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Falling Chimney

  1. Jul 7, 2005 #1
    I'm having trouble with solving this problem:

    Suppose we have a cylindrical brick chimney with height L. It starts to topple over, rotating rigidly about its base until it breaks. Show that it is most likely to break a distance L/3 from the base because the torque is too great.​

    So I'm approaching this by trying to find where [tex]\frac{d \tau}{dx}=0[/tex], where x is the distance along the chimney measured from the base.

    I assume that on an infintesimal piece of the chimney there are two torques. One due to the weight of the piece and one forcing the rigid rotation of the chimney.

    [tex] d \tau_w= -x g \cos \theta dm [/tex]

    [tex] d \tau_r=-\ddot{\theta} dI= -\ddot{\theta} x^2 dm [/tex]

    Then I find [tex]\ddot{\theta}[/tex] to substitute back into the expression above:

    [tex]\tau=I \ddot{\theta}[/tex]

    [tex] -(3/2) (g/L) \cos \theta= (m/3) L^2 \ddot{\theta} [/tex]

    [tex] \ddot{\theta} = -(3/2) (g/L) \cos \theta [/tex]

    Making this substitution and also [tex]dm=(m/L) dx[/tex] I get:

    [tex] d \tau= ( -xg \cos \theta +\frac{3g}{2L} \cos \theta x^2)\frac{m}{L}dx[/tex]


    [tex] \frac{d \tau}{dx}= (-xg \cos \theta + \frac{3g}{2L} \cos \theta x^2)\frac{m}{L} [/tex]

    Setting this equal to zero and solving for x I get that x= (2/3)L. This seems like I'm close, but I'm not entirely sure that my approach makes sense. Did I inadvertently miss a factor of two somewhere? Or did I implicitly use the top of the chimney as the origin instead of the base? Or is this answer just coincidentally close to the right one?

  2. jcsd
  3. Jul 12, 2005 #2
    Falling Chimney Solution

    I've got a solution to this problem now that matches the solution given by the problem statement, but I use a different approach than the one I used above.

    Now I start by writing down the total angular momentum of the chimney as the angular momentum of part of the chimney below a point that is a distance x from the base, and the angular momentum of part of the chimney that is above it.

    [tex] L(x)=L_{above}+L_{below} [/tex]

    [tex] L(x)=\frac{1}{6}\frac{m}{L} \dot{\theta}^2 (L-x)^3+ \frac{1}{3}m x^2 \dot{\theta}^2[/tex]

    Since [tex] \tau=\dot{L}[/tex], then [tex]\frac{d \tau}{dx}=\frac{d}{dx}\dot{L}[/tex]. The torque will be an extreme value when this is equal to zero, and if it's a maximum then we expect the chimney to break.

    [tex] \dot{L}(x)= \frac{1}{3}\frac{m}{L}\dot{\theta}\ddot{\theta}(L-x)^3 + \frac{2}{3} m x^2 \dot{\theta}\ddot{\theta}[/tex]

    [tex]\frac{d}{dx}\dot{L}= -\frac{m}{L}\dot{\theta}\ddot{\theta}(L-x)^2 + \frac{4}{3} m x \dot{\theta}\ddot{\theta}[/tex]

    Setting the above equal to zero and solving for x gives the roots 3L and L/3. The first root is unphysical and is discarded, but the second root is the answer we are looking for.

    So that's the answer I was supposed to get, but I still can't see exactly where the previous approach was wrong. Any ideas on how to fix up what I did above so I get the same answer as I did here?
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