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rigid body in Software
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DataMatrix Reader In Visual Basic .NET Using Barcode scanner for VS .NET Control to read, scan read, scan image in .NET applications. Barcode Generation In Java Using Barcode creation for Android Control to generate, create barcode image in Android applications. extremely good reason to do this in terms of the kinetic energy The coordinates of the mass center are described by xCR = 1 1 body xdm = m rigid body xrdV m rigid ycg = 1 1 body ydm = m rigid body yrdV m rigid (1123) Code 39 Printer In Objective-C Using Barcode printer for iPhone Control to generate, create ANSI/AIM Code 39 image in iPhone applications. USS Code 39 Creation In None Using Barcode encoder for Microsoft Excel Control to generate, create Code 39 image in Excel applications. for any coordinate system (x,y) in the plane of motion If the object has constant density, this can be expressed solely in terms of the volume V xcg = 1 V rigid body
ycg =
rigid body
(1124) in which case the mass center is said to coincide with the volume centroid of the body Rearranging Eq 1123, rigid body
( x - x )dm = ( y - y )dm = 0
cg cg rigid body
(1125) from which it follows that
rigid body
rcg dm =
( x - xcg ) ( y - y ) dm = 0 cg rigid body
(1126) where the vector rcg represents the vector from the center of mass to an arbitrary point on the body This relationship, in turn, enables the kinetic energy for the general case of translation and rotation in a plane (Fig 115) to be simpli ed by splitting it into three terms: V = Vcg + w rcg Vcg w rcg Vcg rcg w rcg
FIGURE 115
Rigid body planar motion: translation with rotation
CAM SYSTEM MODELING
KE = = = =
1 (v cg + w rcg ) (v cg + w rcg ) dm 2 rigid body 1 [(v cg vcg ) + ( w rcg ) (w rcg ) + 2(vcg (w rcg ))] dm 2 rigid body 1 2 1 mvcg + Jcgw 2 + v cg (w rcg ) dm 2 2 rigid body 1 2 1 mvcg + Jcgw 2 + 0 2 2 (1127) The rst term is simply the translational kinetic energy of the body, the second term is the rotational kinetic energy of the body and the third term goes to zero because of the property of the center of mass given in Eq 1126 As a result, the kinetic energy of a body can be described in terms of the mass of the body, the velocity at the center of mass, the moment of inertia about the center of mass and the angular velocity In modeling, this result provides justi cation for describing a rigid body or a collection of rigidly connected rigid bodies by a single point mass at the center of mass and a moment of inertia about the center of mass In terms of Newtonian mechanics, mass and inertia play integral roles in the formation of equations of motion For translation, mass gives the relationship F = ma = m r (1128) between the force vector F applied to the rigid body and the acceleration vector a located at the body s center of gravity The acceleration can also be expressed in terms of the second derivative of the vector r, which represents the vector from any xed inertial frame of reference to the center of mass of the body For rotation in the plane, the equivalent expression is derived from the rate of change of angular momentum, giving Tcg = Jcgw (1129) where Tcg is the net torque about the center of mass and w is the angular velocity Since motion is restricted to the plane, rotation can occur only about one axis, so this is a scalar equation in contrast to Eq 1128 When motion of the body does not lie in a single plane, the situation becomes a bit more complicated, since the proper representation of inertia is then a tensor, not a scalar 1131 Finding Mass and Moment of Inertia Finding the mass of a component is straightforward The element may either be weighed directly on a scale or the mass can be determined from knowledge of the element s density and geometry In this latter case, the mass is given simply by integrating the density over the volume of the body: m=
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