Gregory Diamond
Professional
You are wrong. I`ll give you the best explanation. Imagine we placed the racket vertically on the table(without connecting the handle with the table) and directed fast ball towards the strings. The ball will bounce off the strings(if the speed of the ball is not too great) and the racket fall in the opposite direction. Now we repeat the same but we connect the handle firmly with the table. Ball with the same initial speed will bounce back with much greater speed. There is no doubt. The racket at the end will be in the same place. Racket absorbed the kinetic energy of the ball and gave it back to the ball(not all but a large part of it). In the first example a great part of energy the racket obtained from the ball was wasted. This energy caused the racket to fall back. If we observed what happens during the collision of the ball with the racket in a real stroke but we did it from the point of view of observer moving forward with the speed of the handle we would see the same difference between reaction of the ball and the racket after the stroke. It is easy to explain. The time the ball touches the strings is not the same. It may be short but it differs significantly. Change of momentum = force x time. This formula can be derived from Newton`s rules. It all means that if you want to obtain the same change of momentum of the ball throwing racket at the ball you should do it with much greater speed what has to cause loss of control.I'll put it once, in casual manner to stay within messageboard format. If you need deeper dive into strict description, I'll leave that to you.
1. Your force formula is for force applied to an object. Consider ball mass and ball acceleration to use it. If you want a formula for racquet-ball interaction, it's energy transfer and E=mc^2. However, in a simple application it doesn't take into account object shapes, elasticity, etc, which are a part of equation for actual physics of tennis stroke. If you had a steel sphere hitting a tennis ball directly center-towards-center, you could more or less consider that sphere mass and approach velocity to calculate E. And compare those of different spheres moving with different velocities.
2. For rotational motions it's the relation between "distance of contact point to pivot point" and "distance of mass location to pivot point" which determines the mass "piece" contribution to the interaction energy. Think baseball bat: hit the ball conventional, or flip the bat upside-down and hit the ball with the handle. Moreover, distance is squared for this equation. Which makes mass located around the contact significantly more contributing. You asked about cutting the racquet - yes, a racquet with lighter handle, but same mass in the hoop, will deliver nearly same power for the hit. So even if we you somehow managed to ensure non-elastic wrist, contribution of arm mass would be diminishing with the distance, significantly.
3. You just cannot ensure the wrist to be firm enough to create a monolith arm-racquet structure. Check it - "lock" your wrist as hard as you can with a racquet and make someone throw a ball onto it. It will give way due to joints and muscle elasticity and inconsistent nature of muscle deformation - muscle fibers are never truly static. Now, there'd be some point of limit for the wrist extra lay-back under pressure of collision with incoming ball, but the ball would be gone by that moment. If 3-4ms figure tells you nothing, whatch slow-mo of shanks by pro-players: even though the racquet twists badly, the ball departs with almost the same direction as for sweetspot hit. All visual twisting happens later. Same for the wrist - it still absorbs the collision, tendons stretching, while the ball is already gone. No "feel of entire body", or even arm, through contact.
4. Once again to the "even if we you somehow managed to ensure non-elastic wrist", there's an arm pivoting at shoulder joint before, during and after contact, which separates it from the torso. Actually, if you want entire body mass to contribute to the contact, you should grip the racquet head with both arms at 9 and 3 o'clock and place it facing the incoming ball at belly level, arms bent 90 deg and elbows fixed against pelvis bones. This would arguably eliminate most elastic joints.
5. Even for linear motions, like straight punch in karate, E=mc^2 works for the fastest moving part. The mass of forearm. The velocity of forearm. All body contributes to the punch with proper technic, but its via building up the speed - either via kinetic chain transfer, or via compound, simultaneous activation. This might be a tad simplified, as joints are not exactly hinges, but still a decent model. While actual physics of tennis strokes still hasn't been fully described and modeled, neither the swingpath, nor the sringbed-ball collision. To many moving parts, biomechanical power sources and elastic interactions.
Disclaimer: I'm sorry for all the impresise terms, being neither English native speaker, nor having studied physics in English. Still tried to cover key ideas in a popular manner.
So you can ask what is the best solution. You just have to accelerate the racket(speed of the racket is important factor) but you cant surpass the speed when you cant keep the right wrist laid back.You should be near that point but you cant accelerate the racket more because you will lose control. You can be sure that your results will be the best if you play that way.
Last edited: