- #1
gulsen
- 217
- 0
Suppose, there's an inclined surface, and a sphere, with radius [tex]R[/tex], is rolling without slipping. The Lagrangian is
[tex]L = I \frac{\dot \theta ^2}{2} + m \frac{\dot s ^2}{2} - V[/tex]
where [tex]\theta[/tex] is the angle of rotation of sphere and [tex]s[/tex] is the curve length from top, and V is a potential which depends on coordinates. But coordinates [tex]\theta[/tex] and [tex]s[/tex] are interralated with [tex]s = R\theta[/tex] so Lagrangian can be equivalenty written as
[tex]L = \left ( \frac{ds}{dt} \right )^2 \left ( \frac{I}{mR^2} + 1 \right ) - V/m[/tex]
I'd expect the solution will be reduced to point particle solution in the limit [tex]R \to 0[/tex] but...
The point particle Lagrangian is
[tex]L = \left ( \frac{ds}{dt} \right )^2 - V/m[/tex]
so I assume this's what the solution will be reduced, but when I put [tex]I=(2/5)mR^2[/tex] in the previous Lagrangian, there are no longer any [tex]R[/tex]s, i.e. the Lagrangian is independent of radius and
[tex]L = \left ( \frac{ds}{dt} \right )^2 \left ( \frac{7}{5} \right ) - V/m[/tex]
becomes. What's more painful is, the same weirdness occurs in any circular rolling object; if my object was a rotating disk in the first place, my Lagrangian would become
[tex]L = \left ( \frac{ds}{dt} \right )^2 \left ( \frac{3}{2} \right ) - V/m[/tex]
and in the point particle limit, where I imagine the point particle as a disk with infinitesimal radius and thickness...
1. Can someone shed light on this "paradox"; what's going on here?!?
2. And how do i get point particle limit? Or is there a way to get point particle limit, which has a delta function mass density, starting from a continuous, smooth mass distribution??
[tex]L = I \frac{\dot \theta ^2}{2} + m \frac{\dot s ^2}{2} - V[/tex]
where [tex]\theta[/tex] is the angle of rotation of sphere and [tex]s[/tex] is the curve length from top, and V is a potential which depends on coordinates. But coordinates [tex]\theta[/tex] and [tex]s[/tex] are interralated with [tex]s = R\theta[/tex] so Lagrangian can be equivalenty written as
[tex]L = \left ( \frac{ds}{dt} \right )^2 \left ( \frac{I}{mR^2} + 1 \right ) - V/m[/tex]
I'd expect the solution will be reduced to point particle solution in the limit [tex]R \to 0[/tex] but...
The point particle Lagrangian is
[tex]L = \left ( \frac{ds}{dt} \right )^2 - V/m[/tex]
so I assume this's what the solution will be reduced, but when I put [tex]I=(2/5)mR^2[/tex] in the previous Lagrangian, there are no longer any [tex]R[/tex]s, i.e. the Lagrangian is independent of radius and
[tex]L = \left ( \frac{ds}{dt} \right )^2 \left ( \frac{7}{5} \right ) - V/m[/tex]
becomes. What's more painful is, the same weirdness occurs in any circular rolling object; if my object was a rotating disk in the first place, my Lagrangian would become
[tex]L = \left ( \frac{ds}{dt} \right )^2 \left ( \frac{3}{2} \right ) - V/m[/tex]
and in the point particle limit, where I imagine the point particle as a disk with infinitesimal radius and thickness...
1. Can someone shed light on this "paradox"; what's going on here?!?
2. And how do i get point particle limit? Or is there a way to get point particle limit, which has a delta function mass density, starting from a continuous, smooth mass distribution??
Last edited: