The average force needed to throw a baseball 90mph?

[Phillipv2004]

Homework Statement

The average force needed to throw a baseball 90mph. I tried to find the force at 90mph on release and 90mph at the plate and got two wildly different answers and it seems off that they vary soo much.

Homework Equations

· Velocity- 90 mph or 40.23 m/s measured at release point.

· Mass- Objectives of the Game. Rules of Baseball. Major League Baseball Enterprises, 1998."It shall weigh not less than five nor more than 5 ¼ ounces avoirdupois" – 142g-149g so 145g=.145kg

· Distance- Ball accelerated over a distance of 1.75m arm length

Work= Force x Distance
Force= Work/Distance
KE=.5mv²

The Attempt at a Solution

KE=1/2(.145kg)(40.23m/s)²
KE=117.34J

F=117.34J/1.75m=67.05N

I then attempted to determine a 90mph baseball at the plate.

a=ΔV/Δt

Major League Baseball the mound and plate are 18.44m apart
t=18.44m/40.23m/s=.46s

a=(40.23m/s-0)/.46s
a=87.46m/s²

F=ma
F=(.145kg)(87.46m/s²)=12.68N

 

[PhanthomJay]

If the ball leaves the pitcher’s hand at 90 mph and reaches home plate at 90 mph, what is it’s acceleration during that journey from release of ball to home?

 

[haruspex]

Work= Force x Distance

This is fine for finding a constant force to achieve a certain KE over a certain distance, but it will not give the average of a varying force. The average force is defined as the change in momentum divided by the elapsed time.
Muscle power is typically limited in three ways: maximum force, maximum power and maximum rate of contraction. In throwing a baseball I suggest that at the start of the action it's the force that's the limit, but pretty soon it is the power; as the speed increases greater power is needed to maintain the same force. By the time of release, rate of contraction may be the limit.
My guess would be that the most accurate of those three models is constant power. You could calculate the time to reach the required speed on that basis and hence deduce the average force.

 

[jbriggs444]

I tried to find the force at 90mph on release and 90mph at the plate and got two wildly different answers and it seems off that they vary soo much.
[...]
Work= Force x Distance
Force= Work/Distance
[...]
F=117.34J/1.75m=67.05N

This is the average force over the path from start of throw (as the pitcher's hand begins moving) to end of throw (when the ball leaves the pitcher's hand).

[@haruspex makes the point that the standard assumption is that "average force" means an average over time. This is an average over a directed displacement instead. One could make additional technical points about dot products and averages over paths, but that would ignore the elephant in the room]

If we make the crude assumption that the throw is at constant acceleration, we are talking about 40 meters per second top speed and 20 meters-per-second average speed during the throw. It will take about 1.75 meters / 20 meters-per-second = 87 milliseconds for the throw.

Roughly speaking, you have obtained the average force over the 87 millisecond throw.

I then attempted to determine a 90mph baseball at the plate.

a=ΔV/Δt

Major League Baseball the mound and plate are 18.44m apart
t=18.44m/40.23m/s=.46s

a=(40.23m/s-0)/.46s
a=87.46m/s²

F=ma
F=(.145kg)(87.46m/s²)=12.68N

This time it seems that you have attempted to compute the average force during the interval between the ball starting to move in the pitcher's hand to the instant before it contacts the catcher's glove.

You've included the 460 milliseconds during which the ball moves from pitcher to catcher and ignored the [estimated] 87 milliseconds of the throw itself.

Roughly speaking, you've obtained average force over an estimated 460 millisecond throw.

The elephant in the room is the fact that you have averaged over different intervals.

It is no surprise that computing average force for the same delta v over different delta t's will produce different results.

 

[ZapperZ]

I think this is simpler than what the OP is thinking of.

For example, to find the average velocity, we simply find the change in displacement, and divide that over the time taken, i.e.
<v> = \frac{\Delta x}{\Delta t}

So the same should apply here, where
<F> = \frac{\Delta p}{\Delta t}

where Δp is the change in momentum. Presuming that this is to throw the ball from rest up to 90 mph, then the change in momentum is rather straight-forward. The average force is then just the final momentum over the period of time that that the ball is being pushed to get to that speed.

Zz.

 

[haruspex]

I think this is simpler than what the OP is thinking of.

For example, to find the average velocity, we simply find the change in displacement, and divide that over the time taken, i.e.
<v> = \frac{\Delta x}{\Delta t}

So the same should apply here, where
<F> = \frac{\Delta p}{\Delta t}

where Δp is the change in momentum. Presuming that this is to throw the ball from rest up to 90 mph, then the change in momentum is rather straight-forward. The average force is then just the final momentum over the period of time that that the ball is being pushed to get to that speed.

Zz.

Yes, it is easy if the time for which the ball is accelerated is known, but it seems it is not.

The OP mentions a distance - the arm length - but it is unclear whether this is a fact provided as part of the task or is @Phillipv2004's idea of additional info that is relevant.
As I have posted, it is useful, but it needs to be combined with a model for the profile of the force against time.
If the model is constant force we get one answer, F_c; constant power gives an average force 4/3 times as great, while the model of a spring gives $\frac 4{\pi}F_c$.