Homework 2: Kinematics and Dynamics of Particles Due Friday Feb 8, 2019

Transcription

1 EN4: Dynamics and Vibraions Homework : Kinemaics and Dynamics of Paricles Due Friday Feb 8, 19 School of Engineering Brown Universiy 1. Sraigh Line Moion wih consan acceleraion. Virgin Hyperloop One is a prooype rapid ransporaion sysem. I consiss of an elecric-powered vehicle ha ravels hrough a sealed ube in which he air pressure has been reduced o minimize air drag. Is developers anicipae ha he vehicle will cruise a 67mph (18 km/hr). 1.1 If he acceleraion of he vehicle may no exceed.5 m/s (a ypical discomfor hreshold), how long will i ake he vehicle o reach cruise speed? Using he sraigh-line moion formula v = v + a = = 1sec.5 36 [1 POINT] 1. Wha disance is raveled during acceleraion? The sraigh line moion formula gives x = x + v + a x = = km [1 POINT] 1.3 How long would i ake for he vehicle o ravel beween Providence and Boson (41.5 miles)? (Assume ha boh acceleraion and deceleraion will be a.5 m/s ). You can compare your esimae wih one made by he company here) miles is 66.4 km. I will ake mins o accelerae, and anoher min o decelerae. The oal disance raveled during acceleraion and deceleraion is 36km, leaving 3.4km a cruise speed. 18 km/hr is 3 m/s, so i will ake 34/3 = 11sec o over he remaining 3.4km. The oal ime is 4+11=341sec = 5.7 mins. Virgin hyperloop esimaes 1 mins perhaps hey plan o accelerae less aggressively; and don anicipae being able o ravel in a sraigh line

2 . Calculus review (apologies for inflicing his on you bu hopefully i will be helpful. We sugges doing he problems by hand raher han MATLAB; he problems are mean o help you brush up on calculus ha someimes appears on exams): A paricle a posiion x = and has speed V a ime =. For each case below, please find formulas for he speed v and posiion x of he paricle as funcions of ime..1 The acceleraion of he paricle depends on ime a ( ) = Aexp( / ), where A, are consans v dv a = = Aexp( / ) dv = Aexp( / ) d d V [ 1 exp( / )] [ 1 exp( / )] v V = A v= V + A ( ) [ exp( / ) 1] x dx = V + A [ 1 exp( / )] dx = ( V + A [ 1 exp( / )]) d d x= V + A + A. The acceleraion of he paricle depends on is speed a() v = cv (e.g. air drag), where c is a consan v dv dv a = = cv cd d = v V 1 1 V + = c v = v V 1 + cv x V V dx = dx = d d 1+ cv 1+ cv 1 x = log(1 + cv ) c.3 The acceleraion of he paricle depends on is posiion ax ( ) = ω xwhere ω is a consan (e.g. he acceleraion of a mass on a spring). (You could solve his problem by finding v as a funcion of x firs, hen find x as a funcion of, and finally use your answer o find v as a funcion of. There are oher ways oo). v x dv a = v = ω x vdv = ω xdx dx V 1 1 ( v V ) = ω x v= V ω x x arcsin( ωxv / ) dx 1 cosθ = V ω x dx d d d d = θ = V ω x ω 1 sin θ arcsin( ωx/ V ) = ω x= ( V / ω)sinω v= Vcos ω [4 POINTS]

3 3. Using MATLAB o calculae posiion and acceleraion from velociy measuremens CRS-8 was a 16 Space-X cargo resupply mission o he inernaional space saion. This file conains elemery daa from he launch (in csv forma). Your goal in his problem is o calculae he acceleraion and aliude of he launch vehicle from he daa, and hence o esimae he rocke moor hrus. Wrie a MATLAB scrip ha accomplishes he following asks (Your MATLAB code should be uploaded o CANVAS as a submission o his problem): 3.1 Read he daa ino a marix using he MATLAB csvread command. 3. Plo he speed (in m/s) as a funcion of ime (in sec). [ POINTS] Speed (m/s) [ 1 POINT] 3.3 Calculae he acceleraion of he vehicle as a funcion of ime. You can do his by calculaing he change in speed beween wo successive readings, and dividing by he ime difference beween hem, e.g. if vi () denoes he ih value of x, hen he i componen of acceleraion is ai () = ( vi ( + 1) vi ())/(( i+ 1) i ()) You can consruc he vecor a(i) using a loop (noe ha i will have one fewer rows han he vecor v soring speed). Alernaively, you could use he buil-in MATLAB diff funcion (see he MATLAB manuals for deails). 1 Raw acceleraion 1 8 ) 6 Acceleraion (m/s [ POINTS]

4 3.4 You will find he resuls of 3.3 are very noisy. You can smooh he daa using a simple firs-order filer, as follows. Using a loop, creae a new vecor as(i) wih as(1) = a(1) as(i) = z as(i-1) + (1-z) a(i), i=,3,4 where <z<1 is a number you can use o conrol how much smoohing is applied. You could ry z=.8. Plo he smoohed acceleraion as a funcion of ime. 5 Smoohed acceleraion 4 ) 3 Acceleraion (m/s [ POINTS] 3.5 Use he MATLAB cumrapz funcion o inegrae he speed wih respec o ime and plo he resuls. 3 5 Aliude (km) [ POINTS] 4. Simple circular moion problem The roors on he fields poin wind urbines have a diameer of 9m. Measure heir roaion rae (radians/sec) on a suiably windy day (you can see hem ou of he windows on he 7 h floor of Barus-Holley). Hence, calculae he speed and he magniude of he acceleraion of he roor ips. Be sure o explain your working, since everyone will ge differen numbers. The circular moion formula gives V = Rω. A ypical angular speed is abou rad/sec, which gives a ip speed of 9m/s. The acceleraion of he roor ips is Rω 184ms (his is large!)

5 5. Normal-Tangenial Coordinaes. An aircraf in a dive follows a rajecory 3 x= ( 1) /1 ( ) y = wih x and y in meers, and in seconds. Use a MATLAB live scrip o: 5.1 Plo he rajecory (y v- x) for a ime inerval <<s j V i n 1 Trajecory y (m) (This is jus a simple parameric plo; no explanaion required) 5. Plo he speed as a funcion of ime, for a ime inerval <<s x (m) Speed 6 [ POINTS] 55 5 Speed (m/s) The plo is generaed using V = ( dx / d) + ( dy / d) [ POINTS] 5.3 Plo he normal and angenial componens of acceleraion (on he same plo), for a ime inerval <<s. To calculae he normal-angenial componens, you can (i) find he acceleraion in i,j componens (differeniae he velociy wih respec o ime); (ii) find a uni vecor angen o he pah you could use he velociy vecor, eg, and se V = v = v / V ; (iii) Find a second uni vecor normal o he pah (use he fac ha he normal and angen are muually perpendicular which means - n = or you can also use n= k, boh of hese will ell you ha n= ( dy / di+ dx / dj )/ V ); and (iv) recall ha you can use a do produc o find he componen of a vecor parallel o a uni vecor, so he angenial and normal acceleraions are a = a an = an respecively. Anoher way o find he angenial acceleraion is o use he formula a = dv / d and have malab crank hrough he ime derivaive of V from 5.. There are various oher exoic formulas for normal acceleraion ha you could use if you happen o know hem, for example a n = a (can you prove his?), or you could use V / R if you

6 know how o calculae he radius of curvaure. Bu hese exoic mehods weren covered in ENGN4 (you migh see hem in MA/MA18) and we expeced he do produc approach..5 Acceleraion.4 Tangenial Normal.3 ). Acceleraion (m/s The lif force on he aircraf acs in he n direcion. The load facor is defined as he raio of he lif force o he aircraf weigh. Plo he load facor as a funcion of ime for <<s. (You will need o (i) draw an FBD for he aircraf; and (ii) use Newon s law and he calculaions in 5.3 o find he lif force). The figure shows a FBD. We can wrie Newon s law as ( FT FD) + FLn mgj= ma Take he do produc wih n FL an an 1 dx FL mgjn = man = + jn = + mg g g V d Where a n is from n Load facor.94 F D F L.9 j i mg F T GRADERS: Live scrips were no required for his problem; please grade he plos and any brief descripion of he mehod ha has been provided wih he plos.

7 1 e e r.5 r y Polar Coordinaes: The figure shows he rajecory of a paricle wih mass m in an elecric field. The polar coordinaes of he paricle vary wih ime according o he formulas r = e θ = e Find a formula for he velociy vecor of he paricle as a funcion of ime, as componens in he { er, e θ } basis x dr dθ v= er + r eθ = e er + e e θ d d [1 POINT] 6. Find a formula for he acceleraion vecor of he paricle as a funcion of ime, as componens in he { er, e θ } basis d r dθ d θ dr dθ a= r er + r + e d θ d d d d ( ) ( 4 e e e er e e e e ) 3 ( e 4e ) er = + = eθ [ POINTS]

8 6.3 Show ha he force on he paricle acs in he e r direcion, and show ha he magniude of he force as a funcion of r and m is F = m r 4 3 r F= ma shows 3 ( ) 4 F() = me 4e er = m r 3 e r r This is in he e r because here is no e θ componen. [ POINTS]