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Question

If a car goes through a curve too fast, the car tends to slide out of the curve. For a banked curve with friction, a frictional force acts on a fast car to oppose the tendency to slide out of the curve; the force is directed down the bank. Consider a circular curve of radius r=200m and a bank angle theta, where the coefficient of static friction between tires and pavement is u. In kilometers per hour, evaluate the maximum velocity for a bank angle of 10 and for u=0.6, and again for u=0.05

Answer 1

For a flat curve, the maximum force on the tires before slippage is the static coefficient of friction times the normal force, or weight of the car. f=u*m*g.
This is modeled as the centripetal force holding the car in the curve

The force pulling the car away is the centrifugal force, which is m*v^2/r.

Since the curve is banked, a component of the force adds to the weight of the car, and subtracts from the part pulling the car off the track. Also, gravity acts to pull the car into the curve, or down the bank.

Let's look at gravity first

f=mg
Perpendicular to the surface of the road,
f=cos(theta)*m*g
parallel to the surface of the curve
f=sin(theta)*m*g towards the center.

The component of centrifugal force parallel to the road is
cos(theta)*1/2*m*v^2/r
And the part adding to the perpendicular force is
sin(theta)*1/2*m*v^2/r

using the coefficient of friction u,
the sliding force versus the part pushing downward is
cos(theta)*1/2*m*v^2/r - sin(theta)*m*g
=u(sin(theta)*1/2*m*v^2/r+ cos(theta)*m*g)

Note that the mass divides out
for simplicity , I will type cos for cos(theta)
and sin for sin(theta)
1/2*cos*v^2/r-sin*g=
u*1/2*sin v^2/r +u*cos*g

1/2*v^2/r * (cos-u*sin)
=g*(u*cos+sin)
v=sqrt
(g*2*r*(u*cos+*sin)/
(cos-u*sin))

This is in m/s
for km/hr:

1m/1000km 3600secons/hr
multiply by 3.6

For u=.6, vmax = 210km/hr
for u=.05, vmax=108km/hr

sorry for the delay with the correct answer, Yahoo! answers was down for 40 minutes

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