PHYSICS Cars and Their Cornering Ability
Where ac
equals the centripetal acceleration, v equals the car's velocity around
the curve, and R represents the radius of the curve. A curve that as a
posted speed limit of 25 m/s
and has a radius of 50 m requires a centripetal acceleration of 6.5 m/s2.
In terms of g's this is 0.66 g's, (6.5/9.8
= 0.66 ).
This means that a car can
travel around the curve without slipping if the tires exert a frictional
force of 3.11 m/s2, 0.318 g's. In car and
driver this frictional force is described as the "lateral acceleration."
The limiting factor is 0.317 gs for this curve.
3.36 gs! There are no
passenger cars that do this. The car will not make it around the curve.
The driver will have the wheels turned but the car will continue to slide.
If the curve were banked, the car would slide to the outside of the curve
until the centripetal acceleration needed to negotiate the curve equals
the centripetal acceleration the car can exert. Suppose the car can exert
a centripetal acceleration of 0.88 gs. Then;
Therefore R = 72.47m. The
car will slide outwards along the banked curve until the radius of the
turn equals 72.47 meters. Or until a fence stops the car. Click here for some physics problems relating to cars and their ability to corner. A new window will open when this link is clicked. Close it to get back to this page. |
by Tony Wayne ...(If you are a teacher, please feel free to use these resources in your teaching.)
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