A baseball player holds a bat with a centroidal moment of inertia \overline{I} a distance a from the bats mass center. His “bat speed” is the angular velocity with which he swings the bat. The pitched ball is a fastball which reaches the batter with a velocity v. Assuming his swing is a rigid-body rotation about an axis perpendicular to his hands, where should the batter hit the ball to minimize the impulse felt by his hands?
Chapter 1
Q. 1.11
Step-by-Step
Verified Solution
When the better hits the ball, it exerts an impulse on the bat: call it B. Since the batter is
holding the bat, he feels an impulse as he hits the ball: call it P. The effect of hitting the ball is to change the bat speed from \omega _{1} to \omega _{2}. The impulse momentum diagrams of the bat during the time are shown in Figure 1.26.
Applying the principle of linear impulse and momentum to Figure 1.26 leads to
ma\omega _{1}+P-B=ma\omega _{2} (a)
Application of the principle of angular impulse and angular momentum about an axis
through the batter’s hands yields
\overline{I} \omega _{1} +ma \omega _{1}\left(a\right) -B\left(b\right) = \overline{I} \omega _{2}+ma\omega _{2}\left(a\right) (b)
Solving Equation (b) for B, we have
B = \frac{\left(\overline{I}+ma^{2} \right) }{b} \left(\omega _{2}-\omega _{1}\right) (c)
Substituting Equation (c) into Equation (a) and solving for P leads to
P=\left(\omega _{2}-\omega _{1}\right)\left(\frac{\overline{I}+ma^{2}}{b}- ma \right) (d)
Thus, P = 0 if
b= a +\frac{\overline{I} }{ma} (e)
Thus, the angular impulse felt by the batter is zero if b satisfies Equation (e). The location of b is called the center of percussion
