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## Q. 1.11

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?

## 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 