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ecall that martial arts involve almost totally 3-D actions. This chapter describes almost all those physical properties and physical quantities that were described in the previous chapter. Every description of course will be related to angular motion.9.1 FORCES
In angular motion, there are many different forces, such as eccentric, concentric, centripetal (radial), tangential, friction force, and others, which can act on a body or a point of mass differently. To these forces the speed/velocity and acceleration are directly related to the object in cause.
These forces act of course through different distances. Distance, displace-ment, speed, velocity, and acceleration have been described at the angular kinematics. We will describe mostly the action of the forces.
When we speak about eccentric or concentric forces, we should not be confused with the external and internal forces. These two forces has noth-ing or little commune with any motion to be translational or rotational.
These two forces have been described earlier.
9.1.1 Couple and Eccentric Force
We know that for any movement to occur, a force must be present. In rotary action also, a force must be present. Basically, any rotation of a mass/body first involves a translatory movement, and then will turn into
a rotary movement. This translatory motion has seemingly an extremely short time. Here is an example of how in fact a rotary motion happens.
Two wrestlers A and B face each other. If wrestler A pushes wrestler B exactly on the middle of the chest, then B will presumably have a direct backward linear motion. If the same wrestler A pushes B again but at the shoulder, then a body rotation will occur in the direction of the pushing force. Most of the time when the force of pushing acts through the CoG or CoM, then a translatory motion will occur. This kind of pushing force almost never happens in martial arts. There is always a possibility that the force will not go through the CoG.
A force whose line of action does not pass through the CoG of the body on which it acts (or through the point at which a body is fixed) is called an eccentric force. This force produces translation and rotation. Its rotatory effect is known as torque. On the contrary, a force that has its line of action passing through the CoG of the body on which it acts is called concentric or direct force.
Let us again use our example of the pushing force by a wrestler. If wres-tler A pushes with his right arm the left shoulder of wreswres-tler B, who in turn pushes back with his right arm wrestler A at his left shoulder, there will be a rotatory force exerted by both wrestlers. The forces are eccentric and presumably equal in magnitude.
Both wrestlers’ forces have the tendency to go linear; however, they will be cancelled because of the equal magnitude of both forces. The remaining tendency of each force to rotate the opponent body in a counterclockwise direction will turn the linear force to rotational force. These equal and opposite parallel forces is called a force couple or just couple. There are many examples in daily life, e.g., pushing the bicycle pedals or turning your car driving wheel.
In martial arts, a force couple happens very seldomly. As we know that in martial arts the majority of techniques executed most of the time involve two persons and they must have contact to each other in order to have a couple.
Recall that every rotational movement involves first a linear movement and at the time when we could speak about a force couple, the physical properties involved are strength, speed, and acceleration. To say it better, a momentum (p = m ⋅ v) is involved. So, in this case, there is never a real force couple involved because the momentum of a person differs from the momentum of the other person. The explanation about momentum will be described later on.
9.1.2 Centripetal and Centrifugal Forces
During a back-fist strike (Uraken-uchi) executed by a karateka, muscles such as pectoralis major, deltoid, teres major, latissimus dorsi, and sub-scapularis exert forces on the arm of the karateka. The resultant of these forces can be resolved into two components: The first one acts along the line of the arm through the shoulder joint, which should be considered as the axis of rotation and is named as the radial component of the force (Frad).
The second component that acts perpendicular to the radial force component is named the tangential force (Ftan) (Figure 9.1). This tangen-tial force is responsible for the angular acceleration that increases the tangential velocity of the striking arm. The radial force acts toward the center (axis of rotation) and this force is referred to as the centripetal force or center-seeking force.
The radial acceleration involves change of direction continuing perpen-dicular to the direction of motion. Here are the equations for radial accel-eration and radial (centripetal) force. Radial accelaccel-eration arad = v2/r = ω2r and radial force Frad = m ⋅ v2/r = m ⋅ r ⋅ ω2. Here, v is the magnitude of the tangential circular velocity or speed and r is the length of the radius of rotation.
Let us take an example and calculate the radial force. A karateka strikes toward the opponent’s face with a technique back-fist strike (Uraken-uchi) as shown in Figure 9.1. The karateka weighs 70 kg, which we do not count at this time; his total arm mass is 3.45 kg; the velocity is 11 m/s; and the total arm length (with the fist closed) is 0.69 m. Frad = m ⋅ v2/r = 3.45 kg × 112/0.69 m = 605 N.
Frad
Ftan Target
r
Portion of the circle g+CoM
FIGURE 9.1 Radial and tangential forces.
Centripetal force coexists with another force named centrifugal force.
The word comes from the Latin word “centrum” (center) and “fugere” to (flee). The centrifugal force has two slightly different manifestation forms:
• Reactive centrifugal force that occurs in reaction to a centripetal accel-eration on a mass. This force is equal in magnitude to the centripetal force and is directed from the center of rotation. We can observe this force when we are sitting in a car. When the car is turning to the left, our body moves to the right. This motion to the right is the centrifu-gal force. In martial arts, an example of the centrifucentrifu-gal force mani-festation is the aikido technique named “entering throw—negative”
execution (Irimi-nage). Here, the executor (attacker) (Shite) guides the opponent in a rotary fashion. The defender (Uke) feels that the guiding force throws him away from the closeness of his attacker.
Besides the attacker’s guiding force, the defender also feels the cen-trifugal force. More description about the centripetal and cencen-trifugal forces can be found under Part IV.
• Pseudocentrifugal force appears when a rotating reference frame is used for analysis. The true frame acceleration is substituted by a pseudocentrifugal force that is exerted on all objects, and directed away from the axis of rotation.
Figure 9.1 is seen from the top of the head of the karateka. As you can see the tangential force is perpendicular to the radial force. The drawing shows that the technique starts with the elbow bent.
However, for the correct calculation of the angular velocity and the radial force, the arm should be completely extended. r represents the radius of the circle and also the total length of the arm.
9.2 CENTER OF MASS
As we know already that every object/matter is comprised of mass. This amount of mass has a volume and density. The amount of mass also has its center, what we call the center of mass (CoM). This CoM represents the average position of the mass distribution of an object and it is considered that the mass of the body lies within the CoG.
The linear and/or rotational movements and their positions of the rigid bodies can be analyzed by finding out the CoG positions of the body in case. There are different methods to find out the CoG of a rigid body. One
method is the reaction board method. We will not describe these methods in this book. The entire mass of a rigid body may be assumed to be at the CoM of the object in case considering that the object is under an action of an external force.
Generally, the CoM of a body coincides with the CoG of that body. The CoM and CoG are located just under the navel at an approximate distance of 4 cm. The CoM and also the CoG have a location not only within the body, but can also be found outside of the moving body.
Analyzing the CoM of a rigid body is much easier than analyzing a non rigid body. Analyzing a rigid body, the researcher uses different geo-metrical shapes, which is similar to the body that will be analyzed later on. Let us consider two examples: A ball that rolls down a path and has an inclination of θ, or a ping-pong racket thrown up and forward in the air will certainly have a different path and also a different CoM.
In the first example, the forces appear to operate on a straight line and the CoM = m ⋅ g sin θ. In this example, the CoM appears to be exactly in the middle of the ball. In the second example of throwing a ping-pong racket, the path will be a parabolic one. Here the CoM = m ⋅ g only. It is possible to be stated that the CoM is a geometric point, which holds together all the mass and external force(s) during motion. To calculate the CoM of a system of particles, we use the following equation.
CoM mass of individual particle position of particle
position
t
= Σ ×
Mootal
Substituting with appropriate symbols, the equation will become rCoM = Σ m r Mi i/ total, where ri is the position of individual particle, mi
is the mass of individual particle, and rCoM is the position of CoM. See Appendix C where two karateka CoM is described with figures and equations.
The above equations are useful for the CoM of individual particles that are rotating in the same path. However, if the mass of the individual particle(s) is rotating by moving in different paths or this moving from one path to another will take time, then the original equation will be different, and vector components must be used for the above calculations. Let us pre-suppose that two objects are located on the same axis (X) and both objects are moving from left to right and the pivot point also moves to the right, then the equation for the particle masses will be ΔxCoM = m1Δx1 + m2Δx2/ m1 + m2, where xCoM is the pivot point, m are the point masses, and x is the
moving distance of the point masses. If we also take into consideration the time spent (velocity) by moving the point masses, then the equation would be VCoM = m1 ⋅ v1 + m2 ⋅ v2/m1 + m2, where v is the velocity; however, this equation is not realistic because calculating the CoM should not involve time spending, rather it should involve the distance all the time.
9.3 EQUILIBRIUM
Equilibrium is strongly related to the CoM. When the body loses its equi-librium, the CoM is out of its initial place. When an object is in equilib-rium, its state of position or motion is not changing. An obvious condition for equilibrium is that the net force acting on the object must be zero.
From Newton’s second law, Fnet = m ⋅ a, then we have that the accelera-tion = 0 and the velocity is constant. This requirement for a point mass to be in equilibrium is not a guarantee of a rigid body to be in equilibrium.
There are several basic conditions for equilibrium:
• Fnet = 0, which mean that for a 3-D movement of x, y, and z, compo-nents of forces may be separately set = 0.
• F = 0, which mean that forces left = forces right and forces up = forces down.
• During a perfect rotary equilibrium where the rotational condition is not changed, the net torque must be zero. This condition is a require-ment that the sum of all clockwise (CW) torques must be equal to the sum of all counterclockwise (CCW) torques. For a rigid body to be in a complete state of equilibrium, it must first be in a state of translational equilibrium where the sum of all the forces equal zero.
The conditions for a body to be in equilibrium under angular motion have been stated earlier. Now, how can we calculate equilibrium? The majority of motions involve torque. But torque involves moment of inertia, lever(s), and angular acceleration. The torque is apparent when the body has contact with a force and also has contact with the center of the rotation that is connected to the line of action where the force is activated. In this case, we speak about torque, which we will describe later.
But what about a body that has no torque, for example, a man running with a constant speed in a circle, which, let us say, has a diameter of 20 m.
In this case, how do we know that he lost equilibrium? Just by watching