Kinetic Energy and the Human Body - Shot Put Illustration

In a previous post (Remembering Gravity), I was estimating the change to my vertical jump if my strength, speed, & power capabilities remained constant and my weight decreased. I mentioned that one of my assumptions was that the amount of change in the weight involved was probably such that the force production of my legs was the same on the light me as the heavy me. I also suggested that if the amount of difference in weight was significant, this assumption would start to unravel. My example to motivate this at an intuitive level was to compare shot putting with a 12 lb shot with shot putting a softball. Intuitively, you can probably recognize, that producing the same "power" with the softball as with the shot would involve an incredible amount of speed and that the result of shot putting a softball would be governed by maximum speed production rather than maximum force production. I thought it would be interesting to put numbers to this illustration.

When I jump or throw or "put", I am imparting force over distance. That force may not be constant and the direction may not be either but at the end of force application I have imparted a net amount of work to an object (my own body, ball, shot). In the end, the object has an amount of kinetic energy
$$W_{NET} = K = \frac{1}{2} m v^2.$$
So, to state my illustration more technically, I am saying that, using an identical movement pattern, allowing only changes in speed of movement of my body, I cannot impart as much kinetic energy to a very light object as I can to a heavier, but still manageable object.

So, to establish the general invalidity of the assumption of equivalent force production under different loading we can put a 7.260 kg shot and a 190 g softball (or 146 g baseball) and determine the approximate amount of kinetic energy imparted. A little hitch would be the need to know the release angle. An interesting post on Brunel University's site (Shot Put Projection Angle) highlights that in practice the optimum release angle is not 45° but may be between 26° and 38° and and this is due to a biomechanical advantage favoring superior speed production in the horizontal direction.

Let's Pretend

Suppose we take a world record shot put of 23.12 m with a 7.260 kg shot, assume a 30° release angle and determine the kinetic energy imparted to the shot by the putter. Then we will make the very erroneous assumption that the same person would have been capable of imparting the same kinetic energy to a softball or baseball, using the same biomechanical movement pattern. We are expecting to get a ridiculous answer which will satisfy us that the assumption that we can do this is bogus.

To determine the kinetic energy of the shot, I need to find the initial velocity. We use a few common kinematic equations. We can use the relationship between the vertical and horizontal velocity which is determined by the release angle, so that $\tan \theta = v_y / v_x$, where $\theta$ is the release angle. We neglect the effect of drag from the air. Taking $X$ as the horizontal distance traveled by the shot, we have
$$X = v_x t_t = \frac{v_y}{\tan \theta} t_t$$
and so
$$t_t = \tan \theta \frac{X}{v_y}.$$
Using the vertical displacement equation, we have
$$0 = d_R + v_y t_t + \frac{1}{2} g t_t^2$$
$$0 = d_R + v_y \Big(\frac{\tan \theta}{v_y} X\Big) + \frac{1}{2} g \Big(\frac{\tan \theta}{v_y} X\Big)^2$$
$$0 = d_R + X \tan \theta + \frac{g X^2 \tan^2 \theta}{2 v_y^2}$$
$$\frac{X^2 g \tan^2 \theta}{2 v_y^2} = -d_R - X \tan \theta$$
$$2 v_y^2 = - \frac{X^2 g \tan^2 \theta}{d_R + X \tan \theta}$$
$$v_y = X \tan \theta \sqrt{\frac{-g}{2 (d_R + X \tan \theta)}}$$
We have our velocity, $v_p$ of the shot put, then as $$v_p = \frac{v_y}{\cos \theta}.$$
Here's where we make our faulty assumption. (Faulty, because we are comparing objects of very different masses.) We take the kinetic energy the putter could generate on the shot and assume he can generate the same kinetic energy with a much smaller mass. Hence,
$$\frac{1}{2} m_p v_p^2 = \frac{1}{2} m_s v_s^2$$
and so
$$v_s = v_p \sqrt{\frac{m_p}{m_s}}.$$
Since we are assuming biomechanical identity, we use the same release angle giving us the vertical initial velocity of the softball, $v_{sy}$
$$v_{sy} = v_s \sin \theta \sqrt{\frac{m_p}{m_s}}.$$
and horizontal
$$v_{sx} = v_s \cos \theta \sqrt{\frac{m_p}{m_s}}.$$
From here we proceed to determine how long the ball will be in the air. There will be a time to the peak $t_p$ and a time to drop $t_d$ in the total time $t_t$. For the time to peak we have
$$0 = v_{sy} + g t_p$$
$$t_p = -\frac{v_{sy}}{g}$$
From this we get our maximum height, $h_p$
$$h_p = d_R + v_{sy} t_p + \frac{1}{2} g t_p^2$$
$$= d_R - \frac{v_{sy}^2}{2 g}$$
This allows us to find our time to drop from the displacement equation as
$$0 = h_p + \frac{1}{2} g t_d^2$$
$$t_d = \sqrt{\frac{2 h_p}{-g}}$$
So, we can give an expression of our bogus displacement of the softball as a shot put by
$$X_{sb} = v_{sx} (t_p + t_d).$$
We can follow through these expressions to give output without spending the time to make a monster expression out of it. I looked at doing that briefly and it didn't look like there was much for cancellations that made the expression more tidy. So, there is no computation savings by combining the expression into one monster formula. Therefore, since I want a number, I will just put these expressions into maxima and crunch.

Here's the sequence of expressions in maxima:

theta: float(30 * %pi/180);
m_p: 7.260;
m_s: 0.190;
d_R: 2.0;
X_sp: 23.12;
g: -9.81;

v_y: X_sp * tan(theta) * sqrt(-g/(2*(d_R+X_sp*tan(theta))));
v_p: v_y / cos(theta);

v_s: v_p * sqrt(m_p/m_s);
v_sx: v_s * cos(theta);
v_sy: v_s * sin(theta);
t_p: -v_sy/g;
h_p: d_R - v_sy^2/(2*g);
t_d: sqrt(2*h_p/(-g));

X_sb: v_sx* (t_p + t_d);

The final answer comes out at about 259.52 m for a softball. Changing the mass to a baseball, gives 336.71 m.

Athletes who use biomechanically superior form to throw a baseball actually come in more about 136 m or so (Glen Gorbous).

So, if you're trying to use the human body to generate kinetic energy, there's a mass sweet spot somewhere. Too little mass and you just can't generate the speed. Too much mass and you can't move it at all. Somewhere in between is the optimum, if, for reasons unknown, you care about generating a lot of kinetic energy.