Chapter 5, Problem 29P

To determine

The slowest speed the car can go around the car can go around the curve without sliding down the bank.

Answer to Problem 29P

The slowest speed the car can go around the car can go around the curve without sliding down the bank is $11.5\text{m}/\text{s}$

Explanation of Solution

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

$\begin{array}{c}{\displaystyle \sum F_y}=N_y-mg+f_y\\ =0\end{array}$ (I)

Here, $F_y$ is the net force acting along $y$ direction, $N_y$ is the normal force, $m$ is the mass of the car, $g$ is the acceleration due to gravity, $f_y$ is frictional force acting along y direction.

Write the expression for the net force acting along $x$ direction.

${\displaystyle \sum F_x}=N_x-f_x$ (II)

Here, $F_x$ is the force acting along x direction, $f_x$ is the frictional force acting along $x$ direction.

Write the expression for the net force acting along $x$ direction.

$\begin{array}{c}{\displaystyle \sum F_x}=m{a}_r\\ =m\frac{v^2}{r}\end{array}$ (III)

Here, $v$ is the speed, and $r$ is the radius

Equate equation II and equation III, and resolve the normal vector and force in the x and y directions,

$\begin{array}{c}m\frac{v^2}{r}=N_x-f_x\\ =N\sin \theta -{\mu }_{s}N\cos \theta \end{array}$ (IV)

Rearrange the above equation to find $v$.

$v=\sqrt{\frac{rN}{n}\left(\sin \theta -{\mu }_s\cos \theta \right)}$ (V)

Write the expression for the frictional force along $x$ and $y$ direction,

$\begin{array}{l}{f}_x={\mu }_{s}N\cos \theta \\ f_y={\mu }_{s}N\sin \theta \end{array}$ (VI)

Here, ${\mu }_s$ is the coefficient of static friction, $\theta$ is the banking angle.

Substitute equation IV in equation I to find $N$.

$\begin{array}{c}{N}_y-mg+f_y=N\cos \theta -mg+{\mu }_{s}N\sin \theta \\ =0\end{array}$

Rearrange the above equation to find $N$.

$\begin{array}{c}N\left(\cos \theta +{\mu }_s\sin \theta \right)=mg\\ N=\frac{mg}{\cos \theta +{\mu }_s\sin \theta }\end{array}$ (VII)

Substitute $N$ from equation VII to find $v$.

$\begin{array}{c}v=\sqrt{\frac{r}{m}\left(\frac{mg}{\cos \theta +{\mu }_s\sin \theta }\right)\left(\sin \theta -{\mu }_s\cos \theta \right)}\\ =\sqrt{\frac{gr\left(\sin \theta -{\mu }_s\cos \theta \right)}{\cos \theta +{\mu }_s\sin \theta }}\\ =\sqrt{\frac{gr\left(\tan \theta -{\mu }_s\right)}{1+{\mu }_s\tan \theta }}\end{array}$                                                                VIII

Write the expression for the banking angle.

$\theta ={\tan }^{-1}\left(\frac{v_b^2}{rg}\right)$ (IX)

Here, $\theta$ is the banking angle, $v_b$ is the speed that a car navigate without friction.

Substitute equation IX in equation VIII to find $v$.

$\begin{array}{c}v=\sqrt{\frac{gr\left[\tan {\tan }^{-1}{v}_b^2/rg-{\mu }_s\right]}{1+{\mu }_s\tan {\tan }^{-1}{v}_b^2/\left(rg\right)}}\\ =\sqrt{\frac{gr\left[\frac{v_b^2}{r_g}-{\mu }_s\right]}{1+{\mu }_s\left[\frac{v_b^2}{\left(rg\right)}\right]}}\\ =\sqrt{\frac{v_b^2-{\mu }_{s}gr}{1+{\mu }_s\frac{v_b^2}{rg}}}\end{array}$ (X)

Conclusion:

Substitute $15.0\text{m}/{\text{s}}^2$ for $v$ . $0.120$ for ${\mu }_s$ , $9.80\text{m}/{\text{s}}^2$ for $g$, $75.0\text{m}$ for $r$ in equation x  to find $v$.

$\begin{array}{c}v=\sqrt{\frac{\left(15.0\text{m}/{\text{s}}^2\right)-0.120\left(9.80\text{m}/{\text{s}}^2\right)\left(75.0\text{m}\right)}{1+\frac{0.120\left(15.0\text{m}/{\text{s}}^2\right)}{9.80\text{m}/{\text{s}}^2\left(75.0\text{m}\right)}}}\\ =11.5\text{m}/\text{s}\end{array}$

Therefore, the slowest speed the car can go around the car can go around the curve without sliding down the bank is $11.5\text{m}/\text{s}$