What is the rotational acceleration of a falling chimney?

Click For Summary
SUMMARY

The discussion centers on the calculation of rotational acceleration for a falling chimney using energy conservation principles. The original equation presented was incorrect due to the use of sin(55) instead of cos(35) for the angle in the energy conservation equation. The correct formulation involves recognizing the trigonometric identity sin(x) = cos(90-x), which clarifies the relationship between the angles involved. The participants concluded that the mistake was due to a misunderstanding of trigonometric identities rather than a fundamental error in the energy conservation approach.

PREREQUISITES
  • Understanding of energy conservation principles in physics
  • Familiarity with rotational dynamics and angular acceleration
  • Knowledge of trigonometric identities, specifically sin and cos functions
  • Basic algebra skills for solving equations
NEXT STEPS
  • Study the derivation of energy conservation equations in rotational motion
  • Learn about the application of trigonometric identities in physics problems
  • Explore examples of rotational dynamics involving falling objects
  • Practice solving problems involving angular acceleration and energy conservation
USEFUL FOR

Students of physics, educators teaching rotational dynamics, and anyone interested in understanding energy conservation in mechanical systems.

spsch
Messages
111
Reaction score
21
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
 
Physics news on Phys.org
90-55=35
sin(x) = cos(90-x)
 
  • Like
Likes   Reactions: spsch
ω
Orodruin said:
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!
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
13
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K