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Chapter 2 Kinematics of a Particle

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Chapter 2 Kinematics of a Particle CHAPTER 2 KINEMATICS OF A PARTICLE Kinematics: It is the study of the geometry of motion of particles, rigid bodies, etc., disregarding the forces associated with these motions. Kinematics of a particle motion of a point in space 1 • Interest is on defining quantities such as position, velocity, and acceleration. • Need to specify a reference frame (and a coordinate system in it to actually write the vector expressions). • Velocity and acceleration depend on the choice of the reference frame. • Only when we go to laws of motion, the reference frame needs to be the inertial frame. 2 • From the point of view of kinematics, no reference frame is more fundamental or absolute. 2.1 Position, velocity, acceleration path followed by the z object. vP rOP P o O - origin of a y coordinate system in x the reference frame. 3 rOP - position vector (specifies position, given the choice of the origin O). Clearly, rOP changes with time rOP(t) z velocity vector: P r d rOP P’ vP r OP ( t ) lim . rOP t 0 rOP’ dt t O acceleration vector: dd2 x y a v( t ) r ( t ). Pdt P dt 2 OP 4 • speed: vPPPP v v v • magnitude of acceleration: aPPPP a a a Important: the time derivatives or changes in time have been considered relative to (or with respect to) a reference frame. Description in various coordinate systems (slightly different from the text) • Cartesian coordinates, cylindrical coordinates etc. 5 Let i , j , k be the unit vectors k z P i j z Cartesian coordinate system: O The reference frame is x y x y - it is fixed. P’ i j k , j k i , k i j etc. i , j , k are an orthogonal set. Then, position of P is: rOP=x(t) i+y(t) j + z(t) k 6 The time derivative of position is velocity: dr dx()()() t dy t dz t vOP i j k P dt dt dt dt d idj d k x()()() t y t z t dt dt dt If considering rate of change in a frame in didj dk which i , j , k are fixed, 0 dt dt dt dx()()() t dy t dz t v i j k P dt dt dt velocity vector Similarily, acceleration vector aP x()()() t i y t j z t k 7 Cylindrical Coordinates: z er- unit vector in xy k plane in radial direction. P O e - unit vector in xy x z r r plane to er in the x y direction of increasing y e P’ e k- unit vector in z. Then, by definition r r( x2 y 2 ) 1/2 ; tan 1 ( y / x ). 8 The position is: rOP=x(t) i+y(t) j + z(t) k Also, rOP=r() er + z(t) k and rOP= r() cos i+ r() sin j + z(t) k rr ecos i sin j r rr Also, rr r ebut r ( sin i cos j ) r andr e sin i cos j • Imp. to Note: e r and e change with position (). 9 • position: rOP=r() er + z(t) k or rOP= r() cos i+ r() sin j + z(t) k • velocity: vP dr OP dt re r r de r dt zk z dk dt z- direction ( k ) fixed dk dt 0 Thus dedtr( ded r )( ddt ) ( ded r / ) e or vPr re r e zk =radial comp+transverse comp+axial comp 10 •acceleration: dv d aP () re r e zk Prdt dt rerr re re re re zk Now, err e; e e 2 aPr( r r ) e (2 r r ) e zk =radial comp+transverse comp+axial comp Spherical coordinates: reading suggested later 11 Tangential and Normal Components: (intrinsic description) z s r P path P‟ O‟ r OP s O y x s - scalar parameter defining distance along the path from some landmark O . - called the path variable. 12 Note: s s(t) (it depends on time). Suppose that the path is fixed, a given highway for example. Then, rOP= rOP (s) is known. Different vehicles travel at different rates - speeds, changes in speeds. Properties of the highway, a planar or space curve are distinct from the motion s(t). Ex: automobile traveling along a circular race track. 13 • O and O are distinct. north v Position is from O, s is P P being measured from . s(t) rOP O O‟ Now: east a rOP r OP ( s ); s s ( t ) Then drOP dr OPds ds r P v P lim dt ds dt dts 0 s 14 s s e (s) P‟ t P r P‟ r rOP rOP‟ P O P Consider rrPPr P lim lim .lim 1.ett ( s ) e ( s ) s 0 ssr P • et depends in orientation on s (location), its magnitude is always one. vPt se() s velocity vector 15 • velocity is always tangent to path with magnitude (speed) =v Pt se() s s To find expression for acceleration: d a( se ( s )) se ( s ) se ( s ) Pdt t t t d ds de() s se( s ) s ( e ( s )) se ( s ) s2 t tds t dt t ds det To find , consider ett( s ) e ( s ) 1 ds (unit vector at every s) d de() s e( s ) e ( s ) 0 2 e ( s ) t 0 dst t t ds 16 de() s es( ) isr to t t ds de( s ) 1 Let: t e e() s ds nn where de()() s de s es( ) tt / isa normalvector n ds ds - curvature of the path at the location „s‟. - radius of curvature at P (at location „s‟). sv2 2 a se()()() s e s v e s P e P t n P t n ora a e ( s ) a e ( s ) P t t n n 17 z en et s en vP O‟ et rOP eb O x y Now define: e b e t ()() s e n s binormal vector Note that the vectorset , e n ,and e b satisfy et e t e n e n e b e b 1; et e n e t e b e n e b 0. 18 Rate of change of unit vectors along the path: One can show that: de e tn e (1) ds n de e bn (2) - torsion ds de e e n t b (3) ds Frenet‟s formulas (in differential geometry) 19 de e Ex (2): Suppose we want to show: bn ds d de de Consider (e e ) 0 tb e e 0 dst b ds b t ds de e de e Now t n ee b n 0 dstb ds de e r b t ds d de Also (e e 1) b e 0 dsb b ds b de Thus e r b b ds 20 de b is only along e ds n de e Let bn e ds wn 1 - torsion , - twist w Torsion and twist are like radius of curvature and curvature. 21 Ex 3: A rocket lifts-off v, a straight up. A radar station is located L distance away. At height H, the rocket has speed v, and rate of change of speed v . R h Find: R ,R , R , , L the variables measured by the tracking station. 22 et We start with velocity: e er v ve Re R e v, a P t r etrcos e sin e vetr v(cos e sin e ) Re R e R h r y Comparing on two sides: e: v cos R v cos / R L O x er : v sin R ; Also, R L / cos vsin R v cos2 / L 23 Similarly: v2 a ve e ( R R2 ) e ( R 2 R ) e P t n r v(sin er cos e ) comparing on the two sides : 2 er : v sin R R or R vsin v23 cos / L e : v cos R 2 R orcos [ v cos 2 v sin cos2 v / L ]/ L 24 Ex 4: A block C slides along the horizontal rod, while a pendulum attached to the block can swing in the vertical plane x(t) Find: The acceleration of the pendulum mass D. B O,A C Using (x,y,z) coord. system, i j rC x() t i rD [ x ( t ) l sin ] i l cos j D rD [ x ( t ) l cos ] i l sin j 2 rD [ x ( t ) l cos l sin ] i 2 lj[ sin cos ] 25 Ex 5: A slider S is constrained to follow the fixed surface defined by the curve BCD: r a/(1 ); r isin meters, isin radians. Find: v , a B S S r=a/(1+ ) y A r S C O a x D 26 Consider the solution using the cylindrical coordinate system: the unit vectors are eer and The position is: rSr re The velocity is vSr re r e ; Nowra /(1 ), y et e csin( t ), c cos( t ) er dr dr a S r ; dd (1 )2 r aa v e e O Sr(1 )2 (1 ) x 27 Now we consider the acceleration of the block: The expression is: 2 aSr( r r ) e ( r 2 r ) e The various termsin thisexpression are: c cos( t ); c 2 sin( t ) d d a a 2 a rr () dt dt (1 )2 (1 ) 2 (1 ) 3 28 2.2 ANGULAR VELOCITY: It defines the rate of change of orientation of a rigid body - or, a coordinate frame with respect to another. Consider z displacement P” in time t. (displ.
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