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Home , Acceleration, Kinematics

CHAPTER 2 OF A PARTICLE Kinematics: It is the study of the of of particles, rigid bodies, etc., disregarding the associated with these .

Kinematics of a particle  motion of a point in

1 • Interest is on defining quantities such as , , and . • Need to specify a reference frame (and a 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  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 • : vPPPP v  v  v • of acceleration:

aPPPP a  a  a Important: the time 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 of position is velocity: dr dx()()() t dy t dz t vOP  i  j  k P dt dt dt dt d idj d k x()()() t  y t  z t dt dt dt If considering of change in a frame in didj 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  ecos i  sin j  r rr Also, rr r ebut  r (  sin i  cos j )    r andr  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 aP () 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 - parameter defining 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 travel at different rates - , 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 dts 0  s 14 s s e (s) P‟ t P r P‟ r rOP rOP‟ P

O P Consider rrPPr P lim lim .lim  1.ett ( s )  e ( s ) s 0 ssr P • et depends in on s (location), its magnitude is always one.

 vPt se() s velocity vector 15 • velocity is always 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( ) isr 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  - 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 dstb 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 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    etrcos 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  R2 ) 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  orcos  [ v cos  2 v sin  cos2  v / L ]/ L 24 Ex 4: A block C slides along the horizontal rod, while a attached to the block can swing in the vertical plane x(t) Find: The acceleration of the pendulum 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 . 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 : It defines the rate of change of orientation of a - or, a coordinate frame with respect to another.  Consider z P” in time t.  (displ. + rot.) P‟ O P Shown is an x y s A‟ displacement of A a rigid body Chasles‟ Theorem Base Point 29    lim defines the angular velocity  t 0 t

• The angular velocity does not depend on the base point A . Rather, it is a property for the whole body.

• The angular velocity vector will usually change both its magnitude  and direction

e  / continuously with time.

30 2.3 Rigid Body Motion about a Fixed Point: The rigid body rotates  about point O (fixed base Z v point). P - a point fixed in the body. P  r  - angular velocity of the O body relative to XYZ axes. X Y Direction of - instant. axis of . Speed of P srsin .  velocity vr   (along tangent to circle)

31 The acceleration is now calculated, using the definition that it is the time-derivative of velocity: a d()() v dt  d  r dt a  r   r ora  r    (   r ) (Recall: these rates of change are  . r . t . XYZ). • ( (  r))is directed towards the instantan axis from P - centripetal acceleration. •   r - tangential acceleration (not really tangent to the path of P).

32 2.4 The Derivative of a Unit Vector:

Z Let e 1 , e 2 , e 3 be an independent set of  unit vectors attached e2 to a rigid body rotating e with angular velocity 3 O  . The body rotates Y X e1 relative to the reference frame XYZ. Thus, for the unit vectors:

e1  e 1,, e 2    e 2 e 3    e 3 33 Assume that the set ( e 1 , e 2 , e 3 ) is - orthonormal r r r Thus, e 1  e 2; e 1  e 3 and e 2  e 3 This can also be stated as: e1 e 2 e 3, etc. Let    1 e 1   2 e 2   3 e 3 (expressedin moving basis)

e1 () 1 e1   2 e 2   3 e 3  e 1   3 e 2   2 e 3

e2() 1 e1   2 e 2   3 e 3  e 2   1 e 3   3 e 1

e3() 1 e1   2 e 2   3 e 3  e 3   2 e 1   1 e 2

Ex : e1 i,, e 2  j e 3  k

  xij   y   z k 34 Then, d i dt  i  zy j   k

dj dt  j  xz k   i

dk dt  k  yx i   j IMPORTANT: • The rates of change of unit vectors have been calculated with respect to the (X,Z,Y) system - also called “relative to XYZ”. • These rates (vectors) have been expressed in terms of the unit vectors moving with the body.

35 z e 2.6 EXAMPLES: z R ii) Helical Motion:  A particle moves along P a helical path. The helix is defined in terms of the Cylindrical O z Coordinates:  e y rR (constant) x  z kR, where k tan , er  helix

36 Clearly, as changes with time, so does z. So, rr0,  0,   ,    z kR, z  kR   kR 

• vP= rer re  ze z  Re   kRe  z 2 • aP= R erz  R  e  kR  e 2 2 2 • speed s vP ( R )  ( kR  )  R  1  k • constant or uniform speed  s 0, 0 22 a   R eRn  (/) s e •  Rk(12 ) - radius of curvature of the path of the particle 37 iii) Harmonic motion: (Reading assignment)

k x(t) m m - mass, k - stiffness of the spring key point: the is directly proportional to the distance of the particle from some point  xx2 ; x- displacement 2  0 - a constant ( of natural freq.) Solution: Let x(t) = A cos(t + )

38 Here, A - amplitude,  - phase angle (these are determined by initial conditions x, x at t = 0). If  = 0, i.e., x ( t )  0 at t = 0, x(t) = A cos t

x simple A T harmonic t motion

39 T = 2/ - time period of harmonic motion  - circular

• x(t) = A cos t  x( t )   A sin  t  A  cos (  t  ) 2 x( t )   A22 cos  t  A  cos (  t   )  In , extreme values of position and acceleration occur when the velocity vanishes. Also, the velocity is out of phase with position by /2, and the acceleration is out of phase by . 40 Two-dimensional harmonic motion: y • Consider the spring-mass  system shown: k x 22 x; y   y x x( t ) A cos( t ); m k y( t ) B cos( t ) • Choose reference time such that  = 0  r()()() t x t i y t j Acos t i  B cos(  t   ) j v( t )  A sin  t i   B sin(  t   ) j

22 a( t )  A cos  t i   B cos(  t   ) j 41 Now: cos t = x/A, and cos (t + ) = y/B or, cos t cos- sintsin = y/B Using expression for x/A  y/B= (x/A) cos- sint sin  sint =[(x/A) cos - y/B] sin Since cos2 t + sin2 t =1 

(sin)-2[(x/A)2+ (y/B)2-2 (x/A)(y/B) cos]=1 (equation of an ellipse)

42 y P v r 2B O x

2A Harmonic motion in two dimensions (a plane)

43 2.7 Velocity and Acceleration of a Point in a Rigid Body

Z  v P rigid body rotating at the rate of  relative P  to XYZ. A O vA X Y A, P are two points on the same rigid body

44  Z vP  - position of P as measured from A. P  A O Now, rrOP OA  vA So, vPA v d dt X Y vvPA    (since |  |= const.,  changes only in orientation)

• vPAPA v    v / velocity of P .r.t. Point A, as viewed in the reference frame XYZ. 45 Now, consider the acceleration:

aPPA dv dt  d() v   dt

a A        or aaPA      ()   

This is the acceleration of point P in a rigid body as viewed .r.t the frame XYZ; the point P is in the rigid body which is rotating at angular velocity  relative to XYZ, and this rotation rate is changing at the rate . 46 2.8 Vector Derivative in Rotating Systems

• O - a fixed point in Z A the body  • e1,e2,e3 - triad of unit vectors in the e3 e2 body O •  - angular velocity X of the body Y e1 • Consider now an arbitrary vectorA It can be represented as A  A e  A e  A e 1 1 2 2 473 3 There are R stationary with XYZ two observers - S Tmoving with the body (). dA  can be with respect to XYZ (XYZ dA dt) Then,    dt can be with respect to the moving body ( dA dt) (depends on the observer) Let A  XYZ dA dt  rate of change .r.t. XYZ Then A Ae11  Ae 22  Ae 33  Ae 11  Ae 22  Ae 33 e ()A r- the rate of change w.r.t. the body in which i are fixed 48 Now e1  e 1;; e 2    e 2 e 3    e 3

A1 e 1  A 2 e 2  A 3 e 3   A

AAA ()r   In a more general , let A and B be two bodies;  A / B - angular velocity of A as viewed (by an observer) from B;

(Note  B / A    A / B - angular velocity of B as viewed from A). Then

()()AAAABBA  /  49 2.9 Motion of a Particle in a Moving Coordinate System

P  • XYZ - fixed reference frame (really, a given Z  frame) r OP z • xyz - a frame denoting a moving body with O‟ y angular velocity O R  x relative to XYZ. X Y 50 O- origin of coordinate system in P xyz  R- position of Z  r - position of point r OP OP z P (moving object) ρ- position of P .r.t. O‟ y R Then, the position of O x the particle is X Y rOP =R+ρ

51 Then, the velocity with respect to the XYZ is

rOP  vP  R   but () r     (rate of change of  ... rt the rotating frame (the rotating body))

 vRP  ()r     - velocity of P .r.t. XYZ.

() r - velocity of P P  in xyz. R - velocity of O   . r . t . XYZ. R - velocity of a point P  in the rotating body which is coincident with P at this instant. 52 rOP  dvPP dt a - acceleration in XYZ frame

oraP  d [ R  ( )r     ] dt

R  d()r dt        d  dt

Nowd ( )r dt  (  ) r    (  ) r andd dt  (  )r    

aRP     (  )rr  2   (  )    (    )

coriolis centripetal This is the most general expression for accel 53 Ex: Motion of a Particle Relative to Rotating We now consider one of the application of the general formulation for acceleration when a is quite natural. Here, an object is moving and its motion is observed by someone moving with earth. Assumptions: • Earth rotates about the sun. • The acceleration of center of earth is, however, relatively small compared to and acceleration on the earth‟s surface due to

earth‟s spin, especially away from poles. 54 Z O-center   z O‟-location of observer on earth‟s surface  y O‟ parallel  R m O  x 

Y X meridian circle 55 Basic definitions: x - local south (tangent to the meridian circle) y - local east (tangent to a parallel)  - longitude (defines location of a meridian plane relative to plane through Greenwich)  - latitude (defines location of a parallel relative to the Equator) z - local vertical XYZ - Fixed Frame located at O - center of the earth

56 More definitions: OXY - Equatorial plane OZ - axis of earth‟s rotation xyz - attached to the surface of earth at O (at latitude -  ; longitude - )  - angular velocity of the =  cos i  sin k Note: is constant    0 (for a vector to be constant, both its magnitude and the direction must remain constant w.r.t. the reference frame) 57 Now, we use the notation already established to define the kinematics: R - position vector of to O.  - position of the mass particle relative to O (the point on earth‟s surface) = xi  yj  zk ()()d dt  x i  y j  z k r rel to xyz () x i  y j  z k   (  cos i   sin  k )  ( xi  y j  zk )  [ y sin i  ( z cos   x sin  ) j  y cos  k ]

58 (    )  2 [  sin  (z cos   x sin  ) i y j cos ( z cos   x sin  ) k ] (centripetal acceleration)

2 (  )r  2(  cos i   sin  k )  ( xi  y j  zk ) 2  [ y sin i  ( z cos   x sin  ) j  y cos  k ] (coriolis acceleration)

59 Finally, 0 Now: aRP     (  )rr  2   (  )    (    ) R Rk (constant w.r.t. xyz)

R() dR dtrel to XYZ   R RR ()   22R(  cos sin  i  cos  k )

60 Note: 7.2910-5 rad/sec 22 aP   R(  cos sin  i  cos  k )  0 2  [ y sin i  ( z cos   x sin  ) j  ykcos ]  2[  sin (z cos   x sin  ) i  y j cos (z cos  x sin  ) k ] x i  y j  z k

• close to earth‟s surface  22x R,. etc

61 22 aP   R(  cos sin  i  cos  k ) 2  [ y sin i  ( z cos   x sin  ) j  ykcos ] x i  y j  z k (for motions near earth‟s surface). • Since   7.29  10  52 ,  terms are also neglected in study of most motions close to earth‟s surface. Note that this depends also on the latitude  of the point O‟.

62 Ex: Motion of a Particle in Near Earth‟s Surface z (local north) - Consider a particle moving close to the m earth‟s surfare; mg - We will write the O‟ y (local east) using the co- ordinate system attached to the Uniform gravitational surface of moving O field W=-mgk earth; 63 ‟s 2nd Law: F  maP (in an inertial frame)

mgk  maPP   gk  a Neglecting  2 terms and air  i: x 2  y sin  0 (1) j: y 2  ( x sin  z cos )  0 (2) k: z 2  y cos  g  0 (3)

Note: 2 terms also need to be always

neglected in calculations to follow. 64 Initial conditions are: x(0) =0, y(0) =0, z(0) =h; x(0)  y(0)  z(0)  0 (1)x  2  y sin  cons tan t  0 (4) (3)z  2  y cos  gt  cons tan t  0 (5) (4),(5)in (2) y2  (2  y sin22  2  y cos   gt cos  )  0 or y2  gt cos  0 y   gt3 cos / 3 (6)

65 (4), (6) x (2 3) 23 gt sin cos 0  xt( ) 0 (7) (5), (6) z (2 3) 2 gt 3 cos 2   gt  gt z( t )  h  gt 2 / 2 (8)

Time to reach earth‟s surface: z0  t  2 h / g Coordinate of the landing point: x0, y  (2/3)  h 2/ h g cos, z  0

66 Schematics

z (local north) h

O‟ y (local east)

y(2/ 3) h 2 h / g cos O

67 2.10 Plane Motion (motion of a 3-D rigid body in a plane) • There exists a plane such that every point has velocity and acceleration parallel to this one fixed plane. lamina of motion All planes are moving parallel to the the plane colored green. This lamina contains the centroid of the rigid

body. 68 Consider motion in the vB  plane called the „lamina v B A of motion‟. Let  - angular velocity AB of the body (same for every plane in the body A v A and  R to the lamina of motion).

LetvvAB , velocitiesof Aand B, twopoints

in thelamina.Then vvA A   AB This relates of two points on the same rigid body. 69 Question: Does there exist a point such that its velocity is zero, even if only instantaneously ?

• Suppose that C be such a point: vC=0. Then, considering points A and B, velocities are: r vA v C    CA  v A   CA and r vB v C    CB  v B   CB (Assuming thatv  0) Using these, we canC construct and find the point C, as well as the angular velocity  , given the

velocities vv AB and for points A and B. 70 C Consider the construction on v  B the left. Points A and B are given with their velocities. B Then, one can follow the

CA construction and note that vA CA and A v A vB CB

(Assuming thatvC  0) C - instantaneous center (of zero velocity)

vvACABCB//  

71 Ex: Rolling Motion: Consider a wheel moving  on the fixed ground. r O‟ - center of wheel i O‟ Let: v - velocity of O‟ vO O‟ j  - angular velocity C ‟ W C of the wheel They are not independent CG in rolling. • Two physical points C W , C G ; one belongs to the wheel, the other to the ground on which it is rolling  vv CCWG

• Ground fixed  vv0   0. 72 CCGW • wheel is one rigid body vC  0 is CG W the instant center (of zero velocity).

• Then vO'' v C   r CO  ()  rj

r i  vOO'' i  v  r (thisisalways valid) differentiating with respect to time 

vOO'' a r • Let   = of the wheel.

aOO''  ror a   ri   ri

73 Now, consider acceleration of the wheel center. The relation is:

aaO''' C    CO   ()    CO

 CO' rj;

 CO'  k ()  rj   r i

r i  aC   r i   ()    rj

2 aC   k   r i    rj (it is directed towards the wheel center).

74 Reading assignment: Examples 2.3, 2.4-2.5, 2.7 Example 2.6 Consider a wheel of radius r2, rolling inside the fixed track of radius r 1 . The arm OO ' rotates at a constant angular velocity  about the fixed point O’.

75 Find: The velocity vP and acceleration aP of a point P on the wheel, specified by angle  w.r.t. line OO’ . First Method: O - origin of the fixed frame; Arm is the moving frame; O‟- origin of the e moving coordinate system. n (e , e , e ) - a unit vector triad et t n b 76 vP  R ()()r      v O'   r    

where R vO';  rOP'

NowR rOO'   Re n   ( r 1  r 2 ) e n

SoR  ( r1  r 2 ) enn   ( r 1  r 2 )  e

 (r1  r 2 ) eb  e n  ( r 1  r 2 ) e t  v O '

Alsor2 (cos  ent sin  e )

( )r r2  (  sin  e n  cos  e t )

77    erb 2 (cos  e n  sin  e t )

 rr22cos  etn   sin  e

vP ( r1  r 2 ) e t  r 2  (sine   n  cose)  t

r2 (  cos  etn  sin  e )

r2 Rolling Constraint :v C  0, andPC when =

vC () r1  r 2 e t  r 2 e t   r 2 e t  0

or(r1  r 2     r 1 / r 2 always valid )

78 So,vPt [( r1  r 2 )  r 1 cos  r 2 cos  ]  e

[ r12 sin  r sin  ]  e n

orvP ( r12  r ) {[1  cos  ] e t  sin  e n }

Also, rr12 /    0 Acceleration:

aRP      (    )  (  ) r  2   (  ) r 2 R()()() r1  r 2  eb  e t  r 1  r 2  e n 0 79 Consider theother terms :

(    )  eb  [   r22 cos  e t   r sin  e n ] 2 r2 (  cos  ent  sin  e )

( )rr2  (  sin  e n  cos  e t )

r2(   cos  ent   sin  e ) 22 r12(  cos  ent  sin  e ) / r

2 (  )r  2 e b   r1 [  sin  e n  cos  e t ] 2  2re1 (cos nt sin e ) 80 Thus, the acceleration is

22 aP( r1  r 2 ) e n   r 2 (  cos  e n  sin  e t ) 22 r12(  cos  ent  sin  e ) / r 2 2r1 (cos  ent sin  e )

2 2 2 aP [( r1  r 2 )   ( r 1  r 2 ) cos  / r 2 ]  22 [ (r1 r 2 ) sin / r 2 ] e t

81 Method: Let wheel serve as the moving frame; O  - origin of the moving coordinate system, attached to the wheel.

Then, ()   e b

()()   eebb     ee   0 bb en

   ()    e b et 82 R rOO'   Re n ( Note :O'ison thearm)

R eb   Re n  Re t () R  r12  r

r2 (cos  ent sin  e )

( )r  0 (moving frameattached to the wheel)

  (    )eb  r2 (cos  e n  sin  e t )

(   )r2 (sin  ent  cos  e )

vP ( r1  r 2 ) e t  (    ) r 2 (sin  e n  cos et ) Now, theconstraint is Rolling

 vC 0 and P  C when   83 Imposing constraint vC=0 

 r12/ r    0 (sin ce   0)

vPt( r12  r ){1  cos } e  (r12 r ) sin e n Acceleration: 2 ( )rn =0; R= (r12 r ) e . ..etc

84 Rate of Change of a vector in a Rotating Reference Frame • Let A and B be two reference frames, B moves relative to A P

• {E1,E2,E3} – orthonormal basis in A E3 b • {e1,e2,e3} – orthonormal e3 Reference frame B basis in B Reference

We can express: frame A e2 O b=b1e1+b2e2+b3e3 e1 in the basis {e1,e2,e3} E 1 E 2 85