# Work and Rotational Kinetic Energy (Falling Chimney)

Tags:
1. Nov 17, 2014

### maxhersch

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

a2r=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.

2. Nov 17, 2014

### ehild

What is h? Recall that the potential energy is as if all the mass of the rod was concentrated in the centre of mass.

Correct the expression of the potential energy. And you also miss some parentheses.
What is cos(0)? You ignored both cos(0) and the minus sign in front of cos(θ)