Tension for the crown

Here is my "energy-based solution" (see Steven Chase's post for a "force-based" solution).

The potential energy of the wire is given by $E = mgh$ (where $h$ is the height at which the wire lies; here I set the height to be 0 at the center of the sphere). This may be re-expressed as a function of the length $\ell$ of the wire. Use $h^2 + (\frac{\ell}{2 \pi})^2 = r^2$ to get: $E = mg\sqrt{ r^2 - (\tfrac{\ell}{2 \pi})^2 }$ Now imagine the wire would increase its length by $\text{d}\ell$ , then the energy would change as $\text{d}E = -F \text{d}\ell$ where $F$ is the tension (the change in energy corresponds to a negative work by the tension in the wire). Hence $\begin{array}{rl} F =& -\dfrac{ \text{d}E }{ \text{d}\ell }\\ = & mg \dfrac{1}{ \sqrt{ r^2 -(\tfrac{\ell}{2 \pi})^2 } } \dfrac{\ell}{4 \pi^2}\\ =& \dfrac{mg \ell}{4\pi^2 \sqrt{ r^2 -(\tfrac{\ell}{2 \pi})^2 } } \end{array}$ Pluging in the values give $\approx \fbox{0.75 N}$ .

Suppose the ring is divided into many little pieces of angle $d \theta$ . Suppose also that each piece is acted on by a force $d F = \alpha \, d \theta$ , which points radially away from the sphere. The vectorial force components are:

$d F_x = \alpha \, d \theta \, cos \theta \, sin \phi \\ d F_y = \alpha \, d \theta \, sin \theta \, sin \phi \\ d F_z = \alpha \, d \theta \, cos \phi$

The angle $\phi$ is a fixed quantity determined by the parameters of the problem. We know that the integral of $d F_z$ must equal the weight.

$F_z = \alpha \, cos \phi \int_0^{2 \pi} d \theta = 2 \pi \, cos \phi \, \alpha = m g \\ \alpha = \frac{m g}{2 \pi \, cos \phi }$

Calculate the $x$ force associated with an arbitrarily defined angle window $-\theta_f \leq \theta \leq \theta_f$ :

$F_x = \alpha \, sin \phi \int_{-\theta_f}^{\theta_f} cos \theta \, d \theta \\ = \frac{m g \, tan \phi}{2 \pi} \int_{-\theta_f}^{\theta_f} cos \theta \, d \theta = \frac{m g \, tan \phi}{\pi} \, sin \theta_f$

We know that this $x$ force is countered by the twice the corresponding tension component, evaluated at $\theta_f$ . This is because the tension pulls on the partial ring segment equally at both ends.

$\frac{m g \, tan \phi}{\pi} \, sin \theta_f = 2 \, T \, sin \theta_f \\ T = \frac{m g \, tan \phi}{2 \pi} \approx 0.75$

Here's a "force-based" solution using the non-standard analysis of infinitesimals. I don't mean this to be taken as a formal solution; rather, I show it to give insight into how some analysts think about this type of problem.

Warning: While the theorems of nonstandard analysis are consistent (under the same assumptions on the consistency of set theory in standard mathematics), your instructors will probably give you no points for using this method in a class for the standard methods.

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Let $O$ denote the center of the spherical head, let $C$ denote the center of the circle formed by the crown, let $Q$ denote an arbitrary point on the crown, and let $T(Q)$ denote the tension at $Q$ . We note that the "weight density per radian" of the crown is $\frac{mg}{2\pi}$ , the radius of the crown is $CQ = \frac{\ell}{2\pi}$ , and the height of the crown above $O$ is $OC = \sqrt{r^2 - \left(\frac{\ell}{2\pi}\right)^2}.$

Now let $d\theta$ denote an infinitesimal. We consider a section of the crown centered at $Q$ and including $d\theta$ radians to either side. There is a tension force $T(Q)$ at each end of this section acting in the same direction as the ends. The net effect is a force acting in the direction of $\vec{QC}$ with magnitude $2T(Q)\sin(d\theta)$ .

We now consider the forces acting on this small segment:

• Its weight acts in the direction of $\vec{CO}$ with magnitude $\frac{mg}{2\pi} \cdot 2\,d\theta = \frac{mg}{\pi}d\theta$
• The net tension acts in the direction of $\vec{QC}$ with magnitude $2T(Q)\sin(d\theta)$ .
• The normal force acts in the direction of $\vec{OQ}$ .

In order for these to balance, the sum of the weight and net tension vectors acting on the segment must act in the direction of $\vec{QO}$ , so by similar triangles, $\frac{2T(Q)\sin(d\theta)}{\frac{mg}{\pi}d\theta} = \frac{CQ}{OC} = \frac{\frac{\ell}{2\pi}}{\sqrt{r^2 - \left(\frac{\ell}{2\pi}\right)^2}}$ Here, we note that $\frac{\sin(d\theta)}{d\theta} = 1$ , so applying this and solving for the tension: $T(Q) = \frac{mg\ell}{4\pi^2\sqrt{r^2 - \left(\frac{\ell}{2\pi}\right)^2}}$

Using the given values yields $\boxed{T(Q) \approx 0.75 \mbox{ N}}$