# Homework Help: Estimating the force of a baseball swing

1. Sep 4, 2006

### danago

Hey. I did some calculations, and estimated that an average person hitting a ball with a baseball bat would apply approximately 600-800N to the ball. Does this sound approximately right, or am i way off?

2. Sep 4, 2006

### Clef

what was the given information? and that seems like a fair bit....

3. Sep 4, 2006

### danago

Thats the thing...we were given no information at all. We are supposed to calculate the force of the average persons baseball swing, but nobody can seem to be able to do it.

What we did was use a 1kg mass, resting on a surface, and hit it with a baseball bat. We then timed the time it took to reach the 1m mark, and then found an average velocity. Now, my partner did the trials while i wasnt there, and he said that he calculated the velocty to be around 6m/s. I though this sounded a bit high, but we used it for our calculations.

From that average velocity, i calculated the change in momentum, which would be 6kg m/s, since its a 1kg mass. Then divided by an estimated contact time of 0.01s, which gave us the force.

Im thinking that my mistake was underestimating the contact time. I really wouldnt have a clue. I know its less than 1 second, but thats about all i know about the contact time :s

But other than that, does the rest of it sound ok? If i changed my working, and used a contact time of 0.1s, or even 0.05, id get a force of 60-120N. Does that sound better?

Im not sure if im even close to producing valid results, but its an attempt i guess.

4. Sep 4, 2006

### big man

I would have thought that it was a fair estimate. However, I think that is a fair estimate for a moving ball. I'd say that the average pitcher (I'm not American and I don't play baseball so I could be wrong) would pitch at maybe 90 - 100 km/h. I think that when the ball is hit you are going to have the ball going in the opposite direction at a higher speed than the initial pitch I'd say. Maybe about 120 or more km/h. Momentum has to be conserved so you solve as follows:

mv1 + Fdt = mv2

This is also assuming that the ball is thrown perfectly flat.
You can solve for the force in the x-direction. This is obviously assuming that the ball is hit absolutely horizontally. If you have a vertical component (you hit the ball so that it makes an angle with the horizontal) you can estimate the force by considering the x and y components of velocity and force.

The total force will be given by:

Ftotal = sqrt (Fx^2 +Fy^2)

This is what I think, but I could be wrong. There are probably other people here who know more and can be more informative.

Remember to draw a diagram as well and convert all units to SI units.

5. Sep 4, 2006

### Farsight

Your numbers are a little low, danago!

http://www.kettering.edu/~drussell/bats-new/impulse.htm [Broken]

Last edited by a moderator: May 2, 2017