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Quoted post said:Michael Press said:"[email hidden]" <[email hidden]> wrote:
Angular momentum is conserved.
Linear momentum is conserved.
Energy is conserved.
You are trying to argue that angular momentum converts to
linear momentum. In the world of physics it does not.
Look at the calculation I presented. I assume half the
energy of the turning wheels is converted into heating of
the floor and tires. The other half is converted into
kinetic energy of linear motion. I get a resultant speed
of ~ 5 km/hr for two 0.5 kg wheels. Angular momentum and
linear momentum are transferred to the motion of matter
associated with the earth to conserve these quantities.
Where did you get that factor of one-half of the energy going
into heat? Hand waving. When friction plays a large part in
a problem - as it must here, to decelerate the tire on contact
with the floor - arguing from energy considerations can easily
lead you astray. If you try figuring out how much energy goes
into friction in your solution, you'll see it is inconsistent.
There is a physical model behind my approach to the problem,
which I will now explain in tedious detail.
As I said in the quoted paragraph, I _assumed_ it.
Quoted message said:
Before I get even more boring, let me say: I'm an observer,
not a theorist. I didn't try bunny hopping off rollers (I have
downstairs neighbors), but anyone can try this experiment
without rollers. Lift up the rear wheel of your bike. Put it in
high gear and crank the pedal as fast as you possibly can,
so the rear wheel is really going. Now firmly set the rear wheel
down on the floor (use an old rug), putting your weight on the
bike. You might get a little skid mark, but the forward motion
is trivial. It's not like the bike breaks out of your hand and tries
to run across the room.
Okay, here is the full model of the problem. This may seem
overly mathematical for rbr, but it's just multiplication. I'm going
to assume the rider, unlike Mad "Cuddles" Dog, stops pedaling.
Consider one wheel, with mass m1 = 0.8 kg at the rim, rotating at
speed v1=14 m/s. The instant it hits the floor, half the bike+rider
weight is on it, call this m2=40 kg. (Double the quantities to
do two wheels).
As the wheel hits the floor, the contact patch is moving at
14 m/s backward relative to the floor. The friction force of the floor
on the tire is F_fric = mu*m2*g, where g=9.8 m/s^2 and mu=0.7 is the
coefficient of friction (0.7 is about right for rubber on concrete).
Now what happens is the frictional force pushes forward on the
tire, both decelerating the wheel and pushing the bike forward.
This happens for a very short time (the skid) until the rotation
speed is decelerated to match the forward speed of the bike.
Once the speeds match, the skidding ends and the bike rolls
forward normally.
The skid time is dt and the impulse (change in momentum)
delivered to the wheel is P=F_fric * dt = mu*m2*g*dt. This
actually puts a torque on the wheel to change its angular momentum,
but since the torque and the wheel speed are both measured at
the rim, the factor of radius is the same for both. The
wheel is decelerated by P/m1, so the final wheel velocity
is v1_final = v1 - P/m1. The force puts the same change in
momentum P into the (half-weight) bike+rider, so it accelerates
the bike from 0 to v2_final = P/m2.
The skidding stops when the bike speed and wheel speed match,
so v2_final = v1_final. This means mu*g*dt = v1 - mu*g*dt*(m2/m1).
Simplifying, v1 = (1 + m2/m1) * v2_final. This is almost the
same as what I derived earlier from conservation of momentum
(it's different becase earlier I neglected the small angular momentum
remaining in the rotating wheels at the end of the skid). When I
substitute in the numbers:
final speed of bike, v2_final = 0.275 m/sec (a whopping 0.6 mph)
time of skid dt = 0.04 seconds.
Most of the rotating wheel's energy has been dissipated during the
skid as friction slows the wheel. During the skid the bike speed goes
only from 0 to 0.275 m/s, so the skid mark is very short. This
calculation gives 0.5 cm, but a real tire's contact patch is longer
than that - the skid mark has to be at least as long as the contact
patch. BTW, notice that the final speed doesn't depend on the
coefficient of friction, but the time of skid does.
My work here is done. By now you should all be bored enough
to be looking forward to Laff@me's reports from the libel trial.
OK. I ran a linear model and got the same final speed.
Define the quantities
M mass of bicycle and rider
m mass of wheels
r radius of wheel
I moment of inertia
u coefficient of friction
w_0 initial angular speed of wheels
w angular speed of wheels
F frictional force
t elapsed time from initial contact of the wheel with the pavement
T time constant
D differentiation operator wrt time
Assume a model where the frictional force is proportional
to the relative speed of tire and pavement.
F = (wr - v) uMg
After the dismount from the roller, no force is applied to the wheel
through the chain.
This is a linear model.
The time constant for typical values is ~ 0.005 sec.
This means that with this model the tire skids
in the real worls for about 0.02 sec.
The final speed is mrw_0/(M+m).
For a typical situation the final speed is ~0.5 m /sec = 1.8 km / hr.
Now assume a model where the rider continues to turn the crank.
F = (w_0 r - v) uMg
In this case the time constant is 1/(ug) ~ 0.15 sec. for u = 0.7,
and ~ 0.3 for u = 0.3.
Were the rider to continue to spin the wheel
as it was spinning at dismount
for the length of time of the time constant
the speed of the bicycle would be about 0.6 times
the virtual speed in the rollers.
M mass of bicycle and rider
m mass of wheels
r radius of wheel
I moment of inertia
u coefficient of friction
w_0 initial angular speed of wheels
w angular speed of wheels
F frictional force
t elapsed time from initial contact of the wheel with the pavement
T time constant
D differentiation operator wrt time
Expression for I
I = mr^2
Assume the frictional force is proportional to the relative speed of tire and pavement.
F = (wr - v) uMg
Acceleration of the wheel
Dw = -Fr/I
Acceleration of the bicycle
Dv = F/M
Differentiate the expression for the force
DF = (r Dw - Dv) uMg = (-Fr^2/I - F/M) uMg = (-F/m - F/M) uMg
= (-1/m - 1/M) uMg F
A solution is
F = A exp(-t/T),
where A = ruMgw_0
and T = -1/[(-1/m - 1/M)uMg] = m/[(M+m)ug]
This solution implies an infinite time for spinning the wheel.
We can still use it if we take a finiite cut off time such as 2T.
Integrating the expression for the acceleration of the bicycle we get
v(t) = AT/M (1 - exp(-t/T) = mrw_0/(M+m) (1 - exp(-t/T)
The final speed of the bicycle is mrw_0/(M+m).
The final speed is independent of u, the coefficient of friction,
but if we integrate the expression for v(t) we will see that the
length of the skid mark increases with decreasing coefficient of friction.
The numbers are still small however.
For a typical situation
M = 80 kg
m = 2 kg
w_0 = 70 radian/sec
r = 0.35 m
A = 9.8 x 10^3
T = 5 x 10^(-3) sec
This gives a final speed of ~ 0.5 km /hr.
--
Michael Press
