EN40: Dynamcs and Vbratons School of Engneerng Brown Unversty Homework : Knematcs and Dynamcs of Partcles Due Frday Feb 7, 014 Max Score 45 Ponts + 8 Extra Credt 1. An expermental mcro-robot (see a descrpton here most people would call the robot a speck of drt ) s moved around on a surface by subectng t to a constant magnetc force P that acts tangent to the surface. In addton, an electrostatc force wth magntude P can be appled to attract the mcro-robot to the surface. Frcton wth coeffcent acts between the robot and the surface, so the partcle can be moved around by turnng the electrostatc force on and off. Gravty may be neglected. Suppose that the mcro-robot has mass m, and s at rest at x=0 at tme t=0, wth the electrostatc force turned on. To move the robot through a dstance d, the electrostatc force s turned off for a tme 0 t t 1, and then turned back on agan. The goal of ths problem s to derve a formula relatng t 1 to d. x P P P d Electrostatc force off Electrostatc force on 1.1 Use Newton s law to fnd the acceleraton of the mcro-robot durng the tme when the electrostatc force s swtched off 0 t t1, and hence fnd formulas that gve the speed and poston of the mcrorobot as functons of tme for 0 t t1, n terms of P and m. Durng ths phase P s the only force actng on the robot. Its acceleraton s constant and has magntude P/m. The straght-lne moton formulas gve v ( P / m) t x ( P / m) t / [ POINTS] 1. Smlarly, fnd formulas that predct the speed and poston of the mcro-robot for tme t t1 A free body dagram s shown n the fgure to the rght. Newton s law shows that ( P N) ( N P) ma The and components show that N P a ( 1)P/ m N P N P
At tme t 1 the poston and velocty of the partcle are v ( P / m) t1 x ( P / m) t1 / The straght-lne moton formulas then gve v v a( t t ) 0 0 ( P / m) t1 ( 1) P( t t1) / m x x0 v0 ( t t0) a( t t0) / ( P / m) t1 / ( P / m) t1( t t1) ( 1) P( t t1) / m [3 POINTS] 1.3 Use the condtons that the robot comes to rest at poston d to determne t 1 n terms of P, d, and m. Check the dmensons (unts) of your soluton. When moton stops we know that v 0 ( t t ) t / ( 1) 1 1 t1 t d ( P / m) 1 t1 ( 1) ( 1) ( 1) Pt1 ( 1) 1 d m 1 md( 1) t1 P [ POINTS]. Ths webpage shows some hgh speed vdeos of three dfferent bow desgns as they fre an arrow (Courtesy of Ahyoung Choe, 15). Poston-v-tme data for the arrow has been extracted from the mages, usng the MATLAB mage processng toolbox. Your msson s to wrte a MATLAB scrpt that wll analyze ths data to plot graphs showng the velocty and acceleraton of the arrow. The poston-v-tme data are stored n three.csv ( comma separated value ) fles that can be downloaded from ths webpage. You can read the data fles drectly nto MATLAB usng the csvread command for example compoundbow_data = csvread('compound_data.csv'); The varable compoundbow_data s a matrx, wth the frst column (compoundbow_data(:,1)) contanng tme values, n seconds, and the second column (compoundbow_data(:,)) contanng the dstance moved by the arrow, n m. You can also open the fles wth excel or any text edtor. The easest approach to calculatng velocty and acceleraton s to numercally dfferentate the data. For example, let the N tme and
poston values be t, x wth =1 N. You can estmate the velocty at the mdpont of a tme nterval between two successve data ponts as v x / t, gvng t ( t t 1) / v ( x 1 x ) / ( t 1 t) 1... N 1 You can calculate vectors of tme and velocty usng these formulas and a loop, and then plot them. You can then calculate the acceleratons by dfferentatng the velocty a v / t. The acceleraton wll be qute nosy. If you would lke to try somethng more sophstcated you can explore MATLAB s Curve Fttng Toolbox, whch has a lot of capabltes for curve-fttng, smoothng, and computng numercal dervatves. 1.4 Submt graphs that show the velocty and acceleraton of at least ONE bow as a functon of tme (you can plot them all f you are curous but ths s not requred). Be sure to specfy whch bow you analyzed. Also nclude a bref descrpton that wll help the grader understand how you processed the data (numercal dfferentaton, curve fttng, etc). You do not need to submt MATLAB code. Velocty-v-tme s shown below Acceleraton results computed by drect numercal dfferentaton are shown below [3 POINTS] Acceleraton data smoothed usng the MATLAB cft functon (the dots show the raw data) are shown below [3 POINTS any approach s fne]
1.5 Estmate the maxmum force exerted by the bow on the arrow as t s fred. For comparson, the statc force-v-draw curves (.e. the force requred to draw the bow through a dstance d) for the three bows are shown on the webpage. Why do you thnk the dynamc force dffers from the statc force? Would you expect the average dynamc force as the arrow s fred to be greater, or less than the statc force? The maxmum dynamc forces are the arrow mass multpled by the maxmum acceleraton: Longbow approx. 50N Compound bow approx. 400N Recurve bow approx. 150N The dynamc force (roughly speakng) dffers from the statc force because under statc loadng the nternal force n the bow balances the external force exerted on the bow to draw t Fnt Fext. Under dynamc condtons the nternal forces n the bow must accelerate the bow Fnt Fext mbow abow (t s not clear qute what a bow and m bow represent because the bow s not a partcle). The average dynamc force must be less than the statc force. 1.6 Optonal for extra credt Plot graphs that compare the statc force exerted by the bow as a functon of draw dstance d to the actual, dynamc force exerted on the arrow whle t s fred (plot draw dstance on the horzontal axs and the force on the vertcal axs) [5 POINTS]
Normal accel (a n ) (m/s ) Speed (m/s) 3. The path and speed of a vehcle drvng around a sharp bend s shown n the fgure below (the vehcle s at pont A at tme t=0). 3.1 Sketch graphs of the normal and tangental acceleraton of the vehcle. Explan brefly how you calculated relevant quanttes. D 35 30 50m 5 C 0 A B n V t 15 10 5 7.85s 5.3s 5.3s Tangental accel (at) m/s 5 4 3 1 0-1 - -3-4 100m -5 0 5 10 15 0 5 Tme (s) 0 0 5 10 15 0 5 10 8 6 4 Tme (sec) 0 0 5 10 15 0 5 Tme (s) Ths problem nvolves applyng and nterpretng the formula for acceleraton n normal-tangental coordnates dv V a t n dt R Durng the frst 5 sec the car travels wth average speed 0 m/s, and therefore travels a total dstance 100m and ust reaches B. Durng the frst 5 sec the tangental acceleraton s -0/5=-4m/s^ The dstance traveled durng the subsequent 7.85 sec s 78.5m the angle BOC s therefore 78.5/50=1.57 radans (approxmately /). The car reaches C at the end of ths tme. [1 POINT] Durng ths perod the tangental acceleraton s zero (speed s constant). The normal acceleraton s 100/50=m/s^ Durng the subsequent 5.3sec, the average speed s 15m/s, and the car therefore travels 78.5m and reaches D at the end of the tme perod. The tangental acceleraton s equal to 10/5.3 = 1.91m/s^ (for 10.46 sec) The speed durng ths phase vares lnearly wth tme the normal acceleraton vares quadratcally. At D, the speed s 0 m/s, so the normal acceleraton s 400/50=8 m/s^. [1 POINT] Thereafter, the normal acceleraton drops to zero (the radus of curvature s nfnty).
3. Optonal extra credt Calculate the mnmum frcton coeffcent requred to prevent the vehcle from slppng (Draw a FBD; use Newton s laws to calculate the reactons; and use the frcton law to fnd ) A FBD s shown. Newton s law gves V ( TAt TBt ) t ( TAn TBn ) n ( NA NB mg) k mat t m n R V Thus TAt TBt at TAn TBn NA NB mg R For no slp TAt TBt TAn TBn ( N A NB ) V at R To fnd the crtcal frcton coeffcent we need to dentfy the maxmum value of g V at R - from the graphs t s clear that the max value occurs at pont D, where at 1.9 and V / R =8.. The frcton coeffcent must exceed 8. 0.83 9.81 [3 POINTS EXTRA CREDIT] 4. The fgure shows an arcraft ust startng ts take-off roll. The engnes provde a total thrust F T that act a heght h below the center of mass, producng an acceleraton ax ( g/ ). Snce the arcraft s not yet movng lft and drag forces are zero. L x C T At n t T An k N A mg T Bt T Bn N B h y C 4.1 Draw a free body dagram showng the forces actng on the arcraft.
N A F T mg N B [ POINTS] 4. Wrte down Newton s law of moton and the equaton of rotatonal moton for the arcraft (assume straght lne moton wthout rotaton) g F ma FT ( NA NB mg) M N ( L x ) F h x N k 0 C A c T c B [ POINTS] 4.3 Hence, fnd formulas for the reacton forces on the wheels (for the rear wheels, calculate the total force). The prevous problem gves three equatons, whch can be solved for NA, NB, F T wth the results x / c h h x 1 c FT mg NA mg NB mg L L L L [ POINTS] 4.4 Hence, show that the front wheel wll lose contact wth the ground f h exceeds a crtcal value, and fnd a formula for ths crtcal value of h. xc h Notce that NA 0 0 h xc L L The crtcal value of h s x c [ POINTS] 5. The fgure shows an unbalanced rotor that spns at constant angular speed d / dt 40 rad/s. The center of mass of the rotor s a dstance L from the axle. As a result, large fluctuatng reacton forces develop at the axle. The horzontal and vertcal reacton forces actng on the wheel are plotted n the graphs shown (the forces act n the postve and drectons). At tme t=0 the angle s zero. L LAPHROAIG 18 h A 100 gram mass s to be added to the dsk to balance t. goal of ths problem s to calculate the poston. The
5.1 As a prelmnary step, consder a mass m that rotates at constant angular rate d at the end of a massless lnk wth radus R (see the fgure). dt Fnd the acceleraton vector for the mass, expressng your answer as components n the, bass. O R m The poston vector s easy to wrte down and can then be dfferentated. Or you can ust use the standard formulas for crcular moton f you prefer. r R cos sn d v R sn cos dt d a R dt cos sn R cos sn [ POINTS] 5. Draw a free body dagram showng the forces and moments actng on the lnk and mass together (nclude reacton forces/moments at O and gravty) [3 POINTS] 5.3 Use Newton s laws to calculate a formula for the reacton forces actng on the lnk, n terms of,, R, m, g. x y cos y cos sn F R R mg ma mr R mr R mr sn mg x [ POINTS] 5.4 Hence, fnd the dstance R for whch the 100 gram mass wll produce the same horzontal reacton force as the dsk (see the fgure) when rotatng at the same speed as the dsk. R x M R R y m mg
The out-of-balance dsk has a reacton force that vares harmoncally wth tme, wth frequency equal to the angular speed of the dsk (see the fgure the perod s 0.05 sec, whch s 40 rad/s Notce that the mass produces a harmonc reacton force exactly the same as what s measured on the out-of-balance rotor. We can make the mass produce a reacton force that s equal and opposte to what s measured on the dsk by choosng mr 300N R 300 / (0.1 (40 ) ) 0.19m N [ POINTS] 5.5 Draw a sketch showng where the mass m should be located on the dsk (show the locaton of the center of mass of the dsk, together wth the added mass). L LAPHROAIG 18 h (any sketch wth the added mass dametrcally opposte the COM s fne). 5.6 What are the reacton forces actng on the dsk after the dsk has been balanced? The fluctuatng (harmonc) forces cancel, but the average vertcal reactons add. Notce that the average vertcal reacton force on the unbalanced dsk s 00 N so after addng the mass the total vertcal reacton s R 00 0.1g 01N y