Chapter 18. Kinetics of Rigid Bodies in Three Dimensions Introduction

Chapter 18. Kinetics of Rigid Bodies in Three Dimensions

Introduction

Rigid Body Angular Momentum in Three Dimensions Principle of Impulse and Momentum Kinetic Energy Motion of a Rigid Body in Three Dimensions Euler’s Equations of Motion and D’Alembert’s Principle Motion About a Fixed Point or a Fixed Axis Motion of a Gyroscope. Eulerian Angles Steady Precession of a Gyroscope Motion of an Axisymmetrical Body Under No Force

Three dimensional analyses are needed to determine the forces and moments on the gimbals of gyroscopes, the joints of robotic welders, and the supports of radio telescopes.

Introduction • The fundamental relations developed for the plane motion of rigid bodies may also be applied to the general motion of three dimensional bodies.   H = Iω   • The relation G which was used F = ma ∑ to determine the angular momentum of a   ∑ M G = HG rigid slab is not valid for general three 18.1 dimensional bodies and motion. The current chapter is concerned with evaluation of the angular momentum and its rate of change for three dimensional motion and application to effective forces, the impulse-momentum and the work- energy principles. 18.1A Angular Momentum of a Rigid Body in Three Dimensions Angular momentum of a body about its mass center,  n   n    HG = ∑(ri′×vi∆mi ) = ∑[ri′×(ω × ri′)Δmi ] i=1 i=1 The x component of the angular momentum, n  ′  ′ H x = ∑[yi (ω × ri )z − zi (ω × ri )y ]Δmi i=1 n = ∑[yi (ω x yi −ω y xi )− zi (ω z xi −ω x zi )]Δmi i=1 n n n 2 2 = ω x ∑ (yi + zi )Δmi −ω y ∑ xi yiΔmi −ω z ∑ zi xiΔmi i=1 i=1 i=1 =ω22 +− ωω − Hxx∫( y z) dmy ∫∫ xy dm z zx dm

=+−−IIxωωω x xy y I xz z

HIy=−+− yxωω x I y y I yz ω z

HIz=−−+ zxω x I zy ωω y I z z  ω Transformation of into is characterized by the inertia HG tensor for the body,  + I − I − I   x xy xz  − I yx + I y − I yz     − I zx − I zy + I z 

• With respect to the principal axes of inertia,  I 0 0   x′   0 I y′ 0  HIx′= xx ′′ωωω HI y ′ = yy ′′ HI z ′ = zz ′′    0 0 I ′  z  The angular momentum HG of a rigid body and its  angular velocity ω have the same direction if, and only if,  ω is directed along a principal axis of inertia. The momenta of the particles of a rigid body can be reduced to:  L = linear momentum  =  mv HG = angular momentum about G H x = +I xω x − I xyω y − I xzω z

H y = −I yxω x + I yω y − I yzω z H = −I ω − I ω + I ω z zx x zy y z z

• The angular momentum about any other given point O is     HO = r × mv + HG

The angular momentum of a body constrained to rotate about a fixed point may be calculated from • Or, the     angular momentum may be HO = r × mv + HG computed directly from the moments and products of inertia with respect to the Oxyz frame.

 n   HO = ∑(ri ×viΔm) i=1 n    = ∑[ri ×(ω × ri )Δmi ] i=1

HIx=+−− xωωω x I xy y I xz z

HIy=−+− yxωω x I y y I yz ω z

HIz=−−+ zxω x I zy ωω y I z z

18.1B Principle of Impulse and Momentum

• The principle of impulse and momentum can be applied directly to the three-dimensional motion of a rigid body,

Syst Momenta1 + Syst Ext Imp1-2 = Syst Momenta2 • The free-body diagram equation is used to develop component and moment equations. • For bodies rotating about a fixed point, eliminate the impulse of the reactions at O by writing equation for moments of momenta and impulses about O.

18.1C Kinetic Energy Kinetic energy of particles forming rigid body, n = 1 2 + 1 ′2 T 2 mv 2 ∑Δmivi i=1 n = 1 2 + 1 ω × ′2 2 mv 2 ∑ ri Δmi i=1 = 1 2 + 1 ω 2 + ω 2 + ω 2 − ω ω 2 mv 2 (Ix x I y y Iz z 2Ixy x y − 2I ω ω − 2I ω ω ) yz y z zx z x • If the axes correspond instantaneously with the principle axes, = 1 2 + 1 ω 2 + ω 2 + ω 2 T 2 mv 2 (I x′ x′ I y′ y′ I z′ z′ )

• With these results, the principles of work and energy and conservation of energy may be applied to the three-dimensional motion of a rigid body.

• Kinetic energy of a rigid body with a fixed point, = 1 ω 2 + ω 2 + ω 2 − ω ω T 2 (I x x I y y I z z 2I xy x y − 2I ω ω − 2I ω ω ) yz y z zx z x • If the axes Oxyz correspond instantaneously with the principle axes Ox’y’z’, T = 1 (I ω 2 + I ω 2 + I ω 2 ) 2 x′ x′ y′ y′ z′ z′

Sample Problem 18.1 Rectangular plate of mass m that is suspended from two wires is hit at D in a direction perpendicular to the plate. Immediately after the impact, determine a) the velocity of the mass center G, and b) the angular velocity of the plate.

STRATEGY: • Apply the principle of impulse and momentum. Since the initial momenta is zero, the system of impulses must be equivalent to the final system of momenta. • Assume that the supporting cables remain taut such that the vertical velocity and the rotation about an axis normal to the plate is zero. • Principle of impulse and momentum yields to two equations for linear momentum and two equations for angular momentum. • Solve for the two horizontal components of the linear and angular velocity vectors.

MODELING and ANALYSIS: • Apply the principle of impulse and momentum. Since the initial momenta is zero, the system of impulses must be equivalent to the final system of momenta. • Assume that the supporting cables remain taut such that the vertical velocity and the rotation about an axis normal to the plate is zero.       ω = ω i + ω j v = vxi + vzk x y Since the x, y, and z axes are principal axes of inertia,      = ω + ω = 1 2ω + 1 2ω HG Ix xi I y y j 12 mb xi 12 ma y j

• Principle of impulse and momentum yields two equations for linear momentum and two equations for angular momentum. • Solve for the two horizontal components of the linear and angular velocity vectors. 1 −=1 2 bFΔ t= H x 2 aFΔ t H y 0 = mvx −=FΔ t mvz 1 2 1 2 = 12 mb ωx = 12 ma ωy vx = 0 vz = − FΔ tm

  ω = 6FΔ t mb ωy = −(6FΔ t ma) v= −( FΔ tmk) x

 6FtΔ  ω =(ai + b j) mab

REFLECT and THINK: • Equating the y components of the impulses and

momenta and their

moments about the z axis,

you can obtain two additional equations that   − yield TA = TB = ½W.v (FΔt m)k • This verifies that the wires remain taut and that the initial assumption was correct. If the impulse were  6FΔt   at G,ω this= would(a ireduce+ b j ) to a  mab  two-dimensional1122 problem. HGx=12 mbωω i + 12 ma y j

Sample Problem 18.2 A homogeneous disk of mass m is mounted on an axle OG of negligible mass. The disk rotates counter-clockwise

at the rate w1 about OG. Determine: a) the angular velocity of the disk, b) its angular momentum about O, c) its kinetic energy, and d) the vector and couple at G equivalent to the momenta of the particles of the disk. STRATEGY: • The disk rotates about the vertical axis through O as well as about OG. Combine the rotation components for the angular velocity of the disk. • Compute the angular momentum of the disk using principle axes of inertia and noting that O is a fixed point. • The kinetic energy is computed from the angular velocity and moments of inertia. • The vector and couple at G are also computed from the angular velocity and moments of inertia. MODELING and ANALYSIS: • The disk rotates about the vertical axis through O as well as about OG. Combine the rotation components for the angular velocity of the disk.    ω = ω1i + ω2 j Noting that the velocity at C is zero,    vC = ω × rC = 0     0 = (ω1i + ω2 j )× (Li − rj )  = (Lω − rω )k 2 1 ω2 = rω1 L   ωω=i − ( r ω Lj) 11

• Compute the angular momentum of the disk using principle axes of inertia and noting that O is a fixed point.     = ω + ω + ω HO Ix xi I y y j Iz zk 1 2 Hx= I xxωω = ( 2 mr ) 1

221 Hy==+− I yyωω( mL4 mr)( r1 L)    ω = ω i − (rω L) j 221 1 1 Hz==+= I zzω ( mL4 mr )00   112 22 HO =24 mrωω11 i −+ m( L r)( r L) j

• The kinetic energy is computed from the angular velocity and moments of inertia.

= 1 ω 2 + ω 2 + ω 2 T 2 (Ix x I y y Iz z )

= 1 [mr2ω 2 + m(L2 + 1 r2 )(− rω L)2 ] 2 1 4 1

r 2 =1 22 + ω T8 mr 6 2 1 L The vector and couple at G are also computed from the angular velocity and moments of inertia.   mv = mrω1k   HG=++ I xx′′ωωω iI yy jI zz ′ k  1122 =24mrωω1 i +− mr( r L) j     r ω = ω − ω 1 2 1i (r 1 L) j HG =2 mrω1  i − j 2L

REFLECT and THINK: • If the mass of the axle were not negligible and it was instead

modeled as a slender rod with a mass Maxle, it would also contribute to the kinetic energy 2 2 Taxle = ½(1/3 MaxleL ) ω2 and the momenta 2 Haxle = ½(1/3 MaxleL ) ω2 of the system.

18.2 Motion of a Rigid Body in Three Dimensions

  ∑ F = ma   ∑ M = HG

18.2A Rate of Change of Angular Momentum Angular momentum and its rate of change are taken with respect to centroidal axes GX’Y’Z’ of fixed orientation.   Transformation of ω into HG is independent of the system of coordinate axes. Convenient to use body fixed axes Gxyz where moments and products of inertia are not time dependent.  • Define rate of change of change of HG with respect to the rotating frame,         (HG )Gxyz = H xi + H y j + H zk

Then,       HG = (HG )Gxyz + Ω × HG Ω = ω

18.2B Euler’s Eqs of Motion   With Ω = ω and Gxyz chosen to correspond to the principal axes of inertia,     ∑ M = (H ) + Ω × H G G Gxyz G Euler’s Equations: ∑ M x = I xω x − (I y − I z )ω yω z

∑ M y = I yω y − (I z − I x )ω zω x M = I ω − I − I ω ω ∑ z z  z ( x y ) x y • System of external forces and effective forces are equivalent for general three dimensional motion. • System of external forces are equivalent to the vector and couple,   ma and HG .

18.2C,D Motion About a Fixed Point or a Fixed Axis For a rigid body rotation around a fixed point,   ∑ M O = HO    = (H ) + Ω × H O Oxyz O For a rigid body rotation around a fixed axis,

HIx= − xzω HIy= − yzω HIzz= ω

   ∑ MHOO= +×ω H O ( )Oxyz  =−−+( Ixz i I yz j Ik z )ω 

+ωωk ×−( Ixz i − I yz j + Ik z )

 2 =−( IiIjIkxz − yz + z )αω +−( IjIixz + yz )

2 ∑ MIIx=−+ xzαω yz 2 ∑ MIIy=−+ yzαω xz

∑ MIzz= α

If symmetrical with respect to the xy plane, M = 0 M = 0 M = I α ∑ x ∑ y ∑ z z • If not symmetrical, the sum of external moments will not be zero, even if α = 0, M = I ω 2 M = I ω 2 M = 0 ∑ x yz ∑ y xz ∑ z

• A rotating shaft requires both static (ω = 0)and dynamic (ω ≠ 0) balancing to avoid excessive vibration and bearing reactions.

Sample Problem 18.3 Rod AB with mass m = 20 kg is pinned at A to a vertical axle which rotates with constant angular velocity ω = 15 rad/s. The rod position is maintained by a horizontal wire BC. Determine the tension in the wire and the reaction at A.

STRATEGY:  • Evaluate the system of effective forces by reducing them to a vector ma attached at G  and couple HG . • Expressing that the system of external forces is equivalent to the system of effective forces, write vector expressions for the sum of moments about A and the summation of forces. • Solve for the wire tension and the reactions at A. MODELING and ANALYSIS: • Evaluate the system of effective forces by reducing them to   a vector ma attached at G and couple HG.

  HG=++ I xxωωω iI yy jI zz k 1122 Ix= 12 mL I yz= 0 I= 12 mL

ωx=−== ωcos βω yz ω sin βω 0  1 2  HG = − 12 mLωβcos i

Expressing that the system of external forces is equivalent to the system of effective forces, write vector expressions for the sum of moments about A and the summation of forces.   M = (M ) ∑ A ∑ A eff       1.732 J×−( TI) + 0.5 I ×−( 196.2 J) = 0.866 J ×−( 2250 I) + 649.5 K  (1.732TK−=+ 98.1) ( 1948.5 649.5) K

 T =1557 N ∑∑FF= ( ) eff     AXY I++ AJ AK Z −1613 I − 196.2 J =− 2250 I   AI=−+697 N 196.2 N J ( ) ( ) •

Gyroscopes are used in the navigation system of the Hubble telescope, and can also be used as sensors . Modern gyroscopes can also be MEMS (Micro Electro- Mechanical System) devices, or based on fiber optic technology.

* Motion of a Gyroscope. Eulerian Angles * A gyroscope consists of a rotor with its mass center fixed in space but which can spin freely about its geometric axis and assume any orientation. From a reference position with gimbals and a reference diameter of the rotor aligned, the gyroscope may be brought to any orientation through a succession of three steps:

a) rotation of outer gimbal through φ about AA’, b) rotation of inner gimbal through θ about BB’, c) rotation of the rotor through ψ about CC’. • φ θ ψ : Eulerian Angles

φ = rate of precession θ = rate of nutation Ψ = rate of spin

The angular velocity of the gyroscope,       ω = φK +θ j +Ψ k    with K=−+ sinθθ ik cos   ω =− φθθ + + Ψ+ φ  θ sin ij ( cos ) k Equation of motion,     ∑ M = (H ) + Ω × H O O Oxyz O    ′′  HO =− Iφθsin iIjI + θ +( Ψ+ φcos θ) k  Ω=φθKj +

′      MIx =−(φθθφθθsin + 2 cos ) +I( Ψ+ φθcos ) ∑ MI=′ θφ −  2 sincos θ θ+I φ sin θ Ψ+ φ  cos θ ∑ y ( ) ( ) d   ∑ MIz =( Ψ+φθcos ) dt

Steady Precession of a Gyroscope Steady precession,

θ ,φ,ψ are constant    ω = −φsinθ i + ωzk    HO = −I′φsinθ i + Iωzk    Ω = −φsinθ i + φcosθ k

   ∑ MO = Ω × HO  ′   = (Iωz − I φ cosθ )φ sinθ j Couple is applied about an axis perpendicular to the precession and spin axes

When the precession and spin axis are at a right angle, θ = 90°     ∑ M O = IΨφ j Gyroscope will precess about an axis perpendicular to both the spin axis and couple axis.

Motion of an Axisymmetrical Body Under No Force Consider motion about its mass center of an axisymmetrical body under no force but its own weight, e.g., projectiles, satellites, and space craft.   H = 0 H = constant G G  Define the Z axis to be aligned with HG and z in a rotating axes system along the axis of symmetry. The x axis is chosen to lie in the Zz plane. HG sinθ ′ ω x = − H x = −HG sinθ = I ω x I′ ′ HIyy=0 = ω ωy = 0 θ HH=cosθω = I HG cos zG z ωz = • θ = constant and body is in steady precession. I ω I − x = tanγ = tanθ ω I′ • Note: z • Two cases of motion of an axisymmetrical body which under no force which involve no precession: • Body set to spin about its axis of symmetry, ω x = H x = 0   and body keeps spinning ω and HG are aligned about its axis of symmetry.

• Body is set to spin about its transverse axis, ω z = H z = 0   ω and HG are aligned and body keeps spinning about the given transverse axis.

The motion of a body about a fixed point (or its mass center) can be represented by the motion of a body cone rolling on a space cone. In the case of steady precession the two cones are circular.

• I < I’. Case of an elongated body. γθ< and the vector ω lies inside the angle ZGz. The space cone and body cone are tangent externally; the spin and precession are both counterclockwise from the positive z axis. The precession is said to be direct. • I > I’. Case of a flattened body. γθ< and the vector ω lies outside the angle ZGz. The space cone is inside the body cone; the spin and precession have opposite senses. The precession is said to be retrograde.