How Do You Calculate the Speed of a Falling Chimney's Center of Mass?

[NasuSama]

2

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 vcm, the linear speed of the center of mass of the chimney just as it hits the ground.

Homework Equations

τ = Iα
τ = rF
ω_f² = ω_0² + 2aθ
v = ωr

The Attempt at a Solution

τ = mgL/2
τ = Iα

mgL/2 = 1/12 * mL² * α
α = 6g/L

ω = √(2αθ) [I was thinking that θ = π/2, but this gives incorrect answer]

v = ωr
= √(2αθ)r

But the whole answer is wrong

 

[Doc Al]

The acceleration is not constant, so you cannot use those kinematic formulas for constant acceleration.

Instead of kinematics, think in terms of energy.

 

[NasuSama]

The acceleration is not constant, so you cannot use those kinematic formulas for constant acceleration.

Instead of kinematics, think in terms of energy.

Then, I believe it is...

PE = KE

mgL/2 = 1/2 * Iω²

 

[NasuSama]

Actually, I used that form and tried to use kinematic as before, but as you just said it's not right.

mgL/2 = ½ * 1/12 * mL²ω²
g = 1/12 * L * ω²
ω = √(12g/L)

v = ωr
= √(12g/L) * L/2

Wait.. As you just said before, I can't use kinematics! But how?

 

[Doc Al]

Then, I believe it is...

PE = KE

mgL/2 = 1/2 * Iω²

Good!

 

[Doc Al]

Actually, I used that form and tried to use kinematic as before, but as you just said it's not right.

mgL/2 = ½ * 1/12 * mL²ω²

What's the rotational inertia about the pivot point?

 

[NasuSama]

What's the rotational inertia about the pivot point?

For the chimney acting like the "thin rod", that is 1/3 * mL². Then, we obtain...

mgL/2 = ½ * 1/3 * mL² * ω² [since the rotational axis is at one end of the chimney]

...so ...

g = 1/3 * L * ω²
ω = √(3g/L)

 

[Doc Al]

For the chimney acting like the "thin rod", that is 1/3 * mL². Then, we obtain...

mgL/2 = ½ * 1/3 * mL² * ω² [since the rotational axis is at one end of the chimney]

Good!

 

[NasuSama]

Good!

Continuing from here...

g = 1/3 * L * ω²
ω = √(3g/L)

Then...

v = ωr

But you said that I can't use kinematics. I though that ω is the angular speed, so I need to multiply that value by L/2 to get the result.

 

[NasuSama]

I don't understand. I though that ω is the angular speed, right? Then, v is the linear speed.

Which is the key to find? ω or v? I don't even get the point here.

 

[Doc Al]

Continuing from here...

g = 1/3 * L * ω²
ω = √(3g/L)

Then...

v = ωr

Good.

But you said that I can't use kinematics. I though that ω is the angular speed, so I need to multiply that value by L/2 to get the result.

I meant that you couldn't use kinematic formulas for constant acceleration. Of course you can use some kinematics, such as v = ωr.

 

[NasuSama]

I don't understand. I though that ω is the angular speed, right? Then, v is the linear speed.

Which is the key to find? ω or v? I don't even get the point here.

Nvm. I believe I get your point when you made the new post.

 

[Doc Al]

I don't understand. I though that ω is the angular speed, right? Then, v is the linear speed.

Right.

Which is the key to find? ω or v? I don't even get the point here.

You are asked to find the linear speed of the center of mass.

 

[NasuSama]

Then, when I substitute all the values in, I get...

v = √(3g/L) * L/2
= √(3 * 9.81/82.6) * 82.6/2
≈ 24.7

 

[Doc Al]

Then, when I substitute all the values in, I get...

v = √(3g/L) * L/2
= √(3 * 9.81/82.6) * 82.6/2
≈ 24.7

Looks good to me.

 

[NasuSama]

Finally, I got it right. Thanks. I checked with the solution key, and I'm actually correct.

 

[Doc Al]

Finally, I got it right. Thanks. I checked with the solution key, and I'm actually correct.

Yay!