Planning the Motion of a Sliding and Rolling Sphere

Size: px
Start display at page:

Download "Planning the Motion of a Sliding and Rolling Sphere"

Transcription

1 2015 IEEE Internatinal Cnference n Rbtics and Autmatin (ICRA) Washingtn State Cnventin Center Seattle, Washingtn, May 26-30, 2015 Planning the Mtin f a Sliding and Rlling Sphere Yan-Bin Jia Department f Cmputer Science Iwa State University Ames, IA 50010, USA jia@iastate.edu Abstract This paper investigates the free mtin f a sphere with initial velcity and angular velcity n a plane under sliding and rlling frictin. The sphere will first slide alng a parablic trajectry (with a cnstant directin f its cntact velcity), and then rll alng a straight trajectry. Such a curved trajectry can be utilized fr bstacle avidance in path planning. A ne-tne crrespndence exists between trajectries cnnecting tw specified lcatins and pairs f sliding and rlling directins within sme 2-dimensinal regin, called the trajectry space, which is then used fr planning. A plane sweep algrithm perating in this space is presented t find all cllisin-free trajectries in the presence f cylindrical bstacles that are vertically riented. I. INTRODUCTION A sphere n a plane can achieve re-rientatin, that is, change the rientatin f its bdy frame, by simply rlling n the plane. Mtin planning fr the sphere, called the ball-plate prblem, explits the nnhlnmic rlling cnstraint t cnstruct a rlling path (desirably clsed) that results in sme prescribed change in the sphere s rientatin. A number f re-rientatin algrithms [5], [7], [8], [11] have been designed using techniues frm plane trignmetry r differential gemetry. Spherical rbts [1], [2] have als been designed t perfrm rlling and spinning peratins in a cntrlled manner. While rlling has received much attentin in path planning, mbile rbts, grasping, and dexterus manipulatin, sliding has largely been regarded as a negative phenmenn t avid, especially in the last tw areas abve. In this paper, we are ging t demnstrate that sliding ffers the benefit f curving the trajectry f a mving sphere, which can be utilized t avid bstacles in path planning. One mtivatin cmes frm billiard sprts. The cue ball after a massé sht will initially slide alng a curve, making slight nise frm scratching the felt, befre rlling straight ahead. Supprt fr this research was prvided in part by Iwa State University (ISU). II. MOVING SPHERE ON A PANE Figure 1 illustrates the general case f a sphere with radius ρ mving n a hrizntal plane P. The sphere P trajectry f ρ Fig. 1. Cntact velcity s f a sphere and cntact frictinal frce f. has n actuatin r cntrl during its mtin, which is cmpletely influenced by frictin. Des there exist initial velcity v 0 and angular velcity ω 0 that will take the sphere t sme lcatin while aviding bstacles? Withut lss f generality, we let the rigin be at the sphere s starting lcatin, the x-axis pint tward, and the xy-plane cincide with P. The fllwing tw assumptins are made: 1) There exists at least ne bstacle at distance ρ r less frm the line segment. 2) v 0 and ω 0 have n cmpnents in the z-directin. If the first assumptin des nt hld, the line segment wuld be a cllisin-free trajectry, making the prblem a trivial ne. The z-cmpnent f v 0 is assumed t be zer t ignre buncing f the sphere. The z-cmpnent f ω 0 plays n part in the initial velcity at the sphere s cntact pint with the plane, referred t as the sliding velcity: s 0 = v 0 + ρẑ ω 0 (1) due t the crss prduct with ẑ = (0, 0, 1) T. We shall see later in this sectin that s 0 and v 0 determine the sphere trajectry. The assumptin abut ω 0 will merely make ur presentatin clean. A. Sliding Sliding happens when s 0 is nt zer. et the sphere have mass m. Its angular inertia matrix is eual t the s ω v /15/$ IEEE 2396

2 2 5 mρ2 times the 3 3 identity matrix. Its velcity v and angular velcity ω yield the fllwing sliding velcity x r at t r sliding s 0 s = v + ρẑ ω. (2) The sphere is subject t a frictinal frce f = µ s mgŝ, where µ s is the cefficient f sliding frictin, g > 0 the magnitude f the gravitatinal acceleratin, and ŝ = s/ s, the sliding directin. Wrking ut the dynamics, we btain the sphere s acceleratin v = µ s gŝ and angular acceleratin ω = 5 2ρ µ sgẑ ŝ. We derive the cntact acceleratin by differentiating (2) and substituting the expressins f v and ω in: ṡ = µ s gŝ µ sgẑ (ẑ ŝ) = 7 2 µ sgŝ, (3) using ẑ ŝ = 0. The abve implies that ṡ is ppsite t s 0, which further implies that s will nt change its directin until it becmes t zer. We refer t ŝ = s 0 / s 0 as the sliding directin. With ŝ being cnstant, we can integrate ṡ, v, and ω t btain s, v, and ω, as derived earlier in [6, pp ]. The sliding trajectry is btained frm ne mre rund f integratin this time f v: x = (x, y) T = v 0 t 1 2 µ sgŝt 2, 0 t t r. (4) Sliding ends (i.e., s = 0) at time t r = 2 s 0 /(7µ s g) with velcity and lcatin given by v r = v 0 µ s g t r ŝ = v s 0, (5) x r = x(t r ) = ( v s 0) tr. (6) B. Pure Rlling At time t r, v r is fund rthgnal t the angular velcity ω r, after taking crss prduct f v with (2) under s = 0. The frictinal frce ppses the directin f v r. Neither ω nr v will change its directin frm t r n. The sphere starts pure rlling withut slip (s = 0). The rlling trajectry is a straight line segment. T derive the deceleratin f the sphere, we adpt Marlw s treatment [6, p. 12] based n the principle f energy dissipatin. Because the sphere rlls n a line, we can nw dente v and ω by their magnitudes v and ω with n ambiguity. Since v = ρω under pure rlling, the kinetic energy f the sphere is E = 1 2 mv (2 5 mρ2) ω 2 = 7 10 mv2. Its dissipatin rate Ė under rlling frictin euals µ r mgv. Frm this, we can shw that the sphere will cme t stp at time t e = t r +7 v r /(5µ r g), alng the fllwing line trajectry trajectry f rlling: x = x r + v r (t t r ) 5 14 µ rgˆr(t t r ) 2, (7) x e at t e rlling Fig. 2. Trajectry f a sphere with unit radius, initial velcities v 0 = (4, 3) T and ω 0 = (14, 25) T, which yield initial sliding velcity s 0 = (29, 17) T. The velcities are drawn as arrws f lengths prprtinal t their magnitudes. Rlling starts at x r = ( , ) T at time t r = and ends at x e = ( , ) T at time t e = The cefficients f frictin are µ s = 0.3 and µ r = 0.5. t r < t t e, where ˆr = v r / v r is the rlling directin. The sphere will be resting at the lcatin x e = x(t e ) = x r + (7 v r /(10µ r g))v r. (8) Euatins (4) and (7) tgether describe the sphere trajectry. An example is shwn in Fig III. TRAJECTORIES TO DESTINATION It is nt difficult t shw that v 0 ω 0 = 0 results in a straight sliding trajectry, using (1) and (4). Under Assumptin 1, a straight trajectry wuld result in cllisin with sme bstacle. Therefre, the case v 0 ω 0 = 0 will nt be cnsidered frm nw n. Therem 1: When v 0 ω 0 0, the sphere slides alng ( a parabla generated frm first rtating y = 1 2 µs g/ v 0 ŝ 2) x 2 abut the rigin thrugh φ+π/2, where ŝ = ( cs φ sin φ), and then translating it by x = (v 0 ŝ/(µ s g)) ( v (v 0 ŝ)ŝ ). Due t limited space, the prf f the therem is mitted, as will thse f ther prpsitin and therems in this paper. As an example, the sphere trajectry in Fig. 2 is a parabla y = x 2 underging a rtatin thrugh the angle and then a translatin by ( , ) T. We wuld like t describe the set f velcities (v 0, ω 0 ) that will take the sphere t rest at a given lcatin x e =. By rewriting (1) and (5) int a matrix euatin, the pair (s 0, v r ) f initial sliding and rlling velcities has a ne-t-ne crrespndence t (v 0, ρẑ ω 0 ), the secnd element f which in turn has a ne-t-ne crrespndence t ω 0 under Assumptin 2 and v 0 ω 0 0. Thus, we need nly cnsider (s 0, v r ). 1 Frm nw n, the sliding segment f a trajectry will be drawn in blue, and the rlling segment will be drawn in red. v 0 ω

3 Next, plug v 0 = 2 7 s 0 + v r, btained frm (5), int (6) and the result int (8) t rewrite = c s ŝ + c rˆr, where c s and c r are nn-negative expressins in terms f s 0 and v r. Prpsitin 2: There exist v 0 and ω 0 with v 0 ω 0 0 that will take the sphere t a stp at if and nly if the cne spanned by ŝ and ˆr cntains in its interir. π β π π Fig. 3. C-space f trajectries T (α, β) that take the sphere frm the rigin t a pint n the psitive x- axis. The space cnsists f tw pen triangular regins bunded by the tw axes and the lines α β = π and β α = π. π α The unit vectrs ŝ and ˆr are respectively determined by their plar angles α and β, referred t as the sliding angle and rlling angle. We can recnstruct the initial velcities v 0 and ω 0 (and thus the trajectry) frm α and β when the cne spanned by ŝ and ˆv cntains in the interir. It can be shwn that the crrespndence thus defined between (α, β) and (v 0, ω 0 ) is ne-t-ne. Nevertheless, the frms f v 0 and ω 0 in terms f α and β are t cmplicated with sines, csines, and suare rts t be up fr an analysis. Dente by T (α, β) the trajectry frm t determined by α and β. Since the trajectry is a parablic segment fllwed by a tangent line segment, it is cnvex and thus either abve the x-axis r belw it except at and. Under Prpsitin 2, we infer that t reach all feasible (α, β) frm the fllwing set: D = (0, π) (α π, 0) ( π, 0) (0, α + π). (9) Pltted in Fig. 3, D is the cnfiguratin space (r C- space) f trajectries, referred t as the trajectry space. Therem 3: Fr any (α, β) D, T (α, β) and T ( α, β) are reflectins f each ther in the x-axis. et R and R be the tw rays riginating respectively frm and in the directin ŝ, and R and R be the tw rays riginating respectively frm and in the directin ˆr. Therem 4: Fr a fixed sliding angle α (0, π), the fllwing statements hld: i) Fr any tw rlling angles β 1, β 2 (α π, 0) with β 2 < β 1, except at and, T (α, β 1 ) lies inside the regin bunded by T (α, β 2 ) and. ii) Every pint p in the interir f the regin bunded by, R, and R lies n exactly ne trajectry T (α, β), β (α π, 0). Therem 5: Fr a fixed rlling angle β ( π, 0), the fllwing statements hld: i) Fr any tw sliding angles α 1, α 2 (0, β + π) with α 1 < α 2, except at and, T (α 1, β) lies inside the regin bunded by T (α 2, β) and. ii) Every pint p in the interir f the regin bunded by, R, and R lies n exactly ne trajectry T (α, β), α (0, β + π). Fig. 4 plts tw grups f trajectries respectively with the same sliding and rlling angles. The values T (α10, π/6) T (π/6, β 1) (a) (b) T (α1, π/6) T (π/6, β 10) Fig. 4. Tw grups f 10 trajectries each frm the rigin t (3, 0) with (a) the same sliding angle α = π/6 but different rlling angles β i = i (π α), 1 i 10; and (b) the same rlling angle 11 β = π/6 but different sliding angles α i = i (β+π), 1 i µ s = 0.3 and µ r = 0.15 f the cefficients f frictin will als be used in the remaining examples in the paper. IV. MOTION PANNING Accrding t Therem 3, we need nly cnsider hw t find a cllisin-free path frm t that ges abve the x-axis. Such a path is represented by a pint in the lwer triangular regin D in Fig. 3. This implies that we need nly cnsider bstacles that are abve r intersected by the x-axis. T find the trajectry belw the x-axis, simply reflect all bstacles in the x-axis, invke the algrithm and then reflect the fund path in the x- axis again. T simplify, bstacles are vertical line segments standing n the xy-plane at the pints b 1,...,b n and having length at least the radius ρ f the sphere. In the C-space fr path planning, the sphere shrinks t a pint, and every bstacle induces a disk C i at b i f radius ρ. The disk is a C-space bstacle referred t as a C-circle

4 A. Cnstraint Curves in the C-Space f Trajectries Every C-circle C i defines sme regin in the trajectry space D that includes all the trajectries which wuld cllide with the bstacle b i. Such a regin is bunded by curves n which every pint (α, β) yields a trajectry T (α, β) tangent t C i. The bstacles nt belw the x-axis are either A) abve the x-axis; r intersecting the x-axis B) between and, C) t the left f, r D) t the right f. 1) Type A Obstacle: T simplify ntatin, we refer t b i as b and C i as C. There exist tw grups f tangent trajectries: thse that bund C frm abve and thse that bund it frm belw. T characterize them, we nte that there are tw tangent lines frm t C at p + frm the abve and at p frm the belw. Similarly, the tw tangent lines frm t C has tangencies p + and p. Every upper-bunding tangent trajectry has a tangency pint between p + and p+ n C. As illustrated in Fig. 5(a), the maximum β value, dented β max, + is the plar angle f the vectr p +. Frm Fig. 3, α β max + π α max. Meanwhile, the α value f the trajectry must be bunded frm belw by the plar angle f p +, dented α+ min. As α increases frm α+ min (exclusive), the tangent trajectry T (α, β) will have β increase frm α + min π (exclusive). This is the case with T 1 in Fig. 5(a). Eventually, α will increase t a value α + mid where the sphere starts rlling right at p+, as in the case f T 2 = T (α + mid, β+ mid ) in the figure. As α cntinues t increase, the β value will stay at β + mid, which is the case f T 3. All (α, β) values, where T (α, β) is an upperbunding tangent trajectry frm the curve γ + in D pltted in Fig. 5(c). The interir f the regin bunded by the curve and the line α β = π includes all trajectries that will enclse the bstacle b. The secnd grup f tangent trajectries bund the bstacle frm belw, as illustrated in Fig. 5(b). All such tangent trajectries are represented by the curve γ in Fig. 5(c). The interir f the regin in D bunded by the curve and the α- and β-axes represent all cllisinfree trajectries belw the bstacle. The shaded regin in Fig. 5(c) includes all trajectries that will cllide with b. 2) Type B Obstacle: The prtin f the C-circle abve the x-axis intrduces a cnstraint curve in D like γ + in Fig. 5(c). Cllisin-free trajectries are represented by the pen regin enclsed by the curve and the line α β = π. 3) Type C and D Obstacles: Cnstraint curves induced by bstacles f these tw types are very similar with differences in their lcatins in D. We discuss a type D bstacle but will present a type C ne later. b3 b1 b2 b5 (a) b4 γ 5 γ 3 (b) γ 2 γ 1 γ 4 α β = π Fig. 6. (a) Wrk space fr a unit sphere mving frm t = (8, 0) with five bstacles: b 1 = (3, 3), b 2 = (4, 1/2), b 3 = ( 12, 3/10), b 4 = (19/2, 1/2), and b 5 = (4, 3), which are shwn as the centers f C-space disks f radius 1. (b) Cnfiguratin space including five cnstraint curves induced by b k, 1 k 4. The blue curves γ 1 and γ 3 allw pints α, β (i.e., T (α, β) nt clliding their defining bstacles) belw them. The red curves γ 2, γ 4, and γ 3 allw pints abve them. Als shwn in (b) is a sweep line. Fig. 5(d) shws the same destinatin with a type D bstacle at ( 19 2, 1 2 ). Here, p is the pint f tangency frm n the C-circle that is abve the x-axis. It is easy t see that any tangent trajectry T (α, β) must have β β max, the plar angle f p. As α increases frm 0 +, the pint f tangency mves frm infinitely clse t p x, where the circle intersects the x-axis, t p (e.g., the trajectry T 1 ). The trajectry T 2 = T (α min, β min ) starts rlling at p. Frm that pint, β stps changing as fr the trajectry T 3, while α increases t β max + π. The cnstraint curve is shwn in Fig. 5(e). Therem 6: A cnstraint curve starts and ends n the line α β = π with α increasing mntnically and β nn-decreasing. In the trajectry space D, let us refer left, right, belw, and abve t the directins f the vectrs (1, 1), ( 1, 1), ( 1, 1), and (1, 1), respectively. B. Cnfiguratin Space Fig. 6(a) shws an example with five bstacles. The bstacles b 1 and b 4 are exactly the same as in Fig. 5(a) and (d). The bstacle b 5 is belw the rigin and eliminated when planning trajectries abve the x-axis. The ther fur bstacles are f types A, B, C, and D, respectively. Fig. 6(b) plts the cnstraint curves γ j, 1 j 5. The first tw f which are induced by b 1, and fr 2 k 4, γ k+1 is induced by b k. The cnstraint curves partitin D int pen regins. Every pint (α, β) frm the same regin vilate the same set f cnstraints (i.e., T (α, β) cllides with the same set f bstacles). The planning prblem has a slutin if and nly if there exists at least ne pen regin that vilates n cnstraint at all

5 T 3 T 1 T 2 p 1 p + p + b T 3 b C T 2 p p 1 p β β + max + π α + min (a) β + max β max + π T 1 α min (b) β max α T 3 D (α mid, β mid ) (α max, β max) (α + mid, β+ mid ) (α + max, β + max) γ γ + (α min, β min ) (α + min, β+ min ) α β = π T 1 T 2 p p 1 p x b (α mid, β mid) (α max, β max) γ β max (α min, β min) (c) (d) (e) Fig. 5. Tangent trajectries and C-space bstacles (ρ = 1). (a) Three tangent trajectries enclsing an bstacle disk f type A at (3, 3). (b) Three tangent trajectries belw the disk. (c) In the C-space, the curve γ + represents the set f all tangent and enclsing trajectries and the curve γ represents the set f all tangent and nn-enclsing trajectries. Nte that β + mid = β+ max and β mid = β max. (d) Tangent trajectries t an bstacle f Type D at (19/2, 1/2). (e) Cnstraint curve induced by the bstacle in (d) in the dmain D. The shaded regins in (c) and (e) represent all trajectries respectively clliding with the tw bstacles. C. Plane Sweeping We emply a plane sweep techniue frm cmputatinal gemetry [3]. A line initially aligned with α β = π will sweep acrss D, handling events at stps during the sweep. At an intermediate psitin, the line is partitined int segments by the cnstraint curves. Its status is a list f cnstraint curves rdered alng the sweep line, starting with the β-axis and ending with the α-axis. Each curve has a superscript r + t indicate that pints t the left r right f the curve yield trajectries nt clliding with the curve s defining bstacle. When intersected by, a cnstraint curve appears twice in the status with different superscripts. The sweep line status is stred in a data structure such as a red-black tree [4] r a splay tree [10]. Dente by s i the ith segment frm left t right n the sweep line. An array c[ ] is used such that the entry c[i] cunts the bstacles clliding with the trajectry defined by any pint in the regin immediately abve s i. D. Event Handling An event happens when the sweep line passes by either an intersectin pint f tw cnstraint curves r a pint n sme cnstraint curve that is the clsest t the rigin. A dynamically updated event ueue Q rders events in the increasing distance f the event pint t the line α β = π. Suppse the segment s k n at its current psitin is bunded by the curve γ a l n the left and the curve γ b r n the right. After the intersectin, the segment s k will be bunded by γ b r n the left and γ a l n the right. One f the fllwing updates is perfrmed based n (a, b): (, +): c[k] c[k] 2. (+, ): c[k] c[k] + 2. (, ): n change t c[k]. (+, +): n change t c[k]

6 e k e k e k γ 2 b2 γ 1 (a) (b) (c) e k (d) γ i ek Fig. 7. Fur types f intersectin pints: (a) (, +), (b) (+, ), (c) (, ), and (d) (+, +), and (e) an extremum pint. A slid arrw shws the sweeping directin, while a hllw arrw marks the side where pints define trajectries nt in cllisin with the bstacle assciated with the curve. See Fig. 7 (a) (d). Since γ a l and γ b r have switched their rder in the sweep line status, their neighbring curves in the status change. We need t check whether will intersect its new preceding curve and whether will intersect its new succeeding curve in the sweep line status. Whenever an intersectin exists, add its cmputed psitin (alng with the tw curves) t the event ueue Q. A segment s k n bunded by the same curve γ i may shrink t ne extreme pint n it. When this happens, we lcate the tw adjacent segments s k 1 and s k+1, and merge them tgether with s k int ne bunded by the left bunding curve f s k 1 and the right bunding curve f s k+1. N update n c[k 1] needs t be dne. Check if these tw curves will intersect. If s, add the intersectin t the ueue Q. If c[k] = 0, the regin immediately abve s k will be a cllisin-free ne. The entire regin can be traced ut as the sweep cntinues using a dubly-cnnected edge list [3, pp ]. E. Initializatin At the beginning, the sweep line cincides with the line α β = π. Fr each cnstraint curve, we cmpute its extremum pint and insert it int the event ueue Q. Every cnstraint curve yields tw entries in the sweep line status respectively assciated with its tw endpints. In the rdering, cmplicatins arise with multiple bstacles f type C r multiple bstacles f type D. The curves defined by multiple bstacles f type D all start at (0, π), as in Fig. 8 which shws an instance with tw bstacles defining the cnstraint curves γ 1 and γ 2. These curves can be rdered by their intersectins with a hrizntal line β = π +ǫ fr small enugh ǫ > (e) b1 Fig. 8. Tw cnstraint curves γ 1 and γ 2, due t bstacles b 1 and b 2, respectively, start at the lwest pint (0, π) f D. γ 5 e 6 e 5 γ + 3 e 9 γ 2 e 4 e 8 γ + γ e 7 e 1 γ + 4 e γ 3 e γ γ γ 4 1 Fig. 9. Shaded regins in the C-space represent all cllisin-free paths fr the envirnment in Fig. 6(a). 0. Apply Therem 5, we infer that the α value increase with the x-crdinates f the bstacles. Cnseuently, the curve γ i lies t the right f the curve γ j in D if b i has a larger x-crdinate than b j des. In Fig. 8, γ 2 lies t the right f γ 1. All the curves f type D are thus rdered. The cnstraint curves induced by type C bstacles all end at the pint (π, 0). Similarly, these curves appear in the status in the decreasing rder f the x-crdinates f the bstacles. Up t 2n, where n is the number f bstacles, cnstraint curves with superscripts, tgether with the α- and β-axes, are rdered. The array c[ ] can be initialized with a left-t-right scan alng the line α β = π as fllws. We let c[0] t be the number f bstacles f type B. These bstacles are the nly nes that define curves t exclude the pint (0, π). If the kth scanned curve has superscript, set c[k] c[k 1]+1. If the curve has superscript +, set c[k] c[k 1] 1. Table I details sweeping f the C-space in Fig. 9. Intervals n the sweep line are represented by their left bunding curves. The initial sweep line status is described in the secnd rw. Frm c[6] = 0 we see that the regin right abve the 6-th segment is cllisin free. Every rw starting at the third ne represents the updated status immediately after an event. T illustrate, the event e 1 is due t an intersectin between γ

7 i s i, c[i] y-axis, 1 γ 5, 2 γ+ 3, 1 γ 2, 2 γ+ 5,1 γ+ 1,0 γ 1, 1 γ 4,2 γ+ 2,1 γ 3,2 γ+ 4, 1 e 1 γ + 4,0 γ 3, 1 e 2 γ + 2,0 γ 4, 1 e 3 γ + 2,0 e 4 γ + 5,0 γ 2,1 e 5 γ + 3,0 γ 5,1 e 6 γ + 3, 0 e 7 γ 2,1 e 8 γ + 3,0 e 9 y-axis, 1 TABE I EXECUTION OVER THE C-SPACE IN FIG. 9. BANK ENTRIES ARE THE SAME AS IN THE ROWS IMMEDIATEY ABOVE. and γ + 4, which bund the tenth segment. After the intersectin, the tw curves switching their psitins. The bstacles b 3 and b 1 will nt cllide with paths in the regin bunded by γ + 2 and γ 4. The event e 3 happens at an extremum pint n the curve γ 4. In the previus rw labeled by e 2, γ 4 and γ + 4 are next t each ther. At e 3, the tw intervals leftbunded by them alng with the preceding interval leftbunded by γ + 2 are merged int ne labeled by γ+ 2. The regin immediately abve the segment is cllisin free. When the sweep finishes, the set f cllisin-free paths fr the sphere is the unin f the tw shaded pen regins shwn in Fig. 9. The plane sweep algrithm des nt explicitly cnstruct the cnstraint curves which d nt have algebraic frms. Efficient numerical subrutines have been designed t a) lcate a pint n a cnstraint curve given an α r β value; b) cmpute the starting, ending, and cnnectin pints n a cnstraint curve; c) find the extremum pint n a cnstraint curve; and d) intersect tw cnstraint curves. These subrutines emply bisectin r gld sectin search [9], by expliting the mntnicity f trajectries as stated in Therems 4 and 5. Under a specified accuracy, a call f any numerical subrutine abve takes cnstant time. Then, the running time f the sweep line algrithm fr n bstacles is easily analyzed t be O((n+n e )lg n), where n e is the number f events (which is O(n 2 ) in the wrst case). F. Discussin T have a gd clearance frm the bstacles, we may pick a pint (α, β) that lies clse t the center f a regin f cllisin-free trajectries in D. Cmpute the velcities (v 0, ω 0 ) frm α and β. Thugh presented fr vertical line bstacles, the algrithm trivially extends t vertical cylindrical bstacles. Extensin f the algrithm t plyhedral r ther generally shaped bstacles needs further investigatin. The presented wrk intrduces a new way f path planning which takes advantage f sliding under frictin. T apply this techniue t navigatin by spherical rbts, tw issues await investigatin. First, nn-unifrm mass distributin f such a rbt cmplicates the angular inertial matrix t change dynamic euatins significantly. Secnd, actuatin t generate a specified initial mtin can be uite challenging. REFERENCES [1] S. Bhattacharya and S. K. Agrawal. Spherical rlling rbt: a design and mtin planning studies, IEEE Trans. n Rbt. Autmat., vl. 16, pp , [2] A. Bicchi, A. Balluchi, D. Prattichizz, and A. Grelli. Intrducing the spherical: An experimental testbed fr research and teaching in nn-hlnmy, in Prc. IEEE Int. Cnf. Rbt. Autm., pp , [3] M. de Berg, M. van Kreveld, M. Overmars, and O. Schwarzkpf. Cmputatinal Gemetry: Algrithms and Applicatins, 3rd editin, Springer-Verlag, [4] T. H. Crmen, C. E. eisersn, R.. Rivest, and C. Stein. Intrductin t Algrithms, 3rd editin, The MIT Press, [5] Z. i and J. Canny. Mtin f tw rigid bdies with rlling cnstraint, IEEE Trans. n Rbt. Autmat., vl. 6, pp. 72, [6] W. C. Marlw. The Physics f Pcket Billiards, Marlw Advanced Systems Technlgies, [7] R. Mukherjee, M. A. Minr, and J. T. Pukrushpan. Mtin planning fr a spherical mbile rbt: Revising the classcial ballplate prblem, Trans. f ASME, vl. 124, pp , [8] A. Nakashima, K. Nagase, and Y. Hayakawa. Cntrl f a sphere rlling n a plane with cnstrained rlling mtin, in Prc. IEEE Cnf. Decisin and Cntrl, 2005, pp [9] W. H. Press, S. A. Teuklsky, W. T. Vetterling, and B. P. Flannery. Numerical Recipes in C++: The Art f Scientific Cmputing, 2nd editin, Cambridge University Press, [10] D. D. Sleatr and R. E. Tarjan. Self-adjusting binary search trees, J. ACM, vl. 32, pp , [11] M. Svinin and S. Hse. On mtin planning fr ballplate systems with limited cntact area, in Prc. IEEE Int. Cnf. Rbt. Autm., 2007, pp

Figure 1a. A planar mechanism.

Figure 1a. A planar mechanism. ME 5 - Machine Design I Fall Semester 0 Name f Student Lab Sectin Number EXAM. OPEN BOOK AND CLOSED NOTES. Mnday, September rd, 0 Write n ne side nly f the paper prvided fr yur slutins. Where necessary,

More information

Homology groups of disks with holes

Homology groups of disks with holes Hmlgy grups f disks with hles THEOREM. Let p 1,, p k } be a sequence f distinct pints in the interir unit disk D n where n 2, and suppse that fr all j the sets E j Int D n are clsed, pairwise disjint subdisks.

More information

Phys101 Final Code: 1 Term: 132 Wednesday, May 21, 2014 Page: 1

Phys101 Final Code: 1 Term: 132 Wednesday, May 21, 2014 Page: 1 Phys101 Final Cde: 1 Term: 1 Wednesday, May 1, 014 Page: 1 Q1. A car accelerates at.0 m/s alng a straight rad. It passes tw marks that are 0 m apart at times t = 4.0 s and t = 5.0 s. Find the car s velcity

More information

Building to Transformations on Coordinate Axis Grade 5: Geometry Graph points on the coordinate plane to solve real-world and mathematical problems.

Building to Transformations on Coordinate Axis Grade 5: Geometry Graph points on the coordinate plane to solve real-world and mathematical problems. Building t Transfrmatins n Crdinate Axis Grade 5: Gemetry Graph pints n the crdinate plane t slve real-wrld and mathematical prblems. 5.G.1. Use a pair f perpendicular number lines, called axes, t define

More information

14. Which shows the direction of the centripetal force acting on a mass spun in a vertical circle?

14. Which shows the direction of the centripetal force acting on a mass spun in a vertical circle? Physics 0 Public Exam Questins Unit 1: Circular Mtin NAME: August 009---------------------------------------------------------------------------------------------------------------------- 1. Which describes

More information

14. Which shows the direction of the centripetal force acting on a mass spun in a vertical circle?

14. Which shows the direction of the centripetal force acting on a mass spun in a vertical circle? Physics 3204 Public Exam Questins Unit 1: Circular Mtin NAME: August 2009---------------------------------------------------------------------------------------------------------------------- 12. Which

More information

MODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b

MODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b . REVIEW OF SOME BASIC ALGEBRA MODULE () Slving Equatins Yu shuld be able t slve fr x: a + b = c a d + e x + c and get x = e(ba +) b(c a) d(ba +) c Cmmn mistakes and strategies:. a b + c a b + a c, but

More information

ENGI 4430 Parametric Vector Functions Page 2-01

ENGI 4430 Parametric Vector Functions Page 2-01 ENGI 4430 Parametric Vectr Functins Page -01. Parametric Vectr Functins (cntinued) Any nn-zer vectr r can be decmpsed int its magnitude r and its directin: r rrˆ, where r r 0 Tangent Vectr: dx dy dz dr

More information

SPH3U1 Lesson 06 Kinematics

SPH3U1 Lesson 06 Kinematics PROJECTILE MOTION LEARNING GOALS Students will: Describe the mtin f an bject thrwn at arbitrary angles thrugh the air. Describe the hrizntal and vertical mtins f a prjectile. Slve prjectile mtin prblems.

More information

Phys102 Final-061 Zero Version Coordinator: Nasser Wednesday, January 24, 2007 Page: 1

Phys102 Final-061 Zero Version Coordinator: Nasser Wednesday, January 24, 2007 Page: 1 Crdinatr: Nasser Wednesday, January 4, 007 Page: 1 Q1. Tw transmitters, S 1 and S shwn in the figure, emit identical sund waves f wavelength λ. The transmitters are separated by a distance λ /. Cnsider

More information

Differentiation Applications 1: Related Rates

Differentiation Applications 1: Related Rates Differentiatin Applicatins 1: Related Rates 151 Differentiatin Applicatins 1: Related Rates Mdel 1: Sliding Ladder 10 ladder y 10 ladder 10 ladder A 10 ft ladder is leaning against a wall when the bttm

More information

Kinetics of Particles. Chapter 3

Kinetics of Particles. Chapter 3 Kinetics f Particles Chapter 3 1 Kinetics f Particles It is the study f the relatins existing between the frces acting n bdy, the mass f the bdy, and the mtin f the bdy. It is the study f the relatin between

More information

AP Physics Kinematic Wrap Up

AP Physics Kinematic Wrap Up AP Physics Kinematic Wrap Up S what d yu need t knw abut this mtin in tw-dimensin stuff t get a gd scre n the ld AP Physics Test? First ff, here are the equatins that yu ll have t wrk with: v v at x x

More information

37 Maxwell s Equations

37 Maxwell s Equations 37 Maxwell s quatins In this chapter, the plan is t summarize much f what we knw abut electricity and magnetism in a manner similar t the way in which James Clerk Maxwell summarized what was knwn abut

More information

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System Flipping Physics Lecture Ntes: Simple Harmnic Mtin Intrductin via a Hrizntal Mass-Spring System A Hrizntal Mass-Spring System is where a mass is attached t a spring, riented hrizntally, and then placed

More information

Chapter 2 GAUSS LAW Recommended Problems:

Chapter 2 GAUSS LAW Recommended Problems: Chapter GAUSS LAW Recmmended Prblems: 1,4,5,6,7,9,11,13,15,18,19,1,7,9,31,35,37,39,41,43,45,47,49,51,55,57,61,6,69. LCTRIC FLUX lectric flux is a measure f the number f electric filed lines penetrating

More information

Chapter 3 Kinematics in Two Dimensions; Vectors

Chapter 3 Kinematics in Two Dimensions; Vectors Chapter 3 Kinematics in Tw Dimensins; Vectrs Vectrs and Scalars Additin f Vectrs Graphical Methds (One and Tw- Dimensin) Multiplicatin f a Vectr b a Scalar Subtractin f Vectrs Graphical Methds Adding Vectrs

More information

Rigid Body Dynamics (continued)

Rigid Body Dynamics (continued) Last time: Rigid dy Dynamics (cntinued) Discussin f pint mass, rigid bdy as useful abstractins f reality Many-particle apprach t rigid bdy mdeling: Newtn s Secnd Law, Euler s Law Cntinuus bdy apprach t

More information

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System Flipping Physics Lecture Ntes: Simple Harmnic Mtin Intrductin via a Hrizntal Mass-Spring System A Hrizntal Mass-Spring System is where a mass is attached t a spring, riented hrizntally, and then placed

More information

CHAPTER 8b Static Equilibrium Units

CHAPTER 8b Static Equilibrium Units CHAPTER 8b Static Equilibrium Units The Cnditins fr Equilibrium Slving Statics Prblems Stability and Balance Elasticity; Stress and Strain The Cnditins fr Equilibrium An bject with frces acting n it, but

More information

Q1. In figure 1, Q = 60 µc, q = 20 µc, a = 3.0 m, and b = 4.0 m. Calculate the total electric force on q due to the other 2 charges.

Q1. In figure 1, Q = 60 µc, q = 20 µc, a = 3.0 m, and b = 4.0 m. Calculate the total electric force on q due to the other 2 charges. Phys10 Secnd Majr-08 Zer Versin Crdinatr: Dr. I. M. Nasser Saturday, May 3, 009 Page: 1 Q1. In figure 1, Q = 60 µc, q = 0 µc, a = 3.0 m, and b = 4.0 m. Calculate the ttal electric frce n q due t the ther

More information

Chapter 9 Vector Differential Calculus, Grad, Div, Curl

Chapter 9 Vector Differential Calculus, Grad, Div, Curl Chapter 9 Vectr Differential Calculus, Grad, Div, Curl 9.1 Vectrs in 2-Space and 3-Space 9.2 Inner Prduct (Dt Prduct) 9.3 Vectr Prduct (Crss Prduct, Outer Prduct) 9.4 Vectr and Scalar Functins and Fields

More information

Chapter 5: Force and Motion I-a

Chapter 5: Force and Motion I-a Chapter 5: rce and Mtin I-a rce is the interactin between bjects is a vectr causes acceleratin Net frce: vectr sum f all the frces n an bject. v v N v v v v v ttal net = i = + + 3 + 4 i= Envirnment respnse

More information

Phys101 Second Major-061 Zero Version Coordinator: AbdelMonem Saturday, December 09, 2006 Page: 1

Phys101 Second Major-061 Zero Version Coordinator: AbdelMonem Saturday, December 09, 2006 Page: 1 Crdinatr: AbdelMnem Saturday, December 09, 006 Page: Q. A 6 kg crate falls frm rest frm a height f.0 m nt a spring scale with a spring cnstant f.74 0 3 N/m. Find the maximum distance the spring is cmpressed.

More information

Kinematic transformation of mechanical behavior Neville Hogan

Kinematic transformation of mechanical behavior Neville Hogan inematic transfrmatin f mechanical behavir Neville Hgan Generalized crdinates are fundamental If we assume that a linkage may accurately be described as a cllectin f linked rigid bdies, their generalized

More information

Introduction: A Generalized approach for computing the trajectories associated with the Newtonian N Body Problem

Introduction: A Generalized approach for computing the trajectories associated with the Newtonian N Body Problem A Generalized apprach fr cmputing the trajectries assciated with the Newtnian N Bdy Prblem AbuBar Mehmd, Syed Umer Abbas Shah and Ghulam Shabbir Faculty f Engineering Sciences, GIK Institute f Engineering

More information

GAUSS' LAW E. A. surface

GAUSS' LAW E. A. surface Prf. Dr. I. M. A. Nasser GAUSS' LAW 08.11.017 GAUSS' LAW Intrductin: The electric field f a given charge distributin can in principle be calculated using Culmb's law. The examples discussed in electric

More information

Conceptual Dynamics SDC. An Interactive Text and Workbook. Kirstie Plantenberg Richard Hill. Better Textbooks. Lower Prices.

Conceptual Dynamics SDC. An Interactive Text and Workbook. Kirstie Plantenberg Richard Hill. Better Textbooks. Lower Prices. Cnceptual Dynamics An Interactive Text and Wrkbk Kirstie Plantenberg Richard Hill SDC P U B L I C AT I O N S Better Textbks. Lwer Prices. www.sdcpublicatins.cm Pwered by TCPDF (www.tcpdf.rg) Visit the

More information

Yeu-Sheng Paul Shiue, Ph.D 薛宇盛 Professor and Chair Mechanical Engineering Department Christian Brothers University 650 East Parkway South Memphis, TN

Yeu-Sheng Paul Shiue, Ph.D 薛宇盛 Professor and Chair Mechanical Engineering Department Christian Brothers University 650 East Parkway South Memphis, TN Yeu-Sheng Paul Shiue, Ph.D 薛宇盛 Prfessr and Chair Mechanical Engineering Department Christian Brthers University 650 East Parkway Suth Memphis, TN 38104 Office: (901) 321-3424 Rm: N-110 Fax : (901) 321-3402

More information

Equilibrium of Stress

Equilibrium of Stress Equilibrium f Stress Cnsider tw perpendicular planes passing thrugh a pint p. The stress cmpnents acting n these planes are as shwn in ig. 3.4.1a. These stresses are usuall shwn tgether acting n a small

More information

205MPa and a modulus of elasticity E 207 GPa. The critical load 75kN. Gravity is vertically downward and the weight of link 3 is W3

205MPa and a modulus of elasticity E 207 GPa. The critical load 75kN. Gravity is vertically downward and the weight of link 3 is W3 ME 5 - Machine Design I Fall Semester 06 Name f Student: Lab Sectin Number: Final Exam. Open bk clsed ntes. Friday, December 6th, 06 ur name lab sectin number must be included in the spaces prvided at

More information

making triangle (ie same reference angle) ). This is a standard form that will allow us all to have the X= y=

making triangle (ie same reference angle) ). This is a standard form that will allow us all to have the X= y= Intrductin t Vectrs I 21 Intrductin t Vectrs I 22 I. Determine the hrizntal and vertical cmpnents f the resultant vectr by cunting n the grid. X= y= J. Draw a mangle with hrizntal and vertical cmpnents

More information

Plan o o. I(t) Divide problem into sub-problems Modify schematic and coordinate system (if needed) Write general equations

Plan o o. I(t) Divide problem into sub-problems Modify schematic and coordinate system (if needed) Write general equations STAPLE Physics 201 Name Final Exam May 14, 2013 This is a clsed bk examinatin but during the exam yu may refer t a 5 x7 nte card with wrds f wisdm yu have written n it. There is extra scratch paper available.

More information

Example 1. A robot has a mass of 60 kg. How much does that robot weigh sitting on the earth at sea level? Given: m. Find: Relationships: W

Example 1. A robot has a mass of 60 kg. How much does that robot weigh sitting on the earth at sea level? Given: m. Find: Relationships: W Eample 1 rbt has a mass f 60 kg. Hw much des that rbt weigh sitting n the earth at sea level? Given: m Rbt = 60 kg ind: Rbt Relatinships: Slutin: Rbt =589 N = mg, g = 9.81 m/s Rbt = mrbt g = 60 9. 81 =

More information

Q1. A) 48 m/s B) 17 m/s C) 22 m/s D) 66 m/s E) 53 m/s. Ans: = 84.0 Q2.

Q1. A) 48 m/s B) 17 m/s C) 22 m/s D) 66 m/s E) 53 m/s. Ans: = 84.0 Q2. Phys10 Final-133 Zer Versin Crdinatr: A.A.Naqvi Wednesday, August 13, 014 Page: 1 Q1. A string, f length 0.75 m and fixed at bth ends, is vibrating in its fundamental mde. The maximum transverse speed

More information

Fall 2013 Physics 172 Recitation 3 Momentum and Springs

Fall 2013 Physics 172 Recitation 3 Momentum and Springs Fall 03 Physics 7 Recitatin 3 Mmentum and Springs Purpse: The purpse f this recitatin is t give yu experience wrking with mmentum and the mmentum update frmula. Readings: Chapter.3-.5 Learning Objectives:.3.

More information

Sections 15.1 to 15.12, 16.1 and 16.2 of the textbook (Robbins-Miller) cover the materials required for this topic.

Sections 15.1 to 15.12, 16.1 and 16.2 of the textbook (Robbins-Miller) cover the materials required for this topic. Tpic : AC Fundamentals, Sinusidal Wavefrm, and Phasrs Sectins 5. t 5., 6. and 6. f the textbk (Rbbins-Miller) cver the materials required fr this tpic.. Wavefrms in electrical systems are current r vltage

More information

. (7.1.1) This centripetal acceleration is provided by centripetal force. It is directed towards the center of the circle and has a magnitude

. (7.1.1) This centripetal acceleration is provided by centripetal force. It is directed towards the center of the circle and has a magnitude Lecture #7-1 Dynamics f Rtatin, Trque, Static Equilirium We have already studied kinematics f rtatinal mtin We discussed unifrm as well as nnunifrm rtatin Hwever, when we mved n dynamics f rtatin, the

More information

CHAPTER 6 -- ENERGY. Approach #2: Using the component of mg along the line of d:

CHAPTER 6 -- ENERGY. Approach #2: Using the component of mg along the line of d: Slutins--Ch. 6 (Energy) CHAPTER 6 -- ENERGY 6.) The f.b.d. shwn t the right has been prvided t identify all the frces acting n the bdy as it mves up the incline. a.) T determine the wrk dne by gravity

More information

Pattern Recognition 2014 Support Vector Machines

Pattern Recognition 2014 Support Vector Machines Pattern Recgnitin 2014 Supprt Vectr Machines Ad Feelders Universiteit Utrecht Ad Feelders ( Universiteit Utrecht ) Pattern Recgnitin 1 / 55 Overview 1 Separable Case 2 Kernel Functins 3 Allwing Errrs (Sft

More information

Information for Physics 1201 Midterm I Wednesday, February 20

Information for Physics 1201 Midterm I Wednesday, February 20 My lecture slides are psted at http://www.physics.hi-state.edu/~humanic/ Infrmatin fr Physics 1201 Midterm I Wednesday, February 20 1) Frmat: 10 multiple chice questins (each wrth 5 pints) and tw shw-wrk

More information

and the Doppler frequency rate f R , can be related to the coefficients of this polynomial. The relationships are:

and the Doppler frequency rate f R , can be related to the coefficients of this polynomial. The relationships are: Algrithm fr Estimating R and R - (David Sandwell, SIO, August 4, 2006) Azimith cmpressin invlves the alignment f successive eches t be fcused n a pint target Let s be the slw time alng the satellite track

More information

Cambridge Assessment International Education Cambridge Ordinary Level. Published

Cambridge Assessment International Education Cambridge Ordinary Level. Published Cambridge Assessment Internatinal Educatin Cambridge Ordinary Level ADDITIONAL MATHEMATICS 4037/1 Paper 1 Octber/Nvember 017 MARK SCHEME Maximum Mark: 80 Published This mark scheme is published as an aid

More information

Department of Economics, University of California, Davis Ecn 200C Micro Theory Professor Giacomo Bonanno. Insurance Markets

Department of Economics, University of California, Davis Ecn 200C Micro Theory Professor Giacomo Bonanno. Insurance Markets Department f Ecnmics, University f alifrnia, Davis Ecn 200 Micr Thery Prfessr Giacm Bnann Insurance Markets nsider an individual wh has an initial wealth f. ith sme prbability p he faces a lss f x (0

More information

Lecture 5: Equilibrium and Oscillations

Lecture 5: Equilibrium and Oscillations Lecture 5: Equilibrium and Oscillatins Energy and Mtin Last time, we fund that fr a system with energy cnserved, v = ± E U m ( ) ( ) One result we see immediately is that there is n slutin fr velcity if

More information

CHAPTER 3 INEQUALITIES. Copyright -The Institute of Chartered Accountants of India

CHAPTER 3 INEQUALITIES. Copyright -The Institute of Chartered Accountants of India CHAPTER 3 INEQUALITIES Cpyright -The Institute f Chartered Accuntants f India INEQUALITIES LEARNING OBJECTIVES One f the widely used decisin making prblems, nwadays, is t decide n the ptimal mix f scarce

More information

Electric Current and Resistance

Electric Current and Resistance Electric Current and Resistance Electric Current Electric current is the rate f flw f charge thrugh sme regin f space The SI unit f current is the ampere (A) 1 A = 1 C / s The symbl fr electric current

More information

Module 4: General Formulation of Electric Circuit Theory

Module 4: General Formulation of Electric Circuit Theory Mdule 4: General Frmulatin f Electric Circuit Thery 4. General Frmulatin f Electric Circuit Thery All electrmagnetic phenmena are described at a fundamental level by Maxwell's equatins and the assciated

More information

Surface and Contact Stress

Surface and Contact Stress Surface and Cntact Stress The cncept f the frce is fundamental t mechanics and many imprtant prblems can be cast in terms f frces nly, fr example the prblems cnsidered in Chapter. Hwever, mre sphisticated

More information

Faculty of Engineering and Department of Physics Engineering Physics 131 Midterm Examination February 27, 2006; 7:00 pm 8:30 pm

Faculty of Engineering and Department of Physics Engineering Physics 131 Midterm Examination February 27, 2006; 7:00 pm 8:30 pm Faculty f Engineering and Department f Physics Engineering Physics 131 Midterm Examinatin February 27, 2006; 7:00 pm 8:30 pm N ntes r textbks allwed. Frmula sheet is n the last page (may be remved). Calculatrs

More information

Lyapunov Stability Stability of Equilibrium Points

Lyapunov Stability Stability of Equilibrium Points Lyapunv Stability Stability f Equilibrium Pints 1. Stability f Equilibrium Pints - Definitins In this sectin we cnsider n-th rder nnlinear time varying cntinuus time (C) systems f the frm x = f ( t, x),

More information

Q1. A string of length L is fixed at both ends. Which one of the following is NOT a possible wavelength for standing waves on this string?

Q1. A string of length L is fixed at both ends. Which one of the following is NOT a possible wavelength for standing waves on this string? Term: 111 Thursday, January 05, 2012 Page: 1 Q1. A string f length L is fixed at bth ends. Which ne f the fllwing is NOT a pssible wavelength fr standing waves n this string? Q2. λ n = 2L n = A) 4L B)

More information

Introduction to Smith Charts

Introduction to Smith Charts Intrductin t Smith Charts Dr. Russell P. Jedlicka Klipsch Schl f Electrical and Cmputer Engineering New Mexic State University as Cruces, NM 88003 September 2002 EE521 ecture 3 08/22/02 Smith Chart Summary

More information

3D FE Modeling Simulation of Cold Rotary Forging with Double Symmetry Rolls X. H. Han 1, a, L. Hua 1, b, Y. M. Zhao 1, c

3D FE Modeling Simulation of Cold Rotary Forging with Double Symmetry Rolls X. H. Han 1, a, L. Hua 1, b, Y. M. Zhao 1, c Materials Science Frum Online: 2009-08-31 ISSN: 1662-9752, Vls. 628-629, pp 623-628 di:10.4028/www.scientific.net/msf.628-629.623 2009 Trans Tech Publicatins, Switzerland 3D FE Mdeling Simulatin f Cld

More information

Work, Energy, and Power

Work, Energy, and Power rk, Energy, and Pwer Physics 1 There are many different TYPES f Energy. Energy is expressed in JOULES (J 419J 4.19 1 calrie Energy can be expressed mre specifically by using the term ORK( rk The Scalar

More information

39th International Physics Olympiad - Hanoi - Vietnam Theoretical Problem No. 1 /Solution. Solution

39th International Physics Olympiad - Hanoi - Vietnam Theoretical Problem No. 1 /Solution. Solution 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin Slutin. The structure f the mrtar.. Calculating the distance TG The vlume f water in the bucket is V = = 3 3 3 cm m.

More information

The Sputtering Problem James A Glackin, James V. Matheson

The Sputtering Problem James A Glackin, James V. Matheson The Sputtering Prblem James A Glackin, James V. Mathesn I prpse t cnsider first the varius elements f the subject, next its varius parts r sectins, and finally the whle in its internal structure. In ther

More information

Q x = cos 1 30 = 53.1 South

Q x = cos 1 30 = 53.1 South Crdinatr: Dr. G. Khattak Thursday, August 0, 01 Page 1 Q1. A particle mves in ne dimensin such that its psitin x(t) as a functin f time t is given by x(t) =.0 + 7 t t, where t is in secnds and x(t) is

More information

I. Analytical Potential and Field of a Uniform Rod. V E d. The definition of electric potential difference is

I. Analytical Potential and Field of a Uniform Rod. V E d. The definition of electric potential difference is Length L>>a,b,c Phys 232 Lab 4 Ch 17 Electric Ptential Difference Materials: whitebards & pens, cmputers with VPythn, pwer supply & cables, multimeter, crkbard, thumbtacks, individual prbes and jined prbes,

More information

February 28, 2013 COMMENTS ON DIFFUSION, DIFFUSIVITY AND DERIVATION OF HYPERBOLIC EQUATIONS DESCRIBING THE DIFFUSION PHENOMENA

February 28, 2013 COMMENTS ON DIFFUSION, DIFFUSIVITY AND DERIVATION OF HYPERBOLIC EQUATIONS DESCRIBING THE DIFFUSION PHENOMENA February 28, 2013 COMMENTS ON DIFFUSION, DIFFUSIVITY AND DERIVATION OF HYPERBOLIC EQUATIONS DESCRIBING THE DIFFUSION PHENOMENA Mental Experiment regarding 1D randm walk Cnsider a cntainer f gas in thermal

More information

Physics 212. Lecture 12. Today's Concept: Magnetic Force on moving charges. Physics 212 Lecture 12, Slide 1

Physics 212. Lecture 12. Today's Concept: Magnetic Force on moving charges. Physics 212 Lecture 12, Slide 1 Physics 1 Lecture 1 Tday's Cncept: Magnetic Frce n mving charges F qv Physics 1 Lecture 1, Slide 1 Music Wh is the Artist? A) The Meters ) The Neville rthers C) Trmbne Shrty D) Michael Franti E) Radiatrs

More information

Support-Vector Machines

Support-Vector Machines Supprt-Vectr Machines Intrductin Supprt vectr machine is a linear machine with sme very nice prperties. Haykin chapter 6. See Alpaydin chapter 13 fr similar cntent. Nte: Part f this lecture drew material

More information

THE TOPOLOGY OF SURFACE SKIN FRICTION AND VORTICITY FIELDS IN WALL-BOUNDED FLOWS

THE TOPOLOGY OF SURFACE SKIN FRICTION AND VORTICITY FIELDS IN WALL-BOUNDED FLOWS THE TOPOLOGY OF SURFACE SKIN FRICTION AND VORTICITY FIELDS IN WALL-BOUNDED FLOWS M.S. Chng Department f Mechanical Engineering The University f Melburne Victria 3010 AUSTRALIA min@unimelb.edu.au J.P. Mnty

More information

Three charges, all with a charge of 10 C are situated as shown (each grid line is separated by 1 meter).

Three charges, all with a charge of 10 C are situated as shown (each grid line is separated by 1 meter). Three charges, all with a charge f 0 are situated as shwn (each grid line is separated by meter). ) What is the net wrk needed t assemble this charge distributin? a) +0.5 J b) +0.8 J c) 0 J d) -0.8 J e)

More information

SAFE HANDS & IIT-ian's PACE EDT-04 (JEE) Solutions

SAFE HANDS & IIT-ian's PACE EDT-04 (JEE) Solutions ED- (JEE) Slutins Answer : Optin () ass f the remved part will be / I Answer : Optin () r L m (u csθ) (H) Answer : Optin () P 5 rad/s ms - because f translatin ωr ms - because f rtatin Cnsider a thin shell

More information

1.2.1 Vectors. 1 P age. Examples What is the reference vector angle for a vector that points 50 degrees east of south?

1.2.1 Vectors. 1 P age. Examples What is the reference vector angle for a vector that points 50 degrees east of south? 1.2.1 Vectrs Definitins Vectrs are represented n paper by arrws directin = magnitude = Examples f vectrs: Examples What is the reference vectr angle fr a vectr that pints 50 degrees east f suth? What is

More information

Math 302 Learning Objectives

Math 302 Learning Objectives Multivariable Calculus (Part I) 13.1 Vectrs in Three-Dimensinal Space Math 302 Learning Objectives Plt pints in three-dimensinal space. Find the distance between tw pints in three-dimensinal space. Write

More information

Admissibility Conditions and Asymptotic Behavior of Strongly Regular Graphs

Admissibility Conditions and Asymptotic Behavior of Strongly Regular Graphs Admissibility Cnditins and Asympttic Behavir f Strngly Regular Graphs VASCO MOÇO MANO Department f Mathematics University f Prt Oprt PORTUGAL vascmcman@gmailcm LUÍS ANTÓNIO DE ALMEIDA VIEIRA Department

More information

AP Statistics Notes Unit Two: The Normal Distributions

AP Statistics Notes Unit Two: The Normal Distributions AP Statistics Ntes Unit Tw: The Nrmal Distributins Syllabus Objectives: 1.5 The student will summarize distributins f data measuring the psitin using quartiles, percentiles, and standardized scres (z-scres).

More information

Finding the Earth s magnetic field

Finding the Earth s magnetic field Labratry #6 Name: Phys 1402 - Dr. Cristian Bahrim Finding the Earth s magnetic field The thery accepted tday fr the rigin f the Earth s magnetic field is based n the mtin f the plasma (a miture f electrns

More information

Physics 321 Solutions for Final Exam

Physics 321 Solutions for Final Exam Page f 8 Physics 3 Slutins fr inal Exa ) A sall blb f clay with ass is drpped fr a height h abve a thin rd f length L and ass M which can pivt frictinlessly abut its center. The initial situatin is shwn

More information

AP Physics. Summer Assignment 2012 Date. Name. F m = = + What is due the first day of school? a. T. b. = ( )( ) =

AP Physics. Summer Assignment 2012 Date. Name. F m = = + What is due the first day of school? a. T. b. = ( )( ) = P Physics Name Summer ssignment 0 Date I. The P curriculum is extensive!! This means we have t wrk at a fast pace. This summer hmewrk will allw us t start n new Physics subject matter immediately when

More information

UNIT 1 COPLANAR AND NON-COPLANAR FORCES

UNIT 1 COPLANAR AND NON-COPLANAR FORCES UNIT 1 COPLANA AND NON-COPLANA FOCES Cplanar and Nn-Cplanar Frces Structure 1.1 Intrductin Objectives 1. System f Frces 1.3 Cplanar Frce 1.3.1 Law f Parallelgram f Frces 1.3. Law f Plygn f Frces 1.3.3

More information

CESAR Science Case The differential rotation of the Sun and its Chromosphere. Introduction. Material that is necessary during the laboratory

CESAR Science Case The differential rotation of the Sun and its Chromosphere. Introduction. Material that is necessary during the laboratory Teacher s guide CESAR Science Case The differential rtatin f the Sun and its Chrmsphere Material that is necessary during the labratry CESAR Astrnmical wrd list CESAR Bklet CESAR Frmula sheet CESAR Student

More information

Physics 2010 Motion with Constant Acceleration Experiment 1

Physics 2010 Motion with Constant Acceleration Experiment 1 . Physics 00 Mtin with Cnstant Acceleratin Experiment In this lab, we will study the mtin f a glider as it accelerates dwnhill n a tilted air track. The glider is supprted ver the air track by a cushin

More information

CHAPTER 6 WORK AND ENERGY

CHAPTER 6 WORK AND ENERGY CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS 16. REASONING AND SOLUTION A trapeze artist, starting rm rest, swings dwnward n the bar, lets g at the bttm the swing, and alls reely t the net. An assistant,

More information

NGSS High School Physics Domain Model

NGSS High School Physics Domain Model NGSS High Schl Physics Dmain Mdel Mtin and Stability: Frces and Interactins HS-PS2-1: Students will be able t analyze data t supprt the claim that Newtn s secnd law f mtin describes the mathematical relatinship

More information

MATHEMATICS SYLLABUS SECONDARY 5th YEAR

MATHEMATICS SYLLABUS SECONDARY 5th YEAR Eurpean Schls Office f the Secretary-General Pedaggical Develpment Unit Ref. : 011-01-D-8-en- Orig. : EN MATHEMATICS SYLLABUS SECONDARY 5th YEAR 6 perid/week curse APPROVED BY THE JOINT TEACHING COMMITTEE

More information

ASTRODYNAMICS. o o o. Early Space Exploration. Kepler's Laws. Nicolaus Copernicus ( ) Placed Sun at center of solar system

ASTRODYNAMICS. o o o. Early Space Exploration. Kepler's Laws. Nicolaus Copernicus ( ) Placed Sun at center of solar system ASTRODYNAMICS Early Space Explratin Niclaus Cpernicus (1473-1543) Placed Sun at center f slar system Shwed Earth rtates n its axis nce a day Thught planets rbit in unifrm circles (wrng!) Jhannes Kepler

More information

NWACC Dept of Mathematics Dept Final Exam Review for Trig - Part 3 Trigonometry, 9th Edition; Lial, Hornsby, Schneider Fall 2008

NWACC Dept of Mathematics Dept Final Exam Review for Trig - Part 3 Trigonometry, 9th Edition; Lial, Hornsby, Schneider Fall 2008 NWACC Dept f Mathematics Dept Final Exam Review fr Trig - Part Trignmetry, 9th Editin; Lial, Hrnsby, Schneider Fall 008 Departmental Objectives: Departmental Final Exam Review fr Trignmetry Part : Chapters

More information

(1.1) V which contains charges. If a charge density ρ, is defined as the limit of the ratio of the charge contained. 0, and if a force density f

(1.1) V which contains charges. If a charge density ρ, is defined as the limit of the ratio of the charge contained. 0, and if a force density f 1.0 Review f Electrmagnetic Field Thery Selected aspects f electrmagnetic thery are reviewed in this sectin, with emphasis n cncepts which are useful in understanding magnet design. Detailed, rigrus treatments

More information

A Simple Set of Test Matrices for Eigenvalue Programs*

A Simple Set of Test Matrices for Eigenvalue Programs* Simple Set f Test Matrices fr Eigenvalue Prgrams* By C. W. Gear** bstract. Sets f simple matrices f rder N are given, tgether with all f their eigenvalues and right eigenvectrs, and simple rules fr generating

More information

Phys102 Second Major-102 Zero Version Coordinator: Al-Shukri Thursday, May 05, 2011 Page: 1

Phys102 Second Major-102 Zero Version Coordinator: Al-Shukri Thursday, May 05, 2011 Page: 1 Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 1 1. Particles A and B are electrically neutral and are separated by 5.0 μm. If 5.0 x 10 6 electrns are transferred frm particle A t particle B, the magnitude

More information

Discrete Actuator Array Vectoreld Design for Distributed Manipulation Jonathan E. Luntz, William Messner, and Howie Choset Department of Mechanical En

Discrete Actuator Array Vectoreld Design for Distributed Manipulation Jonathan E. Luntz, William Messner, and Howie Choset Department of Mechanical En Discrete Actuatr Array Vectreld Design fr Distributed Manipulatin Jnathan E. Luntz, William Messner, and Hwie Chset Department f Mechanical Engineering Carnegie Melln University Pittsburgh, PA 11 jl9f@cmu.edu,

More information

Synchronous Motor V-Curves

Synchronous Motor V-Curves Synchrnus Mtr V-Curves 1 Synchrnus Mtr V-Curves Intrductin Synchrnus mtrs are used in applicatins such as textile mills where cnstant speed peratin is critical. Mst small synchrnus mtrs cntain squirrel

More information

Examiner: Dr. Mohamed Elsharnoby Time: 180 min. Attempt all the following questions Solve the following five questions, and assume any missing data

Examiner: Dr. Mohamed Elsharnoby Time: 180 min. Attempt all the following questions Solve the following five questions, and assume any missing data Benha University Cllege f Engineering at Banha Department f Mechanical Eng. First Year Mechanical Subject : Fluid Mechanics M111 Date:4/5/016 Questins Fr Final Crrective Examinatin Examiner: Dr. Mhamed

More information

Subject: KINEMATICS OF MACHINES Topic: VELOCITY AND ACCELERATION Session I

Subject: KINEMATICS OF MACHINES Topic: VELOCITY AND ACCELERATION Session I Subject: KINEMTIS OF MHINES Tpic: VELOITY ND ELERTION Sessin I Intrductin Kinematics deals with study f relative mtin between the varius parts f the machines. Kinematics des nt invlve study f frces. Thus

More information

Momentum 1. MOMENTUM. An object of mass m traveling at velocity v has a linear momentum (or just B. momentum) p, given by. p m B v. (6.

Momentum 1. MOMENTUM. An object of mass m traveling at velocity v has a linear momentum (or just B. momentum) p, given by. p m B v. (6. Mmentum 6 In this chapter we begin ur study f mre realistic systems in which the bjects are n lnger pint particles but have extensin in space. Up until nw we ve generally limited urselves t the dynamics

More information

Introduction to Spacetime Geometry

Introduction to Spacetime Geometry Intrductin t Spacetime Gemetry Let s start with a review f a basic feature f Euclidean gemetry, the Pythagrean therem. In a twdimensinal crdinate system we can relate the length f a line segment t the

More information

= m. Suppose the speed of a wave on a string is given by v = Κ τμ

= m. Suppose the speed of a wave on a string is given by v = Κ τμ Phys101 First Majr-11 Zer Versin Sunday, Octber 07, 01 Page: 1 Q1. Find the mass f a slid cylinder f cpper with a radius f 5.00 cm and a height f 10.0 inches if the density f cpper is 8.90 g/cm 3 (1 inch

More information

5 th grade Common Core Standards

5 th grade Common Core Standards 5 th grade Cmmn Cre Standards In Grade 5, instructinal time shuld fcus n three critical areas: (1) develping fluency with additin and subtractin f fractins, and develping understanding f the multiplicatin

More information

The ultra-high energy cosmic rays image of Virgo A

The ultra-high energy cosmic rays image of Virgo A The ultra-high energy csmic rays image f Virg A Radmír Šmída Karlsruhe Institute f Technlgy, Germany E-mail: radmir.smida@kit.edu Ralph Engel Karlsruhe Institute f Technlgy, Germany Arrival directins f

More information

Physics 101 Math Review. Solutions

Physics 101 Math Review. Solutions Physics 0 Math eview Slutins . The fllwing are rdinary physics prblems. Place the answer in scientific ntatin when apprpriate and simplify the units (Scientific ntatin is used when it takes less time t

More information

CLASS XI SET A PHYSICS

CLASS XI SET A PHYSICS PHYSIS. If the acceleratin f wedge in the shwn arrangement is a twards left then at this instant acceleratin f the blck wuld be, (assume all surfaces t be frictinless) a () ( cs )a () a () cs a If the

More information

NUMBERS, MATHEMATICS AND EQUATIONS

NUMBERS, MATHEMATICS AND EQUATIONS AUSTRALIAN CURRICULUM PHYSICS GETTING STARTED WITH PHYSICS NUMBERS, MATHEMATICS AND EQUATIONS An integral part t the understanding f ur physical wrld is the use f mathematical mdels which can be used t

More information

22.54 Neutron Interactions and Applications (Spring 2004) Chapter 11 (3/11/04) Neutron Diffusion

22.54 Neutron Interactions and Applications (Spring 2004) Chapter 11 (3/11/04) Neutron Diffusion .54 Neutrn Interactins and Applicatins (Spring 004) Chapter (3//04) Neutrn Diffusin References -- J. R. Lamarsh, Intrductin t Nuclear Reactr Thery (Addisn-Wesley, Reading, 966) T study neutrn diffusin

More information

Trigonometric Ratios Unit 5 Tentative TEST date

Trigonometric Ratios Unit 5 Tentative TEST date 1 U n i t 5 11U Date: Name: Trignmetric Ratis Unit 5 Tentative TEST date Big idea/learning Gals In this unit yu will extend yur knwledge f SOH CAH TOA t wrk with btuse and reflex angles. This extensin

More information

1 Course Notes in Introductory Physics Jeffrey Seguritan

1 Course Notes in Introductory Physics Jeffrey Seguritan Intrductin & Kinematics I Intrductin Quickie Cncepts Units SI is standard system f units used t measure physical quantities. Base units that we use: meter (m) is standard unit f length kilgram (kg) is

More information

Lecture 17: Free Energy of Multi-phase Solutions at Equilibrium

Lecture 17: Free Energy of Multi-phase Solutions at Equilibrium Lecture 17: 11.07.05 Free Energy f Multi-phase Slutins at Equilibrium Tday: LAST TIME...2 FREE ENERGY DIAGRAMS OF MULTI-PHASE SOLUTIONS 1...3 The cmmn tangent cnstructin and the lever rule...3 Practical

More information