Dennis writes:
<snip>
>Given that acceleration = force/mass, and given that Car B has
2f and 2m, we
>determine that f/m = 2f/2m So in other words, both cars
accelerate at the same
>rate.
So far, so good.
>At the end of the straight, both cars are travelling at an
identical velocity V.
Right on.
>Immediately following the straight is a turn with radius R.
>
>Now both cars may be travelling at the same speed, but Car A has
a momentum of
>mV, whereas Car B has a momentum of 2mV. In order for Car B to
maintain the same
>speed through the turn as Car A, it must either be capable of
maintaining a
>higher lateral acceleration.
Nope.
> (Or is the lateral acceleration value uneffected,
>but Car B needs more lateral grip to maintain that
acceleration?)
Yep.
>How much higher?
Well, f = m*asubc = m*v^2/r, where asubc = centripetal
acceleration, m = mass, v = linear (i.e., tangential component)
velocity, and r = turn radius. Doubling the mass requires
doubling the lateral force to maintain the same speed through the
same radius turn. A "g" is a unit of acceleration, not force.
>If both cars are capable of maintaining 1G laterally, and both
cars have the
>same amount of grip,
Both can't be true. If they're both capable of 1g, then car B, by
definition, must have twice the available absolute grip (in units
of force) of car A.
> then Car B must slow down.
No. 1g = 1g = 1g, and that's got nothing to do with mass. 1g =~
32.2ft/sec^2.
Jay
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