Chapter 6, Problem 59PQ

Banked curves are designed so that the radial component of the normal force on the car rounding the curve provides the centripetal force required to execute uniform circular motion and safely negotiate the curve. A car rounds a banked curve with banking angle θ = 22.0° and radius of curvature 150 m. a. If the coefficient of static friction between the car’s tires and the road is μ_s = 0.400, what is the range of speeds for which the car can safely negotiate the turn without slipping? b. What is the minimum value of μ_s for which the car’s minimum safe speed is zero? Note that friction points up the incline here.

To determine

The magnitude of range of velocities.

Answer to Problem 59PQ

The range of velocities are $2.26\text{m}/\text{s}$ and $37.6\text{m}/\text{s}$.

Explanation of Solution

Case 1:

When the car slips down towards the inclined plane, then the frictional force is directed up from the inclined plane.

Write the expression for the y-component force on the car.

  $F_N\cos \theta =mg-f\sin \theta$ (I)

Here, $F_N$ is the normal force, m is the mass of the car, g is the acceleration due to gravity , f is the static friction, and $\theta$ is the direction of inclined plane.

Write the expression to calculate the static frictional force on the car.

  $f={\mu }_s{F}_N$

Here, ${\mu }_s$ is the co-efficient of static friction.

Substitute the above equation in (I) to calculate $F_N$.

  $\begin{array}{c}{F}_N\cos \theta =mg-\left({\mu }_s{F}_N\right)\sin \theta \\ F_N\left(\cos \theta +{\mu }_s\sin \theta \right)=mg\\ F_N=\frac{mg}{\cos \theta \left(1+{\mu }_s\tan \theta \right)}\end{array}$

Write the expression for the x-component of force on the car.

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

Here, $v_{\min }$ is the minimum speed of the car required and R is the radius of curvature.

Substitute the expression for $F_N$ and f in the above equation to rewrite in terms of $v_{\min }$.

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

Substitute $9.80\text{m}/{\text{s}}^{\text{2}}$ for g, $22.0°$ for $\theta$, $0.400$ for ${\mu }_s$ and $150\text{m}$ for R in the above equation to calculate $v_{\min }$.

  $\begin{array}{c}{v}_{\min }=\sqrt{9.80\text{m}/{\text{s}}^{\text{2}}\left(150\text{m}\right)\left(\frac{\tan 22.0°-0.400}{1+\left(0.400\right)\tan 22.0°}\right)}\\ =2.26\text{m}/\text{s}\end{array}$

Case 2:

When the car is slip up the incline plane the frictional force is directed towards down the incline plane.

Write the expression for the y-component force on the car.

  $F_N\cos \theta =mg+f\sin \theta$ (II)

Substitute the equation for f in (II) to calculate $F_N$.

  $\begin{array}{c}{F}_N\cos \theta =mg+\left({\mu }_s{F}_N\right)\sin \theta \\ F_N\left(\cos \theta -{\mu }_s\sin \theta \right)=mg\\ F_N=\frac{mg}{\cos \theta \left(1-{\mu }_s\tan \theta \right)}\end{array}$

Write the expression for the x-component of force on the car.

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

Here, $v_{\max }$ is the maximum speed of the car required and R is the radius of curvature.

Substitute the expression for $F_N$ and f in the above equation to rewrite in terms of $v_{\min }$.

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

Substitute $9.80\text{m}/{\text{s}}^{\text{2}}$ for g, $22.0°$ for $\theta$, $0.400$ for ${\mu }_s$ and $150\text{m}$ for R in the above equation to calculate $v_{\min }$.

  $\begin{array}{c}{v}_{\max }=\sqrt{9.80\text{m}/{\text{s}}^{\text{2}}\left(150\text{m}\right)\left(\frac{\tan 22.0°+0.400}{1-\left(0.400\right)\tan 22.0°}\right)}\\ =37.6\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the range of velocities are $2.26\text{m}/\text{s}$ and $37.6\text{m}/\text{s}$.

To determine

The minimum value for ${\mu }_s$ so that car is moving with zero minimum speed.

Answer to Problem 59PQ

The minimum value for ${\mu }_s$ is $0.404$.

Explanation of Solution

Write the expression for minimum speed for the car.

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

Equate the above equation to zero to calculate the expression for ${\mu }_s$.

    $\begin{array}{c}\sqrt{gR\left(\frac{\tan \theta -{\mu }_s}{1+{\mu }_s\tan \theta }\right)}=0\\ \frac{\tan \theta -{\mu }_s}{1+{\mu }_s\tan \theta }=0\\ \tan \theta -{\mu }_s=0\\ {\mu }_s=\tan \theta \end{array}$

Substitute $22.0°$ for $\theta$ in the above equation to calculate ${\mu }_s$.

    $\begin{array}{c}{\mu }_s=\tan 22.0°\\ =0.404\end{array}$

Conclusion:

Therefore, the minimum value for ${\mu }_s$ is $0.404$.