Rope in a circular tube

The rope is a semi-circle of radius R. Suppose the midpoint of the rope segment is offset from the vertical by an angle $\delta$ . The mass associated with an infinitesimal portion of the rope is:

$\large{dm = \frac{d\theta}{\pi} M}$

The height ( $y$ ) above the lowest part of the tube is (where $\theta$ is the angle between the infinitesimal and the vertical):

$\large{y = R(1-cos\theta)}$

The gravitational potential energy of the infinitesimal is:

$\large{dU = dm \, g \, y = \frac{d\theta}{\pi} M g R(1-cos\theta) = \frac{MgR}{\pi} (1-cos\theta) d\theta }$

Recalling that the midpoint of the rope segment is offset from the vertical by an angle $\delta$ , the gravitational potential energy of the entire rope is:

$\large{U = \frac{MgR}{\pi} \int_{\frac{-\pi}{2} + \delta}^{\frac{\pi}{2} + \delta} (1-cos\theta) d\theta = \frac{MgR}{\pi} (\pi - 2cos\delta) }$

As the rope segment oscillates, the offset $\delta$ will vary with time. The associated kinetic energy is:

$\large{KE = \frac{1}{2} M v^2 = \frac{1}{2} M (R \dot{\delta})^2 = \frac{1}{2} M R^2 \dot{\delta}^2}$

We can solve for the dynamics using Lagrangian Mechanics . We have one state variable ( $\delta)$ . The system Lagrangian is:

$\large{L = KE - U = \frac{1}{2} M R^2 \dot{\delta}^2 - \frac{MgR}{\pi} (\pi - 2cos\delta)}$

The Euler-Lagrange equation is:

$\large{\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\delta}}} = \frac{\partial{L}}{\partial{\delta}}}$

Evaluating the left-hand side yields:

$\large{\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\delta}}} = \frac{d}{dt} (M R^2 \dot{\delta}) = M R^2 \ddot{\delta} }$

Evaluating the right-hand side yields:

$\large{\frac{\partial{L}}{\partial{\delta}} = \frac{d}{d\delta} \Big(\frac{2MgR}{\pi} cos\delta\Big) = -\frac{2MgR}{\pi} sin\delta }$

Equating the two yields:

$\large{M R^2 \ddot{\delta} = -\frac{2MgR}{\pi} sin\delta \\ \ddot{\delta} = -\frac{2g}{\pi R} sin\delta}$

Assuming the initial angular displacement is small, the following approximation can be made:

$\large{\ddot{\delta} \approx -\frac{2g}{\pi R} \delta}$

This differential equation clearly corresponds to simple harmonic motion with angular speed as shown below:

$\large{\omega^2 = \frac{2g}{\pi R} = \frac{4 \pi^2}{T^2} \\ T = \sqrt{\frac{4 \pi^3 R}{2g}}}$

We are given the value of R:

$\large{R = \frac{20}{\pi^3}}$

Substituting in for $R$ and $(g = 10)$ yields:

$\large{T = \sqrt{\frac{4 \pi^3 20}{20 \pi^3}} = \sqrt{4} = \boxed{2}}$

u can do it using taylor theorem ,

by calculating potential at general displacement and double differentiate it to get the value at the eqil. point ,

hope i'm right and u liked it.

we can easily do it with com concept. i.e distance of com of ring from centre is 2R/pi .Mg as restoring torque acts there is displced at small angle theta.mow use T=I*alpha .here I is MR^2

I will attempt to draw an analogue to the spring harmonic oscillator to solve this problem.

-Given small displacement $\theta$ on one side, we will have additional mass $\frac{2\theta}{\pi}m$ on that side as well and restoring force

$F = \frac{2\theta}{\pi}mg$

-Using the middle of the rope as a point of reference, the linear displacement $d$ follows the relation relation with $d = r\theta$ which we can substitute into the previous equation to rewrite the restoring force

$F = \frac{2gm}{\pi r}d$

-Recall the spring harmonic oscillator which had restoring force $kd$ and period

$T = 2\pi\sqrt{\frac{m}{k}}$ .

Our restoring force has the exact same form, just with a different k! Substituting in we get

$T = 2\pi\sqrt{\frac{\pi r}{2g}}$

Plugging in radius and gravity we get

$T = 2\pi\sqrt{\frac{\pi 20}{\pi^{3}2 * 10 }}\ = \boxed{2}$