Physics 5A Review 1 Eric Reichwein Deparmen of Physics Universiy of California, Sana Cruz Ocober 31, 2012 Conens 1 Error, Sig Figs, and Dimensional Analysis 1 2 Vecor Review 2 2.1 Adding/Subracing Vecors............................. 2 2.2 Muliplying Vecors: Do and Cross Producs................... 3 3 One Dimensional Kinemaics 3 3.1 Displacemen vs. Disance.............................. 3 3.2 Velociy vs. Speed.................................. 3 3.3 Consan Acceleraion Kinemaics.......................... 4 3.4 Example: Train Disance............................... 4 4 Two Dimensional Kinemaics 6 4.1 X-Direcion Equaions................................ 7 4.2 Y-Direcion Equaions................................ 8 4.3 Cool Facs....................................... 9 5 Dynamics 9 5.1 Coordinae axes.................................... 10 5.2 Free-Body-Diagrams................................. 10 5.3 Uniform Circular Moion............................... 12 6 Graviaion 13 1 Error, Sig Figs, and Dimensional Analysis If we have measuremens 8.31, 8.35, and 8.41 we ge an average of 8.36 wih an error of ±0.05. The basic rule of humb is firs ake ino accoun he error in he measuremens, hen look a he significan figures. For significan figures all we do is firs look a each measured values ha are given, bu no he given consans. Then we find he value wih he leas significan figure and use ha 1
as how many sig figs you will quoe. If we have he measured values: 320, 20.1, 4.1213, and 0.0013345.The leas sig fig ou of hese numbers is 320 wih wo sig figs. Our final value would be 1500 or 32 or 0.00000095 or any number wih wo sig figs. Dimensional analysis is a way of changing unis. Therefore we mus have some conversion raios o acually use dimensional analysis. The way we do his is by aking our unis and muliplying i by he conversion raio. As an example I will conver from kilograms o Eric s. 1kg 2 Vecor Review 10F reds 1kg 20Jams 10F reds 1Eric 2Jams = 10Erics A vecor is a mahemaical ool used o represen physical quaniies. Vecors are exremely useful for physics (and necessary!) o describe naural phenomena. For example, if I say here is wo forces acing on a paricle and hen ask you o calculae he resuling acceleraion, you won be able o. Why? Because he acceleraion is dependen on he direcion of he ne force and on is direcion. To overcome his problem we use vecors. A vecor jus gives us a direcion and magniude of he quaniy in quesion. I can represen a vecor in muliple ways. Firs I could say he firs force is direcly upwards, and he second force is direcly downwards, and ha hey have equal magniudes. Now you could represen hese vecors as F 1 = ( ) 0 î + F ĵ F 2 = ( ) 0 î F ĵ To calculae he acceleraion we need o find he sum of he forces on he body and hen divide by mass (see secion 5, or chaper 4 in Giancoli, for furher deails). So now we need o know how o add vecors. 2.1 Adding/Subracing Vecors To add wo vecors ogeher we jus add each corresponding componen of he vecor ogeher. Using he vecors from he example we will obain F 1 + F 2 = ( ) ( ) ( ) ( ) 0 î + F ĵ + 0 î F ĵ = 0 î + 0 î + F ĵ + F ĵ = 0 î + 0ĵ If we waned o subrac o vecors all we would do is muliply each componen of he vecor being subraced hen add hem ogeher. The same process is used for adding vecors and subracing vecors. We can characerize his process o as many vecors as we wan using summaion noaion as ( n n ) n n v 1 + v 2 + v 3 + + v n = v m = v mx î + v my ĵ + v mzˆk (1) m=0 m=0 m=0 m=0 2
2.2 Muliplying Vecors: Do and Cross Producs There are wo operaions we use when muliplying vecors ogeher. I will omi all heory and in-deph geomeric discussion abou hese wo ypes of vecor muliplicaion. However, I will discuss heir relaions o physics. 2.2.1 Do Produc or Scalar Produc Do produc produces a scalar value. I represens he amoun of one vecor ha is in he direcion of anoher vecor. For example, if we waned o calculae he work done on a paricle by moving i hrough some force field we would use he do produc. Why you ask? Because work is done only when he force is he direcion of moion. Is like a air hockey able, he puck moves around wihou any push from you because he air blowing up on i effecively ges rid of fricion, and he force of graviy only acs downward. Hence, he work needed o move he puck is zero since is direcion of moion is perpendicular o he direcion of he force. The do produc is compued by muliplying he corresponding componens of he wo vecors ogeher. ( ) v 1 v 2 = v 1x v 2x î î + v 1y v 2y ĵ ĵ + v 1z v 2zˆk ˆk = v 1x v 2x + v 1y v 2y + v 1z v 2z (2) 2.2.2 Cross Produc or Vecor Produc r F î ĵ ˆk = r x r y r z F x F y F z = î (F z r y F y r z ) ĵ (F z r x F x r z ) + ˆk (F y r x F x r y ) The magniude of a cross produc is defined as r F = r F sinθ 3 One Dimensional Kinemaics 3.1 Displacemen vs. Disance Disance is a scalar measure of he inerval beween wo locaions measured along he acual pah connecing hem. Displacemen is a vecor measure of he inerval beween wo locaions measured along he shores pah connecing hem. If you walk o he lef 20 meers hen reurn o he posiion you sared a your displacemen is 0 meers, and he disance raveled is 40 meers. 3.2 Velociy vs. Speed Velociy is he rae of change of displacemen per uni ime. This implies ha velociy is a vecor. There can be a change in velociy wihou a change in speed, simply by changing direcion. 3
Speed is he rae of change of he disance. This implies ha speed is a scalar. 3.3 Consan Acceleraion Kinemaics We live in graviaional field produced by he mass of earh. We assume ha i is consan 9.8 m near he surface of he earh. We define his acceleraion as he consan g. We say ha s 2 he acceleraion in he graviaional field is he rae of change of he velociy, a = dv = 9.8 m. d s 2 Now we jus inegrae o obain he velociy equaion of moion v() Following he same process, wih v() = dx d x() 3.3.1 Key Kinemaic Equaions 1. a() = a 2. v() = v 0 + a 3. x() = x 0 + v 0 + a2 2 4. v 2 () = v 2 0 + 2a (x() x 0 ) 3.4 Example: Train Disance dv = a d dv = ad v 0 0 v() v 0 = a 0 v() = v 0 + a (3) dx = v d 0 + a we obain he posiion as a funcion of ime as dx = v 0 + a d x 0 0 x() x 0 = v 0 + 1 2 a 2 0 x() = x 0 + v 0 + 1 2 a2 (4) A rain a saion A sars from res a = 0 and acceleraes a 5 m for 10 seconds. I remains a s 2 his velociy for one hour hen deceleraes a -4 m. (a) Wha is he maximum velociy reached? s 2 (b) How long does i ake o come o a complee sop, and how far does i ravel in ha ime? (c) Wha is he oal displacemen of he rain? (d) Graph he posiion, velociy, acceleraion as a funcion of ime. 4
3.4.1 Par A Since we are given he acceleraion of he rain we jus inegrae acceleraion funcion wih respec o ime (acceleraion as a funcion of ime is consan). The final velociy afer his acceleraion would give us he maximum velociy since here is no more posiive acceleraion afer = 10s. We would obain Equaion 1 wih v 0 = 0 hence v( = 10s) = v 0 + a v( = 10s) = 0 m s + 5 m s 2 10s v( = 10s) = 50 m s The maximum velociy (and speed) is 50 m s. I will remain a 50 m s for he nex hour. 3.4.2 Par B For he rain o come o a complee sop means v() = 0. We know he iniial velociy o be 50 m as found in par A. We inegrae he acceleraion funcion (a consan) which is resuls in s Equaion 1 wih v() = 0 and v 0 = 50 m. s v() = v 0 + a 0 = 50 m s 4 m s 2 4 m s 2 = 50 m s = 50 m s 4 m s 2 = 12.5s Now ha we know how long i ook o come o a complee sop we can deermine how far i raveled during ha ime. Inegraing our velociy funcion we obain Equaion 2 wih x 0 = 0, since we are only ineresed in he disance i raveled from when i began deceleraing. x( = 12.5s) = 0 + 50m/s12.5s + 1 2 ( 4 m s 2 )(12.5s) 2 x( = 12.5s) = 625m + 312.5m x( = 12.5s) = 937.5m The rain ravels 937.5m during he ime ime i akes o sop. 3.4.3 Par C We have o break up he rip ino hree segmens since acceleraion is no consan hroughou he rip. The segmens are as follows: firs where he rain acceleraes, second where he rain moves wih consan velociy, hird where he rain deceleraes o res. The firs disance is given by Equaion 2 as x( = 10s) = 0m + 0 m s 10s + 1 2 (5 m s 2 )(10s) 2 x( = 10s) = 250m = 0.25km 5
The second disance is given by inegraing Equaion 1, or Equaion 2 wih a = 0, since here is no acceleraion. x( = 3600s) = 0m + 50 m s 3600s + 1 2 (0 m s 2 )(3600s) 2 x( = 3600s) = 180, 000m = 180km The hird disance will be deermined in he same way as he firs disance. We have already deermined how far he rain will ravel during he deceleraion in Par B, which was 937.5m. Now all we do is add all he disances ogeher o deermine he oal disance. x oal = 250m + 180, 000m + 937.5m = 181, 187.5m = 181.1875km 3.4.4 Par D x v a 10s 1hr 3612.5s 10s 1hr 3612.5s 10s 1hr 3612.5 4 Two Dimensional Kinemaics The same rules apply for kinemaics in wo dimensions (and hree dimensions). We generally break kinemaic problems apar ino componens, he x-direcion and y-direcion (and he z- direcion). For projecile moion here is no acceleraion in he x-direcion, unless you consider air resisance 1. From he figure below we see ha only he verical componen of velociy changes magniude, hence here is only verical acceleraion (or deceleraion). This makes sense because graviy is he only force acing on he projecile (excluding air resisance) and i always poins downward. 1 For his class and almos all inroducory physics courses we consider air resisance o be negligible. 6
Figure 1: The componens of velociy of a projecile. The moion is parabolic. Air resisance is negligible. To deermine he velociy or posiion as a funcion of ime we mus inegrae he acceleraion funcion. For mos kinemaics problems he acceleraion is jus graviy which is a consan (near he earhs surface). The velociy funcion and posiion funcion, respecively, are v() = a()d + v 0 x() = [ a()d + v 0 ] d + x 0 Where he consans of inegraion, v 0 and x 0, are he iniial velociy and iniial posiion, respecively. 4.1 X-Direcion Equaions 4.1.1 X-Acceleraion Unless here is air resisance or some kind of oher exernal force acing on he projecile here is no acceleraion in he x-direcion. 4.1.2 X-Velociy The velociy in he x-direcion is, by definiion of no acceleraion, consan during he enire rajecory. The equaion for velociy in he x-direcion as a funcion of ime v x () = v 0 cos(θ) Where θ is he angle from he horizonal ha he projecile is launched from, and v 0 is he iniial velociy. 4.1.3 X-Posiion Since he velociy in he x-direcion is consan he posiion of he projecile increases proporionally o ime. Depending on how much iniial velociy he projecile is given will deermine how rapidly i gains disance. Therefore he x-posiion as a funcion of ime is given by x() = v x 7
4.1.4 X-Graphs x v x a x 4.2 Y-Direcion Equaions 4.2.1 Y-Acceleraion For all projecile moion problems here will be graviy acing on he objec during is enire rajecory. Graviy always poins downward and is consan hroughou he duraion of moion. I is expressed as a = g = 9.8 m s 2 This value will only change when on a differen planes surface ( i.e. he moon, mars, venus, ec) or when very far from he earhs surface. However, here is a differen approach o solving problems very far from he earhs surface. These are ofen rockes and saellies. 4.2.2 Y-Velociy The velociy in he y-direcion sars from a maximum (if we assume moion is launched above he horizonal) hen decreases unil i reaches he apex of he rajecory. A he apex i has zero velociy in he verical direcion. This is where he velociy changes from posiive o negaive 2. Remember ha acceleraion is consan, i is he same when i has maximum posiive velociy as when i has no velociy or negaive velociy. To describe he velociy as a funcion of ime we use Equaion 1 v() = v 0 + a We see ha his is a sraigh line wih v 0 as he y-inercep and he acceleraion, usually g, is he slope of he line. This equaion works for any siuaion of projecile moion. 4.2.3 Y-Posiion The verical posiion of he projecile increases o a maximum hen decreases. I is parabolic moion, jus like he equaion describing i y() = y 0 + v 0 sin(θ) + 1 2 a y 2 2 Or i changes from negaive o posiive depending on he coordinae axes you chose. 8
4.2.4 Y-Graphs y v y a y 4.3 Cool Facs A ball hrown upward wih iniial velociy v 0 reaches a max heigh x = v2 0 2g velociy when i reaches he ground of v 0. and has a final The same ball will ake a ime = v 0 g o reach max velociy. This found by knowing ha a he velociy is zero and using our velociy equaion. A bulle sho horizonally a 1000 m akes he same amoun of ime o hi he ground as s he shell casing. If we neglec air resisance. If you dropped a ball from a cliff, of heigh H and a he same ime your friend hrough a ball upwards wih iniial velociy v 0, he balls would collide (if hey did a all) a ime = H v 0. This makes sense if you pu yourself in acceleraing reference frame... 4.3.1 Monkey and he Huner Remember he demonsraion of he monkey dropping and he huner (you blowing a ball hrough a ube), he monkey and he ball fell downward a he same rae even hough he ball had a large x-velociy. Here s a picure of he seup. 4.3.2 Relaive Velociy If a rain falls down a 1 m and you are driving in he posiive x direcion a 5 m s (. The rain on s your window will make an angle wih he negaive x-axis θ = an 1 v rain V car ). Using his idea we can find he velociy of he rain if we measure he angle θ and know how fas he car is going. From his we find ha he velociy of rain is given by v rain = v car anθ 5 Dynamics We explained how objecs move using kinemaics, now we will describe why objecs move using dynamics. There are hree laws of dynamics which were posulaed by Isaac Newon. The laws are as follows 9
1. An objec in moion ends o say in moion unless oherwise aced upon by an exernal force. Similarly an objec a res ends o remain a res unless aced upon by an exernal force. 2. An exernal force acing upon an objec is proporional o he acceleraion and is dependen on he mass. The force is also equivalen o he insananeous rae of change of he momenum. Force is also equivalen o he negaive of he insananeous rae of change of poenial energy. 3. Every force (acion) has an equal and opposie force (reacion). Basically, all of classical mechanics is wrapped up in hese hree laws (mainly he second law). There are few imporan echniques and conceps ha will be applied in every mechanics problem. 5.1 Coordinae axes We do no necessarily have o choose he x-axis going righ and y-axis up. In fac i is ofen easier o no choose he sandard coordinae sysem. For inclined planes we end o pu he y-axis in he direcion of he normal force, as seen in he figure below. This would pu he x-axis enirely in he direcion of moion, and hence he direcion of acceleraion. By puing one of he coordinae axes in he direcion of acceleraion. This makes he equaions simpler. +y N T +x T α f R α Mg mg + Figure 2: We choose wo differen coordinae axes for each box in his sysem. The box on he inclined plane has a iled coordinae axes, while he hanging mass has posiive downward. I is imporan o noe ha for wo body sysems we can can choose wo differen coordinae sysems, we jus have o say consisen wih our equaions. Please noe ha i is very convenien o pu one axis in he direcion of acceleraion. This will give you wo equaions of Newons second law (one for x-direcion and one for y-direcion) in which one of hem is zero. 5.2 Free-Body-Diagrams The free body diagram is easily he mos imporan idea semming from Newons Laws. I allows us o clearly label all he forces on he objec in quesion. This in urn gives us an easy way o wrie our equaions describing moion. There is no difference beween he equaions and he free body diagrams (FBD). The FBD shows us he equaions o use. They are merely 10
wo differen ways of expressing physical phenomena. A free body diagram is shown in Figure 2 above (middle and lef objecs). Figure 3: A free body diagram. There are forces in he x-direcion and he y-direcion. We look a each direcion separaely when using Newons second law. To use a FBD we firs deermine wha forces are acing on he objec. We ask ourselves: Is here fricion? Force of graviy? Tension? ec. We hen label hese forces on he FBD in he direcion ha he force is being applied. I am siing in a chair righ now. The chair is pushing up wih a normal force ha is couner acing he force due o my mass in earhs graviaional field. The FBD of siuaion would be a box, circle, do -somehing simple o represen myself. I would hen draw an arrow upwards represening he force applied by he chair called he normal force. Then I would draw an arrow downward represening he force due o my weigh. These arrows should be equal and opposie if I were no acceleraing. If I was in an adjusable heigh chair and slowly decreasing he heigh hen he arrows would no be equal. Wha arrow would longer? The force due o my weigh would keep he same magniude, hence he arrow would say he same lengh. However, he he normal force would decrease. I would decrease because I am acceleraing downward. This means here is a ne force on my body downwards. This is he process for every problem encounered for his secion. There are wo main forces we will look a 5.2.1 Weigh and Normal Force Weigh and normal force are an example of Newons hird law. The weigh of an objec is an acion and he normal force is he reacion. Every objec ha is a res relaive o he earhs surface has a normal force. If i did no have a normal force i would no be a res. The normal force holds your exbook on he shelf, phone on he able car on he ground,ec. Please noe ha he normal force is no always equal in magniude o he weigh of he objec. I also conrols how much your car grips he ground hrough he force of fricion. 5.2.2 Fricion Fricion is always a rearding force. I acs in opposiion o moion. Fricion is caused by he microscopic irregulariies on wo surfaces ha come ino conac. These forces are ulimaely caused by elecromagneic ineracions beween elecrons of he aoms making up he wo differen objecs. There are wo ypes of fricion saic and kineic. Saic fricion is he fricion keeping an objec from sliding around. I is sronger han kineic energy. Tha is why i is harder o push a heavy box saring a res han afer i has begun moving. The amoun of saic fricion is deermined experimenally for each maerial. Bu is deermined by a coefficien and he normal force. The amoun of fricion, saic or kineic, is direcly proporional o hese 11
wo quaniies. F s µ s N (5) Where N is normal force and µ s is he coefficien of saic fricion. here is a les han or equal o sign because fricion is only as big as i needs o be. Oherwise, objecs would fly off ables lef and righ due o fricion! Kineic fricion is always less han saic fricion for he same maerial. The equaion for kineic fricion is similar o saic fricion excep here is no less han par. The coefficien of kineic fricion is a consan. F k = µ k N (6) Where N is normal force and µ k is he coefficien of kineic fricion. The graph as a funcion of applied force o an objec looks like he following funcion of fricional force vs applied force. Figure 4: During he saic region he objec your applying a force has no moved. Once i reaches he boundary beween saic and kineic (sliding) i overcomes he elecromagneic forces beween elecrons 5.3 Uniform Circular Moion Uniform circular moion is weird because he paricle/objec has a consan angenial velociy, and consan angular velociy. Since velociy is a vecor he objec is acually acceleraing. No because is increasing is speed bu because is changing is direcion. We call his ype of acceleraion cenripeal acceleraion. I will omi he derivaion because he ex has a very nice derivaion. The cenripeal acceleraion is given by a c = v2 r = ω2 r = vω (7) Where we have used he relaions and can use he following relaions o obain differen quaniies v =ωr (8) ω = 2π = 2πf T (9) v = 2πr = 2πrf T (10) 12
5.3.1 Uniform Circular Moion and Newon s Second Law We can use Newon s second law o deermine he moion of an objec by using our cenripeal acceleraion in he second law F = mac = m v2 (11) r Then we can carry ou our calculaions as in oher dynamics problem. Please noe ha here mus be some NET force (ension, graviy, fricion, ec.) keeping he objec moving in he circular. And his equaion only works for uniform circular moion, where uniform jus means consan speed. 6 Graviaion Everyhing ha has mass has a graviaional field. You have a graviaional field because you have mass. The earh, moon, all planes, sars galaxies all have mass. Graviy is he weakes of he four fundamenal forces, ye i is he dominan force in he universe for shaping he large scale srucure of galaxies, sars, ec. The graviaional force beween wo masses m 1 and m 2 is given by he relaionship: F g = G m 1m 2 r 2 ˆr (12) 11 Nm2 Where G = 6.67 10 and is called universal graviaional consan. This is ofen kg 2 called he universal law of graviaion and G he universal graviaion consan. I is an example of an inverse square law force. The force is always aracive and acs along he line joining he ceners of mass of he wo masses. The forces on he wo masses are equal in size bu opposie in direcion, obeying Newon s hird law. Viewed as an exchange force, he mass-less exchange paricle is called he gravion. Any poin source which spreads is influence equally in all direcions wihou a limi o is range will obey he inverse square law. This comes from sricly geomerical consideraions. As one of he fields which obey he general inverse square law, he graviy field can be pu in he form shown below, showing ha he acceleraion of graviy, g, is an expression of he inensiy of he graviy field. 13
Figure 5: This is a graphical represenaion of how graviy permeaes space ime. 14