A stunt driver for a movie needs to make a 2545kg car skid on a large, flat, parking lot surface. The force of friction between the tires and the concrete surface is 1.75x10^4N and he is driving at a speed of 24m/s. As he turns more sharply, what radius of curvature will he reach when the car just begins to skid? (Ans: 83.8m)

Question

anonymous · Answer

The car is negotiating a flat curve. There is a centripetal force, $F_c$ that makes the car to follow a circular path. What is helping the car to negotiate the curve is friction, $f$, otherwise it will be thrown away of the road. So:

$F_c = f$

$\large{\frac{mv^2}{r} = f}$

anonymous · Answer

Now I think you can solve for the radius of curvature (:

anonymous · Answer

To make it clearer, the centripetal force is being provided by friction.