Force of a Bat on a Baseball

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Bibliographic Entry Result
(w/surrounding text)
Standardized
Result
Cutnell, John D. & Kenneth W. Johnson. Physics: 3rd Edition. New York: Wiley, 1995: 198-199. "A force of 3400 N corresponds to 760 lbs, such a large value being necessary to change the ball's momentum during the brief contact time." 3.4 kN
Adair, Robert K. The Physics of Baseball. New York: Harper & Row, 1990: 52. "Very large fores, reaching values as high as 8000 pounds, are required to change the motion of he 5 1/8 ounce ball from a speed of 90 mph toward the plate to a speed of 110 mph toward the center-field bleachers in the 1/1000th of a second of bat-ball contact." 36 kN
The Physics of Baseball. Boston White Sox. "The batter exerts some 6000-8000 pounds of force on the ball. This force is required to change a 5 1/8th-ounce ball from a speed of 90 mph to a speed of 110 mph, this distorts the baseball to half its original diameter and the bat is compressed one fiftieth of it's size." 27–36 kN
Knobler, Mike. Hard Breaking Balls Haven't Thrown this Physicist for a Curve. Jackson Clarion Ledger. "It takes 1/2000th of a second for Cecil Fielder's bat to deliver nearly 10,000 pounds of force." 45 kN

In order for a baseball player to hit the ball with the greatest ability, both the force applied by the bat and the amount of time that the bat is in contact with the ball are important. The hitter's objective is to hit the ball with the part of the bat that will cause the moment of impact to be the longest and the average applied force to be the greatest. I say average, because the force exerted by the bat changes while it's making contact with the ball. This force is zero when the ball reaches the bat, at a maximum for an instant, and then returns to zero just when the ball leaves the bat.

The product of force and time is a quantity called impulse (measured in newton seconds). Newton's second law states that average net force equals mass times acceleration. We also know that acceleration is equal to the change in velocity, divided by the time interval. Applying these two formulas, we see that the average force on an object is equal to its change of momentum over time.

F = ma
a = (vfvi)/Δt
F = (mvfmvi)/Δt

Knowing his, all we need to determine the average force applied on the ball by the bat (which is equal to the force applied on the bat by the ball, from Newton's law of action and reaction) is the velocity (prior to, and after, contact) and mass of the ball, as well as the contact time between the bat and the ball. These numbers will vary, depending on the situation. A rather slow speed for a baseball with a mass of 0.14 kg that's approaching a hitter is -85 miles/hour, which is equivalent to -38 m/s.

-85 miles x 1609 meters x   1 hour   = -38 m/s hour     mile   3600 seconds

A common speed leaving the bat is +130 miles/hour or +58 meters/second. If the contact time is 0.004 seconds (which is longer than usual, since the original pitch was not very fast), we plug these numbers into our formula for the average force.

F = (mvfmvi)/Δt
F = [(0.14 kg)(+58 m/s)–(0.14 kg)(-38 m/s)]/0.004 s
F = 3360 kg*m/s
F = 3400 N

Albert Klyachko -- 2000

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