DG wrote in part:
> 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. (Or is the lateral acceleration value
uneffected,
First of all this is all idealized physics and does not apply to the real
world (we are not taking aerodynamics into account for example)
To go around the same corner at the same speed means that both cars must
pull the same number of lateral G's or have the same lateral acceleration.
But since one car weighs twice as much that car must exert twice as much
force to follow its path. How does a car generate lateral force? By pressing
sideways with its microscopically uneven tire surface across the
microscopically uneven pavement. How much force can it generate? That
depends proportionally on the weight of the car and the coefficient of
friction for the two surfaces. Assuming the same coefficient of friction for
both cars the car that weighs twice as much can generate twice as much
lateral force.
In other words, both cars can go just as fast around the corner!!!??
But, according to the great god of auto-x ( Colin Chapman ), lighter is
faster. Why? The non-idealized model must take into account frictional
losses, bend and stretching losses etc etc. One example is the heat they
produce. The heavier car will heat its tire patch more. Once the tire patch
heats above a certain point its coefficient of friction drops and the
heavier car must slow down.
All this is IMHO :-)
Regards,
Alan
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