The caber toss is a traditional Scottish sport that involves hurling a caber, which is essentially a large piece of a tree: a caber is a 6 m long, 80 kg tree trunk. Large strong people hurl these and the goal is to get the caber to land as far away as possible and rotate in the air, so what was the highest part of the trunk initially is actually the part that hits the gound first. See this clip to understand how the caber rotates.

It's HARD to do this. To see how hard, consider a perfectly vertical caber with one end on the ground. You then launch the caber vertically with some speed v0 and give it a rotation. What is the minimum kinetic energy in Joules you need to give the caber so that when it lands the caber is perfectly vertical again, but the OTHER end of the caber hits the ground?

Details and assumptions
The acceleration of gravity is −9.8 m/s2.
The caber can be modeled as a uniform rod.

To find the minimum kinetic energy required to ensure that the caber is perfectly vertical when the other end strikes the ground, we can consider the principle of conservation of mechanical energy.

The initial mechanical energy of the caber can be given as:

E_initial = K_initial + U_initial

where:
K_initial is the initial kinetic energy
U_initial is the initial gravitational potential energy

At the moment when the caber is perfectly vertical again and the other end hits the ground, all of its initial potential energy is converted into kinetic energy, and the potential energy becomes zero. Therefore, the final mechanical energy is only kinetic energy:

E_final = K_final

Since energy is conserved, we have:

E_initial = E_final

Now, let's break down the initial and final mechanical energies.

For the initial mechanical energy (E_initial), the kinetic energy (K_initial) comes from both linear and rotational motions:

K_initial = (1/2) * m * v0^2 + (1/2) * I * ω^2

where:
m is the mass of the caber (80 kg)
v0 is the initial speed of the caber
I is the moment of inertia of the caber, which can be calculated as (1/3) * m * L^2 for a uniform rod (L = length of the caber)

The rotational speed ω can be related to the linear speed v using ω = v / r, where r is the radius of rotation. In this case, r is half the length of the caber, so:

ω = v0 / (L/2)

Simplifying further, we get:

ω^2 = (4 * v0^2) / L^2

Substituting these equations back into the initial mechanical energy expression, we have:

E_initial = (1/2) * m * v0^2 + (1/2) * (1/3) * m * L^2 * (4 * v0^2) / L^2

E_initial = (1/2) * m * v0^2 + (2/6) * m * v0^2

E_initial = (4/6) * m * v0^2

E_initial = (2/3) * m * v0^2

For the final mechanical energy (E_final), only the kinetic energy is considered:

E_final = K_final = (1/2) * m * vf^2

Since the height is the same and the other end of the caber hits the ground, the final velocity vf can be related to the initial velocity v0 using the conservation of linear momentum:

m * v0 = m * vf

vf = v0

E_final = (1/2) * m * vf^2 = (1/2) * m * v0^2

Finally, to satisfy the conservation of energy, we set E_initial equal to E_final:

(2/3) * m * v0^2 = (1/2) * m * v0^2

Simplifying and solving for v0^2, we get:

(2/3) * v0^2 = (1/2) * v0^2

Multiplying both sides by 6/2:

4 * v0^2 = 3 * v0^2

v0^2 = 0

This means that the minimum kinetic energy required to achieve the desired outcome is zero. In other words, no initial kinetic energy is necessary to get the caber to land perfectly vertical with the other end hitting the ground.