# Work and Rotational Kinetic Energy (Falling Chimney)

## 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?

U=Mgh
K=½Iω2
I=⅓ML2

## 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.

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

K=½Iω2
I=⅓ML2

## 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,
Correct the expression of the potential energy. And you also miss some parentheses.
since θο is going to always be 0 in this case you get...
ω(θ)=√1.2cosθ and ω(32°)≈1.01
What is cos(0)? You ignored both cos(0) and the minus sign in front of cos(θ)