- #1

rc75

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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.