smple enough to analyze by hand. So let s assume that the path s crcular, and try to fnd the necessary ntal conons. If you stll remember the crcular moton formulas, you could use them to do ths. But only morons use formulas here we wll derve M the soluton from scratch. Note that, for a crcular path (a) the partcle s radus r=constant. In fact, we now r=r, from the poston at tme t=0. (b) The satellte must move at constant speed, and the angle must ncrease lnearly wth tme,.e. t where s a constant (see secton to revew moton at constant speed around a crcle). Wth ths nformaton we can solve the equatons of moton. Recall that the poston, velocty and acceleraton vectors for a partcle travelng at constant speed around a crcle are

44 We now that r R cos ( t) Rsn ( t) v R( sn ( t) cos ( t) ) a R cos ( t) Rsn ( t) v V from the ntal conons, and v s constant. Ths tells us that V Fnally, we can substtute ths nto Newton s law R GMm r F ma ma GMm (cos sn ) m V (cos sn ) R R R R Both components of the equaton of moton are satsfed f we choose GM V R R So, f we choose ntal values of GM, V, R satsfyng ths equaton, the orbt wll be crcular. In fact, our orgnal choce, GM 1, V 1, R 1 should have gven a crcular orbt. It dd not. Agan, ths means our computer generated soluton s totally wrong. Fxng the problem: In general, when computer predctons are suspect, we need to chec the followng 1. Is there an error n our MATLAB program? Ths s nearly always the cause of the problem. In ths case, however, the program s correct (t s too smple to get wrong, even for me).. There may be somethng wrong wth our equatons of moton (because we made a mstae n the dervaton). Ths would not explan the dscrepancy between the crcular orbt we predct and the smulaton, snce we used the same equatons n both cases. 3. Is the MATLAB soluton suffcently accurate? Remember that by default the ODE solver tres to gve a soluton that has 0.1% error. Ths may not be good enough. So we can try solvng the problem agan, but wth better accuracy. We can do ths by modfyng the MATLAB call to the equaton solver as follows optons = odeset('reltol', ); [t_values,w_vlues] = ode45(@odefunc,[0,tme],w0,optons); 4. Is there some feature of the equaton of moton that maes them especally dffcult to solve? In ths case we mght have to try a dfferent equaton solver, or try a dfferent way to set up the problem. The fgure on the rght shows the orbt predcted wth the better accuracy. You can see there s no longer any problem the orbt s perfectly crcular. The fgures below plot the energy and angular momentum predcted by the computer.

45 There s a small change n energy and angular momentum but the rate of change has been reduced dramatcally. We can mae the error smaller stll by usng mprovng the tolerances further, f ths s needed. But the changes n energy and angular momentum are only of order 0.01% over a large number of orbts: ths would be suffcently accurate for most practcal applcatons. Most ODE solvers are purposely desgned to lose a small amount of energy as the smulaton proceeds, because ths helps to mae them more stable. But n some applcatons ths s unacceptable for example n a molecular dynamc smulaton where we are tryng to predct the entropc response of a polymer, or a free vbraton problem where we need to run the smulaton for an extended perod of tme. There are specal ODE solvers that wll conserve energy exactly. Example 3: Earthquae response of a -storey buldng The fgure shows a very smple dealzaton of a -storey buldng. The roof and second floor are dealzed as blocs wth mass m. They are supported by structural columns, whch can be dealzed as sprngs wth stffness and unstretched length L. At tme t=0 the floors are at rest and the columns have lengths l1 L mg / l L mg / (can you show ths?). We wll neglect the thcness of the floors themselves, to eep thngs smple. For tme t>0, an earthquae maes the ground vbrate vertcally. The ground moton can be descrbed usng the equaton d d 0 snt. d 0 snt Horzontal moton may be neglected. Our goal s to calculate the moton of the frst and second floor of the buldng. l l 1 m m y y 1 It s worth notng a few ponts about ths problem: 1. You may be septcal that the floor of a buldng can be dealzed as a partcle (then agan, maybe you couldn t care less ). If so, you are rght t certanly s not a `small obect. However, because the floors move vertcally wthout rotaton, the rgd body equatons of moton smply reduce to F ma and M=0, where the moments are taen about the center of mass of the bloc. The floors behave as though they are partcles, even though they are very large.. Real earthquaes nvolve predomnantly horzontal, not vertcal moton of the ground. In adon, structural columns resst extensonal loadng much more strongly than transverse loadng. So we should really be analyzng horzontal moton of the buldng rather than vertcal

46 moton. However, the free body dagrams for horzontal moton are messy (see f you can draw them) and the equatons of moton for vertcal and horzontal moton turn out to be the same, so we consder vertcal moton to eep thngs smple. 3. Ths problem could be solved analytcally (e.g. usng the `dsolve feature of MAPLE) a numercal soluton s not necessary. Try ths for yourself. 1. Introduce varables to descrbe the moton: We wll use the heght of each floor ( y1, y ) as the varables.. Wrte down the poston vector n terms of these varables: We now have to worry about two masses, and must wrte down the poston vector of both r1 y1 r y Note that we must measure the poston of each mass from a fxed pont. 3. Dfferentate the poston vector wth respect to tme to fnd the acceleraton. dy1 dy 1 v1 1 a dy dy v a 4. Draw a free body dagram. We must draw a free body dagram for each mass. The resultant force actng on the bottom and top masses, respectvely, are F mg F F F mg F 1 S1 S S where FS1, F S are the forces n the two sprngs (note that we assume that all the sprngs are n tenson ths maes the calculaton easer). The sprng forces are equal to the stffness multpled by the ncrease n length of the sprngs F ( l L) F ( l L) S1 1 S F S F S1 mg F S mg F S1 We wll have to fnd the sprng lengths l 1, l n terms of our coordnates y1, y to solve the problem. Geometry shows that l y y 1 l 1 y 1 d 0 sn t So, fnally F mg ( y d sn t L) ( y y L) F mg ( y y1 L) 5. Wrte down Newton s laws of moton. F=ma for each mass gves d y mg ( l 1 1 L) ( l L) m d y mg ( l L) m Ths s two equatons of moton we can substtute for l 1, l and rearrange them as

47 dy1 g ( y1 d0 sn t L) ( y y1 L) m m dy g ( y y1 L) m 6. Elmnate reactons ths s not needed n ths example. 7. Identfy ntal conons. We now that, at tme t=0 dy1 dy y1 L mg / 0 y L 3 mg / 0 8. Solve the equatons of moton. We need to () reduce the equatons to the standard MATLAB form and () wrte a MATLAB scrpt to solve them. Convertng the equatons. We now need to do two thngs: (a) remove the second dervatves wth respect to tme, by ntroducng new varables; and (b) rearrange the equatons nto the form dy / f( t, y ). We dl remove the dervatves by ntroducng 1 dl v 1 v as adonal unnown varables, n the usual way. Our equatons of moton can then be expressed as dy1 v1 dy v dv1 ( g y1 d0 sn t L) ( y y1 L) m m dv g ( y y1 L) m We can now code MATLAB to solve these equatons drectly for dy/. A scrpt (whch plots the poston of each floor as a functon of tme) s shown below. functon buldng % =100; m=1; omega=9; d=0.1; L=10; tme=0; g = 9.81; w0 = [L-m*g/,*L-3*m*g/(*),0,0]; [t_values,w_values] = ode45(@eom,[0,tme],w0); plot(t_values,w_values(:,1:)); functon dw = eom(t,w) y1=w(1); y=w(); v1=w(3); v=w(4);

48 end end dw = [v1;v;... -**(y1-d*sn(omega*t)-l)/m+**(y-y1-l)/m;... -g-**(y-y1-l)/m] The fgures below plot the heght of each floor as a functon of tme, for varous earthquae frequences. For specal earthquae frequences (near the two resonant frequences of the structure) the buldng vbratons are very severe. As long as the structure s desgned so that ts resonant frequences are well away from the frequency of a typcal earthquae, t wll be safe. We wll dscuss vbratons n much more detal later n ths course. Ths problem llustrates a shortcomng of solvng problems on the computer wthout thnng too much about them. The way we set up the problem, t loos as though the soluton depends on g, and the unstretched sprng length L, but n fact ths s not the case. Not beng aware of ths maes the equatons of moton much more complcated than they really are, and maes t harder to nterpret the results. Of course we could learn by tral and error that the soluton s ndependent of L and g, but a better approach s to elmnate these varables from the equatons of moton altogether. There s a standard way to do ths nstead of solvng for the lengths of the sprngs y, we solve for the deflecton of the masses from ther statc equlbrum postons. We wll dscuss ths n more general terms when we dscuss vbraton problems later n the course. But we ll wor through the process here, because t s useful to use the same approach n the mass launcher desgn proect.