Question

A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 59.9 m. At the instant it makes an angle of 39.3° with the vertical as it falls, what are:

(a) the radial acceleration of the top
(b) the tangential acceleration of the top. (Hint: Use energy considerations, not a torque.)
(c) At what angle θ is the tangential acceleration equal to g? (Assume free-fall acceleration to be equal to 9.81 m/s^2.)

Answer 1

The answer for (a) is 6.6491 m/s^2

Answer 2

I obtained that by using conservation of kinetic and potential energy.
K_i + U_i = K_f + U_f
K = (1/2)(I)(omega^2)
U = mgy
I of a rod rotating around its center of mass is (1/12)mL^2. Using the parallel axis theorem, I found that I for the chimney (which is rotating around its base) is (1/3)mL^2
You need to know the height of the center of mass while the chimney is falling as well. That is (1/2)(L)(cos theta)

When you put the conservation equation together, you get:
u_i = u_f + k_f
(1/2)mgL = (1/2)mgL(cos theta) + (1/6)mL(omega^2)

You can solve this for omega^2:

\[\omega^2 = \frac{ 3g }{ L }(1 - \cos \theta)\]

Radial acceleration is omega^2 * r, r here being equal to L, so it gets canceled out.