- #1
z_offer09
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Anyone familiar with "centrifugal potential" and "brachistochrone" in polar coords?
Hi there,
The issue appears within this problem:
a bead is sliding without friction on a straight wire; the wire rotates in a plane:
\omega= const
No external fields, no gravity.
*******************Which is true:
1) bead's velocity **IS** v = \omega r, because in a rotating frame the kinetic energy = centrifugal energy:
(1/2) m v^2 = (1/2) m (r^2) (\omega^2)
2) bead's velocity **IS NOT** v = \omega r, because expressing v in polar coordinates r,phi:
v = \sqrt(\dot{r}^2 + (r^2)*(\dot{phi}^2))
The bead turns with the wire, so \dot{phi} = \omega
Then v = \sqrt(\dot{r}^2 + (r^2)*(\omega^2)) > \omega r,
because thoughout the motion \dot{r} is never zero.
Any ideas?
---------------------------
Once we know the answer, let's try to build the functional
T = \int (ds/v) for the above problem. Try to express it in polar coordinates. Any luck?
It is really useful if the integral can be Euler-Lagrange minimized for any constraint (not just a straight line). For instance, how do I write T = \int (ds/v) when the bead is constrained to slide on a string shaped as an Archimedean spiral r = \phi \prime?
When this one is minimized with Euler-Lagrange, it should give a straight-line trajectory seen from the inertial frame of reference.
Any input is appreciated.
Hi there,
The issue appears within this problem:
a bead is sliding without friction on a straight wire; the wire rotates in a plane:
\omega= const
No external fields, no gravity.
*******************Which is true:
1) bead's velocity **IS** v = \omega r, because in a rotating frame the kinetic energy = centrifugal energy:
(1/2) m v^2 = (1/2) m (r^2) (\omega^2)
2) bead's velocity **IS NOT** v = \omega r, because expressing v in polar coordinates r,phi:
v = \sqrt(\dot{r}^2 + (r^2)*(\dot{phi}^2))
The bead turns with the wire, so \dot{phi} = \omega
Then v = \sqrt(\dot{r}^2 + (r^2)*(\omega^2)) > \omega r,
because thoughout the motion \dot{r} is never zero.
Any ideas?
---------------------------
Once we know the answer, let's try to build the functional
T = \int (ds/v) for the above problem. Try to express it in polar coordinates. Any luck?
It is really useful if the integral can be Euler-Lagrange minimized for any constraint (not just a straight line). For instance, how do I write T = \int (ds/v) when the bead is constrained to slide on a string shaped as an Archimedean spiral r = \phi \prime?
When this one is minimized with Euler-Lagrange, it should give a straight-line trajectory seen from the inertial frame of reference.
Any input is appreciated.