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

[NasuSama]

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.

 

[NasuSama]

Actually, I followed the method from this link: https://www.physicsforums.com/showthread.php?t=647339

This link is the previous thread I just posted. I attempted to use conservation of energy, but that doesn't even seem right (the way I solve that problem)

 

[Doc Al]

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.

 

[NasuSama]

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?

 

[NasuSama]

I mean why is it that there is the same angular speed?

 

[NasuSama]

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.

 

[Doc Al]

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?

 

[NasuSama]

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]

 

[Doc Al]

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.