Rotational problem another time (this time not the center of mass)

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Homework Help Overview

The problem involves a chimney that behaves like a thin rod and topples due to gravity after cracking at the base. Participants are tasked with finding the linear speed of the top of the chimney just before it hits the ground, using principles of conservation of energy and rotational motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants discuss using conservation of energy to relate potential energy and rotational kinetic energy, questioning the change in potential energy as the chimney falls.
  • There are attempts to derive the angular speed and its relationship to the linear speed at the top of the chimney, with some participants expressing confusion about the consistency of angular speed throughout the fall.
  • Participants reference previous discussions and methods, indicating a connection to earlier problems while exploring the nuances of this specific scenario.

Discussion Status

Participants are actively engaging with the problem, attempting various methods and questioning their reasoning. There is recognition of the similarities to previous problems, and some guidance is offered regarding the relationships between angular speed and linear speed. However, there is no explicit consensus on the correct approach or solution yet.

Contextual Notes

Participants note that the chimney is treated as a rigid body and that the problem constraints include the absence of friction and the assumption that the chimney does not break apart as it falls. There is also mention of the previous thread that may contain relevant insights.

NasuSama
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Homework Statement



A chimney (length L = 82.6 m, mass M = 2280 kg) cracks at the base, and topples. Assume:
- the chimney behaves like a thin rod, and it does not break apart as it falls
- only gravity (no friction) acts on the chimney as if falls
- the bottom of the chimney tilts but does not move left or right

Find vtop, the linear speed of a point on the top of the chimney just as it hits the ground.

Homework Equations



→Conservation of energy
→KE = ½Iω²?
→v = ωr?

The Attempt at a Solution



What if I tried this method?

mgL = ½ * 1/3 * mL² * ω²
ω = √(6g/L)

Then...

v = √(6g/L) * L

But the whole answer is wrong.
 
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NasuSama said:
What if I tried this method?

mgL = ½ * 1/3 * mL² * ω²
What's the change in PE for the chimney as it falls? (Nothing has changed from your earlier problem except the point you are finding the speed of.)

And since this problem is an offshoot of your other one, why start a separate thread? You've already done the hard work in your other thread. You've found ω.
 
Doc Al said:
What's the change in PE for the chimney as it falls? (Nothing has changed from your earlier problem except the point you are finding the speed of.)

And since this problem is an offshoot of your other one, why start a separate thread? You've already done the hard work in your other thread. You've found ω.

Why is it that there is the same speed of the chimney on the top as the center of mass?
 
I mean why is it that there is the same angular speed?
 
Sorry for another post.

Then, using the same ω instead of different ω, we have...

v = ωr
= √(3g/L) * L

Instead of L/2, we have L since we want to find the speed of the top of the chimney.

Sorry for another thread. I thought this problem would be totally different from the previous one.
 
NasuSama said:
I mean why is it that there is the same angular speed?
Did anything change? It's the same chimney falling in the same manner. Why would the angular speed be different?
 
Doc Al said:
Did anything change? It's the same chimney falling in the same manner. Why would the angular speed be different?

Nvm. I was stumped and unclear of the situation here. Now, I should get it.

It's just v = √(3g/L) * L [with same ω]
 
NasuSama said:
Then, using the same ω instead of different ω, we have...

v = ωr
= √(3g/L) * L

Instead of L/2, we have L since we want to find the speed of the top of the chimney.
Good.
 

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