Ring Tension

Method 1

Find $\theta$ such that:

$2 \, T \, \cos \theta = \frac{\pi + 2 \theta}{2 \pi} M g \\ T = M g \\ \implies \cos \theta = \frac{\pi + 2 \theta}{4 \pi}$

Method 2

Another approach considers the force balance for an infinitesimal segment:

$T(\theta) \cos(\theta) - T(\theta - d\theta) \cos(\theta - d\theta) = M g \frac{d \theta}{2 \pi}$

The process for each loop iteration is:

1) Store the values for $T(\theta - d\theta)$ and $\theta - d\theta$
2) Update the value of $\theta$
3) Solve for $T(\theta)$

The following code gives the solution for Method 2, and the result is the same

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

import math

M = 2.0
g = 10.0

dtheta = (math.pi/2.0)/(10.0**5.0)

T = 0.25*M*g   # initial conditions
theta = 0.0

while T < M*g:

    T_store = T
    theta_store = theta

    theta = theta + dtheta

    right = T_store*math.cos(theta_store) + M*g*dtheta/(2.0*math.pi)
    T = right/math.cos(theta)


print (theta/math.pi)*180.0
#64.5732