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 make 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 know the orentaton of the satellte, and you know 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 Newtonan physcs we have to measure poston and moton n a so-called Inertal Frame. Ths concept wll be x O y k r P z

2 dscussed n more detal n Secton 3.. For now, suppose that we can dentfy 1. Three, mutually perpendcular, fxed drectons n space: the three drectons are descrbed by unt,, k vectors { }. We choose a convenent statonary (or non-acceleratng) pont to use as orgn. 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 k In dynamcs problems, x,y,z can all be functons of tme, but {,, k} are fxed. 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+ δt) r() t v = lm δ t 0 δt Velocty s a vector, and can therefore be expressed n terms of ts Cartesan components v= vx+ vy+ vzk You can vsualze a velocty vector as follows The drecton of the vector s parallel to the drecton of moton The magntude of the vector example). v= v = vx + vy + vz s the speed of the partcle (n meters/sec, for When both poston and velocty vectors are expressed n terms Cartesan components, t s smple to,, k 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+ vzk = ( x() t + y() t + z() t k) = + + k 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+ δt) v() t a = lm δ t 0 δt The acceleraton s a vector, wth Cartesan representaton a= ax+ ay+ azk. Lke velocty, acceleraton has magntude and drecton. Sometmes t may be possble to vsualze an acceleraton vector for example, f you know 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 work through the math.

3 The relatons between Cartesan components of poston, velocty and acceleraton are dvx d x dvy d y dvz d z ax = = a y = = a z = = Revew of some deas from calculus To analyze moton, we often have to calculate acceleratons gven poston vectors as a functon of tme, or (more commonly) calculate velocty and poston by ntegratng acceleratons. You should have been beaten nto submsson wth ths sort of problem n calculus courses (and maybe some physcs courses as well) so we ll ust revew the most mportant procedures here as a remnder. We wll focus on the generc one-dmensonal problem: gven a, calculate v and x; or vce-versa. Calculatng postons, veloctes and acceleratons graphcally You wll remember that: Speed s the slope of the dstance-v-tme curve Dstance s the area under the speed-v-tme curve Or alternatvely Acceleraton s the slope of the speed-v-tme curve Speed s the area under the acceleraton-v-tme curve These deas are llustrated n the fgure below: f you can sketch a graph of acceleraton, velocty or poston, you can often use geometry to calculate all the other quanttes of nterest.

4 Rules for ntegratng acceleratons and speeds Acceleraton gven as a functon of tme v t dv a() t dv a() t = = v0 0 Example:

5 dv 3/ t v v0 t = = + 3 Acceleraton gven as a functon of speed v t dv dv = gt () f( v) gt () = f () v v0 0 Example: v t dv dv v = cv c log( ) ct = = v v v0 0 Acceleraton gven as a functon of dstance v x dv dv dx dv = f( x) = f( x) v= f( x) vdv f( x) dx dx dx = v0 0 Example v x dv = kx vdv kxdx v v0 kx v v0 kx = = = v0 0 Velocty gven as a functon of tme Example x t v() t dx v() t dx = = x0 0 3/ 4 5/ v= v0 + t x= x0 + vt 0 + t 3 15 Velocty gven as a separable functon of poston and tme v t dx dx = gt () f( x) gt () = f ( x) v0 0 Example x t dx dx = v0 kx = x v0 kx kx 1 kx sn sn 0 t k = v 0 v 0 v 0 1 kx x sn kt sn 0 = + k v 0 0 0

6 3.1.4 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 takeoff 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= ( v0 + at) a= a 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 taken hgh school physcs classes have used these formulas to solve so many problems that they automatcally apply them to everythng ths works for hgh school problems but not always n real lfe! x(t) 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. k,l 0 m For example, the poston vector of the mass mght be gven by r = xt ( ) = ( X0 + Xsn( πt / T) ) Here X 0 s the average length of the sprng, X0 + X 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).

7 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 ω = π /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 π X v= = cos( πt / T) T d xt () 4π X a= = sn( πt / T) T Note that: The velocty and acceleraton are also harmonc, and have the same perod and frequency as the dsplacement. If you know 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 π π π 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 make 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 θ ncreases at constant rate. We can wrte θ = ωt, where ω s the (constant) angular speed of the shaft, n radans/seconds. The speed of the partcle s related to ω by V = Rω. To see ths, notce that the crcumferental dstance traveled by the partcle s s V = ds / = Rdθ / = Rω. For ths example the poston vector s r = Rcosθ+ Rsnθ The velocty can be calculated by dfferentatng the poston vector. dr dθ dθ v= = R snθ+ R cos θ= Rω( snθ+ cos θ ) Here, we have used the chan rule of dfferentaton, and noted that dθ / = ω. R Rcosθ snθ t cosθ n θ=ωt Rsn θ = Rθ. Therefore,

8 The acceleraton vector follows as dv dθ dθ a= = Rω( cosθ sn θ) = Rω (cosθ+ sn θ ) Note that () The magntude of the velocty s V = Rω, and ts drecton s (obvously!) tangent to the path (to see ths, vsualze (usng trg) the drecton of the unt vector t = ( snθ+ cos θ ) () The magntude of the acceleraton s Rω and ts drecton s towards the center of the crcle. To see ths, vsualze (usng trg) the drecton of the unt vector n= (cosθ+ sn θ ) We can wrte these mathematcally as dr dv V v= = Rωt = Vt a= = Rω n= n R Example 4: More general moton around a crcular path We next look at more general crcular moton, where the partcle stll moves around a crcular path, but does not move at constant speed. The angle θ s now a general functon of tme. We can wrte down some useful scalar relatons: dθ Angular rate: ω = dω d θ Angular acceleraton α = = dθ Speed V = R = Rω dv d θ dω Rate of change of speed = R = R = Rα snθ t cosθ R n θ Rsn θ Rcosθ We can now calculate vector veloctes and acceleratons r = Rcosθ+ Rsnθ The velocty can be calculated by dfferentatng the poston vector. dr dθ dθ v= = R snθ+ R cos θ= Rω( snθ+ cos θ ) The acceleraton vector follows as dv dω dθ dθ α= = R ( snθ+ cos θ) + Rω( cosθ sn θ) = Rα( snθ+ cos θ) Rω (cosθ+ sn θ) 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 = ( snθ+ cos θ ), n= (cosθ+ sn θ ). Then dv V v= Rωt = Vt α= Rαt + Rω n= t + n R These are the famous crcular moton formulas that you mght have seen n physcs class.

9 Usng MATLAB lve scrpts If you fnd that your calculus s a bt rusty you can use MATLAB to do the tedous work for you. For example, to dfferentate the vector r = x() t + y() t + z() t k you would type syms x(t) y(t) z(t) r(t) v(t) a(t) r(t) = [x(t),y(t),z(t)] v(t) = dff(r(t),t) a(t) = dff(v(t),t) It s essental to type n the (t) after x,y,and z f you don t do ths, MAPLE assumes that these varables are constants, and takes 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 know that the angle θ s some arbtrary functon of tme, but don t know what the functon s syms R theta(t) r(t) v(t) a(t) r(t) = [R*cos(theta(t)),R*sn(theta(t))] v(t) = smplfy(dff(r(t),t)) a(t) = smplfy(dff(v(t),t)) MATLAB can make 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 know what the program s dong so keep takng those math classes

10 3.1.5 Velocty and acceleraton n normal-tangental coordnates. In some cases t s helpful to use specal bass vectors to wrte down velocty and acceleraton vectors, nstead of a fxed {,,k} bass. If you see that ths approach can be used to quckly solve a problem go ahead and use t. If not, ust use Cartesan coordnates ths wll always work, 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, the formulas for velocty and acceleraton look very complcated n (,) coordnates but become much smpler f they are expressed as components usng a bass orented parallel and perpendcular to the path. These bass vectors are called normaltangental coordnates. For example, normal-tangental coordnates are nearly always used n vehcle dynamcs problems, because the { tnk,, } drectons pont forwards sdeways and vertcally, so t s easy to understand the sgnfcance of acceleratons along { tnk,, }, whle {,,k} components are generally very dffcult to nterpret. r(s) s t R n To use normal-tangental coordnates we Specfy the path by wrtng down the poston vector of a pont on the path n terms of the dstance s travelled along the path. For a D curve: r() s = xs () + ys () 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 (ths sounds a bt scary, but n s ust perpendcular to t, and f the path curves to the rght n ponts to the rght; f t curves to the left, t ponts to the left). Introduce the radus of curvature of the path R (n most problems we solve R s gven, but we ll gve some formulas later). Denote the speed of the partcle by V. The speed can vary wth tme. We then use the followng formulas to calculate speed, velocty and acceleraton v= Vt dv V a= t+ n R We can also use the formula V = ds / to wrte velocty and acceleraton n terms of dstance traveled along the path and ts tme dervatves ds v= t d s 1 ds a= t+ n R In words, these equatons tell us: (1) 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.

11 (3) The speed s the dervatve of dstance traveled wth respect to tme (4) The acceleraton vector can be constructed by addng two components: the component of acceleraton tangent to the partcle s path s equal to dv /. We call ths the tangental component of acceleraton. The component of acceleraton perpendcular to the path (towards the center of curvature) s equal to V / R. We call ths the normal component of acceleraton. Dervng the normal-tangental formulas In Newtonan physcs, we have to start by defnng an nertal frame, whch (mathematcally) s always a non-acceleratng and non-rotatng Cartesan {,,k} bass. The x,y,z coordnates of a partcle and ther dervatves tell us the Newtonan defntons of poston, velocty and acceleraton. r(s) As we wll see wth specfc examples below, some problems can be smplfed by replacng the x,y,z coordnates wth some smpler set of coordnates (these mght be angles, or dstances, or some combnaton of both that descrbe our system). Whenever we use new coordnated, we proceed by wrtng down the x,y,z coordnates n terms of our new ones. We usually also replace the {,,k} bass wth a new set of drectons that are defned by our new coordnate system. As we make small changes to our new coordnates, we wll move around n space the drectons we move usually have some specal sgnfcance, so t s helpful to wrte down all vector quanttes of nterest as components n the bass defned by these new drectons. Ths sounds very abstract, so let s see how t works for normal-tangental coordnates. For problems where a partcle moves along a known path, we can always wrte down the poston of a pont on the path n terms of the dstance travelled along the path. For a D curve: r() s = xs () + ys () Here s s the arc-length traveled along the path. Snce the path s known (x,y are gven functons) we only need one coordnate (s) to specfy where we are. We now generate some bass vectors by workng out how poston vector changes f we make small changes to our new coordnate s. It s convenent to defne dr 1 1 t = n= R R= = ds ds d x d y ds ds + ds ds The vectors { tnk,, } have unt length, are mutually perpendcular, and therefore defne a new Cartesan bass. We can defne any vector n ths bass n the usual way. Wth these defntons we are now ready to derve our formulas for velocty and acceleraton: s t R n

12 dr() s dr() s ds v= = = Vt ds dv dv dv ds dv n dv V a= = t+ V = t+ V = t+ V V = t+ n ds R R These are all ust repeated applcatons of the chan rule. Examples usng normal-tangental coordnates We ll work through a few examples below that show how to work wth normal-tangental coordnates. These are all qute hard: you need to be able to use lots of basc deas from vectors and calculus to be able to answer them. Example: Crcular moton n normal-tangental coordnates We already analyzed crcular moton as a specal case. We ll revst that example as an example of moton along a general curved path. The dstance traveled around the crcle s the arc length s. R θ t n s The arc-length formula gves θ = s/ R s s Therefore r = Rcos + Rsn (ust use trg) R R dr s s The tangent vector s t = = sn + cos (ths agrees wth our earler formula) ds R R The normal vector s d t n= R = cos s sn s (also the same as our earler formula) ds R R ds The formula says velocty vector s v= t (the same as our earler formula) d s 1 ds The formula for acceleraton vector s a= t+ n (also agrees wth the earler result) R Example (a smple exam problem): A vehcle starts at rest at A and travels wth constant tangental acceleraton a t around a crcular path. To avod skddng, the magntude of the acceleraton must not exceed µ g, where µ s the frcton coeffcent between tres and road, and g s the gravtatonal acceleraton. Fnd a formula for the shortest possble tme to reach B. B R We can use the formula: n normal-tangental coordnates n t A

13 dv V V a= t+ n= att+ n R R Snce we are told a t s constant and the car s at rest at t=0 we can use the constant acceleraton formula to fnd V dv = a t V = at t Furthermore ds 1 at t s at t = = The arc-length from A to B s π R so the tme to travel from A to B follows as We can fnd a t from the conon that T = a µ g π R/ at ( at) 4 4 V V t a = att+ n = at + = a t + µ g R R R (notce that we ust used the usual Cartesan formula to fnd the magntude of the vector). The maxmum acceleraton occurs at B, so we can substtute for t and solve for a t Fnally remember T = π R/ at so t 4 t a = a + π a µ g a µ g / 1+ 4π t π R T + µ g ( 1 4π ) 1/4 Example: Desgn speed lmt for a curvy road: As a consultng frm specalzng n hghway desgn, we have been asked 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( π 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 maxmum allowable 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. Α V

14 Calculaton: We can solve ths problem quckly usng normal-tangental coordnates. Snce the speed s constant, the acceleraton vector s V a= n R Our only problem s that we don t know R but we can use what we know about vectors and normaltangental coordnates to fgure t out. The poston vector s r = x+ Asn( π x/ L), where x s some unknown functon of dstance s traveled along the path. We can calculate the tangent from the formula for the tangent (and the chan rule) d r d r dx π t = = = + Acos( π x/ L) dx ds dx ds L ds We know that t must be a unt vector, therefore 4π ( ) dx t t = 1 1+ A cos π x/ L L ds 1/ dx 4π = 1+ A cos ( π x/ L) ds L We could separate varables and ntegrate ths to calculate x(s) f we need t but n practce we don t need to bother. So now we know that π + Acos( π x/ L) L t = 4π 1+ A cos ( π x/ L) L The normal vector n must be perpendcular to both t and k, so we can create t usng π π + Acos( πx/ L) Acos( πx/ L) + L L n=± k t =± k =± 4π 4π 1+ A cos ( πx/ L) 1+ A cos ( πx/ L) L L (the ± s because there are two vectors normal to both t and k). Fnally we know that 4π Asn ( π x/ L) L dx π d 1 n= R = + + Acos( π x/ L) ds 4π ds L ds 4π 1+ A cos ( πx/ L) 1+ A cos ( πx/ L) L L

15 Snce n s a unt vector, we can take the dot product of both sdes of ths expresson wth n (from above) 4π Asn ( π x/ L) L dx n n = 1 =± R 4π ds 1+ A cos ( π x/ L) L Therefore (usng our earler expresson for dx / ds, and notng that R s postve by defnton) 4π Asn ( π x/ L) 1 L = R 3/ 4π 1+ A cos ( π x/ L) L So now we know the acceleraton vector s a A πv L πx L ( / ) sn( / ) = 3/ 4π 1+ A cos ( π x/ L) L We are nterested n the magntude of the acceleraton A( πv / L) sn( πx/ L) a = 3/ 4π 1+ A cos ( π x/ L) L We see from ths that the car has the bggest acceleraton when x = L/. The maxmum acceleraton follows as amax = A( πv / L) The formula for the speed lmt s therefore V < ( L/ π ) 0. g / A Now we send n a bll for a bg consultng fee n Poston, Velocty and Acceleraton n cylndrcal-polar coordnates. 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 θ e r Polar coordnate formulas Polar coordnates are related to x,y coordnates through 1 r = x + y θ = tan ( y/ x) x= rcosθ y = rsnθ We can also specfy heght out of the plane of the pcture usng the usual z coordnate. r θ

16 Suppose that the poston of a partcle s specfed by ts polar coordnates (, r θ ) 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 θ be a unt vector pontng n the tangental drecton,.e er = cosθ+ snθ eθ = snθ+ cosθ The poston, velocty and acceleraton of the partcle can then be expressed as r = rer + zk dr dθ dz v= er + r eθ + k d r dθ d θ dr dθ d z a= r er + r + e θ + k In most problems we solve, we ust substtute known nformaton about (, r θ ) nto these formulas. Dervng the polar coordnate formulas The formulas for polar coordnate can be derved usng the same deas we used to set up normaltangental coordnates. The general procedure to set up any new coordnate system n Newtonan physcs s: (1) Choose an nertal frame ths defnes a statonary Cartesan {, k, } bass () Choose some new coordnates here we use (, r θ,) z and wrte down poston vector n the nertal frame n terms of these new coordnates. For the cylndrcal-polar system r = rcosθ+ rsnθ+ zk (3) We now thnk about makng very small changes to each of r, θ, z. As we do so, r wll change. We defne unt vectors that pont n each of the drecton assocated wth makng changes to r, θ, z: mathematcally ths operaton s 1 dr er = = cosθ+ snθ dr / dr dr 1 dr rsnθ+ rcosθ eθ = = = snθ+ cosθ dr / dθ dθ r 1 dr ez = = k dr / dz dz For the polar coordnate system t turns out that { er, eθ, k } are mutually perpendcular and so are a Cartesan bass. But note that the drectons { er, eθ, k} are functons of r, θ, z. Regardless, we can express any vectors as components n our new bass usng the usual deas the vector wll be made up of contrbutons parallel to each of the new bass vectors. The polar coordnate formulas now follow by smple calculus. Snce we have chosen to work wth r, θ, z nstead of x,y,z, and { er, eθ, k } nstead of {, k, }, we would lke to wrte down poston, velocty and acceleraton n the { er, eθ, k} bass, terms of tme dervatves of r, θ, z. Before we work through the detals we have to do some busy-work. Snce { er, eθ, k} are functons of r, θ, z we wll need to know how to dfferentate them wth respect to r, θ, z. To do ths we go back to ther orgnal defntons:

17 de cos sn r der de e r r = θ+ θ = = 0 = snθ+ cosθ= eθ dr dz dθ de d d sn cos θ eθ e e θ θ = θ+ θ = = 0 = cosθ snθ= er dr dz dθ dez dez de e z z = k = = = 0 dr dθ dz The rest s ust a tedous exercse n usng the chan rule. The poston vector can be wrtten down by nspecton as r = rer + zk. Then r = rer + zk dr dr der dθ dz dr dθ dz v= = er + r + k = er + r eθ + k dθ dv d r dr der dθ dr dθ d θ dr dθ de d d z r r θ θ a= = e + + eθ + e d θ + + k d θ θ d r dθ d θ dr dθ d z = r er + r + e θ + k Examples usng polar coordnates Example The robotc manpulator shown n the fgure rotates wth constant angular speed ω L about the k axs. Fnd a formula for the k maxmum allowable (constant) rate of extenson O dl / f the acceleraton of the grpper may not exceed g. θ We can smply wrte down the acceleraton O vector, usng polar coordnates. We dentfy ω = dθ / and r=l, so that a dl 4 dl dl 1 = ( Lω ) e r + ω θ = L ω + 4 ω < g < g / ω L ω e a 4 e θ L θ e r Example: The poston of a partcle n polar coordnates s gven by θ = t r = t / π (meters). At the nstant when θ = π, calculate the followng quanttes: The poston vector n, components and n e, e components r θ When θ = π, r = 1 so e r e

18 r = ( meters) r = er ( meters) The velocty vector n e, e and, r θ dr dθ 1 1 The polar coordnate formula s v= er + r eθ = er + teθ = er + π e θ m/s π π 1 By nspecton we see that = er, = e θ at the nstant of nterest, so v= π π The acceleraton vector n e, e components r θ d r dθ d θ dr dθ a= r er + r + e θ 1 = 4 t er + ( + 4 t) eθ = 4π er + 6eθ π Unt vectors tn, tangent and normal to the path, n er, e θ. (Choose n to pont towards the center of curvature) m/s We know t s parallel to v so v t = = er + πeθ = er + πe v 4π + 1/ π π 4π + 1 ( ) We can fnd n n three ways: frst, we know n must be perpendcular to t and must le n the, plane (and hence s perpendcular to k). Remember that you can create a vector perpendcular to two others usng a cross product, so n=± k t. The postve choce ponts towards the center of curvature (by nspecton), and note k er = eθ k eθ = e r 1 so n= ( π er + e θ ) 4π + 1 You can also use the conon t n = 0 - f we assume n= nrer + n θ e θ then 1 t n= 0 ( er + π eθ ) ( nrer + nθeθ ) = 0 4π + 1 nr + π nθ = 0 Any nr, n θ that satsfes ths (eg nr = π, n θ = 1 ) s perpendcular to t. But n must be a unt vector, and we know we want the vector to pont towards the orgn (because the center of curvature of the path s nsde the turn). So we have to choose 1 n= ( π er + e θ ) 4π + 1 θ

19 The last (cumbersome, but general) way to do the calculaton s to note that a ( a tt ) must be parallel to n πer + 6eθ ( 4πer + 6eθ ) ( er + πeθ ) ( er + πeθ ) 4π + 1 4π + 1 8π 4 π(3 + 4 π ) (3 + 4 π ) = 4πer + 6eθ ( e ) r + πeθ = e r + e θ 4π + 1 4π + 1 4π + 1 Dvdng by the magntude of ths vector (to create a unt vector) gves the same answer as before. Tangental and normal components of acceleraton at, a n : we know (from ENGN30) that we can fnd the component of a vector n a bass by dottng t wth the bass vectors, so 8π 8π + 6 at = a t = an = a n= m/s 1+ 4π 1+ 4π You can also do ths problem usng the formula dv V dv a= t+ n. Ths shows at =, whch s not too hard to calculate (but s a pan so I R can t be bothered). To fnd an = V / R we would ether have to use our MA000 super-powers dv V to fnd the radus of curvature of the path, or else use a t an =+ n R = n and then take the magntude of the vector on the left to get a n 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 known 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 knd 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 known 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 trackng and a reference pont. Can only measure dstances of mm or less, but precson can be down to mcron accuracy.

20 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 trackng 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 or where a backup s needed for GPS. They are often combned wth GPS recevers as well: accelerometers are very good for measurng changes n velocty and poston over a short tme nterval, and GPS s very good over long tme ntervals, so you can use sensor fuson to make a sensor that uses both sgnals to get the best possble measurement. 3. Calculatng forces requred to cause prescrbed moton of a partcle 3..1 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: (1) We can dentfy some pont n the unverse that s not acceleratng () We can dentfy three mutually perpendcular drectons that are fxed n the sense that they do not rotate. Together, these defne a so-called nertal frame a Cartesan coordnate system n whch moton obeys Newton s laws. For engneerng calculatons, dentfyng a sutable orgn and fxed drectons usually poses no dffculty. If we are solvng problems nvolvng terrestral moton over short dstances compared wth the earth s radus, we smply take a pont on the earth s surface as fxed, and take 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 take 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. Experments show that Newton s laws predct moton suffcently accurately for our needs. But there wll always be some very small error. In 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 gves us a framework that avods havng to choose an nertal frame. More elaborate calculatons show that Newton s laws are rgorous approxmatons to the general equatons (n the sense of a Taylor expanson of the more general equatons for low partcle speeds compared wth the speed of lght), and would also (n prncple) tell us the best choce of drectons to set up a Newtonan frame at any pont n space.

21 3.. Newton s laws of moton for a partcle If we gnore these conceptual dffcultes, 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, n the nertal frame) 3. a denote the acceleraton of the partcle (agan, as a vector n the nertal frame). Then F= ma Occasonally, we use a partcle dealzaton to model systems whch, strctly speakng, 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 unknown 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 We wll see where ths equaton comes from when we analyze rgd body dynamcs, and we ll also understand when t s no longer correct. It s very mportant to take 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). Newton s laws of moton can be used to calculate the forces requred to make a partcle move n a partcular way. We use the followng general procedure to solve problems lke 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 knematcs 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 unknown components of force or acceleraton (ths s ust lke a statcs problem, except the rght hand sde s not zero).

22 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 take off from the deck of an arcraft carrer (the fgure s from We wll estmate the acceleraton requred to reach takeoff speed, assumng the arcraft accelerates from zero speed to takeoff speed along the deck of the carrer, and then use Newton s laws to deduce the force. Data/ Assumptons: 1. The flght deck of a Nmtz class arcraft carrer s about 300m long ( but only a fracton of ths s used for takeoff (the angled runway s used for landng). We wll take the length of the runway to be d=00m. We wll assume that the acceleraton durng takeoff 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 takeoff. We neglect moton to make the calculaton smpler) 4. The FA18 Super Hornet s a typcal arcraft used on a carrer t has max catapult weght of m=15000kg 5. The manufacturers are somewhat retcent about performance specfcatons for the Hornet but v = 150 knots (77 m/s) s a reasonable guess for a mnmum controllable arspeed for ths t arcraft. Calculatons: 1. Idealzaton: We wll dealze the arcraft as a partcle. We can do ths because the arcraft s not rotatng durng takeoff.. FBD: The fgure shows a free body dagram. F T represents the (unknown) force exerted on the arcraft due to ts engnes. F T NA mg N B 3. Knematcs: We must calculate the acceleraton requred to reach takeoff speed. We are gven () the dstance to takeoff d, () the takeoff speed v t and () the arcraft s at rest at the start of the takeoff roll. We can therefore wrte down the poston vector r and velocty v of the arcraft at takeoff, and use the straght lne moton formulas for r and v to calculate the tme t to reach takeoff speed and the acceleraton a. Takng the orgn at the ntal poston of the arcraft, we have that, at the nstant of takeoff 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

23 v F t T = ma= m d 5. Soluton: The component of the equaton of moton gves an equaton for the unknown force n terms of known quanttes v F t T = m d Substtutng numbers gves the magntude of the force as F= kn. Ths s very close, but slghtly greater than, the 00kN (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. Example : Mechancs of Magc! You have no doubt seen the smple `tablecloth trck 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 make the glass move wth the tablecloth (.e. to prevent slp), and see f these forces are bg enough to cause slp. 1. Idealzaton: We wll assume that the glass behaves lke 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 unknown. 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 mg T N

24 T < µ N where µ s the frcton coeffcent. Substtutng for T and N from (5) shows that for no slp a < µ g To do the trck, therefore, the acceleraton must exceed µ g. For a frcton coeffcent of order 0.1, ths gves an acceleraton of order 1 m/ s. There s a specal trck to pullng the tablecloth wth a large acceleraton but that s a secret. Example 3: Bcycle Safety. If a bke rder brakes too hard on the front wheel, hs or her bke wll tp over (the fgure s from In ths example we nvestgate the conons that wll lead the bke to capsze, and dentfy desgn varables that can nfluence these conons. If the bke 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 bke 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 bke racers but that s how we thnk of you ). The partcle s located at the center of mass of the rder. The fgure shows the most mportant desgn h parameters- these are the heght of the rder s COM, the wheelbase L and the dstance of the COM from the rear wheel. b. We assume that the bke s a massless A frame. The wheels are also assumed to d B have no mass. Ths means that the forces L 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 thnk about statcs of wheels, you should reread the notes on ths topc n partcular, make 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 brakes so hard that the front wheel s prevented from rotatng. It must therefore skd over the ground. Frcton wll resst ths sldng. We denote the frcton coeffcent at the contact pont B by µ. 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 bke 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 N A W T B N B

25 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 bke; b. Only a normal force acts at the contact pont on the rear wheel because t s freely rotatng, and behaves lke a -force member. 3. Knematcs The bke s movng n the drecton. As a vector, ts acceleraton s therefore a= a, where a s unknown. 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 take moments about the partcle (.e. the rder s COM). F= ma gves T + ( N + N W ) = ma C = B A B M 0gves N ( L d ) k N dk T hk = 0 B A B The two nonzero components of F= ma and the one nonzero component of M = C 0 gve us three scalar equatons T = ma B ( N + N W) = 0 A B N ( L d) Nd Th= 0 B A B We have four unknowns the reacton components NA, NB, T B and the acceleraton a so we need another equaton. The mssng equaton s the frcton law T = µ N 5. Soluton: (tedous algebra you could avod ths by usng Matlab) The thrd equaton and the frcton law show that N( L d µ h) Nd= 0 B Multply ( N + N W) = 0by ( L d µ h) and subtract t from ths equaton: A B B Nd ( L d µ h)( N W) = 0 A = B L d µ h NA W L µ h We are nterested n fndng what makes the reacton force at A go to zero (that s when the bke s about to tp). So L d µ h NA = W 0 µ ( L d)/ h L µ h A A

26 Ths tells us that the bke wll tp f the frcton coeffcent exceeds a crtcal magntude, whch depends on the geometry of the bke. The smplest way to desgn a tp-resstant bke s to make 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 bke 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. 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 know that both masses must move n the drecton. We also know 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 aa= ab 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 unknowns (the unknowns are aa, a B and T). 5. Soluton: More algebra. We can elmnate a B so that the last two equatons are:

27 ( T mag) = maaa ( T mbg) = mbaa Now we can multply the frst equaton by mb and the second by ( TmA + TmB mambg) = 0 So m A m T = B g ma + mb We pass! ma and add them Example 5: Another stupd FE exam problem: The fgure shows a small block on a rotatng bar. The contact between the block and the bar has frcton coeffcent µ. The bar rotates at constant angular speed ω. Fnd the crtcal angular velocty that wll ust make the block start to slp when θ = 0. Whch way does the block slde? The general approach to ths problem s the same as for the Magc trck example we wll calculate the reacton force exerted by the bar on the block, 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. 1. Idealzaton We wll dealze the block as a partcle. Ths s dangerous, because the block s clearly rotatng. We hope that because t rotates at constant rate, the rotaton wll not have a sgnfcant effect but we can only check ths once we know how to deal wth rotatonal moton.. FBD: The fgure shows a free body dagram for the block. The block 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 θ N slppng, we can choose the drecton of the tangental component of the mg reacton force arbtrarly. The resultant force on the block s F= T cosθ N sn θ + ( N cosθ + T sn θ mg) ( ) 3. Knematcs We can use the crcular moton formula to wrte down the acceleraton of tbe block (see secton 3.1.3) a= rω (cos θ+ sn θ ) 4. EOM: The equaton of moton s T cosθ N sn θ + ( N cos θ mg) = mrω (cosθ+ sn θ ) ( ) 5. Soluton: The and components of the equaton of moton can be solved for N and T. Dong ths by hand s a pan, but Matlab makes t panless syms T N theta m r omega g real eq1 = T*cos(theta)-N*sn(theta) == -m*r*omega^*cos(theta); eq = N*cos(theta)+T*sn(theta)-m*g == -m*r*omega^*sn(theta); [N,T] = (solve([eq1,eq],[n,t])); N = smplfy(n) T = smplfy(t) ω r θ

28 To fnd the pont where the block ust starts to slp, we use the frcton law. Recall that, at the pont of slp T = µ N For the block to slp wth θ = 0 rω = µ g so the crtcal angular velocty s ω = µg / r. Snce the tangental tracton T s negatve, and the frcton force must oppose sldng, the block must slde outwards,.e. r s ncreasng durng slp. 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 block does not slp, t moves wth speed V = ωraround a crcular arc wth radus r. Its acceleraton vector has magntude V / 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 cosθ) ( mg snθ T ) F= t+ n ( cosθ) ( snθ ) ω 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 cosθ T = mg snθ mω r Ths s the same soluton as before. Normal-tangental coordnates makes the equatons and algebra much smpler, however. ω r n θ t n r and T θ N mg

29 Example 6: Wndow blnds. Have you ever wondered how wndow shades work? You gve the shade a lttle downward erk, 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 remarkable 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 bracket 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 counterclockwse, 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 counterclockwse. Statonary Hub Rotatng Drum Statonary Hub Wndow shade To rase the shade, you need to gve the end of the shade a erk downwards, and then release t. When the drum rotates suffcently quckly (we wll calculate how quckly shortly) the 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 lockng 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 ω. 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.. 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 lnk, and () gravty. The resultant force s F= T cos( φ θ) N cos θ + ( N snθ + T sn( φ θ) mg) ( ) 3. Knematcs We can use the crcular moton formula to wrte down the R T θ φ mg θ N

30 acceleraton of the partcle(see secton 3.1.3) a= Rω (cosθ+ sn θ ) 4. EOM: The equaton of moton s T cos( φ θ) N cos θ + ( N snθ + T sn( φ θ) mg) = a= Rω (cosθ+ sn θ ) ( ) 5. Soluton: The and components of the equaton of moton can be solved for N and T MAPLE makes ths panless syms T N theta m R omega g ph real eq1 = -T*cos(ph-theta) - N*cos(theta) == -R*omega^*cos(theta); eq = -N*sn(theta)+T*sn(ph-theta)-m*g == -R*omega^*sn(theta); [T,N] = solve([eq1,eq],[t,n]); T = smplfy(t) N = smplfy(n) The normal reacton force s therefore N = mg cos( θ φ) / snφ + mrω We are lookng for the pont where ths can frst become zero or negatve. Note that max{cos( θ φ)} = 1 at the pont where θ φ =0. The smallest value of N therefore occurs at ths pont, and has magntude Nmn = mg / snφ + mrω The crtcal speed where N=0 follows as ω = g / ( Rsn φ) Changng the angle φ 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 work 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 lnk, and () gravty. The resultant force s F = ( N + T cosφ + mg sn θ) er + ( T snφ mg cos θ) e θ θ T e θ φ mg N e r

31 . Knematcs The acceleraton vector s now a= Rω e r 3. EOM: The equaton of moton s ( N + T cosφ + mg sn θ) e + ( T snφ mg cos θ) e = Rω e r 4. Soluton: The er, eθ components of the equaton of moton can be solved for N and T. If we use polar coordnates we can do ths by hand the e θ component shows that T = mg cos θ / snθ We can substtute ths back nto the e r component to get N = mg cos( θ φ) / snφ + mrω We are lookng for the pont where ths can frst become zero or negatve. Note that max{cos( θ φ)} = 1 at the pont where θ φ =0. The smallest value of N therefore occurs at ths pont, and has magntude Nmn = mg / snφ + mrω The crtcal speed where N=0 follows as ω = g / ( Rsn φ) Changng the angle φ and the radus R gves a convenent way to control the crtcal speed n desgnng an nertal latch. θ r Example 7: Arcraft Dynamcs Arcraft performng certan nstrument approach procedures (such as holdng patterns or procedure turns) are requred to make all turns at a standard rate, so that a complete 360 degree turn takes mnutes. All turns must be made at constant alttude and constant speed, V. People who desgn nstrument approach procedures need to know the radus of the resultng turn, to make 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 bank 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 makes 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 banks 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. k θ R α

32 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. The resultant force s [( FT FD)cosθ FLsnαsn θ] + [( FD FT)snθ FLsnαcosθ] + ( FLcosα mg ) k (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 k drecton as FL = FLsnαRO + FLcosαk. Then note that RO = snθ cosθ.) 3. Knematcs a. The arcraft moves at constant speed around a crcle, so the angle θ = ωt, where ω s the (constant) angular speed of the lne OP. Snce the arcraft completes a turn n two mnutes, we know that ω = π / ( 60) = π / 60 rad/sec b. The poston vector of the plane s r = Rsnωt+ Rcosωt We can dfferentate ths expresson wth respect to tme to fnd the velocty v= Rω(cosωt sn ωt) c. The magntude of the velocty s V = Rω, so f the arcraft fles at speed V, the radus of the turn must be R= V / ω d. Dfferentatng the velocty gves the acceleraton a= Rω (sn ωt+ cos ωt ) 4. EOM: The equaton of moton s [( FT FD)cosθ FLsnαsn θ] + [( FD FT)snθ FLsnαcosθ] + ( FLcosα mg ) k = mrω (snθ+ cos θ) = mvω(snθ+ cos θ) 5. Soluton: The and k components of the equaton of moton gve three equatons that can be solved for F T, F L and α. We assume that the drag force s known, snce ths s a functon of the arcraft s speed. syms FT FD FL alpha theta R m omega g V real eq1 = (FT-FD)*cos(theta)-FL*sn(alpha)*sn(theta)==-m*V*omega*sn(theta); eq = (FT-FD)*cos(theta)-FL*sn(alpha)*cos(theta)==-m*V*omega*cos(theta); eq3 = FL*cos(alpha)-m*g ==0; [FL,alpha,FT] = solve([eq1,eq,eq3],[fl,alpha,ft]); FL = smplfy(fl) alpha = smplfy(alpha) FT = smplfy(ft) k O θ R F T F L α mg F D

33 (The two solutons are a bt werd, but to a mathematcan havng the arplane fly upsde down and generate negatve lft s a perfectly acceptable soluton) 1 = tan (( g V + g ) / V ) FL = mg 1 + V / g FT = FD a ω ω ω We can calculate values of α, R= V / ω and the load factor FL / mg for a few arcraft 0 a. Cessna 150 V=70knots (36 m/s) : α = 11 R=690m, FL / mg = b. Boeng 747: V=00 knots (10 m/s) α = 8 R=1950m, FL / mg = c. F111 V=300 knots (154 m/s) α = 39 R=950m, FL / mg = 1.3 Alternatve soluton usng normal-tangental coordnates Ths problem can also be solved rather more quckly 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= Vωn R where n s a unt vector pontng towards the center of the crcle. k θ R n α k t () Force vector. The force vector can be wrtten n terms of the unt vectors n,t,k as F= ( FT FD) t+ FLsn αn+ ( FLcos α mg) k () Newton s law F= ( FT FD) t+ FLsn αn+ ( FLcos α mg) k = mvωn The n, t and k components of ths equaton gve three equatons that can be solved for F T, We can easly do ths by hand 1 = tan ( V / g) FL = mg 1 + V / g FT = FD a ω ω F L and α.

34 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. 6. If you wsh, you can elmnate any unknown 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 unknown varables, you should look for any constrants that restrct the moton of the partcles. The constrants must be expressed n terms of the unknown acceleratons. 8. Identfy the ntal conons for the varables defned n (1). These are usually the values of the unknown varables, ther tme dervatves, at tme t=0. If you happen to know the values of the varables at some other nstant n tme, you can use that too. If you don t know ther values at all, you should ust ntroduce new (unknown) 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 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.

35 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+ Zk wth ntal velocty vector V0 = Vx+ Vy+ Vzk. Calculate ts traectory as a functon of tme. k X0 V 0 1. Introduce varables to descrbe the moton: We can smply use the Cartesan coordnates of the partcle ( xt ( ), yt ( ), zt ( )). Wrte down the poston vector n terms of these varables: r = x() t + y() t + z() t k 3. Dfferentate the poston vector wth respect to tme to fnd the acceleraton. For ths example, ths s trval dx dy dz d x d y d z v = + + k a= + + k 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 F= mgk. 5. Wrte down Newton s laws of moton. Ths s easy d x d y d z F= ma mgk = m + + k The vector equaton actually represents three separate dfferental equatons of moton d x d y d z = 0 = 0 = g 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= X0 = 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 k mg

36 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 y x t t z t x y z X0 0 Y0 0 Z0 0 ( ) dx = V dy = V dz = V gt 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 k v= V + V + V gt k x y z ( ) 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 work 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 Vx = v0cosθ Vy = 0 Vz = v0snθ H. The components of the poston of the partcle at tme t=0 are X = 0, Y = 0, Z = H k 3. The traectory of the partcle follows as 1 r = ( v0cosθt) + H + v0snθt gt k D 4. When the partcle hts the ground, ts poston vector s r = D. Ths must be on the traectory, so v 0 θ

37 1 v0 ti + H + v0 ti gti k = D ( cosθ ) snθ where t I s the tme of mpact. 5. The two components of ths vector equaton gves us two equatons for the two unknowns { ti, D }, whch can be solved clear all syms v0 theta ti g D H real eq1 = v0*cos(theta)*ti==d; eq = H+v0*sn(theta)*tI - g*ti^/; [D,tI] = solve([eq1,eq],[d,ti]); D = smplfy(d) 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 know - you can probably do better, especally f you are on the baseball/softball teams). 3. Throwng angle 45 degrees. Substtutng numbers gves 36m (118ft). If you want to go wld, you can maxmze D wth respect to θ, 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 kt / m, where m s ther mass, T s the temperature and k= 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 = kt / m 3. For tme t>0 the poston vector of the partcle follows as ( kt / m ) t 1 r = gt k Its velocty s ( kt / m gt ) v= 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 k

38 r = yk s a maxmum, then dr/ = v= ( dy / ) k = 0). Therefore, at the nstant of maxmum heght v= ( kt / m gtmax ) k = 0 5. Ths shows that the nstant of max heght occurs at tme = ( ) tmax kt / m / g 6. Substtutng ths tme back nto the poston vector shows that the poston vector at max heght s kt 1kT kt r = k = k mg mg mg 7. S has a densty of about 330 kg/m^3. At room temperature (93K) we fnd that the dstance s surprsngly large: 10mm or so. Gravty s a very weak 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 k 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 k,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 take the orgn at O as shown n the fgure. The poston vector of the center of mass of the block 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

39 5. Wrte down Newton s laws of moton. Ths s easy d x F= ma ( RAx + RBx ) ( mg + k( x L0 )) = m The and components gve two scalar equatons of moton ( R + R ) = 0 Ax Bx d x k = 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 know 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 Matlab. The equaton of moton s d x k = g ( x L + 0) m We can re-wrte ths n terms of dx vx = Ths gves dvx dvx dx dvx k = = vx = g + ( x L0 ) dx dx m We can separate varables and ntegrate vx x k vxdvx = g + ( x L0 ) dx m 0 L 1 k k vx = g( x L) ( x L ) + L0 ( x L) m m k mg mg vx = L L0 + x L0 + m k k Don t worry f the last lne looks mysterous wrtng the soluton n ths form ust makes the algebra a bt smpler. We can now ntegrate the velocty to fnd x dx k mg mg vx = = L L0 + x L0 + m k k x t dx = L k mg mg 0 L L0 + x L0 + m k k The ntegral on the left can be evaluated usng the substtuton

40 so that ( 0 + / ) ( / ) ( x L0 + mg / k ) ( L L + mg / k ) 0 = cosθ θ0 t snθdθ 1 + = θ0 = cos k cos θ m θ0 = t k m x L mg k k = cos t L L0 + mg k m ( x L0 mg / k ) ( L L + mg / k ) k x = ( L L0+ mg / k ) cos t + L0 mg / k m Here s the Matlab soluton clear all syms g k m L L0 real assume(k>0); assume(m>0); syms x(t) v(t) eq = dff(x(t),t,) == -(g+k*(x(t)-l0)/m); v(t) = dff(x(t),t); IC = [x(0)==l, v(0)==0]; x(t) = smplfy(dsolve([eq,ic])) It doesn t qute look the same as the hand calculaton but of course cosθ = cos( θ) so they really are the same. 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 / k. Here, mg/k 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 / k. Ths corresponds to the dstance of the mass above ts average poston at tme t=0. 3. The perod of oscllaton (the tme taken for one complete cycle of vbraton) s T = π m/ k

41 1 k 4. The frequency of oscllaton (the number of cycles per second) s f = (note f=1/t). π m Frequency s also sometmes quoted as angular frequency, whch s related to f by ω = π f = k / m. Angular frequency s n radans per second. 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 δ = mg / k, you can calculate the (angular) vbraton frequency as ω = g / δ 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 block and rod has frcton coeffcent µ. The rod pushes a block of mass m, whch starts at r=0 wth radal speed V. Fnd an expresson for r(t). ω L r θ e θ e r 1. Introduce varables to descrbe the moton the polar coordnates r, θ work 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= rω e r + ωe θ 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 block wll slde radally outwards, and frcton opposes the slp. 4. Wrte down Newton s law 5. Elmnate reactons d r dr Ter + Neθ = m rω e r + ωe θ e θ e r T N The e θ component of F=ma shows that N = dr ω

42 The frcton law gves T = µ N. Substtutng ths nto the e r component of F=ma and smplfyng shows that d r dr + µω rω = 0 6. Identfy ntal conons Here, r=0 dr/=v at tme t=0. 7. Solve the equaton: If you ve taken AM33 you wll know how to solve ths equaton If not you can use Matlab: clear all syms mu omega V0 real syms r(t) v(t) dffeq = dff(r(t),t,) + *mu*omega*dff(r(t),t) - r(t)*omega^ ==0; v(t) = dff(r(t),t); IC = [r(0)==0,v(0)==v0]; r(t) = smplfy(dsolve(dffeq,ic)) Ths can be smplfed slghtly by hand: V µωt rt ( ) = e snh( 1 + µ ωt) ω 1+ µ 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 work, 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. 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.

43 In ths course we wll use a general purpose program called 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 ( xt ( ), yt ( ), zt ( )) k X0 V 0 F D. Wrte down the poston vector n terms of these varables: r = x() t + y() t + z() t k 3. Dfferentate the poston vector wth respect to tme to fnd the acceleraton. Smple calculus gves mg dx dy dz d x d y d z v = + + k a= + + k 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 = mgk. Ar resstance. 1 The magntude of the ar drag force s gven by FD = ρcddv, where ρ 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 = + +

44 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 1 πd v 1 πd dx dy dz FD = ρcd V = ρcd V + + k 4 V 4 The total force vector s therefore 1 π D dx dy dz F= mgk ρcd V + + k 4 5. Wrte down Newton s laws of moton. F 1 π D dx dy dz d x d y d z = m a mg k ρcd V + + = m k k π It s helpful to smplfy the equaton by defnng a specfc drag coeffcent c = ρ DC D, so that 8m dx dy dz d x d y d z g k cv + + = + + k k 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= X0 = 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 known exact soluton. So nstead, we use MATLAB to generate a numercal soluton. Ths takes 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 unknown varables. In other words, we wll solve for ( xyzv,,, x, vy, v z), where vx = dx v dx dx y = vz =

45 These defntons are three new equatons of moton relatng our unknown 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(@(t,w) eom(t,w,c,g),[0,stop_tme],ntal_w); plot3(sols(:,1),sols(:,),sols(:,3)) % Plot the traectory end functon dw = eom(t,w,c,g) % 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^); 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 Ths produces a plot that looks lke ths (the plot s been eed to add the grd,etc)

46 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 v = V. 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= + 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 F g 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 GMm F= = ( x + y ) r 3 r r 5. Wrte down Newton s laws of moton.

47 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 unknown varables. In other words, we wll solve for ( xyv,, x, v y), where dx dx vx = vy = These defntons are new equatons of moton relatng our unknown 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. 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(@(t,w) odefunc(t,w,gm),[0,tme],w0); plot(w_values(:,1),w_values(:,))

48 end functon dw = odefunc(t,w,gm) 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 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. Look 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 checks here. 1. Conserved quanttes For ths partcular problem, we know 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 take ths as gven). The total energy of the system conssts of the potental energy and knetc energy of the satellte, and can be calculated from the formula GMm 1 GMm 1 E = + mv = + m( vx + 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 ( ) ( x y ) ( y x) H= r mv= x+ y m v + v = m xv yv k H = ( xvy yvx) m (If you don t know 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

49 You can then plot the results (e.g. plot(t_values,energy)). The results are shown below. These results look 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 θ 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 know r=r, from the poston at tme t=0. (b) The satellte must move at constant speed, and the angle θ must ncrease lnearly wth tme,.e. θ = ωt where θ = ω 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 r = Rcos θ() t + Rsn θ() t We know that v = Rω( sn θ() t + cos θ()) t ( cos () sn ()) a= ω θ + θ R t R 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 = Rω GMm r F= ma = ma GMm (cosθ+ sn θ) = m V (cosθ+ sn θ ) 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.

50 Fxng the problem: In general, when computer predctons are suspect, we need to check 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 mstake 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(@(t,w) odefunc(t,w,gm),[0,tme],w0,optons); 4. Is there some feature of the equaton of moton that makes 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. There s a small change n energy and angular momentum but the rate of change has been reduced dramatcally. We can make 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 make 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.

51 Example 3: Earthquake response of a -storey buldng The fgure shows a very smple dealzaton of a -storey buldng. The roof and second floor are dealzed as blocks wth mass m. They are supported by structural columns, whch can be dealzed as sprngs wth stffness k and unstretched length L. At tme t=0 the floors are at rest and the columns have lengths l 1 = L mg / k l = L mg /k (can you show ths?). We wll neglect the thckness of the floors themselves, to keep thngs smple. For tme t>0, an earthquake makes the ground vbrate vertcally. The ground moton can be descrbed usng the equaton d = d 0 snωt. d 0 snω t Horzontal moton may be neglected. Our goal s to calculate the moton of the frst and second floor of the buldng. It s worth notng a few ponts about ths problem: 1. You may be skeptcal 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 taken about the center of mass of the block. The floors behave as though they are partcles, even though they are very large.. Real earthquakes 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 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 keep 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. l l 1 k k m m k k y y 1 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

52 { } F1 = mg kl ( 1 L) + kl ( L) F = mg kl ( L) 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 ωt 5. Wrte down Newton s laws of moton. F=ma for each mass gves d y { mg kl ( } 1 1 L) + kl ( L) = m d y mg k l L = m { ( )} mg k(l -L) mg k(l 1 -L) Ths s two equatons of moton we can substtute for l 1, l and rearrange them as dy1 k k = g ( y1 d0 sn ωt L) + ( y y1 L) m m dy k = g ( y y1 L) m 6. Elmnate reactons ths s not needed n ths example. 7. Identfy ntal conons. We know that, at tme t=0 dy1 dy y1 = L mg / k 0 y L 3 mg /k 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 unknown varables, n the usual way. Our equatons of moton can then be expressed as dy1 = v1 dy = v dv1 k k = ( g y1 d0 sn ωt L) + ( y y1 L) m m dv k = 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.

53 functon buldng % k=100; m=1; omega=9; d=0.1; L=10; tme=0; g = 9.81; w0 = [L-m*g/k,*L-3*m*g/(*k),0,0]; [t_values,w_values] = ode45(@(t,w) eom(k,m,l,d,omega,g),[0,tme],w0); plot(t_values,w_values(:,1:)); end functon dw = eom(t,w,k,m,l,d,omega,g) y1=w(1); y=w(); v1=w(3); v=w(4); end dw = [v1;v;... -*k*(y1-d*sn(omega*t)-l)/m+*k*(y-y1-l)/m;... -g-*k*(y-y1-l)/m]; The fgures below plot the heght of each floor as a functon of tme, for varous earthquake frequences. For specal earthquake 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 earthquake, t wll be safe. We wll dscuss vbratons n much more detal later n ths course.

54 Fnal remark: we made ths calculaton a bt more complcated than necessary by solvng for the heghts y1, y. It s better to solve for the deflectons of the floors nstead of ther heghts. Defne z1 = y1 ( L mg / k) z = y (L 3 mg / k) If we substtute for y1, y n our equatons we fnd that dz1 = v1 dz = v dv1 k k = ( z1 d0sn ωt) + ( z z1) m m dv k = ( z z 1) m These are a lot smpler, and more mportantly, tell us that the moton of the system does not depend on the sprng length L or gravty.