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Ch.15 of Rigid Bodies

Equations Defining the Rotation of a About a Fixed Axis

General Plane Absolute and Relative in Plane Motion Instantaneous Center of Rotation in Plane Motion Absolute and Relative in Plane Motion Analysis of Plane Motion in Terms of a Parameter

Rate of Change With Respect to a Rotating Frame Coriolis Acceleration

Applications The between train wheels is an example of curvilinear – the link stays horizontal as it swings through its motion.

How can we determine the velocity of the tip of a turbine blade?

Introduction *Kinematics of rigid bodies: relations between and the positions, , and of the particles forming a rigid body. *Classification of rigid body : Translation: rectilinear translation curvilinear translation

Rotation about a fixed axis - General plane motion - Motion about a fixed point - - General motion

15.1 Translation and Fixed Axis Rotation 15.1A Translation • Consider rigid body in translation: - direction of any straight inside the body is constant, - all particles forming the body move in parallel lines. • For any two particles in the body,    r = r + r B A B A • Differentiating with respect to time,     rB = rA + rB A = rA   vB = vA All particles have the same velocity. • Differentiating with respect to time again,     rB = rA + rB A = rA   aB = aA All particles have the same acceleration.

15.1B Rotation About a Fixed Axis. • Consider rotation of rigid body about a fixed axis AA’

 =  • Velocity vector v dr dt of the particle P is to

= the path with magnitude v ds dt

∆s = (BP)∆θ = (r sinφ )∆θ ds ∆θ v = = lim (r sinφ ) = rθsinφ dt ∆t→0 ∆t (15.4)

• The same result is obtained from(15.5,15.6)   dr   v = = ω × r dt     ω = ωk = θ k =

Concept Quiz What is the direction of the velocity of point A on the turbine blade?

a) → ω b) ← A c) ↑ y L d) ↓ Sol) =ω ×  ˆ ˆ x  ˆ vrA vA =ω k ×− Li vA = − Ljω answer ; d) ------Acceleration

• Differentiating to determine the acceleration,   dv d  a = = (ω × r ) dt dt   dω   dr = × r +ω × dt dt  dω    = × r +ω ×v dt and  dω  = α = dt     = αk = ωk = θ k Fig.15.9 • Acceleration of P is combination of two vectors,

      a = α × r +ω ×ω × r   α × r = tangential acceleration component    ω ×ω × r = radial acceleration component

Representative Slab

• Consider the motion of a representative slab in a plane perpendicular to the axis of rotation. • Velocity of any point P of the slab,      v = ω × r = ωk × r

v = rω

• Acceleration of any point P of the slab,   a = α × r +ω ×ω × r   2  = αk × r −ω r Resolving the acceleration into tangential and normal components,    at = αk × r at = rα  2  2 an = −ω r an = rω

Concept Quiz What is the direction of the normal acceleration of point A on the turbine blade?

a) → ω b) ← A c) ↑ y L d) ↓ x

Sol) = −ω 2  2 ˆ  2 ˆ arn an =−−ω () Li an = Liω

15.1C Equations Defining the Rotation of a Rigid Body About a Fixed Axis

• Motion of a rigid body rotating around a fixed axis is often specified by the type of angular acceleration. • Recall dθ dθ ω = or dt = dt ω dω d 2θ dω α = = = ω dt dt 2 dθ

• Uniform Rotation (angular acceleration=0 )

θ = θ0 +ωt • Uniformly Accelerated Rotation( angular acceleration = constant): ω = ω0 +αt θ = θ +ω + 1α 2 0 0t 2 t 2 2 ω = ω0 + 2α(θ −θ0 ) Sample Problem 15.3

Cable C has a constant acceleration of 225 mm/s2 and an initial velocity of 300 mm/s, both directed to the right. Determine (a) the number of revolutions of the pulley in 2 s, (b) the velocity and change in of the load B after 2 s, and (c) the acceleration of the point D on the rim of the inner pulley at t = 0.

STRATEGY: • Due to the action of the cable, the tangential velocity and acceleration of D are equal to the velocity and acceleration of C. Calculate the initial angular velocity and acceleration. • Apply the relations for uniformly accelerated rotation to determine the velocity and angular position of the pulley after 2 s. • Evaluate the initial tangential and normal acceleration components of D. • MODELING and ANALYSIS: The tangential velocity and acceleration of D are equal to the velocity and acceleration of C.

• Apply the relations for uniformly accelerated rotation to determine velocity and angular position of pulley after 2 s.

11222 θω=+=0tt22 α (4 rad s)( 2 s) +( 3rad s)( 2 s) =14 rad 1 rev N = ( 14 rad)= number of revs N = 2.23rev 2π rad

• Evaluate the initial tangential and normal acceleration components of D.

Magnitude and direction of the total acceleration,

22 aaa=( ) + ( ) DDtn D 22 =(225) +=( 1200) 1221 mm s2

2 aD =1.221 m s φ = ° 79.4

REFLECT and THINK: • A double pulley acts similarly to a system of gears; for every 75 mm that point C moves to the right, point B moves 125 mm upward. This is also similar to the rear tire of your bicycle. As you apply tension to the chain, the rear sprocket rotates a small amount, causing the rear wheel to rotate through a much larger angle. 15.2 General Plane Motion : Velocity As the man approaches to release the bowling ball, his arm has linear velocity and acceleration from both translation (the man moving forward) as well as rotation (the arm rotating about the shoulder).

15.2A Analyzing General Plane Motion

• General plane motion is neither a translation nor a rotation. • General plane motion can be considered as the sum of a translation and rotation.

of particles A and B to A2 and B 2 can be divided into two parts: B′ • translation to A2 and 1 B′ • rotation of 1 about A2 to B 2

15.2B Absolute and in Plane Motion

• Any plane motion can be replaced by a translation of an arbitrary reference point A and a simultaneous rotation about A.    vB = vA + vB A    vB A = ωk × rB A vB A = rω     v = v +ωk × r B A B A

• Assuming that the velocity v A of end A is known, wish to determine the velocity v B

of end B and the angular velocity w in terms of v A, l, and q.

• The direction of v B and v B/A are known. Complete the velocity diagram.

vB vvAA = tanθ = = cosθ vA vlBA ω

vvBA= tanθ v ω = A l cosθ

• Selecting point B as the reference point and solving for the velocity v A of end A and the angular velocity w leads to an equivalent velocity triangle.

• v A/B has the same magnitude but opposite sense of v B/A. The sense of the relative velocity is dependent on the choice of reference point. • Angular velocity w of the rod in its rotation about B is the same as its rotation about A. Angular velocity is not dependent on the choice of reference point.

Sample Problem 15.6

The double gear rolls on the stationary lower rack: the velocity of its center is 1.2 m/s. Determine (a) the angular velocity of the gear, and (b) the velocities of the upper rack R and point D of the gear.

STRATEGY: • The displacement of the gear center in one revolution is equal to the outer circumference. Relate the translational and angular displacements. Differentiate to relate the translational and angular velocities. • The velocity for any point P on the gear may be written as       v = v + v = v + ωk × r P A P A A P A Evaluate the velocities of points B and D.

MODELING and ANALYSIS • The displacement of the gear center in one revolution is equal to the outer circumference.

For x A > 0 (moves to right), w < 0 (rotates clockwise).

y

x xA θ = − xA = −r1θ 2π r 2π Differentiate to relate the translational and angular velocities. vA = −r1ω v 1.2m s ω = − A = −   ω = ω = − r1 0.150m k (8rad s)k

      v = v + v = v + ωk × r • For any point P on the gear, P A P A A P A

Velocity of the upper rack is equal to velocity of point B:      vR = vB = vA + ωk × rB A    = (1.2m s)i + (8rad s)k × (0.10 m) j    = (1.2m s)i + (0.8m s)i  viR = (2m s)

Velocity of the point D:     vD = vA + ωk × rD A    = ( ) + ( ) × (− )   1.2m s i 8rad s k 0.150 m i vD =(1.2m s) ij + ( 1.2m s)

vD =1.697 m s

REFLECT and THINK: • Note that point A was free to translate, and Point C, since it is in contact with the fixed lower rack, has a velocity of zero. Every point along diameter CAB has a velocity vector directed to the right and the magnitude of the velocity increases linearly as the from point C increases.

 Instantaneous Center of Rotation *Plane motion of all particles in a slab can always be replaced by the translation of an arbitrary point A and a rotation about A with an angular velocity that is independent of the choice of A. *The same translational and rotational velocities at A are obtained by allowing the slab to rotate with the same angular velocity about the point C on a perpendicular to the velocity at A.

*The velocity of all other particles in the slab are the same as originally defined since the angular velocity and translational velocity at A are equivalent.

*As far as the velocities are concerned, the slab seems to rotate about the instantaneous center of rotation C.

*If the velocity at two points A and B are known, the instantaneous center of rotation lies at the intersection of the perpendiculars to the velocity vectors through A and B .

*If the velocity vectors are parallel, the instantaneous center of rotation is at infinity and the angular velocity is zero. *If the velocity vectors at A and B are perpendicular to the line AB, the instantaneous center of rotation lies at the intersection of the line AB with the line joining the extremities of the velocity vectors at A and B.

*If the velocity magnitudes are equal, the instantaneous center of rotation is at infinity and the angular velocity is zero.

*The instantaneous center of rotation lies at the intersection of the perpendiculars to the velocity

vectors through A and B .with vv ω =AA = AC l cosθ

Then, vA vB =( BC)ωθ = ( l sin ) Fig. 15.20 l cosθ

= vA tanθ *The velocities of all particles on the rod are as if they were rotated about C. *The particle at the center of rotation has zero velocity. *The particle coinciding with the center of rotation changes with time and the acceleration of the particle at the instantaneous center of rotation is not zero. • The acceleration of the particles in the slab cannot be determined as if the slab were simply rotating about C.

• The trace of the locus of the center of rotation on the body is the body centrode and in is the space centrode.

*At the instant shown, what is the approximate direction of the velocity of point G, the center of bar AB?

a) b) c) d) G

Sample Problem 15.9 The double gear rolls on the stationary lower rack: the velocity of its center is 1.2 m/s. Determine (a) the angular velocity of the gear, and (b) the velocities of the upper rack R and point D of the gear.

STRATEGY: • The point C is in contact with the stationary lower rack and, instantaneously, has zero velocity. It must be the location of the instantaneous center of rotation. • Determine the angular velocity about C based on the given velocity at A. • Evaluate the velocities at B and D based on their rotation about C.

MODELING and ANALYSIS: • The point C is in contact with the stationary lower rack and, instantaneously, has zero velocity. It must be the location of the instantaneous center of rotation. • Determine the angular velocity about C based on the given velocity at A. vA 1.2m s vA = rAω ω = = = 8rad s rA 0.15 m • Evaluate the velocities at B and D based on their rotation about C.   = = ω = ( )( ) v = (2m s)i vR vB rB 0.25 m 8rad s R

= = rD (0.15 m) 2 0.2121 m

vrDD=ω = (0.2121 m)( 8rad s)

vD =1.697 m s   vD =(1.2 ij + 1.2)( m s) REFLECT and THINK: The results are the same as in Sample Prob. 15.6, as you would expect, but it took much less computation to get them.

Sample Problem 15.10 The crank AB has a constant clockwise angular velocity of 2000 rpm. For the crank position indicated, determine (a) the angular velocity of the connecting rod BD, and (b) the velocity of the piston P. Use method of instantaneous center of rotation STRATEGY: • Determine the velocity at B from the given crank rotation data. • The direction of the velocity vectors at B and D are known. The instantaneous center of rotation is at the intersection of the perpendiculars to the velocities through B and D. • Determine the angular velocity about the center of rotation based on the velocity at B. • Calculate the velocity at D based on its rotation about the instantaneous center of rotation. MODELING and ANALYSIS: • From Sample Problem 15.3,

• The instantaneous center of rotation is at the intersection of the perpendiculars to the velocities through B and D. • Determine the angular velocity about the center of γβ B =40 °+ = 53.95 °

γβD =90 °− = 76.05 ° rotation based on the velocity at B.

BC CD 200 mm ωBD = 62.0rad s = = sin 76.05° sin53.95 °° sin50

BC• = 253.4 mm CD = 211.1 mm Calculate the velocity at D based on its rotation about the instantaneous center of rotation.

vvPD= =13,080mm s= 13.08m s Instantaneous Center of Zero Velocity

REFLECT and THINK: What happens to the location of the instantaneous center of velocity if the crankshaft angular velocity increases from 2000 rpm in the previous problem to 3000 rpm? What happens to the location of the instantaneous center of velocity if the angle b is 0?

15.4 GENERAL PLANE MOTION : ACCELERATION 15.4 A Absolute and Relative Acceleration in Plane Motion

• Absolute acceleration of a particle of the slab, •    a = a + a B A B A (15.21)

 a • Relative acceleration B A associated with rotation about A includes tangential and normal components,

   (a ) = αk × r (a ) = rα B A t B A B A t  2  2 (a ) = −ω r (a ) = rω B A n B A B A n (15.22)

  • Given aA and vA,   determine aB and α.

   aB = aA + aB A    = a + (a ) + (a ) A B A n B A t

 • Vector result depends on sense of aA and the relative magnitudes a and (a ) of A B A n • Must also know angular velocity w.

   aB = aA + aB A • Write in terms of the two component equations, 2 + → x components: 0 = aA + lω sinθ − lα cosθ 2 + ↑ y components: − aB = −lω cosθ − lα sinθ

• Solve for aB and a.

15.4B Analysis of Plane Motion in Terms of a Parameter

• In some cases, it is advantageous to determine the absolute velocity and acceleration of a directly.

yB = l cosθ xA = l sinθ

vyBB=  vxAA=  = − θθ = lθθcos l sin = −lωθsin = lωθcos

axAA=  ayBB=  =−+llθ2 sin θθ  cos θ =−−llθ2 cos θθ  sin θ 2 2 =−+llωsin θα cos θ =−−llωcos θα sin θ

15.5 Analyzing Motion with Respect to a Rotating Frame Rotating coordinate systems are often used to analyze mechanisms (such as amusement park rides) as well as weather patterns.

15.5A Rate of Change With Respect to a Rotating Frame • With respect to the rotating Oxyz frame,     = + + Q Qxi Qy j Qzk    =++  (Q) Qixy Q j Qk z Oxyz • With respect to the fixed OXYZ frame,        Q = Q i + Q j + Q k + Q i + Q j + Q k ( )OXYZ x y z x y z     Q i + Q j + Q k = (Q ) = * x y z Oxyz Rate of change with respect to rotating frame.   • Frame OXYZ is fixed. Q • If Q were fixed within Oxyz then ( )OXYZ is • Frame Oxyz rotates equivalent to velocity of a point in a rigid body attached about fixed axis OA with  to Oxyz and angular velocity Ω       ( ) Q i + Q j + Q k = Ω×Q • Vector Q t x y z varies in direction and • With respect to the fixed OXYZ frame, magnitude.     Q = Q + Ω×Q ( )OXYZ ( )Oxyz 15.5B Plane Motion Relative to a Rotating Frame • Frame OXY is fixed and frame Oxy rotates with angular velocity  Ω.  • Position vector rP for the particle P is the same in both frames but the rate of change depends on the choice of frame. • The absolute velocity of the particle P is     v = (r) = Ω× r + (r) P OXY Oxy

• Imagine a rigid slab attached to the rotating frame Oxy or F for short. Let P’ be a point on the slab which corresponds instantaneously to position of particle P.   v = (r) = P F Oxy velocity of P along its path on the slab  vP' = absolute velocity of point P’ on the slab • Absolute velocity for the particle P may be written as    v = v + v P P′ P F • Absolute acceleration for the particle P is   d   arrP =Ω× +Ω×( ) + ( r) OXY dt Oxy but,   = Ω×  +  (r )OXY r (r )Oxy  d  v =Ω× rr +( )    P Oxy [(r )Oxy ]= (r )Oxy + Ω×(r )Oxy  dt =vv + PP′ F         aP =Ω× r +Ω×( Ω× r) +2 Ω×( rr) +( ) Oxy Oxy • Utilizing the conceptual point P’ on the slab,       aP′ = Ω× r + Ω× (Ω× r )   a = (r) P F Oxy • Absolute acceleration for the particle P becomes      aP = aP′ + aP F + 2Ω×(r)Oxy    = aP′ + aP F + ac     a = 2Ω×(r) = 2Ω×v = c Oxy P F Coriolis acceleration • Consider a collar P which is made to slide at constant relative velocity u along rod OB. The rod is rotating at a constant angular velocity w. The point A on the rod corresponds to the instantaneous position of P. • Absolute acceleration of the collar is     a = a + a + a P A P F c where       2 aA = Ω× r + Ω× (Ω× r ) aA = rω  arP F = = 0 ( )Oxy

 = Ω× = ω ac22 v PcF au • The absolute acceleration consists of the radial and tangential vectors shown

• Change in velocity over dt is represented by the sum of three vectors  ∆v = RR′ +TT ′′ +T ′′T ′ • TT ′′ is due to change in direction of the velocity of point A on the rod,

TT ′′ ∆θ 2 lim = lim vA = rωω = rω = aA ∆t→0 ∆t ∆t→0 ∆t       2 recall, aA = Ω× r + Ω× (Ω× r ) aA = rω   at t, vv=A + u   at t+∆ tvv, ′′ =A′ + u • RR′ and T ′′T ′result from combined effects of relative motion of P and rotation of the rod  RR′ T ′′T ′  ∆θ ∆r  lim  +  = lim u +ω  ∆t→0 ∆t ∆t  ∆t→0 ∆t ∆t  = uω +ωu = 2ωu    a = 2Ω×v a = 2ωu recall, c P F c

Sample Problem 15.19

Disk D of the Geneva mechanism rotates with constant

counterclockwise angular velocity w D = 10 rad/s. At the instant when f = 150o, determine (a) the angular velocity of disk S, and (b) the velocity of pin P relative to disk S.

STRATEGY:    v = v ′ + v • The absolute velocity of the point P may be written as P P P s  • Magnitude and direction of velocity vP of pin P are calculated from the radius and angular velocity of disk D.  • Direction of velocity vP′ of point P’ on S coinciding with P is perpendicular to radius OP.  v • Direction of velocity P s of P with respect to S is parallel to the slot. • Solve the vector triangle for the angular velocity of S and relative velocity of P.

MODELING and ANALYSIS: • The absolute velocity of the point P may be written as    v = v + v P P′ P s • Magnitude and direction of absolute velocity of pin P are calculated from radius and angular velocity of disk D. vP = RωD = (50 mm)(10 rad s) = 500mm s

• Direction of velocity of P with respect to S is parallel to slot. From the law of cosines, 2 2 2 2 r = R + l − 2Rl cos30° = 0.551R r = 37.1 mm From the law of cosines, sinβ sin30° sin30° = sin β = β = 42.4° R r 0.742 The interior angle of the vector triangle is

γ = 90° − 42.4° − 30° = 17.6°

• Direction of velocity of point P’ on S coinciding with P is perpendicular to radius OP. From the velocity triangle, vv=sinγ = 500mm s sin17.6°= 151.2mm s PP′ ( ) 151.2mm s =rωω = ss37.1 mm   ωs =( −4.08rad s)k

vvPs= Pcosγ = ( 500m s) cos17.6°

  vPs =(477 m s)( − cos 42.4 °−ij sin 42.4 °)

vP = 500 mm s REFLECT and THINK: The result of the Geneva mechanism is that disk S rotates ¼ turn each time pin P engages, then it remains motionless while pin P rotates around before entering the next slot. Disk D rotates continuously, but disk S rotates intermittently. An alternative approach to drawing the vector triangle is to use vector algebra.

Sample Problem 15.20 In the Geneva mechanism, disk D rotates with a constant counter-clockwise angular velocity of 10 rad/s. At the instant when j = 150o, determine angular acceleration of disk S.

STRATEGY:     a = a ′ + a + a • The absolute acceleration of the pin P may be expressed as P P P s c • The instantaneous angular velocity of Disk S is determined as in Sample Problem 15.9. • The only unknown involved in the acceleration equation is the instantaneous angular acceleration of Disk S. • Resolve each acceleration term into the component parallel to the slot. Solve for the angular acceleration of Disk S.

MODELING and ANALYSIS: • Absolute acceleration of the pin P may be expressed as   aaaP=++ P′ Ps a c • From Sample Problem 15.9.   β = 42.4° ωS = (− 4.08rad s)k    v = (477mm s)(− cos42.4°i − sin 42.4° j ) P s • Considering each term in the acceleration equation, 2 2 2 aP = RωD = (500mm)(10rad s) = 5000mm s    a = (5000mm s2 )(cos30°i − sin30° j ) P   aP′ = (aP′ )n + (aP′ )t  2   (aP′ )n = (rωS )(− cos42.4°i − sin 42.4° j )    (aP′ )t = (rαS )(− sin 42.4°i + cos42.4° j )    = α − ° + ° (aP′ )t ( S )(37.1mm)( sin 42.4 i cos42.4 j )

note: aS may be positive or negative

• The direction of the Coriolis acceleration is obtained by rotating the direction of the relative velocity  vPs o by 90 in the sense of w S.    ac = (2ωSvP s )(− sin 42.4°i + cos42.4 j )   = 2(4.08rad s)(477mm s)(− sin 42.4°i + cos42.4 j )   = (3890mm s2 )(− sin 42.4°i + cos42.4 j )  a • The relative acceleration P s must be parallel to the slot. • Equating components of the acceleration terms perpendicular to the slot,

37.1αS + 3890 − 5000cos17.7° = 0 α = −233rad s S  α = − S ( 233rad s)k

REFLECT and THINK: • It seems reasonable that, since disk S starts and stops over the very short time intervals when pin P is engaged in the slots, the disk must have a very large angular acceleration. An alternative approach would have been to use the vector algebra approach.

15.6 Motion of a Rigid Body in Space

15.6A Motion About a Fixed Point • The most general displacement of a rigid body with a fixed point O is equivalent to a rotation of the body about an axis through O. • With the instantaneous axis of rotation and angular velocity the velocity   dr   v = = ω × r of a particle P of the body is dt

and the acceleration of the particle P is         dω a = α × r +ω ×(ω × r ) α = . dt  ω • The angular acceleration α represents the velocity of the tip of .  • As the vector ω moves within the body and in space, it generates a body cone and space cone which are tangent along the instantaneous axis of rotation. • Angular velocities have magnitude and direction and obey parallelogram law of addition. They are vectors.

15.6B General Motion • For particles A and B of a rigid body,    v = v + v B A B A • Particle A is fixed within the body and motion of the body relative to AX’Y’Z’ is the motion of a body with a fixed point     v = v +ω × r B A B A • Similarly, the acceleration of the particle P is    aB = aA + aB A       = a +α × r +ω × ω × r A B A ( B A ) • Most general motion of a rigid body is equivalent to: - a translation in which all particles have the same velocity and - acceleration of a reference particle A, and of a motion in - which particle A is assumed fixed.

Sample Problem 15.21

The crane rotates with a constant angular velocity w 1 = 0.30 rad/s and the boom is being raised with a constant angular

velocity w 2 = 0.50 rad/s. The length of the boom is l = 12 m. Determine: • angular velocity of the boom, • angular acceleration of the boom, • velocity of the boom tip, and • acceleration of the boom tip. STRATEGY:  With    ω1 = 0.30 j ω2 = 0.50k    r = 12(cos30°i + sin 30° j )   = 10.39i + 6 j

   • Angular velocity of the boom, ω = ω1 +ω2        α = ω1 +ω2 = ω2 = (ω2 )Oxyz + Ω ×ω2 = ω ×ω • Angular acceleration of the boom, 1 2    • Velocity of boom tip, v = ω × r           • Acceleration of boom tip, a = α × r +ω ×(ω × r ) = α × r +ω ×v

MODELING and ANALYSIS: • Angular velocity of the boom,       ω = (0.30rad s) j + (0.50rad s)k ω = ω1 + ω2

• Angular acceleration of the boom,        α = ω1 + ω2 = ω2 = (ω2 )Oxyz + Ω ×ω2     = ω1 ×ω2 = (0.30rad s) j × (0.50rad s)k  α = 2 (0.15rad s )i

• Velocity of boom tip ,    i j k    v = ω × r = 0 0.3 0.5 10.39 6 0     = −( ) + ( ) − ( )  v 3.54m s i 5.20m s j 3.12m s k = 0.50k • Acceleration of boom tip,           a = α × r + ω × (ω × r ) = α × r + ω × v       i j k i j k  a = 0.15 0 0 + 0 0.30 0.50 10.39 6 0 − 3 5.20 − 3.12      = 0.90k − 0.94i − 2.60i −1.50 j + 0.90k  22 2 a=−−+(3.54m s ) i( 1.50m s) jk( 1.80m s )

  ωω12=0.30jk = 0.50   r=10.39 ij + 6

15.7Motion Relative to A Moving Reference Frame 15.7A Three-Dimensional Motion Relative to a Rotating Frame With respect to the fixed frame OXYZ and rotating frame Oxyz,

    Q = Q + Ω×Q ( )OXYZ ( )Oxyz

Consider motion of particle P relative to a rotating frame Oxyz or F for short. The absolute velocity can be expressed as     vP = Ω× r + (r)Oxyz   = v +v P′ P F The absolute acceleration can be expressed as          aP = Ω× r + Ω× (Ω× r )+ 2Ω×(r)Oxyz + (r)Oxyz    = a p′ + aP F + ac      a = 2Ω×(r) = 2Ω×v = Coriolis acceleration c Oxyz P F

15.7B in General Motion • With respect to OXYZ and AX’Y’Z’,    rP = rA + rP A    vP = vA + vP A    a = a + a P A P A • The velocity and acceleration of P relative to AX’Y’Z’ can be found in terms of the velocity and acceleration of P relative to Axyz.      v = v + Ω× r + (r ) P A P A P A Axyz   = v + v P′ P F

Consider: - fixed frame OXYZ, - translating frame AX’Y’Z’, and - translating and rotating frame Axyz or F.

    aaP= A +Ω× r PA +Ω×( Ω× rPA)    +2 Ω×rrPA + PA ( )Axyz ( )Axyz (15.54)   =++aa′ a PPF c

Sample Problem 15.25 For the disk mounted on the arm, the indicated angular rotation rates are constant. Determine: • the velocity of the point P, • the acceleration of P, and • angular velocity and angular acceleration of the disk.     STRATEGY: aaP= A +Ω× r PA +Ω×( Ω× rPA)    +2 Ω×rrPA + PA ( )Axyz ( )Axyz   =++aaPP′ F a c • Define a fixed reference frame OXYZ at O and a moving reference frame Axyz or F attached to the arm at A. • With P’ of the moving reference frame coinciding with P, the velocity of the point P is found    v = v + v from P P′ P F     a = a ′ + a + a • The acceleration of P is found from P P P F c    • The angular velocity and angular acceleration of the disk are ω = Ω +ωD F  α = (ω ) + Ω ×ω MODELING and ANALYSIS: F • Define a fixed reference frame OXYZ at O and a moving reference frame Axyz or F attached to the arm at A.      r = Li + Rj rP A = Rj     ω = ω k Ω = ω1 j D F 2 • With P’ of the moving reference frame coinciding with P, the velocity of the point P is found from    vP = vP′ + vP F        vP′ = Ω × r = ω1 j ×(Li + Rj ) = −ω1L k       v = ω × r = ω k × Rj = −ω R i P F D F P A 2 2

   = −ω −ω vP 2R i 1L k

• The acceleration of P is found from     a = a + a + a P P′ P F c

  2 aP′ =Ω×( Ω×r) =ωω11 j ×( − Lk) =− ω 1 Li  arPFF=××ωω D( D F PA)

 2 =ωω22k ×−( Ri) =− ω 2 R j  avcP=2 Ω× F   =22ω1j ×−( ω 2 R i) = ωω12 Rk

 22  aP =−−+ω1 L i ω 2 Rj2 ωω 12 Rk

• Angular velocity and acceleration of the disk,      ω = Ω +ω ω = ω +ω D F 1 j 2k  αω= +Ω× ω  ( )F  =×+ω1j( ωω 12 jk)   α= ωω12i