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Need help with the rotational problem!

  1. Oct 27, 2012 #1
    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. jcsd
  3. Oct 27, 2012 #2

    Doc Al

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    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.
     
  4. Oct 27, 2012 #3
    Then, I believe it is...

    PE = KE

    mgL/2 = 1/2 * Iω²
     
    Last edited: Oct 27, 2012
  5. Oct 27, 2012 #4
    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?
     
  6. Oct 27, 2012 #5

    Doc Al

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    Good!
     
  7. Oct 27, 2012 #6

    Doc Al

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    What's the rotational inertia about the pivot point?
     
  8. Oct 27, 2012 #7
    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)
     
  9. Oct 27, 2012 #8

    Doc Al

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    Good!
     
  10. Oct 27, 2012 #9
    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.
     
  11. Oct 27, 2012 #10
    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.
     
  12. Oct 27, 2012 #11

    Doc Al

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    Good.

    I meant that you couldn't use kinematic formulas for constant acceleration. Of course you can use some kinematics, such as v = ωr. :wink:
     
  13. Oct 27, 2012 #12
    Nvm. I believe I get your point when you made the new post.
     
  14. Oct 27, 2012 #13

    Doc Al

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    Right.
    You are asked to find the linear speed of the center of mass.
     
  15. Oct 27, 2012 #14
    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
     
  16. Oct 27, 2012 #15

    Doc Al

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    Looks good to me.
     
  17. Oct 27, 2012 #16
    Finally, I got it right. Thanks. I checked with the solution key, and I'm actually correct.
     
  18. Oct 27, 2012 #17

    Doc Al

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    Yay! :approve:
     
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