what additional radical force is necessary to prevent a race car from drifting on the curve at 175 mi/h?

Question

an engineer wants to design an oval racetrack such that 3.20e3 lb race cars can round the exactly 1000 ft radius turns at 95 mi/h without the aid of friction. she estimates that the cars will wound the turns at a maximum of 175 mi/h. banking angle necessary for race cars to navigate the turns at 95 mi/h without the aid of friction is 31.24º.

what additional radical force (Fr) is necessary to prevent a race car from drifting on the curve at 175 mi/h?

Answer 1

I'm assuming 'radical' force means radial force, which is the same as centripetal force.

Banking angle needed at 95 mi/hr

The banking angle is given by tan(θ) = v²/rg

θ = 31.24° when

r = 1000 ft, g = 32 ft/s² and v = 95 mi/hr × (1/3600) hr/s × 5280 ft/mi = 418/3 ft/s

and, if we calculate θ = tan^–1(v²/rg) using these numbers we get θ = 31.24°

Radial force needed at 95 mi/hr

At 95 mi/hr the centripetal force provided by the banked curve must be mv²/r

mv²/r = (3200 lb)(418/3 ft/s)²/(1000 ft) = 62,124 ft-lb/s² (Is the force unit abbreviated pdl?)

Radial force needed at 175 mi/hr

For the faster speed v = 175 mi/hr × (1/3600) hr/s × 5280 ft/mi = 770/3 ft/s

mv²/r = (3200 lb)(770/3 ft/s)²/(1000 ft) = 210,809 ft-lb/s²

(which also could be calculated as (175/95)² × 62,124 ft-lb/s²)

Additional radial force needed

So the additional centripetal force needed is 210,809 – 62,124 = 148,685 ft-lb/s²

Answer (rounded to 3 significant figures)

The additional radial force (Fr) is necessary to prevent a race car from drifting on the curve at 175 mi/h is

149,000 ft-lb/s²

Note

This additional force would be provided by banking the curve at a steeper angle given by

θ = tan^–1(v²/rg) = tan^–1[(770/3 ft/s)²/(1000 ft)(32 ft/s²)] = tan^–1(2.0587)

θ = 64.09°