Originally Posted by travlerajm
Almost every single pro tennis player brings the racquethead above his head on the backswing. As gravity drops the racquethead from that point, the wrist is free to pivot. How fast the racquethead travels as the racquet travels through a high-to-low-to-high pendulum sweep is largely dependent on the relative weight distributions of the arm and racquet.
The pendulum action is easy to see if you watch these guys in slo-mo, who have pretty good forehands:
So the pendulum action you are referring to is a rotation of the wrist around the axis going through the forarm, and not around the axis perpendicular to that going through the wrist (i.e. from thumb to little finger)?
Because if the latter, the pendulum action would be in the plane of the swing path which is mostly in the horizontal plane, and more so when the player meets the ball higher in the bounce. This means gravity, and hence g, would only play a small role in the pendulum action and g would need to be replaced by a, the accelleration of the racket arm executed by the player.
If the pendulum action you're referring to is a rotation around the forearm axis, as in the modern 'windscreen-wiper' stroke, it would be perpendicular to the plane of the swing path and therefore mostly in the vertical plane. Then gravity does play a role, but not exclusively. When you rotate your forearm to bring the rackethead up during the take-back, you store energy in the forearm by twisting the radius and ulna bones relative to each other. When you then bring arm foreward during the stroke, the radius and ulna are forced back into their original position, releasing this stored energy into a rotation of the wrist and, hence, the racket. This rotational force therefore also affects the pendulum action around the wrist, as well as gravity, creating an angular accelleration on top of g.
Shouldn't this also be accounted for in your formula? How would this affect your conclusions?