- 1,336
- 1,121
Consider a hoop of mass ##m## that is situated on a horizontal plane and rolls forward without slipping. The radius of the hoop, if I may be so bold, is a given function of time ##r=r(t)##. These are the ideal constraints.
Let ##x,y## be the coordinates of the hoop's center ##S## and let ##\psi## be its angle of rotation. I can give a formal definition of what this is if needed. The constraints are ##y=r,\quad \dot x=r\dot\psi##.
The kinetic energy is
$$T=\frac{m}{2}\Big(\dot x^2+\dot r^2\Big)+\frac{m}{2}\Big((r\dot\psi)^2+\dot r^2\Big).$$
Taking ##x,\psi## for generalized coordinates, we have
$$L=\frac{m}{2}\dot x^2+\frac{m}{2}(r\dot\psi)^2,\quad [L]_x\delta x+[L]_\psi\delta\psi=0,$$
where ##\delta x=r\delta\psi.##
my answer is
$$\dot\psi(t)=\frac{C}{r^{3/2}(t)}.$$
Let ##x,y## be the coordinates of the hoop's center ##S## and let ##\psi## be its angle of rotation. I can give a formal definition of what this is if needed. The constraints are ##y=r,\quad \dot x=r\dot\psi##.
The kinetic energy is
$$T=\frac{m}{2}\Big(\dot x^2+\dot r^2\Big)+\frac{m}{2}\Big((r\dot\psi)^2+\dot r^2\Big).$$
Taking ##x,\psi## for generalized coordinates, we have
$$L=\frac{m}{2}\dot x^2+\frac{m}{2}(r\dot\psi)^2,\quad [L]_x\delta x+[L]_\psi\delta\psi=0,$$
where ##\delta x=r\delta\psi.##
my answer is
$$\dot\psi(t)=\frac{C}{r^{3/2}(t)}.$$
Last edited: