Rotational motion

Q. Rotational motion

Question
Bob Bray (mass 100 kg) and Shelly Marvel (mass 50 kg) plan to convert the Toronto Sky-Dumb into the world's largest ca-SIN-oh. They climb a 40 m tall pole pivoted at ground level as shown. Shelly is at the very top, inspecting the cap, and Bob is exactly half way up. The pole falls over, starting from rest in the vertical position and swinging down through 90 degrees about the frictionless fixed pivot point until it crashes into the horizontal ground. Both Shelly and Bob hold on tightly and do not change their positions on the pole. (Politicians worry about poll positions). Assume that the pole has a negligibly small mass but is still strong enough to stay straight as it falls. Calculate the speed of Shelly just before striking the ground. Answer in m/s.

Answer

  
Method 1: 
Translatory motion treating Shelly and Bob as particles. 
  
As both will fall with the same angular velocity, 
if velocity of Shelly = v, then
velocity of Bob = v/2.
When the pole falls, total potential energy of Shelly and Bob is converted into their total kinetic energy
=> (50 * 40 + 100 * 20) * (9.81) 
......= (1/2) * 50 v^2 + (1/2) * 100 (v/2)^2
=> 37.5 v^2 = 39240
=> v = 32.3 m/s.

Method 2.
Rotatory Motion of the Pole.

Centre of mass of the pole
= (50*40 + 100*20) / (50 + 100)
= 80/3 m

Moment of inertia of the pole about the centre of mass
= (50) * (40 - 80/3)^2 + (100) * (80/3 - 20)^2 kgm^2

Moment of inertia of the pole about the pivot 
using perpendicular axis theorem
= its moment of inertia about the centre of mass + (150) * (80/3)^2 
= (50) * (40 - 80/3)^2 + (100) * (80/3 - 20)^2 + (150) * (80/3)^2
= 120000 kgm^2.

When the pole falls, its total potential energy is converted into its total rotational kinetic energy
=> (50 * 40 + 100 * 20) * 9.81 = (1/2) * (120000) * ω^2, where ω = angular velocity of the pole.
=> 39240 = 60000 ω^2
=> ω = 0.808 radian/s.
=> velocity of Shelly
= 40ω 
= 40 * 0.808
= 32.3 m/s.