## Biomechanics of Twisting Somersaults

In this section we present an example from the airborne movement of a human. The only significant external force in this case is the force of gravity. The gravitational force passes through the center of mass and therefore the angular momentum about the center of mass must be constant. Equation 9.14 relating the angular momentum to the mass moment of inertia and the angular velocity is used to determine the changes in the rate of rotation of the body as a result of a sudden change of body shape.

Example 9.8. Three-Dimensional Diving. A diver dives off a platform with an angular velocity R«B = «o e2 (Fig. 9.11). His plane of motion immediately after the jump is the plane containing the unit vectors e1 and Figure 9.11. As soon as a diver dives off a platform, he moves one of his arms to the side. The reference frames E and B are attached to the platform and to the diver, respectively.

e3. As soon as the diver pushes off the platform, he moves his left arm to the side. Determine his angular velocity immediately after raising his arm to the horizontal configuration. The diver weighs 80 kg, and each of his upper limbs weigh 4 kg. His height (H), his width (W), and his depth (D) are 1.80 m, 0.35 m, and 0.20 m, respectively.

Solution: As the diver is on air in free fall, the only force acting on him is gravity and this force always passes through the center of mass. Thus there is no external moment acting on the diver and his angular momentum before he raises his arm must be equal to his angular momentum after he raises his arm. Because his motion is planar before he raises his arm, from Eqn. 9.14 we obtain

in which the subscript b refers to time immediately before the raising of the arm. Because the diver is considered to be a symmetric body before he raises his left arm, the following holds:

We assume that the body of the diver except the arms can be represented by a rectangular prism of mass 72 kg, height 1.80 m, width 0.35 m, and depth 0.20 m. Arms are considered as rods with lump masses of 4 kg positioned at midpoint. We further assume that each upper limb is equal half the body height (0.9 m). Under these conditions:

Let us next consider the mass moment of inertia after the diver raises one arm. Because his arm is much lighter than the rest of his body, this change in shape will perturbate the center of mass only by a small amount. For simplicity we assume that the center of mass of the diver remains at all times at the center of mass of the rectangular prism representing the diver's body minus his arms. With this simplification and using Eqn. 9.17, the mass moment of inertia after the raising of the arm, (Ic!y)a, becomes

(Icn)a = 72(1.82 + 0.352)/12 + 4(0.452 + 0.1752) + 4 [0.92 + (0.175 + 0.45)2] = 25.3 kg-m2

72(1.82 + 0.202)/12 + 4(0.45)2 + 4(0.9)2 = 23.1 kg-m2

72(0.352 + 0.202)/12 + 4(0.175)2 + 4(0.175 + 0.45)2 = 2.7 kg-m2

(Ic23)a

Therefore the angular momentum immediately after the raising of the arm is equal to

Equating the angular momentums before and after the shape change, we find:

Thus,

This exercise shows that just by moving one arm, a diver performing plain somersaults could begin performing twisting rotations too. Note that the angular velocity component (w3)a would have changed sign if the diver were to raise his right arm instead of his left arm. 