Kenneth said:Kenneth said:Terry Morse said:Adding mass to a rotating
system doesn't change the steady state power requirement.
Howdy,
I certainly may have this wrong, but...
I thought that a part rotating at a constant speed is
accelerating because the angular component of its motion
vector was constantly changing.
All the best,
Hi again,
Suppose two cyclists are traveling side by side at the same
speed and the only difference between their bicycles is that
one of them has wheels that are somewhat heavier, say (for
the sake of our thought experiment) one ton each.
Would we suggest that each bike requires the same power to
continue moving at their constant velocity?
The comment about a rotating system (above) would seem to
argue "yes" but that sure defies my intuition (and my
earlier comment about angular momentum.)
All the best,
Dear Kenneth,
Surprisingly, yes, the bikes require the same force to
continue moving at the identical speed, despite one having
much heavier wheels.
A constant force produces a constant acceleration, so even
an ounce of net thrust will eventually accelerate the Queen
Mary to the speed of the space shuttle--
But it has to be an net force. The friction on the Queen
Mary is considerably more than an ounce, so the ship doesn't
move at all, no matter how long I press my little finger
against it.
Given the same force, both bicycles accelerate until the net
force is zero, at which point they continue moving without
any acceleration.
The bicycle with the heavier wheels takes much longer to
reach terminal velocity on a level path (say 20 mph), but it
eventually reaches the same velocity when the force pushing
it forward is balanced by the wind, tire, chain, and bearing
drag.
This is why I can ride much faster with a tail-wind--the
same force applied to the rear tire is not matched by the
wind drag until a much higher ground speed is reached.
(Practically, of course, we can't even start a bicycle with
2 tons of wheels--the frictional drag is enormous--but let's
ignore that. But you can push a railroad car slowly on level
tracks--it just accelerates slowly because its huge mass
resists your puny push, and then even the tiny friction of
metal wheels on a metal track builds up to match your push.)
Angular or rotating momentum is just the same as linear or
straight-line momentum--the motion continues until a force
is applied to change it. A wheel spinning in outer space
will spin in the same place endlessly at the same speed
until some force tilts the plane (the same as changing the
direction of a meteor moving in a straight line) or some
force slows or speeds up the rotation (the same as slowing
or speeding up the straight-line meteor).
So once an idealized wheel is rotating at a steady speed,
the only power required to keep it moving is whatever is
necessary to overcome the frictional drag--which doesn't
even exist for an idealized wheel. Practically speaking,
doubling the weight of a real bicycle wheel adds little
frictional drag, so rec.bicycles.tech is appalled by the
notion that a heavier wheel would require more force to
produce the same steady speed.
To return to the original post's query, it's possible that a
heavier shoe is not a purely rotating mass like a wheel,
since it also rocks back and forth, a hideous complication
and one likely to require extra force.
I, for one, do not intend to start riding in galoshes, even
on idealized steady-speed flat courses.
Carl Fogel
