# Homework Help: Need help with the rotational problem!

1. Oct 27, 2012

### NasuSama

21. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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

2. Oct 27, 2012

### Staff: Mentor

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

Instead of kinematics, think in terms of energy.

3. Oct 27, 2012

### NasuSama

Then, I believe it is...

PE = KE

mgL/2 = 1/2 * Iω²

Last edited: Oct 27, 2012
4. Oct 27, 2012

### 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?

5. Oct 27, 2012

### Staff: Mentor

Good!

6. Oct 27, 2012

### Staff: Mentor

What's the rotational inertia about the pivot point?

7. Oct 27, 2012

### NasuSama

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)

8. Oct 27, 2012

### Staff: Mentor

Good!

9. Oct 27, 2012

### NasuSama

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.

10. Oct 27, 2012

### 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.

11. Oct 27, 2012

### Staff: Mentor

Good.

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

12. Oct 27, 2012

### NasuSama

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

13. Oct 27, 2012

### Staff: Mentor

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

14. Oct 27, 2012

### 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

15. Oct 27, 2012

### Staff: Mentor

Looks good to me.

16. Oct 27, 2012

### NasuSama

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

17. Oct 27, 2012

Yay!