Work and Rotational Kinetic Energy (Falling Chimney)

[maxhersch]

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=½Iω²
I=⅓ML²

The Attempt at a Solution

Mgh_ο=Mgh+½Iω²
ω²=(2Mg(h_ο-h))/(⅓ML²)
h=Lcosθ

so...
ω²=(2MgL(cosθ_ο-cosθ))/(⅓ML²)=(6g/L)cosθ_ο-cosθ
...where g=9.8m/s² 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_⊥=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.

 

[ehild]

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

Mgh_ο=Mgh+½Iω²
ω²=(2Mg(h_ο-h))/(⅓ML²)
h=Lcosθ
so...
ω²=(2MgL(cosθ_ο-cosθ))/(⅓ML²)=(6g/L)cosθ_ο-cosθ
...where g=9.8m/s² and L=49m,

Correct the expression of the potential energy. And you also miss some parentheses.

since θ_ο 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)