Question

A car rounds a banked curve as discussed in Example 6.4 and shown in Figure 6.5. The radius of curvature of the road is $\mathit{R}$, the banking angle is $\mathit{\theta }$, and the coefficient of static friction is ${\mathit{\mu }}_{\mathbf{s}}$. (a) Determine the range of speeds the car can have without slipping up or down the road. (b) Find the minimum value for ${\mathit{\mu }}_{\mathbf{s}}$ such that the minimum speed is zero.

Short answer

(a)The minimum speed of the car is$v_{\min }=\sqrt{\frac{gR\left(\tan \theta -{\mu }_s\right)}{1+{\mu }_s\tan \theta }}$and the maximum speed of the car is $v_{\max }=\sqrt{\frac{2R\left({\mu }_s+\tan \theta \right)}{1-{\mu }_s\tan \theta }}$.

(b)The minimum value for${\mu }_s$ is${\mu }_s=tan\theta$.

Step-by-step solution

Defining static friction

The expression for static friction is given by

$f_s={\mu }_{s}n$

Here$f_s$is the static friction,${\mu }_s$is the coefficient of static friction and n is the normal force.

The expression for the centripetal force is given by

$F=m\frac{v^2}{R}$

Here$m$is the mass of the object, v is the speed and R is the radius, F is the centrifugal force. Centrifugal force is also known by the name of centripetal force.

Determining the range of speeds

(a)

When the car is slipping down the inclined plane, then the free body diagram of the car is shown below.

When the car is sliding down the inclined plane, then the frictional force is acted up the inclined plane.

Apply the particle in equilibrium model in the vertical direction

$\sum F_y=0$

$$\begin{gathered} n\cos \theta +f\sin \theta -mg=0 \\ n\cos \theta +f\sin \theta =mg \\ n\cos \theta +{\mu }_{s}n\sin \theta =mg \\ n+{\mu }_{s}n\frac{\sin \theta }{\cos \theta }=\frac{mg}{\cos \theta } \\ n+{\mu }_{s}n\tan \theta =\frac{mg}{\cos \theta } \end{gathered}$$

On simplifying the above expression

$$\begin{gathered} n\left[1+{\mu }_s\tan \theta \right]=\frac{mg}{\cos \theta } \\ n=\frac{mg}{\cos \theta \left(1+{\mu }_s\tan \theta \right)} \end{gathered}$$

The expression for static friction is given by

$f_s={\mu }_{s}n$

Substitute$\frac{mg}{\cos \theta \left(1+{\mu }_s\tan \theta \right)}$for n in the above equation.

$\therefore f_s=\frac{{\mu }_{s}mg}{\cos \theta \left(1+{\mu }_s\tan \theta \right)}$

Apply the particle in equilibrium model in the horizontal direction.

$\sum F_x=0$;

$n\sin \theta -f_s\cos \theta =\frac{m{v^2}_{\min }}{R}$

Substitute $\frac{mg}{\cos \theta \left(1+{\mu }_s\tan \theta \right)}$for n and $\frac{{\mu }_{s}mg}{\cos \theta \left(1+{\mu }_s\tan \theta \right)}$for$f_s$in the above equation.

$\frac{mg}{\cos \theta \left(1+{\mu }_s\tan \theta \right)}\sin \theta -\frac{{\mu }_{s}mg\cos \theta }{\cos \theta \left(1+{\mu }_s\tan \theta \right)}=\frac{m{v^2}_{\min }}{R}$

On simplifying the above expression

$$\begin{gathered} \frac{g\tan \theta }{1+{\mu }_s\tan \theta }-\frac{{\mu }_{s}g}{1+{\mu }_s\tan \theta }=\frac{{v^2}_{\min }}{R} \\ \frac{g}{1+{\mu }_s\tan \theta }\left(\tan \theta -{\mu }_s\right)=\frac{{v^2}_{\min }}{R} \\ {v^2}_{\min }=\frac{gR\left(\tan \theta -{\mu }_s\right)}{1+{\mu }_s\tan \theta } \end{gathered}$$

$\therefore v_{\min }=\sqrt{\frac{gR\left(\tan \theta -{\mu }_s\right)}{1+{\mu }_s\tan \theta }}$

Therefore, the minimum speed of the car is $v_{\min }=\sqrt{\frac{gR\left(\tan \theta -{\mu }_s\right)}{1+{\mu }_s\tan \theta }}$.

When the car is sliding up the inclined plane then the frictional force is directed down the inclined plane. The free body diagram is shown below.

Apply the particle in equilibrium model in the vertical direction

$\stackrel{}{\sum f_y}=0$

$$\begin{gathered} n\cos \theta -f\sin \theta -mg=0 \\ n\cos \theta -{\mu }_s\sin \theta -mg=0 \\ n-{\mu }_{s}n\frac{\sin \theta }{\cos \theta }-\frac{mg}{\cos \theta }=0 \\ n-{\mu }_s\tan \theta =\frac{mg}{\cos \theta } \\ n\left(1-{\mu }_s\tan \theta \right)=\frac{mg}{\cos \theta } \\ n=\frac{mg}{\cos \theta \left(1-{\mu }_s\tan \theta \right)} \end{gathered}$$

The expression for static friction is given by

$f_s={\mu }_{s}n$

Substitute$\frac{mg}{\cos \theta \left(1-{\mu }_s\tan \theta \right)}$for$n$in the above equation.

$\therefore f_s=\frac{{\mu }_{s}mg}{\cos \theta \left(1-{\mu }_s\tan \theta \right)}$

Apply the particle in equilibrium model in the horizontal direction.

$\sum F_x=0$;

$f\cos \theta +n\sin \theta =\frac{m{v^2}_{\max }}{R}$

Substitute $\frac{{\mu }_{s}mg}{\cos \theta \left(1-{\mu }_s\tan \theta \right)}$for$f_s$and$\frac{mg}{\cos \theta \left(1-{\mu }_s\tan \theta \right)}$for n in the above equation.

$\frac{{\mu }_{s}mg\cos \theta }{\cos \theta \left(1-{\mu }_s\tan \theta \right)}+\frac{mg\sin \theta }{\cos \theta \left(1-{\mu }_s\tan \theta \right)}=\frac{m{v^2}_{\max }}{R}$

On simplifying the above expression

$$\begin{gathered} \frac{{\mu }_{s}g}{1-{\mu }_s\tan \theta }+\frac{g\tan \theta }{1-{\mu }_s\tan \theta }=\frac{{v^2}_{\max }}{R} \\ \frac{g}{1-{\mu }_s\tan \theta }\left({\mu }_s+\tan \theta \right)=\frac{{v^2}_{\max }}{R} \\ {v}_{{\max }^2}=\frac{gR\left({\mu }_s+\tan \theta \right)}{1-{\mu }_s\tan \theta } \end{gathered}$$

$\therefore v_{\max }=\sqrt{\frac{2R\left({\mu }_s+\tan \theta \right)}{1-{\mu }_s\tan \theta }}$

Therefore, the maximum speed of the car is$v_{\max }=\sqrt{\frac{2R\left({\mu }_s+\tan \theta \right)}{1-{\mu }_s\tan \theta }}$

Calculating the minimum value of coefficient of static friction

(b)

The expression for the minimum speed of the car is given by

$v_{\min }=\sqrt{\frac{Rg\left(\tan \theta -{\mu }_s\right)}{1+{\mu }_s\tan \theta }}$

If the minimum speed of the car is zero, then

$$\begin{gathered} \frac{Rg\left(\tan \theta -{\mu }_s\right)}{1+{\mu }_s\tan \theta }=0 \\ Rg\left(\tan \theta -{\mu }_s\right)=0 \\ \tan \theta -{\mu }_s=0 \\ {\mu }_s=\tan \theta \end{gathered}$$

Therefore, the minimum value for${\mu }_s$ is ${\mu }_s=tan\theta$.