Question 14.7: A baseball pitcher throws his curve ball at 80 mph with a ro...

In the coordinates shown in Figure 14.29, the rotation of the ball will cause a force tending to move the ball in the y direction. The equation of motion for the ball in the y direction of break is Ma_y =\sum F_y. The drag on the ball acts in the x direction and tends to slow the ball down slightly during its travel to the plate. We will neglect this effect and assume for the ball a constant velocity of 80 mph. The time of flight of the ball is therefore given by

T=\frac{S}{V_{ball}}                                  (A)

where S is the distance to the plate and V_ball is the speed of the ball. Inserting the data, we calculate a flight time

T=\frac{S}{V_{ball}}=\left(\frac{55\ \mathrm{ft} }{80\ mph} \right)\left(\frac{1\ mph}{1.467\ \mathrm{ft}/s} \right)=0.47\ s

Gravity acts on the ball in the negative z direction and causes the ball to drop as it travels to the plate. We will assume this effect acts independently of the lift force created by the rotation of the ball; hence the drop can be calculated as approximately 2.9 ft. The lift force for a rotating smooth sphere is given by

F_L=\frac{1}{2}\rho V_{ball}^2\frac{πD^2}{4}C_L                            (B)

and acts in the y direction. A baseball has raised stitches that are known to affect the trajectory of a pitch. Since we have lift and drag data only for a smooth sphere as given in Figure 14.29, we will use the smooth sphere data in our calculation. Finally, note that if the rotational speed of the ball is assumed constant, then the lift force is constant during the flight.

We can write the equation of motion for the ball in the y direction as ma_y = m(d²y/dt²) = F_L. Integrating this twice we obtain y(t) = (F_L/m)(t²/2) + C₀t + C₁. To evaluate the constants, note that at time t = 0, the ball is at an initial location y₀, thus C₁ = y₀, and we can write y(t) − y₀ = (F_L/m)(t²/2) + C₀t. The remaining constant is found by assuming that the ball has no velocity component in the y direction when released by the pitcher. This allows us to set C₀ = 0 and obtain y(t) − y₀ = (F_L/m)(t²/2). At time T the break or distance traveled in the y direction is given by

\Delta y=\frac{F_L}{m}=\frac{T^2}{2}                                 (C)

Inserting (A) and (B) we have

\Delta y=\frac{πD^2ρC_LS^2}{16m}                                         (D)

To determine the lift coefficient, we first calculate the speed ratio

\frac{U}{V_{ball}}=\frac{D\omega }{2V_{ball}}=\frac{(2.865\ in.)(\mathrm{ft}/12  in.)(1800\ rpm)(2π)(60\ min/h) }{2(80\ mph)(5280\ \mathrm{ft}/mile)}

= 0.19

then use the chart in Figure 14.28 to find that C_L ≈ 0.05. To calculate the break at the different air temperatures, note that ρ₅₀ = 2.420 × 10⁻³ slug/ft³, and ρ₉₀ = 2.244 × 10⁻³ slug/ft³. The diameter of the ball is found to be D = (9 in./\pi)(1 ft /12 in.) = 0 .2387 ft. Thus from (D) we find the break at 50°F is

\Delta y_{50}=\left(\frac{\pi }{16} \right)\left(\frac{1}{5\ oz} \right)\left(\frac{1\ oz}{1.943×10^{−3} \ slug} \right) (0.2387 ft)²(2.420 × 10⁻³ slug/ft³)(0.05)(55 ft)²

At 90°F the break is

\Delta y_{90}=\left(\frac{\rho _{90}}{\rho _{50}} \right)\Delta y_{90}=\left(\frac{2.244×10^{−3}\ slug/\mathrm{ft}^3 }{2.240×10^{−3}\ slug/\mathrm{ft}^3} \right)0.42\ \mathrm{ft}

= 0.39 ft

or about 8% less in the “lighter” summer air. This break is not sufficient to fool a batter if the curve ball is thrown with the rotation axis vertical, as indicated in Figure 14.29. Instead the pitcher throws the ball so that the break is down, i.e., the rotation axis is nearly horizontal. Pitchers use a variety of spins to induce movement of the ball. A knuckle ball is thrown with no spin and darts erratically owing to flow separation. A discussion of the physics of sports balls can be found in an article by R. D. Mehta entitled “Aerodynamics of Sports Balls,” in Annual Review of Fluid Mechanics, volume 17, pages 151–189, 1985.