In nature,the failure of structural members usually results from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling. In cases of tall cylindrical structures like trees,the torque is caused by their own weight bending the structure,such that the vertical line through the centre of gravity does not fall within the base. The elastic torque caused by this bending about the central axis of the tree is given by $\frac{Y\pi r^4}{4R}$,where $Y$ is the Young's modulus,$r$ is the radius of the trunk,and $R$ is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.

(N/A) The bending torque on the trunk of radius $r$ is given by $\tau = \frac{Y \pi r^4}{4R}$,where $R$ is the radius of curvature.
When the trunk bends,the torque due to its own weight $W$ is $\tau = Wd$,where $d$ is the horizontal displacement of the centre of gravity from the vertical axis passing through the base.
Equating the torques: $Wd = \frac{Y \pi r^4}{4R}$.
Assuming the tree has height $h$,its centre of gravity is at height $h/2$. From the geometry of the bent tree,using the Pythagorean theorem in the triangle formed by the centre of curvature,we have $R^2 = (R-d)^2 + (h/2)^2$.
Expanding this: $R^2 = R^2 - 2Rd + d^2 + h^2/4$. Since $d$ is very small,$d^2 \approx 0$,so $2Rd \approx h^2/4$,which gives $d = h^2 / (8R)$.
Let $\rho$ be the density and $g$ be the acceleration due to gravity. The weight $W = \text{Volume} \times \rho g = (\pi r^2 h) \rho g$.
Substituting $W$ and $d$ into the torque equation: $(\pi r^2 h \rho g) \times (h^2 / 8R) = (Y \pi r^4) / (4R)$.
Simplifying the equation: $(\rho g h^3) / 8 = (Y r^2) / 4$.
Solving for $h$: $h^3 = (2 Y r^2) / (\rho g)$,so the critical height $h = \left( \frac{2 Y r^2}{\rho g} \right)^{1/3}$.