Question

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?

Short answer

The relation between $x,y$and$\theta$ is $\mathrm{R}\dot{\mathrm{\theta }}=\dot{\mathrm{x}}+\dot{\mathrm{y}}$.

The Lagrangian for the system is:

$L=m{\dot{x}}^2+m\dot{x}\dot{y}+m{\dot{y}}^2+mgy$

The Lagrang’s equations are:

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

And the hoop will begin to unwind and start moving towards the table’s edges.

Step-by-step solution

Meaning of Lagrange’s equation and Lagrangian

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.

Given Parameters

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$\theta$.

Relating x,y and θ.

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,

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

 Step 4: Find the kinetic energy and the Potential energy

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

 Step 5: Find the Lagrangian of the system

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 $L=m{\dot{x}}^2+m\dot{x}\dot{y}+m{\dot{y}}^2+mgy$.

Find the Lagrange’s equations

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 $\dot{\mathrm{x}}\left(\mathrm{t}\right)<0$.