ASEN 5022 - Spring 2005Dynamics of Aerospace StructuresLecture 04: January 20Review of Newtonian DynamicsTwo independent variables in EOM:Position vector (or displacement): rLinear momentum vector:p = m v, v =drdt(1)Newton’s second law:F =dpdt=d(mv)dt(2)Moment of a force and angular momentumAngular momentum:Ho= r × p = r × (m˙r)(3)Rate of angular momentum:˙Ho=˙r × p + r ×˙p = r ×˙p= r × F = Mo⇓M =˙Ho(4)Work and EnergyIncrement of work:¯dW = F · dr = m¨r · dr= md˙rdt·˙rdt = d(12m˙r ·˙r) = dT(5)Integrating both sides, we obtain:r2r1F · dr =˙r2˙r1d(12m˙r ·˙r)− (V2− V1) = T2− T1(6)Work and Energy - cont’dDefinition of potential energy:V(r) =rrefrF · dr⇓W =−dV(r)Work and Energy - cont’dExample of potential energy for a spring undergoingδ displacement:V(δ) =0δFdx =−0δkx dx =12kδ2(7)Work and Energy - cont’dLifting gravity force from y1to y2:y2y1F · dy =y2y1(−mgj) · dy(j)= -mg (y2− y1) =−(V2− V1)V2= mgy2, V1= mgy1Systems of discrete bodiesMass center: For a system consisting of N masseswhose position vectors are {ri, i = 1, 2, 3, ...N} ,the mass center is defined asN i=1miri= 0 (8)Momentum of N masses: ptotal= m vCm=Ni=1miKinetic energy of N massesri= rc+ ri, vi= vc+ viT =12N i=1mi(vC+ vi) · (vC+ vi)=12m vC· vC+ vC·12N i=1mivi+12N i=1mivi· vi(9)Kinetic energy of N masses- cont’dSincewehavefrom(8),Ni=1miri= 0, whichleadstoN i=1mivi= 0 (10)so that (9) reduces toT =12m vC· vC+12N i=1mivi· vi(11)Rotational moment of inertiaHO=mr × (dmv), v = ω × r=mr × (ω× r) dm(12)Let r andω be expressed asr = xi + yj + zkω = ωxi + ωyj + ωzk(13)Rotational moment of inertia-cont’dStep 1: Compute ω × rω × r = detijkxyzωxωyωz= (zωy− yωz)i + (xωz− zωx)j + (yωx− xωy)k(14)Step 2: Compute r × (ω × r)r × (ω × r) = [(y2+ z2)ωx− xyωy− xzωz]i+ [−yxωx+ (x2+ z2)ωy− yzωz]j+ [−zxωx− zyωy− (x2+ y2)ωz]k(15)Rotational moment of inertia-cont’dStep 3: Computemr × (ω × r)dmH = Hxi + Hyj + HzkHx= Ixxωx+ Ixyωy+ IxzωzHy= Iyxωx+ Iyyωy+ IyzωzHz= Izxωx+ Izyωy+ IzzωzIxx=m(y2+ z2) dm, Ixy=mxy dm, etc.(16)Other quantities may be computed by changing the subscripts from Ixxand Ixy.Rotational moment of inertia-cont’dStep 4: Compute MO=˙HO˙HO= (˙Hxi +˙Hyj +˙Hzk)+ (Hx˙i + Hy˙j + Hz˙k)(17)Since we have˙i =ω × i,˙j = ω × j,˙k = ω × k (18)˙HObecomesRotational moment of inertia-concludedMO=˙HO= (˙Hxi +˙Hyj +˙Hzk)+ω × (Hxi + Hyj + Hzk)(19)Kinetic energy of general rigid-body systemsKinematics:r = rC+ rv = vC+ ω × r(20)Kinetic energy:T =12m vC· vC+12m(ω ×ri) · (ω ×ri) dm(21)General rigid-body systems-cont’dFor a general point P that is different from point C,wehaveT =12m vP· vP+ vP· (ω × rPC)+12m(ω × r)2dm(22)where (x, y, z)-components of12m(ω × r)2dm are given by12m(ω × r)2dm =˜ω IP{ω}˜ω =0 −ωzωyωz0 −ωx−ωyωx0{ω}=ωxωyωzIP=IxxIxyIxzIxyIyyIyzIxzIyzIxzP(23)where subscript P denotes that the rotatioanl moments of inertia are computedwith respect to point