# Hoop rolling inside a circular hole

A hoop of radius $b$ and mass $m$ rolls without slipping within a stationary circular hole of radius $a > b$ and is subject to gravity. Use the generalized coordinates the rotation angle $\phi$ of the hoop and the angular position of the hoop’s center $\theta$. We have the rolling without slipping constraint $$b\phi - a\theta=0.$$ The Lagrangian of the system is $$L=\frac{1}{2}m(a-b)^2\dot{\theta}^2+\frac{1}{2}mb^2\dot{\phi}^2+mg(a-b)\cos\theta.$$ The Euler-Lagrange equations with Lagrange multiplier are $$m(a-b)^2\ddot{\theta}+mg(a-b)\sin\theta=\lambda a, mb^2\ddot{\phi}=-\lambda b$$ Solving for an equation of motion of $\theta$, we have $$(2a^2-2ab+b^2)\ddot{\theta}+g(a-b)\sin\theta=0.$$ My questions are

1. how to find the generalized constraint force that makes the hoop roll without slipping?
2. how to find the constraint force that keeps the hoop’s CM moving on a circular path?

• Did you make a sketch of the situation? Or is it supplied with the question? Sep 6, 2014 at 5:42
• I have a sketch but don't know how to post pictures. Sep 6, 2014 at 5:48
• It's the 4th question in this problem set: astro.caltech.edu/~golwala/ph106ab/ps04_v2.pdf Sep 6, 2014 at 5:51
• Anyone can help? Sep 9, 2014 at 12:49
• It would help to put the sketch in your question. People generally do not like to follow links. Sep 9, 2014 at 12:57

First of all, the constraint equation is incorrect. The rotation angle of the hoop, $\phi$, must be measured with respect to the vertical. This means that $b\phi$ will not be equal to the rolled arc. Instead, the condition will be $b(\phi+\theta)=a\theta$, see figure.
The Lagrangian is $$L=\frac{1}{2}m(a-b)^2\dot{\theta}^2+\frac{1}{2}mb^2\dot{\phi}^2+mg(a-b)\cos\theta.$$ The Euler-Lagrange equations with Lagrange multiplier are $$m(a-b)^2\ddot{\theta}+mg(a-b)\sin\theta=\lambda (a-b)$$ and $$mb^2\ddot{\phi}=-\lambda b.$$
Your question is about the constraint force that makes the hoop roll without slipping. This is friction, $f$. The torque provided by friction, $fb$, must equal $-mb^2\ddot{\phi}$. From the second equation of motion we see that $f=\lambda$.
Using that $\ddot{\phi}=\frac{(a-b)}{b}\ddot{\theta}$ we get, from the second equation of motion, that $\ddot{\theta}=-\lambda \frac{a}{mb(a-b)}.$ Inserting this into the first equation of motion, we get $$-\lambda \frac{(a-b)a}{b}+mg(a-b)\sin\theta=\lambda(a-b),$$ which gives friction as $$f=\lambda=\frac{mg\sin\theta}{1+a/b}.$$