Chemical Engineering - Apllied Mechanics

Equations of motion of rigid bodies

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1. DYNAMICS OF A RIGID BODY This chapter deals with the dynamic analysis of single rigid bodies. 1.1 Planar motion Let us consider a rigid body subjected to a planar motion. Let us denote by m and J G the mass and the mass moment of inertia evaluated about an axis passing through the centre of mass G of the body. Let us denote by ω =and ω =the absolute angular speed and absolute angular acceleration of the body, while=a G is the absolute acceleration of the barycentre G. If the rigid body is translating, any point of the body has the same acceleration, a P, and velocity, v P. If the rigid body is rotating or roto-translating, each point P of the body is characterized by a specific acceleration vector a P. Therefore, the infinitesimal inertia force associated with the infinitesimal mass V dm d ρ = is: P i d dm = − Fa .= Then, all the infinitesimal inertia forces i dF =form a distribution of vectors that are parallel only in the= case of a translational motion. Therefore, in a general case of planar motion, the distribution of all the= inertia forces acting on the rigid body is equivalent to an inertial=force,= GiF , applied to the centre of mass G, and an inertial torque, Gi M , applied to the body. The barycentrical inertial force is given by: GG i m = − Fa (1.1) The inertial torque, evaluated using a mass moment of inertia, J G, about a barycentrical axis, is given by: GG J i = − M ω (1.2) where  ω is the absolute angular acceleration of the body. Now, let us consider an isolated rigid body, that is a body for which any possible external constraint has been removed, and substituted with the corresponding reaction forces and torques, vj R and vk M , respectively, with j = 1, 2, …, n 1, and k = 1, 2, …, n 2 . Besides, if the rigid body is connected to other bodies, in order to isolate this body it is necessary to remove any possible internal joint and replace it with the corresponding internal forces and torques , sjF and sk M , respectively, with j = 1, 2, …, n 3, and k = 1, 2, …, n 4 . The rigid body can be subjected to driving and resistance external forces and moments, ext jF and ext k M , with j = 1, 2, …, n 5, and k = 1, 2, …, n 6 . Then, the effects of possible fields of forces (gravitational, electromagnetic, etc.) can be simulated by means of a suitable set of forces and moments, fjF and fk M , with j = 1, 2, …, n 7, and k = 1, 2, …, n 8 . Let us denote by P j the point to which each of the above mentioned forces are applied, and by ( ) POj− the position vector of these points with respect to an arbitrary pole O. The equations of motion of the rigid body can be expressed as: 1 35 7 vG 1 11 1 n nn n jsjext jf j i j=j=j=j=       ++ ++=             ∑ ∑∑ ∑ R R F F F0 (1.3) ( ) ( ) ( ) 1 35 v 1 11 POPOPO n nn jjjsjjext j j=j=j=   −∧ + −∧ + −∧ +     ∑ ∑∑ R RF () () 7 G 1 P O GO n jfji j= + − ∧ +− ∧ +  ∑ FF = 2 46 8 vG 1 11 1 n nn n kskext kf ki k=k=k=k=       + + + + +=            ∑ ∑∑ ∑ M M M M M0 =(1.4) = = Eq.(1.3)=expresses=the equilibrium of all the forces acting on the isolated rigid body. This vector equation in the two dimensional space can be decomposed into two scalar equations obtained projecting all the forces into two orthogonal directions.= Eq.(1.4) is the equilibrium of all the torques,=and the moments of all the forces,=about=pole O.= 1.2 3-D motion Consider a rigid body subjected to a roto-translational motion in the three dimensional space. Let us denote by m the rigid body mass, while the Inertia Tensor is: [] JJJ J JJJJJJxx xy xz yx yy yz zx zy zz   =   =(1.5)= The angular velocity vector can be expressed as:= xy z = ω+ ω+ ω i jk ω = (1.6)= where xω , yω , zω , are the angular velocity components about=the axes=x, y, z. The angular acceleration vector can be expressed as: xy z = ω+ ω+ ω i jk    ω = (1.7)= where xω , yω , zω , are the angular acceleration components about =the axes=x, y, z. The effects of all the infinitesimal inertia forces i dF , the rigid body is subjected to,=are equivalent to a single inertial force and an inertial torque. The=inertial force, GiF , applied to=the centre=of mass G=is= given by:= GGi m = − Fa = (1.8)= The inertial torque,= i M , is can be expressed as:= 2 2 2 J J J 0J J JJ J J Μ J J JJ J J Μ ixxx xy xz xzy yz x iiyyx yy yz yzxxz y izzx zy zz zyx xyz      Μ −− ω− ω       = Μ =−− − ω + −ω           Μ −− ω − ω       M    = (1.9)= If a barycentrical principal inertial frame=is used,=all the cross-products J ik are null. That is, the origin of the reference system must coincide with the barycentre G of the body and the axes x, y, z must be oriented as the principal axes. Then, the equations of motion can be written considering the equilibrium of all the forces acting on the isolated rigid body. This vector equation in the three dimensional space can be decomposed into three scalar equations obtained projecting all the forces into three orthogonal directions. Besides, the equilibrium of all the torques and of the moments of all the forces about a pole O can be written. This gives rise to a further vector equation in the three dimensional space that can be decomposed into three scalar equations obtained projecting all the moments into three orthogonal directions. The above mentioned six scalar equations allow the motion of the rigid body, which has six degrees of freedom, to be studied.