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## Homework Statement

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*?

## Homework Equations

U=Mgh

K=½Iω

^{2}

I=⅓ML

^{2}

## The Attempt at a Solution

Mgh

_{ο}=Mgh+½Iω

^{2}

ω

^{2}=(2Mg(h

_{ο}-h))/(⅓ML

^{2})

h=Lcosθ

so...

ω

^{2}=(2MgL(cosθ

_{ο}-cosθ))/(⅓ML

^{2})=(6g/L)cosθ

_{ο}-cosθ

...where g=9.8m/s

^{2}and L=49m,

since θ

_{ο}is going to always be 0 in this case you get...

ω(θ)=√1.2cosθ and ω(32°)≈1.01

a

_{⊥}=ω

^{2}r=1.01

^{2}*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.