## Human Body Dynamics Swinging Steel Ball While Standing On Turntable

Problem 9.1. A figure skater spins about her longitudinal axis b2 with constant angular speed of 5 rad/s (Fig. P.9.1). She then begins to raise her arms over her head at a rate of 2 rad/s. Determine the angular velocity of her arms (A) with respect to the inertial reference frame in E (EwA). Express this angular velocity in the auxiliary coordinate system B shown in the figure. Answer: wA = 5 (rad/s) b2 + 2 (rad/s) b3.

Problem 9.2. The following equation relates the acceleration of two points in a rigid object:

Figure P.9.1. A figure skater spins about her longitudinal axis with constant angular speed of 5 rad/s.

Show that it can be written in the following form for the planar motion:

Problem 9.3. A gyroscope is a wheel mounted in a successive series of rings so that its axis is free to turn in any direction. When the wheel is spun rapidly, it will keep its original plane of rotation no matter which way the rings are turned. Gyroscopes are used to keep moving ships and planes level. The gyroscope shown in Fig. P.9.3 is composed of an outer gimbal ring B, an inner gimbal ring C, and a rotor D. The symbol S denotes the ship to which the gyroscope is attached. The unit vectors b, c, and d are parallel to the axes of B, C, and D, respectively. The angles of rotation shown in the figure were determined to obey the following equations:

Problem 9.4. Determine the mass moment of inertia matrix about the center of mass of the shapes shown in Fig. P.9.4. All rod segments have a mass m and length L. Rings have a radius R and mass M.

Figure P.9.3. A gyroscope attached to a ship.

Figure P.9.3. A gyroscope attached to a ship.

Answer: For the bent rod of length of 4L and mass 4m shown in the figure, the mass moment of inertia matrix is equal to

Icii = (2/3) mL2, Ic22 = (8/3) mL2, Ic33 = (10/3) mL2

Problem 9.5. Determine the mass moment of inertia of the same shapes (Fig. P.9.4) with respect to the point A.

Problem 9.6. A man stands still on a turntable that isolates his body from external torques around the vertical b3 axis (Fig. P.9.6). He then begins to swing his arms from one side to the other, as shown in the figure. The angular momentum of the swinging arms about the center of mass of the man is equal to 8 (kg-m2/s) b3. Determine the angular velocity of the trunk of the man after he begins swinging his arms. The man is 1.80 m tall and weighs 65 kg. His average width is 34 cm and average depth 19 cm.

Problem 9.7. A man standing on a frictionless turntable begins swinging a pendulum of 10 kg (Fig. P.9.7). The length of the pendulum is 0.5 m. Swinging of the pendulum generates angular momentum in b1direction. The man keeps his hands fixed at the center of the body while swinging the pendulum. Which direction does his trunk rotate to balance the angular momentum produced by the swinging of the pendulum?

Problem 9.8. A thin disk D of mass m rolls on a table while at the same time it rotates around a vertical axis that passes through point O (Fig.

Figure P.9.6. A man is swinging his arms while standing on a turntable.

P.9.8). The system is set up so that the cable that connects the center of the disk to point O is horizontal. The angle the rope makes with the ei direction varies with time according to the equation:

Figure P.9.7. A man is swinging a steel ball while standing on a turntable.

Figure P.9.8. A disk rolling on a horizontal table. The center of the disk is connected to a cable that passes through point O.

The length of the cable above the table is 48 cm, and the diameter of the disk is 14 cm. Determine the position, velocity, and the acceleration of the center of mass of the disk at t = 2s.

Answer: r = 0.48 [cos 86° e1 + sin 86° e2], a = —0.06 [cos 86° e1 + sin 86° e2]

Problem 9.9. A diver wants to add spin (twist) to his somersaults during the aerial movement. The desired twist is in the counterclockwise direction with respect to an axis that is directed from his foot toward his head. Propose a number of shape changes involving upper and lower limbs that would cause the desired twist. Problem 9.10. A metal coin of radius R and mass m rolls along a horizontal circle (Fig. P.9.10). The angle 0 between the disk's axis and the vertical remains constant. Show that the magnitude v of the velocity of the center of mass of the coin is related to the angle 0 by the following equation:

v2 = (2/3) g cot 0 (r — R cos 0)2/[r — (5/6) R cos 0]

Hint: Let E denote the reference frame fixed on earth such that e3 points vertical upward. Let D be a coordinate system attached to the center of mass of the coin. The unit vector d1 remains always in the horizontal plane while the coin rolls without slip. The angle between the e1 and d1 is denoted as The unit vector d3 is chosen along the axis of the coin as shown in Fig. P.9.10. The angle 0 between e3 and d3 remains constant during the motion of the coin. The angle 0 measures the counterclockwise rotation of the coin about d3. Using these angles and their time derivatives, one can derive expressions for the angular velocity of D in E (01, 02, 03) and the angular velocity of the coin in E (w1, w2, w3).

The rate at which the unit vector d1 rotates in the horizontal plane is given by the following equation d^/dt = v/(r — R cos 0)

Figure P.9.10a-c. A coin rolls along a horizontal circular path (a) of radius r. The reference frame D is shown in (b). The free-body diagram of the coin is shown in (c).

where the term in the denominator is equal to the radius of the circular path the center of mass moves.

Note also that the no-slip condition essentially determines the spin rate (d\$/ dt). Show that the condition that the velocity of the point of contact with the flat surface is zero leads to the relation

The free-body diagram of the coin is shown in the figure (Fig. P.9.10). use the equation of motion of the center of mass to determine the reaction forces. Then use Euler's equations about the center of mass to express v as a function of R, r, g, and 0.

Problem 9.11. Just before a pitcher throws a baseball, the spatial positions of the ball, ankle, and shoulder were determined experimentally (Fig. P.9.11). The data on the time course of the position of markers were then curve fit by an equation of the form:

in which rP/O denotes the position vector from O to P, and 6, a, and ^ are the angles rP/O make with the unit vectors e1,e2, and e3, fixed on earth. The position vectors of the markers are given as follows:

rElb/Shl = 0.41 [cos(1.1 + 3 t2) e1 + cos (0.2 + 4t) e2 + cos (0.2 + 7.2t2) e3]

rHnd/Elb = 0.47 [cos (5 t2) e1 + cos (0.5 - t2) e2 + cos (0.4 - 0.2t2) e3]

Assume that the upper arm weighs 4.1 kg and the lower arm (forearm, hand, and the ball) 2.9 kg. Assume further that each of these limb segments can be represented as a uniform cylindrical rod. Determine the average radii of the upper and lower arms. Assume that the mass density of the upper limb is 1 g/cm3.

Determine the velocity and acceleration of the markers at the shoulder joint, elbow, and hand by using the parallelogram law and taking the time derivatives in the inertial reference frame E.

Determine the acceleration of the center of mass of the upper arm and the lower arm using the same method.

Figure P.9.11. A pitcher in preparation for a throw.

Determine the force applied to the forearm by the upper arm at the elbow and the force applied to the upper arm by the shoulder at time i = 0 by using Newton's laws of motion for the center of mass of an object.

Determine the angular velocity of the lower arm at time i = 0 by either of the following two methods:

Method i: Use the equation that relates the velocity of two points in a rigid body:

The only unknown in this equation is E«Bu.

Method ii: Construct a Cartesian coordinate system B embedded into the upper arm. Let b1 be the unit vector:

b1 = [cos(1.1 + 3 i2) e1 + cos (0.2 + 4i) e2 + cos (0.2 + 7.2i2) e3]

Let us choose b2 along the axis of flexion/extension of the forearm:

b2 = b1 X [cos(5i2) e1 + cos(0.5 - i2) e2 + cos (0.4 - 0.2i2) e3]

This equation and its time derivative can be used to evaluate b2 and its time derivative db2/di.

The third unit vector b3 and its time derivative can be found by using the expression:

Angular velocity is then determined using the expression:

RWB = [(Rdb1/di) • b2]b3 + [(Rdb2/di) • b3]b1 + [(Rdb3/di) • b1]b2

Note that one has to determine not only angular velocity but also angular acceleration to determine the net moment exerted at the elbow and the shoulder joints. Write down a "road map" for computing angular acceleration and the resultant joint moments.