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Falling Chimney

  • Thread starter rc75
  • Start date
6
0
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]

So:

[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?

Thanks.
 
6
0
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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