P
Phil Hobbs
Guest
George Herold wrote:
I'm not really persuaded by the v**0 argument for frictional damping.
Long years of painstaking research in the field of yo-yo tricks has
convinced me that when you have a string sliding on a roller, once you
break it loose there's very little friction.
I suspect that if you put a load cell on the string, you'd find that the
actual retarding force was concentrated in narrow pulses near the peak
of each oscillation. The work required to break the string loose is
pretty well constant, so you'd lose a fixed amount of energy per half
cycle. The total energy is
I omega**2 k*theta**2
E = ---------- + ------------
2 2
where omega = d/dt(theta). The average energy loss would be linear in
time, so
dE
-- = Qdot = I omega d(omega)/dt + k*theta*d(theta)/dt.
dt
At the extremes of motion, omega = 0, so if dE/dt over one cycle is some
constant B, then
d(theta)/dt = B/(k*theta)
so theta = (2B/k)*sqrt(t0-t),
where t0 is the time where the motion stops. That's the case for car
brakes--you have to lighten up on the pedal as you slow down, to avoid
jerking to a stop.
With the usual coefficient-of-friction approximation, i.e. your v**0
approach, the power consumed by the rotor in overcoming friction is
dE/dt = omega Gamma,
where Gamma is the frictional torque.
At the peak velocity, theta = 0, so
d(omega)/dt = -Gamma/I,
and you get a linear decrease in the amplitude, as you say.
If those were the whole story, I'd expect to see the envelope be convex,
i.e. with a linear slope at high amplitudes where the sliding friction
dominates, and a steeper slope at low amplitude where it's the stiction
that matters most.
Your plot's envelope is slightly concave, which looks like you have some
exponential behaviour in there someplace.
Interesting, anyway.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net
Cute demo.
I'm not really persuaded by the v**0 argument for frictional damping.
Long years of painstaking research in the field of yo-yo tricks has
convinced me that when you have a string sliding on a roller, once you
break it loose there's very little friction.
I suspect that if you put a load cell on the string, you'd find that the
actual retarding force was concentrated in narrow pulses near the peak
of each oscillation. The work required to break the string loose is
pretty well constant, so you'd lose a fixed amount of energy per half
cycle. The total energy is
I omega**2 k*theta**2
E = ---------- + ------------
2 2
where omega = d/dt(theta). The average energy loss would be linear in
time, so
dE
-- = Qdot = I omega d(omega)/dt + k*theta*d(theta)/dt.
dt
At the extremes of motion, omega = 0, so if dE/dt over one cycle is some
constant B, then
d(theta)/dt = B/(k*theta)
so theta = (2B/k)*sqrt(t0-t),
where t0 is the time where the motion stops. That's the case for car
brakes--you have to lighten up on the pedal as you slow down, to avoid
jerking to a stop.
With the usual coefficient-of-friction approximation, i.e. your v**0
approach, the power consumed by the rotor in overcoming friction is
dE/dt = omega Gamma,
where Gamma is the frictional torque.
At the peak velocity, theta = 0, so
d(omega)/dt = -Gamma/I,
and you get a linear decrease in the amplitude, as you say.
If those were the whole story, I'd expect to see the envelope be convex,
i.e. with a linear slope at high amplitudes where the sliding friction
dominates, and a steeper slope at low amplitude where it's the stiction
that matters most.
Your plot's envelope is slightly concave, which looks like you have some
exponential behaviour in there someplace.
Interesting, anyway.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net
