Solving a Cylindrical Brick Chimney Toppling Problem

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SUMMARY

The cylindrical brick chimney toppling problem demonstrates that the chimney is most likely to break at a distance of L/3 from the base due to torque distribution. The initial approach involved calculating torque using the expressions dτw and dτr, but the solution was refined by analyzing the total angular momentum of the chimney. The final calculations confirmed that the maximum torque occurs at L/3, validating the problem statement's solution. Key factors affecting the outcome include the uniformity of mass distribution and external forces such as wind.

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rc75
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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 \frac{d \tau}{dx}=0, 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.

d \tau_w= -x g \cos \theta dm

d \tau_r=-\ddot{\theta} dI= -\ddot{\theta} x^2 dm

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

\tau=I \ddot{\theta}

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

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

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

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

So:

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

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?

Thanks.
 
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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.

L(x)=L_{above}+L_{below}

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

Since \tau=\dot{L}, then \frac{d \tau}{dx}=\frac{d}{dx}\dot{L}. 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.

\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}

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

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?
 


Firstly, great job on trying to approach the problem by finding where the torque is zero. Your method seems sound and your calculations are correct. However, there are a few things to consider when solving this problem.

1. The assumption of two torques acting on an infinitesimal piece of the chimney may not be entirely accurate. In reality, there are many other forces at play, such as wind, the weight distribution of the bricks, and the strength of the mortar holding them together. These may affect the distribution of the torque along the chimney and could potentially change the location where the chimney is most likely to break.

2. Your calculation for the torque due to rigid rotation is correct, but you may have missed a factor of 2 when calculating the torque due to the weight of the piece. The correct expression should be dτw=-2xgcosθdm, since the weight acts on both sides of the piece.

3. It is important to note that your calculation assumes a perfectly cylindrical chimney with a uniform distribution of mass. In reality, the chimney may not be perfectly cylindrical and the mass distribution may not be uniform, which could also affect the location where it is most likely to break.

Overall, your approach is a good starting point, but it may not give an accurate prediction of where the chimney is most likely to break. To get a more accurate answer, you may need to consider the other forces at play and possibly use more advanced techniques such as finite element analysis. Keep in mind that in real-life scenarios, there are always uncertainties and variations that may affect the outcome.
 

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