## Moment of Momentum About the Center of Mass

Moment of momentum of a system of particles with respect to a point O fixed in reference frame E was defined as

Ho = 2 ri/o X mivi where as usual O denotes the origin of E and i represents particle i in the system of particles. In many instances, particularly when we want to investigate locomotion (walking, running, jumping) or movement in air, it is more convenient to consider moment of momentum about the center of mass. Using parallegram law of vector addition:

Substituting this expression into the equation for moment of momentum about point O, we find:

in which L is the linear momentum of the system of particles, as defined by Eqn. 3.5:

The last term on the right-hand side of Eqn. 3.39a represents the moment of momentum of the system about the center of mass, Hc. Thus,

Let us next take the time derivative of Eqn. 3.39b:

The first term on the right-hand side of this equation is equal to zero because vc/o and L = (2m1) vc/o are in the same direction and hence their vector product must be equal to zero. Because the time derivative of the moment of momentum about point O is equal to the resultant moment acting on the system with respect to point O, we find:

rc/o X 2F' + dHc/dt = 2ri/o X F1 = 2(rc/o + ri/c) X F1

According to this equation, the time rate of change of moment of momentum about the center of mass is equal to the sum of the moments of external forces and force couples with respect to the center of mass.

Example 3.10. Arm Swinging During Walking. Determine how rigorously one has to swing the arms to avoid twisting during walking.

Solution: We have previously shown (in Example 3.8) that the moment of momentum from arm swinging is given by the following expression (see Eqn. 3.36):

in which d is half the length between the shoulders, and L is the length of an upper limb. The moment of momentum for the legs is given by a similar expression. Thus, the total moment of momentum with respect to the center of mass can be written as

Hc = 2da La ma (d0a/dt) cos 0a e2 — 2di Ll mi (ddl/dt) cos 0l e2

in which the subscripts a and l refer to the upper and lower limbs, respectively. Several observations can be made with regard to this equation. First, the left arm must rotate in the same direction as the right leg and vice versa to cancel the contributions of arm and leg swing to the moment of momentum of the human body. Second, because the lower limbs are longer and heavier than the upper limbs, the arms must rotate faster to cancel out the twisting effect of the legs. As the arms and the legs have the same period of swing, the way to achieve zero moment of momentum from swing is to increase the amplitude of swing of the arms while also increasing their rate of rotation. ## 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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