Euler's Dynamic Equations - Theory - helix

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4. Differential Forms of Euler’s LawsThe first law is easily expressed as a derivative,P ~F=ddt ~ L ~ L´m~vC (20)where ~L is defined as the linear momentum.The derivative form of the second law more complex. Usingthe derivative product formula,ddt (~a£~ b)= ( d dt ~a£~ b)+(~a£ddt ~ b), and referring to the figure with~r= ~ R¡ ~ R P , the second lawdevelops asP ~ M P= R ~r£~adm= R ~r£ d dt ~vdm= R ddt (~r£~v)¡ d dt (~r)£~vdm= R d(~r£~v)dm¡R ddt (~ R¡ R ~ P)£~vdmR=dtR d (~r£~v)dm¡ (~v¡~vP )£~vdm= d dt ~ H P +~v P £ R ~vdm= d dt ~ H P +~v P £m~v C (21)where the angular momentum is defined by~H P ´R~r£~vdm(22)There are two important special cases that are used almost exclusivelyfor rigid body problems, eitherP =C orP = O is a pointfixed in the inertial frame,P ~ MC = d dt ~ H C (23)P ~ M O = d dt ~ H O (24)Like Euler’s 2 nd law, there are two important special cases thatare used almost exclusively for rigid body problems, eitherP =CorP =O is a point fixed in the inertial frame,~H C =I C ~! (29)~H O =I O ~! (30)6. Acceleration Form of Euler’s LawsEuler’s 1 st law in acceleration form, P F=m~aC ~ , was introducedwhen considering the center-of-mass. For a point P on the body,the 2 nd law is derived by differentiating the angular momentum ofa rigid body and transferring the inertial derivative to the movingbody,PMP ~ = d ~ dtH P +~v P £m~v C= d dt (I P~!+m~r C £~v P )+~v P £m~v C= d0dt (I P~!)+ ~!£(I P ~!)+m~r C £~a P= I P ~®+ ~!£I P ~!+m~r C £~a P (31)where ~®´d ~! =d0~! is the angular acceleration, the prime denotesa derivative with respect to the moving body, and d0dt I P =0.dt dtThe expression simplifies for the two special cases where P becomeseither a pointO fixed in the inertial frame,PM ~ O =I O ~®+ ~!£I O ~! (32)orP becomes the center-of-mass C,P ~ F=m~a C (33)5. Angular Momentum for a Rigid BodyIt is desireable to avoid the explicit integration in applying Euler’s2 nd law. This can be done for a rigid body by selecting point P onthe (extended) body and developing an alternative expression forangular momentum,~H P ´ R~r£~vdm= R ~r£(~v P +~!£~r)dm= R ¡~r£(~r£~!)dm+ R ~r£~v P dm= R (~r£) T (~r£)~! dm+( R ~rdm)£~v P (25)P ~ MC =I C ~®+ ~!£I C ~! (34)where the acceleration forms of the 1 st and 2 nd laws expressed atCare grouped together for comparison.The 2 nd law is a bit more complex due to the inertia matrix. Inthe special case of a sphere, the inertia matrix is a scalar multiple ofthe identity matrixI C =I1 and the takes on the form of the 1 st law,P ~ F = m~a C (35)P ~ MC = I~® (36)= ( R (~r£) T (~r£)dm)~!+m~r C £~v P= I P ~!+m~r C £~v P (26)where the inertia matrix is defined as the integralI P ´R(~r£) T (~r£)dm (27)(~r£)is the skew-symmetric form of the cross product operator (i.e.(~a£) ~ b=~a£ ~ bwith(~a£) T =¡(~a£ ), and~! is the angular velocityof the body with respect to the inertial frame. Often the inertiamatrix is expressed in the formI P = R (~r T ~r1¡~r~r T )dm (28)where1 is the identity matrix, which follows from the derivationusing the vector identity~a£( ~ b£~c)=(~a ~ b T ¡~a T ~ b1)~c.For commonly shaped bodies of uniform density, inertia matricesare listed in tables (using principal axes) and the need forexplicit integration is eliminated.27. Inertia TensorThere are various types of quantities that can be associated withpoints in space. The temperature of a particle of matter is a scalarquantity associated with a point. The acceleration of a particle is avector quantity associated with a point and indicates a scalar magnitudeina specific direction. The inertia of a body is a tensor quantityassociated with a point and indicates a scalar magnitude for everydirection. An inertia matrix is a convenient representation of aninertia tensor, just as x, y, and z components are a convenient arepresentation of a vector.Consider a body that rotates about an axis through the centerof-massin the û direction. The angular velocity and accelerationareP~®=®û and ~! = !û. The applied moment about the axis isM C =û T ( P MC ~ ). Using these with the acceleration form of
the 2 nd law gives a scalar equation,P M C =I C (û)® (37)whereI C (û)´û T I Cûis the scalar inertia about the axisû:Therefore,the inertiaI C (û)is a ‘resistance ’to angular accelerationaboutan axis just as massm is a ‘resistance ’ to linear acceleration.Unlike mass, every direction is associated with another scalarinertiaI C (û). Therefore, the inertia tensor contains the informationfor inertia in all directions. Just as an arrow with magnitude anddirection can be used to visualize a vector, a bundle of arrows canbe used to visualize an inertia tensor. Further, the tips of the arrowslie on the surface of an ellipsoid, so sometimes the inertia tensor isvisualized as an ellipsoid. The major, minor, and intermediate axesof the ellipsoid define the principal axes of inertia which are determinedby a solving an eigenvalue problem. The inertias along thesedirections take on maximum, minimum, and stationary values.8. Principal Axescan be combined into a single equationU¤=I C U, where matrices¤´diag(¸1;¸2;¸3)andU´[û 1 û 2 û 3 ]. However due tothe orthogonality of the û i , the matrixU is also orthonormal soU T U=UU T =1 andU ¡1 =U T . This allows the inertia matrixbe expressed in terms of its eigensystemI C =U¤U T (39)Since the û i are orthonormal,U is the direction cosines of aset of orthogonal axes. Further, if these principal axes are used as acoordinate frame thenU=1 andI C =¤:Inertia matrices are usuallylisted in tables by only giving the principal moments of inertia.The difference in representing inertia by either¤or a general formI C has only to do with the orientation of the coordinate system.9. Parallel Axis TheoremThe inertia matrix is usually tabulated for the center-of-mass. Theinertia matrix for other points on the (extended) body can be determinedby using the parallel axis formulation. Let~r=~r PC +~r C¢where ~r PC points from P to C and~r C¢ points from C to pointson the body. Substituting this into the latter expression forI P , expanding,and using R dm = m, R R~r C¢ dm = ~r CC = ~0, I C =(~rTC¢~r C¢ ¡~r C¢ ~r T C¢)dm yields,I P =I C +m(~r T PC ~r PC1¡~r PC ~r T PC) (40)Introducing the notationI xx ,I xy ,:::for the components ofI C and~r PC =[ ¹x ¹y ¹z ] T , then the parallel axis theorem becomes,2I P = 4 I 3 23xx I xy I xzI yx I yy I yz5+m4 ¹y2 +¹z 2 ¡¹x¹y ¡¹x¹z¡¹y¹x ¹x 2 +¹z 2 ¡¹y¹z 5I zx I zy I zz ¡¹z¹x ¡¹z¹y ¹x 2 +¹y 2 (41)Note since inertia matrices are symmetric that actual computationsare reduced.The inertia matrix has two fundamental properties that follow directlyfrom its definition: symmetry, I C = I T Cand positive definiteness,I C (û) > 0 for all directions û. Additional propertiesfollow from examining the stationary properties of inertia which inturn leads to an eigenvalue problem.It is desired to determine the directions for which the inertiais stationary, i.e. maximal, minimal, and inflectional. For brevitylet ¸ ´ I C (û). Thus it is necessary to get the stationary valuesof ¸ = û T I Cû forû Tû=1. A straightforward way to solve thisconstrained optimization problem is to use the Lagrange multiplertechnique. Another is to multiply the second equation by¸and set itequal to the first,û T I Cû=¸û Tû. Taking the partials of the equationwith respect toû in component form, u i ;i = 1;2;3, setting@¸@u i= 0, and then collecting the equations yields the eigenvalueproblem,¸û=I Cû or (I C ¡¸1)û=~0 (38)The solution method is well-known. Forû6=~0, then(I C ¡¸1)must be singular which means that its determinant vanishes. ExpandingjIC ¡¸1j=0 yields a cubic equation in¸ whose solutionsare ¸i. Backsubstitution into the eigenvalue equation allows solutionfor the eigenvectors û i which turn out to be orthonormal,û T iû j = ± ij . Theû i are the directions of the principal axes of inertiaand the stationary values¸i are called the principal momentsof inertia.The three eigenvalue-eigenvector pairs satisfy¸iû i =I Cû i and310. Kinetic Energy and PowerThe kinetic energyT of a rigid body canbe expressed in terms of thecenter-of-massC by summing the kinetic energies of the individualparticles,T = 1 2R ~v T ~vdm= 1 2R (~vC +~!£~r C¢ ) T (~v C + ~!£~r C¢ )dm= 1 2R (~vTC ~v C +~v T C(~!£~r C¢ )+(~!£~r C¢ ) T ~v C +(~!£~r C¢ ) T (~!£~r C¢ ))dmR= 1 2 (~vTC ~v C +2~v C(~!£~r T C¢ )+ ~! T (~r C¢ £) T (~r C¢ £)~!)dm= 1 2 ~vT C ~v RC dm+~vTC (~!£ R ~r C¢dm)+ 1 2 ~!T ( R (~r C¢ £) T (~r C¢ £)dm)~!
Page 4: = 1 2 m~vT C ~v C + 1 2 ~!T I C ~!

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