I didn't write down all my references, hope that you guys can take me on my word for these
figures. Okay, here goes...
Some values gleaned from these here internets:
mass of Tiger's driver = 68 g
mass of an average human arm = 3.35 kg
speed of Tiger's swing = 125 mph = 55.88 m/s
mass of an average newborn = 3.4 kg
height of Tiger = 1.83 m
length of avg. arm = 1/2 height, so = 0.915 m
length of Tiger's driver = 45" = 1.143 m
length of chain used in hammer throw = 1.22 m
average shoulder height = 4/5 * height = 1.464 m
Just before contact with the ball, Tiger's driver head is travelling at 125 mph.
Presumably, Tiger would hammer throw the baby using one of those attrocious child leashes
which are likely around the same length as the chain used in the hammer throw and adding a
negligible amount of weight. Let us presume that, given the nature of a hammer throw
(wind up time, use of full body weight to generate torque, etc.), Tiger is able to provide
the same amount of rotational energy at the moment just before releasing the baby.
If we assume the rotational energy of his club (arm included) is equal to the rotational
energy of his baby on a leash (arm included), we can say that:
1/2*(m(arm)+m(club))*w(club)^2 = 1/2*(m(arm)+m(baby))*w(baby)^2
where w is angular velocity.
We note that:
w = v/r
where v is the velocity (perpendicular to the rotation) and r is the radius of rotation.
Therefore by substitution and dividing both sides by 1/2 (or multiplying both sides by 2
for the pedants out there):
(m(arm)+m(club))*v(club)^2/(l(arm)+l(driver))^2 =
(m(arm)+m(baby))*v(baby)^2/(l(arm)+l(chain))^2
Adding in the numbers because damn, is this getting hard to read:
(3.35 + 0.068) * 55.88^2 / (0.915 + 1.143)^2 = (3.35 + 3.4) * v(baby)^2 / (0.915 + 1.22)^2
solving for v(baby):
v(baby) = ((3.418 * 2.135^2 * 55.88^2) / (6.75 * 2.058^2))^(1/2)
v(baby) = 41.252 m/s
Now, it may be asked, well, what if the baby's alive and kicking and such? The answer is
simple: the centrifugal force being exerted on the baby will knock it clean out, so he'll
either be swinging one limp baby or one stiff baby (if it was deceased sufficiently long
before the throw attempt). For the purposes of this calculation, we'll be ignoring wind
resistance (it will be negligible for an object traveling at such a slow (comparatively)
speed and with such a small surface area, anyway), so limp or stiff won't matter much. If I find the time later, I'll do the
calculus required to include this in my calculations.
For optimum range, Tiger would want to throw the baby at a 45 degree angle. He'll release
it at the height of his swing, which would make it:
h(baby) = h(shoulder) + l(arm) * sin 45 degrees = 1.464 + (0.915 * sin 45) = 2.111 m
The range, R, can be determined by the following formula:
R = delta x = v(x, initial) * delta t + 1/2 * a(x) * (delta t)^2
since we're assuming no wind resistance (and so no acceleration in the x axis), this
simplifies to:
R = v(x, initial) * delta t
where: v(x, initial) = v(baby) * cos 45 = 29.170 m/s, and delta t is the time the baby
remains in the air.
To determine delta t, we look at the y axis:
delta y = v(y, initial) * delta t + 1/2 * a(y) * (delta t)^2
Now considering that: delta y = h(final) - h(initial) = h(ground) - h(baby) = 0 - 2.111 =
-2.111 m, a(y) = g = -9.8 m/s^2, and v(y, initial) = v(baby) * sin 45 = 29.170 m/s:
-2.111 = 29.170 * delta t + 1/2 * -9.8 * (delta t)^2
using the quadratic:
delta t = 3.012 s
plugging that back in to our first equation:
R = 29.170 * 3.012 = 87.86 m
Now for comparison (and to see how valid our number probably is):
The current world record for the hammer throw is 86.74 m. Oh my, Tiger Woods would set a
world record! He's *so* in the wrong profession.
Okay, not really. The hammer throw event uses a ball with a weight of 7.257 kg, a bit
more than double the weight of the average newborn, so all is still right in the world,
and I think the numbers still make a good amount of sense.
For any that are interested, I put a better typeset version of this at: http://www.llamanation.org/tigerwoods.