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[Phys-L] Re: centripetal force question



Anthony Lapinski wrote:
We know that a motorcycle driver leans inward (toward the center of a
circle) when turning a corner that is UNBANKED. My question is how does
the FBD look for the driver and cycle? Are both behaving like they are on
a BANKED curve, or is N still vertical to balance w, and then fs is the
centripetal force? Or is N slanted inward, and both Nx and fs add to
provide the centripetal?

It makes sense that the driver FEELS heavier during this turn, like he/she
would on a banked curve. This means N is more, and then N would be slanted
inward (like a banked curve), and not upward. However, N is always
perpendicular to the surface (ground), which is flat. There appears to be
a contradiction here. Can anyone help me out with this analysis?

There are at least four different things going on here.

1a) The fact that the motorcycle needs to lean has essentially nothing to
do with whether or not the road is banked. It has to do with the fact
that the motorcycle makes contact with the road at only two points
such that its footprint has essentially zero width transverse to the
direction of motion. These two points determine a line. Treat this
line as an axis. In equilibrium, there must be just the right amount
of torque around this axis. This gives us an equation that fully
constrains the angle of lean.

1b) The rider "FEELS" heavier by a factor of the secant of the angle
of lean. This is another quantity that has nothing to do with
whether the road is banked or not.

2) Most people find this problem incomparably easier to analyze in the
non-inertial frame comoving with the motorcycle. Yes, I know that
non-inertial frames are anathema in most high-school physics classes.
But in the real world, everybody uses them. Certainly motorcycle
riders use them. Any questions about what the rider "FEELS" are
most appropriately formulated in such a frame.

1+2) In the comoving frame, the idea expressed in paragraph (1) is
particularly easy to state: there should be zero torque around
the axis defined by the points of tire-contact. The earth's
gravitational field plus the centrifugal field combine to create
an effective field. According to Einstein's principle of equivalence,
this effective field is locally indistinguishable from a gravitational
field. If we consider the motorcycle to be a rigid object, the result
is a force acting on the center of mass of the object. The lever arm
of this force with respect to the tire-contact axis must be zero;
otherwise there would be a torque tending to tip the motorcycle to
one side or the other.

The rider makes (literally) a seat-of-the-pants estimate of this
tipping torque and adjusts the angle of lean to zero it out.

3) How much does friction contribute to the centripetal force?
There are actually *five* different cases of interest, as diagrammed at
http://www.av8n.com/physics/img48/banked-road.png

-- If the road is horizontal (zero bank), friction contributes 100%
of the centripetal force.
-- If the road has the ideal bank angle, friction contributes 0% of
the centripetal force.
-- If the road is banked the wrong way, friction contributes more than
100%.
-- If the road is banked the right way, but too much, friction makes
a negative contribution to the centripetal force.
-- In the most common case, i.e. an underbanked road, friction contributes
something more than 0% but less than 100%.

4) For a road that is banked the right way by any positive amount, the
centripetal force makes at least some positive contribution to the normal
component of the tire/road force.

As a consequence, the amount of available frictional force is increased.
So in this sense, banking is a win/win proposition. However, the total
normal force is only slightly greater than the natural weight for the
smallish values of bank-angle one finds on ordinary roads.

For an unbanked (i.e. horizontal) road, there is no increase in normal
force. The centripetal force has no projection on the road-normal.

Note: Here when I talk about friction, I am referring to the quasi-static
rolling friction between the tire(s) and the road.

==================================

For what it's worth, if you reeeally want to get a seat-of-the-pants
perception of what "bank" means and what "secant" means, a motorcycle
is not nearly as interesting as an airplane. A coordinated level turn
with a 75 degree bank angle will get your attention. That's just shy of
four Gees. And you can create any bank angle you like, instantly; no
road-building crew is required.
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