Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
584 views
in Physics by (26.9k points)
closed by

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 and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by \(\frac{Y\,{πr^4}}{4\,R}\).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 tree for a given radius of the trunk.

1 Answer

+1 vote
by (30.0k points)
selected by
 
Best answer

Let us consider the following diagram.

From questions,

Torque on trunk of tree = \(\frac{Y\,πr^4}{4\,R}\)

[r = radius of tree, R = radius of curvature of bent surface]

When the tree is about the buckle,

Wd = \(\frac{Y\,πr^4}{4\,R}\)

If R > > h, then the centre of gravity is at a height l = \(\frac{1}{2}\)h from the ground.

From ∆ABC

R2 = (R – d)2 + (\(\frac{1}{2}\)h)2

If d < < R

R2 = R2 – 2Rd + \(\frac{1}{4}\)h2

∴ d = \(\frac{h^2}{8R}\)

If ω0 is the weight/volume

\(\frac{Y\,πr^4}{4\,R}\) = W0(πr2h)\(\frac{h^2}{8R}\)        [∵ Torque is due to weight]

or h = (\(\frac{2Y}{W_0}\))\(\frac{1}{3}\) r\(\frac{2}{3}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...