## Computation of Reaction Forces

To determine the external forces acting on the dancer, we need to consider the motion of her center of mass. The position of the center of mass of the rod is given by the expression:

rc = (L/2) sin 6 (cos \$ e1 + sin \$ e2) + (L/2)cos 6 e3 (9.26a)

Because the angle of inclination of the dancer with the vertical axis (6) remains constant during the rotation, (d6/dt) = 0. Taking the time derivative of Eqn. 9.26a, the velocity of the center of mass can then be written in the form:

vc = (L/2) wo sin 6 (-sin \$ e1 + cos \$ e2) = (-L/2) wo sin 6 b2 (9.26b)

As the rate of rotation wo was specified as constant, d2\$/dt2 = 0. The acceleration of the center of mass is obtained by taking the time derivative of Eqn. 9.26b:

ac = (L/2) wo2 sin 6 (-cos \$ e1 - sin \$ e2) (9.26c)

Substituting Eqn. 9.26c into Newton's law of motion for the center of mass (Eqn. 9.19), and using the free-body diagram shown in Fig. 9.9d, we find:

-mg e3 + T (-cos \$ e1 - sin \$ e2) + V e3 + f (cos \$ e1 + sin \$ e2)

in which T represents the magnitude of the tensile force exerted by the pole on the dancer's hands and V is the magnitude of the vertical force exerted by the ground. The frictional force exerted by the ground is represented by the last term in the left-hand side of Eqn. 9.27.

Writing this vectorial equation in scalar components in —n and e3 directions, we find:

These two equations are not sufficient to solve for the three unknowns: T, V, and f. Thus, we need to consider the principle of conservation of angular momentum to arrive at an additional equation. Because the dancer rotates around a fixed point O, the time rate of change of her angular momentum about O must be equal to the resultant moment with respect to O. Assuming that the arms of the dancer are positioned at a distance (3L/4) from the feet, this resultant moment Mext is given by the expression:

Mext = (L/2) b3 X (—mg e3) + (3L/4) b3 X T (—n)

The resultant moment about O must be equal to the time rate of change of angular momentum Ho about the same point. Angular momentum Ho was defined as follows:

where r is the position vector from the base of the rod to any point on the rod, v is the velocity of the point with respect to the inertial reference frame E, and dm is a small mass element surrounding the point.

Let h denote distance from point O along the bar (0 < h < L); then, a small mass element dm can be created such that dm = (m/L) dh where dh is a small length element. Expressing r and v in Eqn. 9.30 as a function of h, 0, and \$, one obtains:

Ho = (m/L)Jhb3 X (—h «o sin 0 b2) dh = (mL2/3) «o sin 0 b1 (9.31a) = (mL2/3)«o sin 0 [cos 0 (—cos \$ e1 — sin \$ e2) + sin 0 e3] (9.31b)

We can check the validity of this expression by using the mathematical formulation presented in Section 9.3. Let us illustrate this by going over a few mathematical steps. The angular velocity of the slender rod given by Eqn. 9.25b can be expressed in terms of the unit vectors associated with the reference frame B as

Therefore, in reference frame B, «1 = «o sin 0 and «3 = «o cos 0.

The elements of mass moment of inertia of a slender bar with respect to an endpoint are given as

Equation 9.i6 then dictates that

Ho = (mL2/3) wo sin 0 bi which is what we had found using the direct approach. As can be seen from this equation, Ho is not parallel to the angular velocity wo sin 0 b1 + wo cos 0 b3. This is in contrast with the examples of planar motion presented earlier in this book. The discrepancy stems from the fact that the slender rod (dancer) has no horizontal plane of symmetry in this particular movement. If the rod had swept a plane rather than the surface of a cone during motion, the asymmetry would vanish and once again angular momentum and angular velocity would be colinear.

To compute the reaction forces let us take the time derivative of Ho in the inertial reference frame E. In E the unit vectors ei, e2, and e3 remain constant. Using the expressions given in Eqn. 9.25a for the unit vectors bi, one can show that dHo/dt = (mL2/3) wo2 sin 0 cos 0 b2 (9.32)

According to the conservation of angular momentum this time derivative must be equal to the resultant external moment given by Eqn (9.29):

—mg (L/2) sin 0 + (3L/4) T cos 0 = (mL2/3) wo2 sin 0 cos 0

Thus we arrive at an expression for the tensile force T:

To find the ground reaction forces, we combine Eqn. 9.33a with the equation of motion of the center of mass (Eqn. 9.28). The results are

f = (2/3) mg tan 0 — (1/18) (mL) wo2 sin 0 (9.33c)

These equations show that the vertical ground force acting on the dancer is equal to the weight of the dancer. The force with which the dancer pulls on the pole increases and the ground friction force decreases with increasing rate of rotation. 