# What is the rotational acceleration of a falling chimney?

spsch
Homework Statement:
A 55 Meter chimney breaks and falls down to the side. What is it's rotational acceleration when the chimney makes an angle of 35 degrees to the vertical.
We can assume the chimney as a moment of inertia of a long thin rod.
Relevant Equations:
## mgh = \frac {1}{2} I omega^2 ##
Pythagorean Theorem?
Hi all,

I found this problem in a new textbook I'm working through.
And my energy conservation equation was ## mg\frac {h}{2} = \frac {1}{2} I ω^2 + mg \frac {h}{2}*sin(55) ##
My solution was wrong and after checking why I found that they used cos(35) as the angle. The rest was the same.

I'm a little dumbfounded here. It, unfortunately, doesn't explain why they used the cosine of the angle.
I had sin(55) because I subtracted 35 from 55 to get the angle with the horizontal from which I can get h/2*sin(55) of the remaining height?

Thank you, sorry for asking some weird question again

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90-55=35
sin(x) = cos(90-x)

• spsch
spsch
ω
90-55=35
sin(x) = cos(90-x)
Hi Orodruin.

Thank you, this is an identity I didn't know, or remember! I'm glad that this means I must have just made an algebra or calculator mistake.
Thank you very much!

Edit: I've just drawn it out and it's actually so obvious that they are the same! thanks again!