The Physics of Shot Put & Baseball

A Few Estimates

Let's say you and a couple of your physics-nerd friends were sitting around watching the 2016 Olympic shot put and saw Ryan Crouser win with a distance of 22.52m, you might have wondered how much force he exerted to make a 7.26 kg shot put travel that far. Further, let’s say you wanted to compare the force generated to put the shot that far, versus throw the venerated 95 mile-per-hour fastball.

Before we get into it, let me say that for these calculations, I made some estimations based upon analyzing the shot-putting and baseball-throwing motions. Now, I believe these are reasonable, but there is some minor error introduced since I wasn’t using high speed film captures; however, the same error would be present for both the shot put and baseball throwing motions, so I think everything kind of evens out and this is a fair comparison.

Olympic-Record Shot Put

We’ll start with the shot put first. Keep in mind that our primary formula is a = F/m and we are trying to figure out F. Using our equation of a = F/m, we want to know F, so solving for F, we get F = ma. We already know m, the mass of the shot put, is 7.26 kg. I did some figuring and believe that a reasonable acceleration (a) that Ryan Crouser applied to the shot put on his winning throw was 7.21m/s². So, we have m = 7.26kg and a = 7.21m/s^s, so all we need to do is substitute them into the equation to solve for F:

a = F/m
(m)(a) = F
(7.26kg) (7.75m/s²) = F
F = 52.27 N

So, the force he imparted to the shot put to make it fly 22.52m through the air was 52.27N.

95MPH Fastball

We can compare that to the force needed to generate a 95 mile per hour fastball in baseball. 95 mph is 42.47m/s, which we will say is the velocity when the ball leaves the pitcher’s hand. I made some calculations and figure the acceleration achieved to make that velocity is 55.89m/s². Recall the shot put acceleration of 7.75m/s², so the baseball pitcher generates about 7 times the acceleration as the shot putter. That is quite a big difference between the two throwing movements, but if you are thinking that the baseball pitcher must therefore exert more force, we will see how mass is the big equalizer when it comes to acceleration. The mass of the baseball is 145g, which is 0.145kg. Now we know acceleration and mass, so we just need to complete the equation and get the answer.

a = F/m
(m)(a) = F
(0.145kg)(55.89m/s²) = F
F = 8.1N

Results

So, the pitcher gives 8.1N of force to the baseball to accelerate it from 0 to 95 mph (42.47m/s), while the shot putter generates 52.27N on the shot put to make it travel 22.52m. Although the acceleration to the baseball is about 7 times greater than that to the shot put, the shot put has 50 times the mass of the baseball, which is why so much more force is imparted to the shot put as compared to the baseball.

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