Lecture1 - Location & path of a particle in space HES2310 -...

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HES 2310 LECTURE NOTES DEVELOPED BY DR JAMAL NASER FOR HIS LECTURES HES2310 - Dynamics-1 Jamal Naser Senior Lecturer EN216 Ext: 8655 Swinburne University NB: These lecture notes are prepared from “Engineering Mechanics- Dynamics” (the book by J.L. Merium & L.G. Kraige). These notes will help the students to follow lectures in the class. Students should read the book. Students should not depend on these lecture notes only. HES 2310 LECTURE NOTES DEVELOPED BY DR JAMAL NASER FOR HIS LECTURES Location & path of a particle in space • Location can be defined in – rectangular cordinates by x, y, z – cylindrical coordunates by r, θ , z – spherical coordinates by R, θ, φ • Path in space can be – constrained • guided in a defined path – unconstrained • no defined path HES 2310 LECTURE NOTES DEVELOPED BY DR JAMAL NASER FOR HIS LECTURES Rectilinear motion • The dislacement of the particle in time tis s – hence the velocity is v=lim ( t 0) s/ t s s t=0 t=t t=t+ t -s +s ad vd tvo radsd t s == = = / D / DD () 22 2 sd ts / D 1 v ds a dv or vdv ads sds // DD D D = 3 HES 2310 LECTURE NOTES DEVELOPED BY DR JAMAL NASER FOR HIS LECTURES Graphical relationships between s, v, a & t Tangent to curve in Fig(a) gives Velocity= v = ds/dt velocity at all ponnts determined and plotted in Fig (b) Tangent to curve in Fig(b) gives Acceleration= a =dv/dt Area under the v-t curve is: v dt = ds , on integration : ds vdt s s s s v v 1 2 1 2 21 ∫∫ HES 2310 LECTURE NOTES DEVELOPED BY DR JAMAL NASER FOR HIS LECTURES accelerations at all ponnts determined and plotted in Fig (c) Shaded area under the a-t curve is: adt = dv , on integration : Change in velocity dv adt v v v v t t 1 2 1 2 These graphical representations are important for: 1. Visualizing the relationship between s, v, a & t 2. Approximating results by graphical integration or Differencation when mathematical functions & relationships are not available HES 2310 LECTURE NOTES DEVELOPED BY DR JAMAL NASER FOR HIS LECTURES Acceleration Vs distance plot gives ads = vdv = d(v 2 /2) Integrating Net Area under a-s curve vdv ads v v v v s s 1 2 1 2 1 2 2 2 1 2 From the similar triangles in Fig.(b) CB/v = dv/ds CB=vdv/ds = a
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HES 2310 LECTURE NOTES DEVELOPED BY DR JAMAL NASER FOR HIS LECTURES Problem 2/1 The position coordinate of a particle that is confined to move along a straight line is given by s=2t 3 -24t+6, Determine (a) t=? when v=72 m/s, starting from t=0 (b)a=? when v=30 m/s (c)s=? between t=1 and t=4 sec. Soln: (a) v=ds/dt= d(2t 3 -24t+6)/dt= 6t 2 -24 ------------ (i) Now, v=72=6t 2 -24 HES 2310 LECTURE NOTES DEVELOPED BY DR JAMAL NASER FOR HIS LECTURES Solving the quadratic eqn we get t=+4 & t=-4. The -ve value is neglected (b) Substituting v=30 in equn (i) we get t=3 sec now a=dv/dt= d(6t 2 -24)/dt=12t=12*3=36 m/ s 2 (c) s 4 -s 1 = (2*4 3 -24*4+6)- (2*1 3 -24*1+6) = 54 m HES 2310 LECTURE NOTES DEVELOPED BY DR JAMAL NASER FOR HIS LECTURES Prob 2/2 A particle moves along the x-axis with v=50 m/s. crosses the origin at t=0.
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This note was uploaded on 10/25/2011 for the course HES 2340 taught by Professor Tomedwards during the Three '09 term at Swinburne.

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Lecture1 - Location & path of a particle in space HES2310 -...

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