The center of mass draws a circle of radius (L cos a) with speed vo. Therefore the acceleration of the center of mass is pointed toward the center of the circle and has a magnitude of vo2/(L cos a)

Newton's second law written in vertical (e3) and horizontal directions (n) for the free-body diagram shown in Fig. 9.10 is as follows:

fO + fA = -m (vo2 sin a/L)] (cos $ e1 + sin $ e2) (9.36b)

Because the center of mass traverses a circle with constant velocity, both the acceleration and the contact force tangential to the circle must be equal to zero. Thus, the resultant moment with respect to point O is given by

The resultant moment with respect to O must be equal to the time rate of change of angular momentum Ho. The angular momentum is given by the expression

where r is the position vector from point O to any point P on the disk, v is the velocity of point P, and dm is a small mass element surrounding point P. Note that the holders of the abdominal wheel were assumed to be weightless. Therefore they do not contribute to the angular momentum of the wheel. The following relationship is a direct consequence of the parallelogram addition of vectors:

in which p and 0 are the polar coordinates of the point P as shown in Fig. 9.10d.

using the relationship between velocities of two points in a rigid body, the velocity of point P can be written as v = vo b2 + a X p = vo b2 + (vo/R) (-b1 + tan a b3) X p (cos 0 b2 + sin 0 b3) (9.39b)

The first term on the right-hand side is the velocity of point C. The second term on the right-hand side of the equation is simply the vector product of angular velocity and the position vector connecting point P to point C.

Substituting Eqns. 9.39a and 9.39b into the angular momentum equation (Eqn. 9.38), we express angular momentum as the sum of four terms:

Ho = (Ho)1 + (Ho)2 + (Ho)3 + (Ho)4 (Ho)1 = /(Lb1 X Vob2) dm = m L Vob3 (Ho)2 = /[(Lb1 X (w X p)] dm = L b1 X (w X fpdm) = 0

These results follow because both the terms L b1 and w are constant with respect to the variable of integration and can be taken outside the integral sign. Because C is the center of mass, the integral fpdm = 0.

(Ho)4 = f[(p X (w X p)]dm = (mR2/2) (Vo/R) [-b1 + (1/2) tan a b3]

Summing all four terms we obtain the angular momentum of the rotating disk:

Ho = m L Vo b3 + (mR2/2) (Vo/R) [-b1 + (1/2) tan a b3] (9.40)

Note that this equation can be written in the form:

where

Io33 = (mR2/4) + mL2 «1 = -(Vo/R) «3 = (vo/R) tan a

We could have arrived at this result by using the formulation presented in Section 9.3 (Eqns. 9.16 and 9.17) and the tables of moment of inertia presented in the back of this book.

To use the principle of conservation of angular momentum, let us take the time derivative of Ho in the inertial reference frame E:

The first two terms are zero because the expressions in the brackets are independent of time t. The time derivatives of the unit vectors can be obtained by taking the time derivative of Eqn. 9.34, yielding the result:

(dHo/dt) = -(Vo/R)2 tan a {mR2/2 + (mR2/4 + mL2) tan2 a}b2 (9.41)

Note that we could have obtained the same result by using Eqn. 9.22 for the time derivative of Ho:

Using Eqns. 9.5 and 9.34, one can show that

Using this expression and Eqn. 9.40 in Eqn. 9.42, one again arrives at Eqn. 9.41.

Setting the resultant external moment equal to the time rate of change of angular momentum, one finds

VA = mg cos2 a + m (vo2/R) [1.5 sin2 a/cos a + 0.25 sin4 a/cos3 a]

Note that VA, the vertical force exerted by the ground, cannot be larger in magnitude than the weight mg of the abdominal wheel; otherwise, to assure the force balance in the vertical direction, fA would have to be less than zero. That means, however, that the ground would have to pull point O toward the earth. This is impossible because the ground can only push back but not pull in. The maximum value of Vo, the velocity with which the disk rotates, can be found by equating the vertical ground force exerted at point A to the weight of the disk:

g sin2 a = (vo2/R) [1.5 sin2 a/cos a + 0.25 sin4 a/cos3 a]

This means that if the center of mass rotates with a velocity greater than or equal to 0.89 m/s, the motion is unstable, that is, it could rapidly transform into another mode of rotation, such as that of a disk rolling with none of the holders touching the ground.

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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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