6.3 Centripetal force  (Page 2/15)

Strategy

1. We know that $F_{\text{c}}=\frac{m{v}^2}{r}.$ Thus,
   
   $F_{\text{c}}=\frac{m{v}^2}{r}=\frac{(900.0\text{kg}){(25.00\text{m/s})}^2}{(500.0\text{m})}=1125\text{N}\text{.}$
2. [link] 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 ${\mu }_{\text{s}}N,$ where ${\mu }_{\text{s}}$ is the static coefficient of friction and N is the normal force. The normal force equals the car’s weight on level ground, so $N=mg.$ Thus the centripetal force in this situation is
   
   $F_{\text{c}}\equiv f={\mu }_{\text{s}}N={\mu }_{\text{s}}mg.$
   
   Now we have a relationship between centripetal force and the coefficient of friction. Using the equation
   
   $F_{\text{c}}=m\frac{v^2}{r},$
   
   we obtain
   
   $m\frac{v^2}{r}={\mu }_{\text{s}}mg.$
   
   We solve this for ${\mu }_{\text{s}},$ noting that mass cancels, and obtain
   
   ${\mu }_{\text{s}}=\frac{v^2}{rg}.$
   
   Substituting the knowns,
   
   ${\mu }_{\text{s}}=\frac{{(25.00\text{m/s})}^2}{(500.0\text{m})(9.80{\text{m/s}}^2)}=0.13.$
   
   (Because coefficients of friction are approximate, the answer is given to only two digits.)

Significance

Banked curves

Let us now consider banked curve     s , where the slope of the road helps you negotiate the curve ( [link] ). The greater the angle $\theta$ , the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an “ideally banked curve,” the angle $\theta$ is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an expression for $\theta$ for an ideally banked curve and consider an example related to it.

For ideal banking    , the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the vertical and horizontal directions.