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Falling chimney

  1. Apr 24, 2013 #1
    A chimney is demolished by breaking its base in hope that when it falls the top of the chimney will prescribe a circle. If the chimney is broken into two segments in the air, will the top segment reach the ground first?

    I have attached my professor's solution.

    My professor didnt illustrate on from "both gravity and rotation". I think the rotation is just caused by gravity, and gravity only exerts its effect on rotation. Acceleration from both gravity and rotation is not that clear. Can anyone help me with it? If it can be proved by some equations, it will be helpful:) Thank you!
     

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  3. Apr 24, 2013 #2

    mfb

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    That is a strange description, and in general, it is wrong.
    Imagine you break the chimney after an extremely small tilt - the lower part will be nearly stable, while the upper part can fall down (assuming the two parts do not collide).
    Actually, I am wondering if the lower part can reach the floor first at all - it is slowed down by the contact to the ground.
     
  4. Apr 24, 2013 #3
    The lower segment of the chimney can be taken as a rod rotating about its fixed point on the ground. We have Newton's 2nd law for torques:

    ##Ʃτ=Iα##

    ##mgr=(\frac{1}{3}mL^2)α##

    Assuming the center of gravity of the lower segment is taken to be at its geometrical center, ##r=\frac{L}{2}##

    Then ##α=\frac{3g}{2L}##

    The acceleration of any point on the lower segment is ##a=rα##. Thus for the other end (the end in the air) of the lower segment,

    ##a=(L)\frac{3g}{2L}=\frac{3g}{2}##
     
  5. Apr 24, 2013 #4
    Indeed, I did not understand either when your professor said that there are contributions of acceleration from both rotation and gravity. Is rotation in this case not caused by the chimney's weight which is a gravitational force itself?
     
  6. Apr 24, 2013 #5

    mfb

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    The acceleration of 3g/2 applies to a horizontal chimney orientation only. Otherwise, torque is just mL/2*sin(θ) where θ is measured relative to the vertical axis. This reduces the acceleration to ##\frac{3g\sin(\theta)}{2}##.

    Okay, the lower part can touch the ground first, but it does not have to.
     
  7. Apr 24, 2013 #6

    CWatters

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    I think that bit about not colliding might be key. Does it matter at what angle the failure occurs?

    For example imagine a rod hinged at the bottom with a bead on it that is free to slide up and down the rod. Lets say there is a collar at height h just below the top to stop the bead sliding down. Then as the rod falls would there be a position where gravity cannot provide enough centripetal acceleration to stop the bead flying up and off the top end of the rod ?
     
  8. Apr 24, 2013 #7

    NascentOxygen

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    Falling tall brick chimneys is a spectacle that has been a well-studied (by others, not by me. :smile: ) While upright, the gigantic pile of bricks is held together by little more than gravity. Typically, a tall chimney parts company in three (well, from memory I think it's three) pieces at predictable break points, the relatively weak mortar or concrete being unable to transfer the forces needed to otherwise accelerate the upper part of the structure to greater-than-g acceleration. The last to hit the ground is the top.

    I bet there are plenty of youtube videos of falling chimneys. Sometimes you see spectators leaping for their lives when miscalculation has placed them under the falling chimney top even as the lower part is already flat on the ground.

    i.e., tilting under demolition
     
  9. Apr 24, 2013 #8

    mfb

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    Hmm, I think we can modify the original question a bit: "If the chimney breaks apart into two segments in the air, will the top segment reach the ground first?"
    The difference here: the cause for the breakup is internally now, so it happens only if the lower part "tries" to fall quicker than the upper part.
    The lower part will reach the ground first, due to its faster-than-g acceleration after the breakup. This faster-than-g acceleration is a result of the fixed base - the motion is restricted to a rotation, and the lower part of the chimney will both "fall" and "rotate" (as seen from its center of mass).
     
  10. Apr 24, 2013 #9
    Thank you! But why the lower segmant can hit the ground first? I know its acceleration is larger than g. How to calculate the time taken for lower segment to fall to the ground and hence prove that it takes lesser time than the top segment?
     
  11. Apr 24, 2013 #10
    Can you use equations to clarify both “fall" and "rotate”? Thank you:)
     
  12. Apr 24, 2013 #11

    mfb

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    If I remember correctly, it is messy to solve the differential equation. An acceleration larger than g is sufficient to prove this, for some specific setup.
    For a chimney of length L, falling with an angular velocity of w, the outer tip moves with wL and the bottom does not move. This corresponds to a translation with velocity wL/2 and a rotating around the geometric center with angular velocity w.
     
  13. Apr 24, 2013 #12

    haruspex

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    The provided answer is certainly too glib. If the break occurs when the chimney is still near vertical, the top section will obviously hit the ground first. For the opposite extreme it is less clear. Yes, the top end of the lower section will be accelerating at nearly 3g/2, but the top section has acquired a rotation. (Note that the provided diagram overlooks this.)
    Suppose the break occurs when at some small angle θ to the horizontal. The two sides of the break will have the same speed and are the same distance from the ground. The top of the lower section has accn 3g/2, while the other side of the break has accn g, so will arrive later. But what of the top of the upper section? That will be moving faster, but have further to go. Since the extra speed will be in proportion to the extra distance, the extra g/2 accn of the top of the lower section should still mean it arrives sooner. (I did the algebra and got a time of ≈ θ√(L/3g)(1-Lθ/2x) for top of lower section and θ√(L/3g)(1-θ/3) for the top of the upper section, where x is the distance of the break from the base.)
    It follows that there is some critical height γL for which the top of the lower section will arrive at the same time as (whichever part of) the top section.
     
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