A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 49.0 m. Answer the following for the instant it makes an angle of 32.0° with the vertical as it falls. (Hint: Use energy considerations, not a torque.)
(a) What is the radial acceleration of the top?
(b) What is the tangential acceleration of the top?
(c) At what angle θ is the tangential acceleration equal to g?
The Attempt at a Solution
...where g=9.8m/s2 and L=49m,
since θο is going to always be 0 in this case you get...
ω(θ)=√1.2cosθ and ω(32°)≈1.01
This is not accepted as the correct answer nor does it really make sense but I am not sure where I went wrong.