Let us substitute Eqn. 4.12 into Eqn. 4.11 and take the time derivative of the result:

vQ = vP + (dfi/dt) rQ/P sin 6 [-sin fi e1 + cos fi e2] (4.13)

In deriving this equation we have used the fact that the distance between any two points in a rigid body remains constant, and therefore we set drQ/P/dt = 0. Also, because the motion occurs parallel to the (e1,e2) plane, the angle 6 remains constant; therefore, d6/dt = 0.

Using the definition of vector multiplication of two vectors, Eqn. 4.13 can be shown to be equivalent to the following vector equation:

where m is the angular velocity of the rigid body B with respect to the coordinate system E. Note that in a plane parallel to the (e1, e2) plane, angle fi can be any angle, taken counterclockwise from a line element fixed in E to a line element fixed in the object B (Fig. 4.3). This is true because all such angles differ from each other by a constant and therefore have the same time derivative. If (dfi/dt) > 0, then the object B rotates counterclockwise and it is said to have a positive angular velocity. If (dfi/dt) < 0, then the object B rotates clockwise and it is said to have a negative angular velocity. Note also that angular velocity may vary with time but does not vary from point to point in a rigid body. Thus, knowing the velocity of a single point in a rigid body and its angular velocity, we can determine the velocity of any other point in the rigid body.

Example 4.1. Vertical Jumping. In this example we seek to understand the contributions of segmental rotations of body parts to the vertical velocity of the body's mass center during vertical jumping. As shown in Fig 4.4, the jumper keeps his hands on his hips and jumps as high as he can. We will determine the velocity of the center of mass of the athlete 0.2 s after he begins the preparatory countermovement phase of the jump. The dimensions of the athlete are given as follows: Lf (length of the foot) = 27 cm, Li (length from ankle to knee) = 48 cm, Lt (length from knee to hip) = 50 cm, and Lc (length from hip to center of mass) = 28 cm. He weighs 68 kg.

We model the athlete as composed of four segments as shown in Fig. 4.4. Angles between body segments and the horizontal in the fixed reference frame E are indicated by and As usual, the segment orientation angle is positive when taken counterclockwise from horizontal. The counter movement begins at t = 0 when the body segments make the following angles with the horizontal: = 0, = 66°, = —43°, and 4>c = 23°. Thus, at this instant, the feet are flat on the ground, the knee bent, and the upper body bent forward. Employing the inverse dynamics approach and using a videocamera and a computer, these body segment angles were measured as a function of time for t > 0:

Figure 4.3. The rate of rotation of a rigid body in planar motion is equal to the time rate of change of angle The angle need not be uniquely defined

Figure 4.3. The rate of rotation of a rigid body in planar motion is equal to the time rate of change of angle The angle need not be uniquely defined the time derivative of angle The figure shows two angles whose time derivatives give the same value for the angular velocity.

Figure 4.4a-c. The schematic diagram of a professional vertical jumper (a). The jumper is represented by a four-link model (b). The symbols F, A, K, H, and C denote the tip of the foot, ankle, knee, hip, and the center of mass of the athlete, respectively. The angles these links make with the horizontal axis are shown in (c).

Figure 4.4a-c. The schematic diagram of a professional vertical jumper (a). The jumper is represented by a four-link model (b). The symbols F, A, K, H, and C denote the tip of the foot, ankle, knee, hip, and the center of mass of the athlete, respectively. The angles these links make with the horizontal axis are shown in (c).

In these expressions, time t is in seconds and the segment angles of rotation are in radians. Note that although the jump begins from the stand ing upright position, the jumper is not in that configuration at time t = 0. The center of mass of the jumper has already moved downward toward the ground at t = 0 so that this instant does not correspond to the beginning of the preparatory phase of jumping.

Solution: Our aim is to compute the velocity of the center of mass of the athlete.The angular velocities of segments are obtained by taking the time derivatives of the corresponding angles of rotation:

Angular velocities are in the units of rad/s.

At time t = 0.2 s, we have the following values for the rotation angles: 4>f = -23°, 4>l = 43°, ^ = -43°, and 4>c = 52°. We converted the angles in radians to angles in degrees by using the relation v (rad) = 180°.

The angular velocities at t = 0.2 s are computed by inserting the value of t = 0.2 s in Eqn. 4.17:

Substituting these values successively into Eqn. 4.14a, we obtain the following values for the velocities of joints and the mass center:

vA = 0 + (-2 e3) X [(0.27) (-cos 23° e1 + sin 23° e2)] = [0.2 e1 + 0.5 e2]

vK = [0.2 e1 + 0.5 e2] + (2 e3) X [(0.48) (cos 43° e1 + sin 43° e2)] = [-0.5 e1 + 1.2 e2]

vH = [-0.5 e1 + 1.2 e2] + (-3 eg) X [(0.50) (-cos 43° e1 + sin 43° e2)] = [0.6 ei + 2.3 e2]

vC = [0.6 e1 + 2.3 e2] + (2.5 e3) X [(0.28) (cos 52° e1 + sin 52° e2)] = +2.7 e2

where all velocities are in m/s. Note that vertical velocities increase steadily as one goes up from ankle to knee, to the hips, and finally to the center of mass of the trunk. Bobbert and van Ingen Schenau (1988) found that for 10 skilled jumpers the average velocity of the center of mass increased from 0 to approximately 3.5 m/s during the preparatory phase of the jump.

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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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