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. HO = Ixωxi + I yω y j + Izωzk 1 2 Hx= I xxωω = ( 2 mr ) 1 221 Hy==+− I yyωω( mL4 mr)( r1 L) ω = ω i − (rω L) j ==+=ω 221 1 1 Hz I zz ( mL4 mr )00 2 22 11 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 2 2 2 1 2 2 = [mr ω1 + m(L + r )(− rω1 L) ] 2 4 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.