A hoop of mass m in a vertical plane rests on a frictionless table. A thread is wound many times around the circumference of the hoop. The free end of the thread extends from the bottom of the hoop along the table, passes over a pulley (assumed weightless), and then hangs straight down with a mass m (equal to the mass of the hoop) attached to the end of the thread. Let $\mathit{x}$be the length of thread between the bottom of the hoop and the pulley, let$\mathit{y}$be the length of thread between the pulley and the hanging mass, and let$\mathit{\theta }$be the angle of rotation of the hoop about its center if the thread unwinds. What is the relation between$\mathit{x}\mathbf{,}\mathit{y}$, and$\mathit{\theta }$? Find the Lagrangian and Lagrange’s equations for the system. If the system starts from rest, how does the hoop move?

An ordinary first-order differential equation that is linear in the independent variable and unknown function but not solved for the derivative is termed as Lagrange's equation.

A function that equals the difference between potential and kinetic energy and characterises the state of a dynamic system in terms of position coordinates and their time derivatives is termed as Lagrangian.

The length of thread between the bottom of the hoop$x$ and$y$ the pulley is and is the length of thread between the pulley and the hanging mass $m$. Also, the angle of rotation of the hoop about its center, if the thread unwinds, is$\theta$.

Let $\mathrm{dl}=\mathrm{Rd\theta }$be the amount of hoop unwind where $I$is the length of the unwound thread.

Since $\mathrm{l}=\mathrm{x}+\mathrm{y}+\mathrm{c}$($c$is constant)

Thus,

$\begin{array}{c}\mathrm{dl}=\mathrm{Rd\theta }\mathrm{l}=\mathrm{x}+\mathrm{y}+\mathrm{c}\\ \mathrm{Rd\theta }=\mathrm{dx}+\mathrm{dy}\\ \mathrm{R}\dot{\mathrm{\theta }}=\dot{\mathrm{x}}+\dot{\mathrm{y}}\end{array}$

The reason pf the kinetic energy will be the rotation of the hoop, motion of hoop’s center of mass and falling mass.

This implies that

$\mathrm{T}=\frac{1}{2}\mathrm{I}{\dot{\mathrm{\theta }}}^2+\frac{1}{2}\mathrm{m}{\dot{\mathrm{x}}}^2+\frac{1}{2}\mathrm{m}{\dot{\mathrm{y}}}^2$

So, the kinetic energy is $\mathrm{T}=\frac{1}{2}\mathrm{I}{\dot{\mathrm{\theta }}}^2+\frac{1}{2}\mathrm{m}{\dot{\mathrm{x}}}^2+\frac{1}{2}\mathrm{m}{\dot{\mathrm{y}}}^2$.

Now the reason of Potential energy is the falling mass.

This means,

$\mathrm{U}=-\mathrm{mgy}$

So, the potential energy is $\mathrm{U}=-\mathrm{mgy}$.

Moment of inertia will be:

$\begin{array}{c}\mathrm{I}=\int {\mathrm{R}}^2\mathrm{dm}\\ ={\mathrm{R}}^2\int \mathrm{dm}\\ ={\mathrm{mR}}^2\end{array}$

Then the kinetic energy will be:

$\mathrm{T}=\frac{1}{2}{\mathrm{mR}}^2{\dot{\mathrm{\theta }}}^2+\frac{1}{2}\mathrm{m}{\dot{\mathrm{x}}}^2+\frac{1}{2}\mathrm{m}{\dot{\mathrm{y}}}^2$

The Lagrangian of the system will be calculated by $\mathrm{L}=\mathrm{T}-\mathrm{U}$.

Therefore,

$\begin{array}{l}L=\frac{1}{2}\frac{m{R}^2}{R^2}(\dot{x}+\dot{y}{)}^2+\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\dot{y}}^2-(-mgy)\\ L=m{\dot{x}}^2+m\dot{x}\dot{y}+m{\dot{y}}^2+mgy\end{array}$

So, the Lagrangian is $L=m{\dot{x}}^2+m\dot{x}\dot{y}+m{\dot{y}}^2+mgy$.

By the Euler equation,

$\begin{array}{l}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)=\frac{\partial L}{\partial x}\\ \Rightarrow 2m\ddot{x}+m\ddot{y}=0\\ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{y}}\right)=\frac{\partial L}{\partial y}\\ \Rightarrow 2m\ddot{y}+m\ddot{x}=mg\end{array}$

Since $\ddot{x}=-\frac{\ddot{y}}{2}$,

This follows that,

$\begin{array}{l}2\ddot{\mathrm{y}}-\frac{1}{2}\ddot{\mathrm{y}}=\mathrm{g}\Rightarrow \ddot{\mathrm{y}}=\frac{2}{3}\mathrm{g};\ddot{\mathrm{x}}=-\frac{1}{3}\mathrm{g}\\ \mathrm{y}\left(\mathrm{t}\right)=\frac{1}{3}{\mathrm{gt}}^2+{\mathrm{y}}_0;\dot{\mathrm{y}}\left(0\right)=0\\ \mathrm{x}\left(\mathrm{t}\right)=-\frac{1}{6}{\mathrm{gt}}^2+{\mathrm{x}}_0;\dot{\mathrm{x}}\left(0\right)=0\end{array}$

So, the Lagrange’s equation will be:

$\begin{array}{l}\mathrm{y}\left(\mathrm{t}\right)=\frac{1}{3}{\mathrm{gt}}^2+{\mathrm{y}}_0;\dot{\mathrm{y}}\left(0\right)=0\\ \mathrm{x}\left(\mathrm{t}\right)=-\frac{1}{6}{\mathrm{gt}}^2+{\mathrm{x}}_0;\dot{\mathrm{x}}\left(0\right)=0\end{array}$

And from the Euler equations, it has been cleared that the hoop will begin to unwind and start approaching table edge as $\dot{\mathrm{x}}\left(\mathrm{t}\right)<0$.