Question

A thin uniform pole of length 33 m is pivoted at the bottom end. Calculate the most probable point of rupture on the pole as the pole falls.

Please solve this problem by torque, Newton's law, and conservation of energy. Also, please show all the calculations. Thanks.

Answer 1

Assumptions are:
1) The pole is uniform density which means it's centre of gravity is exactly at 15 m
2) The pole is uniform material properties (ie. no defects in atomic structure)
3) Only gravity forces at work - no other external forces on the pole (including at the pivot)

Problem Set Up
1) the Pole is initially vertical, pivoted at the bottom end
2) as the Pole falls to a horizontal position, you want to find the most probable point of rupture.

Now this is why I think point of rupture may occur at 16.5m
1) prior to impact to the horizontal ground, you have a beam which has equal force distribution (due to gravity). It will look something like this reference except no reactions from the pivot points due to assumptions stated above

2) If you plot the shear diagram of this equally distributed load, the point of maximum shear occurs at 16.5 m. And hence at 16.5 m, the pole will have maximum shear force per meter. Hence, greatest probability of rupture will occur at this point since it has maximum shear force.

3) As the pole strikes the ground (exactly horizontal), the normal force of the ground will equal exactly the distributed force of gravity. Also, the Impulse will equal exactly the distributed force (except in opposite direction). Again, the maximum shear will be at 16.5 m.