The moment of momentum of a rigid object B with respect to a point P that belongs to the object can be written as

in which dm is a small mass element, point Q is at the geometric center of the mass element, rQ/P is the position vector from P to Q, vQ is the velocity of Q, and the integration is over all small mass elements of the object B (Fig. 4.7). The relation that was earlier presented in the chapter about the velocity of two points in a rigid object can be used to reduce the moment of momentum expression into the following form:

Note that both vP and c, although they may vary with time, are constants with respect to the integration over the mass elements of object B. Thus we can take these terms outside the integral sign.

Note that the first term on the right-hand side of Eqn. 4.22 is always equal to zero for two cases: if point P is fixed in space then vP = 0, or if point

Figure 4.7. General plane motion of a rigid object B parallel to the (ej, e2) plane. The reference frame E is fixed on earth. The symbol C denotes the center of mass.

P coincides with the center of mass then the integral of position vector with respect to the center of mass must be equal to zero. Under one of these conditions, the moment of momentum expression reduces to

Let us next make the following substitutions into Eqn. 4.23:

in which e3 is the unit vector of the reference frame E that is perpendicular to the plane of motion. Note that these unit vectors can be taken outside the integral because they do not depend on the mass elements of the rigid object B.

The resultant expression for moment of momentum becomes H = /(x1e1 + x2e2 + x3e3) X (m e3) X (x1e1 + x2e2 + x3e3) dm (4.24) = af(—x1 x3) dm e1 + m/(—x2 x3) dm e2 + af (x12 + x22) dm e3

= a (¡13 e1 + ¡23 e2 + ¡33 e3) in which the mass moment of inertia components are defined as

Mass moment of inertia depends strictly on the geometry and the distribution of mass of the rigid object B. Note that if the plane of the motion coincides with a plane of symmetry of the object B, then ¡13 = ¡23 = 0. Figure 4.8 shows two objects where the plane of motion is also a plane of symmetry (a) and two others when the plane of motion is not a plane of symmetry (b). We focus on the former case. Thus, for the rigid bodies undergoing planar motion in the plane of symmetry we have

The parameter ¡33 depends on the geometry and mass distribution characteristics of the rigid object as well as the position of the point with respect to which it is calculated. The parameter ¡33 is denoted as ¡c when it denotes the mass moment of inertia with respect to an axis that is perpendicular to the plane of motion and passes through the center of mass. Moments of inertia of rigid objects of different shapes have been tabulated in Appendix 2 of this book. For a slender circular arc of angle 6 and radius r, the mass moment of inertia ¡c = mr2, where m is the mass of the arc and r is its radius. For a rectangular solid, ¡c = (m/12) (a2 + b2), where a and b denote the lengths of the sides of the solid in the plane of motion.

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The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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