Conservation of Angular Momentum

The angular momentum of a rigid object with respect to point O is defined as

where rP/O is the position vector from point O to a point P in the body B, EvP is the velocity of point P in reference frame E, and dm is a small mass element surrounding point P (Fig. 9.6). According to the parallelogram law rP/O = rC/O + p where p denotes the position vector from C to P. Note that from the definition of the center of mass Jpdm = 0. The integration of velocity over the mass elements of the body is by definition equal to the mass of the body times the velocity of the center of mass. Using these two relationships in Eqn. 9.12 we obtain:

Ho = J(rC/O + p) X (EvP) dm = m rC/O X EvC + Jp X EvP dm in which m denotes the mass of the body B.

The velocity of a point P in body B can be expressed as a function of the velocity of the center of mass and the angular velocity of the body B:

Figure 9.6. A rigid body undergoing three-dimensional motion with respect to the reference frame E that is fixed on earth. The point O is the origin of the Cartesian coordinate E. The rigid body and the reference frame embedded onto it are denoted by B. The symbol P denotes an ordinary point on B whereas C refers to the center of mass.

Figure 9.6. A rigid body undergoing three-dimensional motion with respect to the reference frame E that is fixed on earth. The point O is the origin of the Cartesian coordinate E. The rigid body and the reference frame embedded onto it are denoted by B. The symbol P denotes an ordinary point on B whereas C refers to the center of mass.

The angular momentum about point O can then be written as

where Hc is the angular momentum about the center of mass:

Expressing p and E«B in terms of unit vectors bi associated with rigid body B, the angular momentum with respect to the center of mass can be written as

where b1, b2, and b3 are a set of orthogonal unit vectors defined in B. The scalar components of the angular momentum are given by the following relationships:

where

and the symbols such as Ic13 refer to the components of mass moment of inertia. This matrix is defined as follows:

The integration in this equation is over the mass of the body. The symbol Xi denotes the distance from the center of mass of the body to the center of mass of the small mass element dm along the bi direction. The entity Sij is equal to one when i = j, and is equal to zero otherwise.

Also, summation over i = 1, 2, and 3 is implied in Eqn. 9.15a when an index such as i or j is repeated twice in a term. Thus, for example:

Ic11 = f[X1X1 + X2X2 + X3X3 - X1X1] dm = f[X22 + X32]dm

Overall, there are nine elements of the inertia matrix, but because the matrix is symmetric, only six independent elements need to be determined. The components of mass moment of inertia are functions of geom etry of the body and the mass distribution within the body. They do not depend on the velocity and acceleration of the body. Mass moment of inertia is a measure of a body's resistance to a change in its rate of rotation. The components of mass moment of inertia for simple shapes have been tabulated in the back of this book.

If a point in a rigid body B, say point O, is fixed in the inertial reference frame E, using Eqns. 9.10a and 9.13a we can express the angular momentum about O as follows:

in which Hoi and «i are the components of angular momentum and angular velocity along the unit vectors associated with reference frame B.

Once the inertia matrix Icj is determined, I°ij can be obtained from it using the following equation:

in which m denotes the mass of the body, and hi are the coordinates of the center of mass of the body relative to point O.

Example 9.4. The Mass Moment of Inertia Matrix of Two Slender Rods. Determine the mass moment of inertia matrix of the body shown in Fig. 9.7 about its center of mass C. The body shown in the figure is composed of two slender rods of mass m and length L.

Solution: The mass moment of inertia of the structure shown in the figure is the sum of contributions from each slender rod. The mass moment of inertia of a rod about its center of mass with respect to an axis perpendicular to the long axis of the rod is equal to mL2/12. Its moment of

Figure 9.7. A rigid object composed of two slender rods of length L and mass m. The mass moment of inertia matrix was computed with respect to the reference frame shown in the figure.

b inertia with respect to point C can be computed by using Eqn. 9.17. Let us illustrate this by computing Ic11:

Ic11 = Ic111 + m (h22 + h32) + Ic2n + m (d22 + d32)

where the symbols hi and di denote the coordinates of the center of mass of OA and OB relative to point C. The first two terms on the right-hand side represent the contribution of rod OA, and the last two terms are the contribution of OB. Similarly, we can show that

Getting Started With Dumbbells

Getting Started With Dumbbells

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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