Aha! A good answer.
>> 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.
Yeah, I kinda thought that sounded dodgy...
OK, so here's the essentials
Car A: Mass of m, thrust of F
Car B: Mass of 2m, thrust of 2F
as F/m = 2F/2M, both cars accelerate the same in a straight line
But when it comes time to turn a radius r, the force f to maintain the turn at
the (common) velocity v is
mv^2/r for Car A, and
2mv^2/r for Car B
So we conclude that Car B will require twice the grip to maintain the same speed
through the turn.
Now here's where it gets interesting:
If we assume that both cars have the same level of grip (f), then Car B must
slow down to make the turn.
Solving for v, we get
sqrt(rf/m) = v
Meaning that v varies as the square root of one over the mass. (uhh, right?)
Something doesn't seem right with that conclusion, could someone check it?
DG
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