Heavy elastic string on a cone.

Question

A heavy elastic string, whose natural length is $2\pi a$, is placed round a smooth cone whose axis is vertical and whose semi-vertical angle is $\alpha$. If W be the weight and $\lambda$ the modulus of elasticity of the string, prove that it will be in equilibrium when in the form of a circle whose radius is $$a=(1+\frac{W}{2\pi\lambda}\cot\alpha)$$

Soln:

Say that the radius in equilibrium is $r$ then we write the potential energy of the system as $$E=-Wr\cot\alpha+\frac12\lambda(2\pi r - 2\pi a)^2$$ using this we differentiate and at equilibrium $dE/dr=0$ this gives $$r=\frac{W\cot\alpha}{4\pi^2\lambda}+a$$

Did I go wrong somewhere? it is somewhat matching but not exactly.

Answer 1

In scalar terms, Hooke's law can be written in at least three different ways:

$$ F = k\,\Delta L = \frac{\lambda}{L}\,\Delta L = \lambda\,\epsilon \,; $$

therefore, dividing both sides by the cross-sectional area, at least four ways are obtained:

$$ \sigma = \frac{k}{A}\,\Delta L = \frac{\lambda}{A\,L}\,\Delta L =\frac{\lambda}{A}\,\epsilon = E\,\epsilon \,. $$

So, given a string of weight $W$, of natural length $2\,\pi\,a$ and of elastic modulus $\lambda$, if $r$ is the radius of the circumference in which this string is in static equilibrium on a smooth and vertical cone at a distance $h = r\,\cot\alpha$ from the vertex:

• for the principle of minimum of the total potential energy:

$$ \frac{\text{d}}{\text{d}r}\left[\frac{1}{2}\,\frac{\lambda}{2\,\pi\,a}\left(2\,\pi\,r - 2\,\pi\,a\right)^2 - W\,h\right] = 0 \quad \Leftrightarrow \quad r = a\left(1+\frac{W\cot\alpha}{2\,\pi\,\lambda}\right); $$

• for the principle of virtual works:

$$ \lambda\,\frac{2\,\pi\,r - 2\,\pi\,a}{2\,\pi\,a}\,\delta(2\,\pi\,r) = W\,\delta h \quad \forall \, \delta h \ne 0 \quad \Leftrightarrow \quad r = a\left(1+\frac{W\cot\alpha}{2\,\pi\,\lambda}\right). $$

That's all.