Question

You want to design an oval racetrack such that 3200 lb racecars can round the turns of radius 1000 ft at 96 mi/h without the aid of friction. You estimate that when elements like downforce and grip in the tires are considered the cars will round the turns at a maximum of 175 mi/h.

a) Find the banking angle ? necessary for the racecars to navigate these turns at 96 mi/h and without the aid of friction.

b)  This banking and radius are very close to the actual turn data at Daytona International Speedway where 3200 lb stock cars travel around the turns at about 175 mi/h. What additional radial force is necessary to hold the racecar on the track at 175 mi/h?

Answer 1

You will see that the mass of the cars doesn't matter.

For the case of no friction, the downslope component of the weight is equal to the upslope component of the centripetal force:
m*g*sin? = m(v²/r)*cos? ? mass m cancels; divide by g*cos?
tan? = v² / g*r ? frictionless case

96mi/h * 22ft/s / 15mi/h = 141 ft/s
so
tan? = (141ft/s)² / 32ft/s²*1000ft = 0.6213
? = 31.85º ? "ideal" banking for this radius and velocity

175mi/h * 22ft/s / 15mi/h = 257 ft/s

At this speed (and any speed greater than 96 mi/h on this track) we need a friction force downslope to hold the car on the curve.

normal force Fn = mgcos? + m(v²/r)sin?
so the friction force
f = µ*Fn = µ(mgcos? + m(v²/r)sin?)

Equilibrium parallel to the slope requires that
mgsin? + f = m(v²/r)cos?
3200lb * sin31.85º + f = (3200lb/32ft/s²)*[(257ft/s)² / 1000ft]*cos31.85º
f = 3922 lb ? friction force

I'm not sure what they mean by "radial" force. To me, that's the component of the friction force that's parallel to the ground, or
3922lb * cos31.85º = 3331 lb

Hope this helps! Please rate the answer.