Question 6.4: (a) Calculate the centripetal force exerted on a 900 kg car ...

Strategy and Solution for (a)

We know that F_c = \frac{mv^2}{r} . Thus,

F_c = \frac{mv^2}{r} = \frac{(900 kg)(25.0 m/s)^2}{(500 m)} = 1125 N.                     (6.26)

Strategy for (b)

Figure 6.12 shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is μ_s N , where μ_s is the static coefficient of friction and N is the normal force. The normal force equals the car’s weight on level ground, so that N = mg . Thus the centripetal force in this situation is

F_c = f = μ_sN = μ_smg.                  (6.27)

Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for F_c from the equation

\left\{\begin{matrix} F_c = m\frac{v^2}{r} \\ F_c = mrw^2 \end{matrix} \right\}                     (6.28)

m\frac{v^2}{r} = μ_smg.                  (6.29)

We solve this for μ_s , noting that mass cancels, and  obtain

μ_s = \frac{v^2}{rg} .                  (6.30)

Solution for (b)

Substituting the knowns,

μ_s = \frac{(25.0 m/s)^2}{(500 m)(9.80 m/s^2)} = 0.13.                  (6.31)

(Because coefficients of friction are approximate, the answer is given to only two digits.)

Discussion 

We could also solve part (a) using the first expression in \left\{\begin{matrix} F_c = m\frac{v^2}{r} \\ F_c = mrw^2 \end{matrix} \right\} because m, v, and r are given. The coefficient of friction found inpart (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, because static friction is a responsive force, being. able to assume a value less than but no more than μ_s N. A higher  coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less as will be discussed below.