1. The problem statement, all variables and given/known data 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? 2. Relevant equations U=Mgh K=½Iω2 I=⅓ML2 3. The attempt at a solution Mghο=Mgh+½Iω2 ω2=(2Mg(hο-h))/(⅓ML2) h=Lcosθ so... ω2=(2MgL(cosθο-cosθ))/(⅓ML2)=(6g/L)cosθο-cosθ ...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 a⊥=ω2r=1.012*49=49.98≈50 This is not accepted as the correct answer nor does it really make sense but I am not sure where I went wrong.