Chapter 5, Problem 30P

To determine

The fastest speed of the car that travels through the curve without skidding.

Answer to Problem 30P

The fastest speed of the car that travels through the curve without skidding is $\underset{\_}{61.55\text{m}/\text{s}}$.

Explanation of Solution

Figure 1 represents the free body diagram of the car along the incline.

Write the expression for net force acting on the car along y direction.

    $\begin{array}{c}{N}_{\text{y}}-f_{\text{y}}-mg=0\\ N\cos \theta -f\sin \theta -mg=0\end{array}$

Here, $N$ is the normal force, $\theta$ is the angle made with the horizontal, $f$ is the frictional force, $m$ is the mass of the car, and $g$ is the acceleration due to gravity.

Since the frictional force $f={\mu }_{\text{s}}N$, the above equation is reduced to

    $N\cos \theta -{\mu }_{\text{s}}N\sin \theta -mg=0$        (I)

Here, ${\mu }_{\text{s}}$ is the coefficient of static friction.

Rearrange the above equation to find $N$

  $\begin{array}{c}N\left(\cos \theta -{\mu }_{\text{s}}\sin \theta \right)-mg=0\\ N=\frac{mg}{\cos \theta -{\mu }_{\text{s}}\sin \theta }\end{array}$        (II)

Write the expression for net force acting on the car along x direction.

    $\begin{array}{c}{N}_x-f_x=\frac{m{v}^2}{R}\\ N\sin \theta +f\cos \theta =\frac{m{v}^2}{R}\end{array}$

Since the frictional force $f={\mu }_{\text{s}}N$, the above equation is reduced to

    $\begin{array}{c}N\sin \theta +{\mu }_{\text{s}}N\cos \theta =\frac{m{v}^2}{R}\\ N\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)=\frac{m{v}^2}{R}\\ v=\sqrt{\frac{RN\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)}{m}}\end{array}$        (III)

Use equation (II) in (III).

    $\begin{array}{c}v=\sqrt{\frac{R\left(\frac{mg}{\cos \theta -{\mu }_{\text{s}}\sin \theta }\right)\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)}{m}}\\ =\sqrt{\frac{Rmg\left(\sin \theta +{\mu }_{\text{s}}\cos \theta \right)}{m\left(\cos \theta -{\mu }_{\text{s}}\sin \theta \right)}}\\ =\sqrt{\frac{Rmg\left(\frac{\sin \theta }{\cos \theta }+{\mu }_{\text{s}}\right)}{m\left(1-{\mu }_{\text{s}}\frac{\sin \theta }{\cos \theta }\right)}}\\ =\sqrt{\frac{Rg\left(\tan \theta +{\mu }_{\text{s}}\right)}{\left(1-{\mu }_{\text{s}}\tan \theta \right)}}\end{array}$        (IV)

Conclusion:

Substitute $610\text{m}$ for $R$, $9.8\text{m}/{\text{s}}^2$ for $g$, $5.8°$ for $\theta$, and $0.50$ for ${\mu }_{\text{s}}$ in equation (IV), to find $v$.

    $\begin{array}{c}v=\sqrt{\frac{\left(610\text{m}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(\tan 5.8°+0.50\right)}{\left(1-\left(0.50\right)\tan 5.8°\right)}}\\ =\sqrt{3788.64}\\ =61.55\text{m}/\text{s}\end{array}$

Therefore, the fastest speed of the car that travels through the curve without skidding is $\underset{\_}{61.55\text{m}/\text{s}}$.